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\begin_body
\begin_layout Title
Teorija linearne algebre za ustni izpit —
IŠRM 2023/24
\end_layout
\begin_layout Author
\noun on
Anton Luka Šijanec
\end_layout
\begin_layout Date
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
today
\end_layout
\end_inset
\end_layout
\begin_layout Abstract
Povzeto po zapiskih s predavanj prof.
Cimpriča.
\end_layout
\begin_layout Standard
\begin_inset CommandInset toc
LatexCommand tableofcontents
\end_inset
\end_layout
\begin_layout Part
Teorija
\end_layout
\begin_layout Section
Prvi semester
\end_layout
\begin_layout Subsection
Vektorji v
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
\end_layout
\begin_layout Standard
Identificaramo
\begin_inset Formula $n-$
\end_inset
terice realnih števil,
točke v
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
,
množice paroma enakih geometrijskih vektorjev.
\end_layout
\begin_layout Standard
Osnovne operacije z vektorji:
Vsota (po komponentah) in množenje s skalarjem (po komponentah),
kjer je skalar realno število.
\end_layout
\begin_layout Standard
Lastnosti teh računskih operacij:
asociativnost in komutativnost vsote,
aditivna enota,
\begin_inset Formula $-\vec{a}=\left(-1\right)\cdot\vec{a}$
\end_inset
,
leva in desna distributivnost,
homogenost,
multiplikativna enota.
\end_layout
\begin_layout Subsubsection
Linearna kombinacija vektorjev
\end_layout
\begin_layout Definition*
Linearna kombinacija vektorjev
\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$
\end_inset
je izraz oblike
\begin_inset Formula $\alpha_{1}\vec{v_{1}}+\cdots+\alpha_{n}\vec{v_{n}}$
\end_inset
,
kjer so
\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}$
\end_inset
skalarji.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Množico vseh linearnih kombinacij vektorjev
\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$
\end_inset
označimo z
\begin_inset Formula $\Lin\left\{ \vec{v_{1}},\dots,\vec{v_{n}}\right\} $
\end_inset
in ji pravimo linearna ogrinjača (angl.
span).
\begin_inset Formula $\Lin\left\{ \vec{v_{1}},\dots,\vec{v_{n}}\right\} =\left\{ \alpha_{1}\vec{v_{1}}+\cdots+\alpha_{n}\vec{v_{n}};\forall\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\right\} $
\end_inset
\end_layout
\begin_layout Subsubsection
Linearna neodvisnost vektorjev
\end_layout
\begin_layout Paragraph*
Ideja
\end_layout
\begin_layout Standard
En vektor je linearno neodvisen,
če ni enak
\begin_inset Formula $\vec{0}$
\end_inset
.
Dva,
če ne ležita na isti premici.
Trije,
če ne ležijo na isti ravnini.
\end_layout
\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:odvisni"
\end_inset
Vektorji
\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$
\end_inset
so linearno odvisni,
če se da enega izmed njih izraziti z linearno kombinacijo preostalih
\begin_inset Formula $n-1$
\end_inset
vektorjev.
Vektorji so linearno neodvisni,
če niso linearno odvisni (in obratno).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:vsi0"
\end_inset
Vektorji
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
so linearno neodvisni,
če za vsake skalarje,
ki zadoščajo
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
\end_inset
,
velja
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$
\end_inset
.
ZDB poleg
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$
\end_inset
ne obstajajo nobeni drugi
\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}$
\end_inset
,
kjer bi veljalo
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:kvečjemu1"
\end_inset
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
so linearno neodvisni,
če se da vsak vektor na kvečjemu en način izraziti kot linearno kombinacijo
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
.
\end_layout
\begin_layout Theorem*
Te tri definicije so ekvivalentne.
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\ref{def:odvisni}\Rightarrow\ref{def:vsi0}\right)$
\end_inset
Recimo,
da so
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
linearno odvisni v smislu
\begin_inset CommandInset ref
LatexCommand ref
reference "def:odvisni"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
.
Dokažimo,
da so tedaj linearno odvisni tudi v smislu
\begin_inset Formula $\ref{def:vsi0}$
\end_inset
.
Obstaja tak
\begin_inset Formula $i$
\end_inset
,
da lahko
\begin_inset Formula $v_{i}$
\end_inset
izrazimo z linearno kombinacijo preostalih,
torej
\begin_inset Formula $v_{i}=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$
\end_inset
za neke
\begin_inset Formula $\alpha$
\end_inset
.
Sledi
\begin_inset Formula $0=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\left(-1\right)v_{i}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$
\end_inset
,
kar pomeni,
da obstaja linearna kombinacija,
ki je enaka 0,
toda niso vsi koeficienti 0 (že koeficient pred
\begin_inset Formula $v_{i}$
\end_inset
je
\begin_inset Formula $-1$
\end_inset
),
tedaj so vektorji po definiciji
\begin_inset CommandInset ref
LatexCommand ref
reference "def:vsi0"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
linearno odvisni.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\ref{def:vsi0}\Rightarrow\ref{def:odvisni}\right)$
\end_inset
Recimo,
da so
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
linearno odvisno v smislu
\begin_inset Formula $\ref{def:vsi0}$
\end_inset
.
Tedaj obstajajo
\begin_inset Formula $\alpha$
\end_inset
,
ki niso vse 0,
da velja
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
\end_inset
.
Tedaj
\begin_inset Formula $\exists i\ni:\alpha_{i}\not=0$
\end_inset
in velja
\begin_inset Formula
\[
\alpha_{i}v_{i}=-\alpha_{1}v_{1}-\cdots-\alpha_{i-1}v_{i-1}-\alpha_{i+1}v_{i+1}-\cdots-\alpha_{n}v_{n}\quad\quad\quad\quad/:\alpha_{i}
\]
\end_inset
\begin_inset Formula
\[
v_{i}=-\frac{\alpha_{1}}{\alpha_{i}}v_{i}-\cdots-\frac{\alpha_{i-1}}{\alpha_{i}}v_{i-1}-\frac{\alpha_{i+1}}{\alpha_{i}}v_{i+1}-\cdots-\frac{\alpha_{n}}{\alpha_{i}}v_{n}\text{,}
\]
\end_inset
s čimer smo
\begin_inset Formula $v_{i}$
\end_inset
izrazili kot linearno kombinacijo preostalih vektorjev.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\ref{def:vsi0}\Leftrightarrow\ref{def:kvečjemu1}\right)$
\end_inset
Naj bodo
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
LN.
Recimo,
da obstaja
\begin_inset Formula $v$
\end_inset
,
ki se ga da na dva načina izraziti kot linearno kombinacijo
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
.
Naj bo
\begin_inset Formula $v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=\beta_{1}v_{1}+\cdots+\beta_{n}v_{n}$
\end_inset
.
Sledi
\begin_inset Formula $0=\left(\alpha_{1}-\beta_{1}\right)v_{1}+\cdots+\left(\alpha_{n}-\beta_{n}\right)v_{n}$
\end_inset
.
Po definiciji
\begin_inset CommandInset ref
LatexCommand ref
reference "def:vsi0"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
velja
\begin_inset Formula $\forall i:\alpha_{i}-\beta_{i}=0\Leftrightarrow\alpha_{i}=\beta_{i}$
\end_inset
,
torej sta načina,
s katerima izrazimo
\begin_inset Formula $v$
\end_inset
,
enaka,
torej lahko
\begin_inset Formula $v$
\end_inset
izrazimo na kvečjemu en način z
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
,
kar ustreza definiciji
\begin_inset CommandInset ref
LatexCommand ref
reference "def:kvečjemu1"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsubsection
Ogrodje in baza
\end_layout
\begin_layout Definition*
Vektorji
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
so ogrodje (angl.
span),
če
\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n}\right\} =\mathbb{R}^{n}\Leftrightarrow\forall v\in\mathbb{R}^{n}\exists\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Vektorji
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
so baza,
če so LN in ogrodje
\begin_inset Formula $\Leftrightarrow\forall v\in\mathbb{R}^{n}:\exists!\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
\end_inset
ZDB vsak vektor
\begin_inset Formula $\in\mathbb{R}^{n}$
\end_inset
se da na natanko en način izraziti kot LK
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
.
\end_layout
\begin_layout Example*
Primer baze je standardna baza
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
:
\begin_inset Formula $\left\{ \left(1,0,0,\dots,0\right),\left(0,1,0,\dots,0\right),\left(0,0,1,\dots,0\right),\left(0,0,0,\dots,1\right)\right\} $
\end_inset
.
To pa ni edina baza.
Primer nestandardne baze v
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset
je
\begin_inset Formula $\left\{ \left(1,1,1\right),\left(0,1,1\right),\left(0,0,1\right)\right\} $
\end_inset
.
\end_layout
\begin_layout Subsubsection
Norma in skalarni produkt
\end_layout
\begin_layout Definition*
Norma vektorja
\begin_inset Formula $v=\left(\alpha_{1},\dots,\alpha_{n}\right)$
\end_inset
je definirana z
\begin_inset Formula $\left|\left|v\right|\right|=\sqrt{\alpha_{1}^{2}+\cdots+\alpha_{n}^{2}}$
\end_inset
.
Geometrijski pomen norme je dolžina krajevnega vektorja z glavo v
\begin_inset Formula $v$
\end_inset
.
\end_layout
\begin_layout Standard
Osnovne lastnosti norme:
\begin_inset Formula $\left|\left|v\right|\right|\geq0$
\end_inset
,
\begin_inset Formula $\left|\left|v\right|\right|=0\Rightarrow v=\vec{0}$
\end_inset
,
\begin_inset Formula $\left|\left|\alpha v\right|\right|=\left|\alpha\right|\cdot\left|\left|v\right|\right|$
\end_inset
,
\begin_inset Formula $\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$
\end_inset
(trikotniška neenakost)
\end_layout
\begin_layout Definition*
Skalarni produkt
\begin_inset Formula $u=\left(\alpha_{1},\dots,\alpha_{n}\right),v=\left(\beta_{1},\dots,\beta_{n}\right)$
\end_inset
označimo z
\begin_inset Formula $\left\langle u,v\right\rangle \coloneqq\alpha_{1}\beta_{1}+\cdots+\alpha_{n}\beta_{n}$
\end_inset
.
Obstaja tudi druga oznaka in pripadajoča drugačna definicija
\begin_inset Formula $u\cdot v\coloneqq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi$
\end_inset
,
kjer je
\begin_inset Formula $\varphi$
\end_inset
kot med
\begin_inset Formula $u,v$
\end_inset
.
\end_layout
\begin_layout Claim*
Velja
\begin_inset Formula $\left\langle u,v\right\rangle =u\cdot v$
\end_inset
.
\end_layout
\begin_layout Proof
Uporabimo kosinusni izrek,
ki pravi,
da v trikotniku s stranicami dolžin
\begin_inset Formula $a,b,c$
\end_inset
velja
\begin_inset Formula $c^{2}=a^{2}+b^{2}-2ab\cos\varphi$
\end_inset
,
kjer je
\begin_inset Formula $\varphi$
\end_inset
kot med
\begin_inset Formula $b$
\end_inset
in
\begin_inset Formula $c$
\end_inset
.
Za vektorja
\begin_inset Formula $v$
\end_inset
in
\begin_inset Formula $u$
\end_inset
z vmesnim kotom
\begin_inset Formula $\varphi$
\end_inset
torej velja
\begin_inset Formula
\[
\left|\left|u-v\right|\right|^{2}=\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi.
\]
\end_inset
Obenem velja
\begin_inset Formula $\left|\left|u\right|\right|^{2}=\alpha_{1}^{2}+\cdots+\alpha_{n}^{2}=\left\langle u,u\right\rangle $
\end_inset
,
torej lahko zgornjo enačbo prepišemo v
\begin_inset Formula
\[
\left\langle u-v,u-v\right\rangle =\left\langle u,u\right\rangle +\left\langle v,v\right\rangle -2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi.
\]
\end_inset
Naj bo
\begin_inset Formula $w=u,v$
\end_inset
.
Iz prihodnosti si izposodimo obe linearnosti in simetričnost.
\begin_inset Formula
\[
\left\langle u-v,u-v\right\rangle =\left\langle u-v,w\right\rangle =\left\langle u,w\right\rangle -\left\langle v,w\right\rangle =\left\langle u,u-v\right\rangle -\left\langle v,u-v\right\rangle =\left\langle u,u\right\rangle -\left\langle u,v\right\rangle -\left\langle v,u\right\rangle +\left\langle v,v\right\rangle
\]
\end_inset
Prišli smo do enačbe
\begin_inset Formula
\[
\cancel{\left\langle u,u\right\rangle }-2\left\langle u,v\right\rangle +\cancel{\left\langle v,v\right\rangle }=\cancel{\left\langle u,u\right\rangle }+\cancel{\left\langle v,v\right\rangle }-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi\quad\quad\quad\quad/:-2
\]
\end_inset
\begin_inset Formula
\[
\left\langle u,v\right\rangle =\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi.
\]
\end_inset
\end_layout
\begin_layout Claim*
Paralelogramska identiteta.
\begin_inset Formula $\left|\left|u+v\right|\right|^{2}+\left|\left|u-v\right|\right|^{2}=2\left|\left|u\right|\right|^{2}+2\left|\left|v\right|\right|^{2}$
\end_inset
ZDB vsota kvadratov dolžin obeh diagonal je enota vsoti kvadratov dolžin vseh štirih stranic.
\end_layout
\begin_layout Proof
\begin_inset Formula
\[
\left|\left|u+v\right|\right|^{2}=\left\langle u+v,u+v\right\rangle =\left\langle u,u+v\right\rangle +\left\langle v,u+v\right\rangle =\left\langle u,u\right\rangle +\left\langle u,v\right\rangle +\left\langle v,u\right\rangle +\left\langle v,v\right\rangle
\]
\end_inset
\begin_inset Formula
\[
\left|\left|u-v\right|\right|^{2}=\left\langle u-v,u-v\right\rangle =\left\langle u,u-v\right\rangle -\left\langle v,u-v\right\rangle =\left\langle u,u\right\rangle -\left\langle u,v\right\rangle -\left\langle v,u\right\rangle +\left\langle v,v\right\rangle
\]
\end_inset
\begin_inset Formula
\[
\left|\left|u+v\right|\right|^{2}+\left|\left|u-v\right|\right|^{2}=2\left\langle u,u\right\rangle +2\left\langle v,v\right\rangle =2\left|\left|u\right|\right|^{2}+2\left|\left|v\right|\right|^{2}
\]
\end_inset
\end_layout
\begin_layout Claim*
Cauchy-Schwarzova neenakost.
\begin_inset Formula $\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$
\end_inset
\end_layout
\begin_layout Proof
\begin_inset Formula $\left|\left\langle u,v\right\rangle \right|=\left|\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi\right|=\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\left|\cos\varphi\right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$
\end_inset
,
kajti
\begin_inset Formula $\left|\cos\varphi\right|\in\left[0,1\right]$
\end_inset
.
\end_layout
\begin_layout Claim*
Trikotniška neenakost.
\begin_inset Formula $\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$
\end_inset
\end_layout
\begin_layout Proof
Sledi iz Cauchy-Schwarzove.
Velja
\begin_inset Formula
\[
-\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\leq\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\quad\quad\quad\quad/\cdot2
\]
\end_inset
\begin_inset Formula
\[
-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\leq2\left|\left\langle u,v\right\rangle \right|\leq2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\quad\quad\quad\quad/+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}
\]
\end_inset
\begin_inset Formula
\[
-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}\leq\cancel{2\left|\left\langle u,v\right\rangle \right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}\leq}2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}
\]
\end_inset
uporabimo kosinusni izrek na levi strani enačbe,
desno pa zložimo v kvadrat:
\begin_inset Formula
\[
\left|\left|u+v\right|\right|^{2}\leq\left(\left|\left|u\right|\right|+\left|\left|v\right|\right|\right)^{2}\quad\quad\quad\quad/\sqrt{}
\]
\end_inset
\begin_inset Formula
\[
\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|
\]
\end_inset
\end_layout
\begin_layout Claim*
Za neničelna vektorja velja
\begin_inset Formula $u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$
\end_inset
.
\end_layout
\begin_layout Proof
\begin_inset Formula $\left\langle u,v\right\rangle =u\cdot v=\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi$
\end_inset
,
kar je 0
\begin_inset Formula $\Leftrightarrow\varphi=\pi=90°$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Vektorski in mešani produkt
\end_layout
\begin_layout Standard
Definirana sta le za vektorje v
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset
.
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $u=\left(\alpha_{1},\alpha_{2},\alpha_{3}\right),v=\left(\beta_{1},\beta_{2},\beta_{3}\right)$
\end_inset
.
\begin_inset Formula $u\times v=\left(\alpha_{2}\beta_{3}-\alpha_{3}\beta_{2},\alpha_{3}\beta_{1}-\alpha_{1}\beta_{3},\alpha_{1}\beta_{2}-\alpha_{2}\beta_{1}\right)$
\end_inset
.
\end_layout
\begin_layout Paragraph
Geometrijski pomen
\end_layout
\begin_layout Standard
Vektor
\begin_inset Formula $u\times v$
\end_inset
je pravokoten na
\begin_inset Formula $u$
\end_inset
in
\begin_inset Formula $v$
\end_inset
,
njegova dolžina je
\begin_inset Formula $\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\sin\varphi$
\end_inset
,
kar je ploščina paralelograma,
ki ga oklepata
\begin_inset Formula $u$
\end_inset
in
\begin_inset Formula $v$
\end_inset
.
\end_layout
\begin_layout Standard
Pravilo desnega vijaka nam je v pomoč pri doložanju usmeritve vektorskega produkta.
Če iztegnjen kazalec desne roke predstavlja
\begin_inset Formula $u$
\end_inset
in iztegnjen sredinec
\begin_inset Formula $v$
\end_inset
,
iztegnjen palec kaže v smeri
\begin_inset Formula $u\times v$
\end_inset
.
\end_layout
\begin_layout Claim*
Lagrangeva identiteta.
\begin_inset Formula $\left|\left|u\times v\right|\right|+\left\langle u,v\right\rangle ^{2}=\left|\left|u\right|\right|^{2}\cdot\left|\left|v\right|\right|^{2}$
\end_inset
\begin_inset Note Note
status open
\begin_layout Plain Layout
DOKAZ???????
\end_layout
\end_inset
\end_layout
\begin_layout Definition*
Mešani produkt vektorjev
\begin_inset Formula $u,v,w$
\end_inset
je skalar
\begin_inset Formula $\left\langle u\times v,w\right\rangle $
\end_inset
.
Oznaka:
\begin_inset Formula $\left[u,v,w\right]=\left\langle u\times v,w\right\rangle $
\end_inset
.
\end_layout
\begin_layout Paragraph*
Geometrijski pomen
\end_layout
\begin_layout Standard
Volumen paralelpipeda,
ki ga določajo
\begin_inset Formula $u,v,w$
\end_inset
.
Razlaga:
\begin_inset Formula $\left[u,v,w\right]=\left\langle u\times v,w\right\rangle =\left|\left|u\times v\right|\right|\cdot\left|\left|w\right|\right|\cdot\cos\varphi$
\end_inset
;
\begin_inset Formula $\left|\left|u\times v\right|\right|$
\end_inset
je namreč ploščina osnovne ploskve,
\begin_inset Formula $\left|\left|w\right|\right|\cdot\cos\varphi$
\end_inset
pa je višina paralelpipeda.
\end_layout
\begin_layout Claim*
Osnovne lastnosti vektorskega produkta so
\begin_inset Formula $u\times u=0$
\end_inset
,
\begin_inset Formula $u\times v=-\left(v\times u\right)$
\end_inset
,
\begin_inset Formula $\left(\alpha u+\beta v\right)\times w=\alpha\left(u\times w\right)+\beta\left(v\times w\right)$
\end_inset
(linearnost)
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Claim*
Osnovne lastnosti mešanega produkta so linearnost v vsakem faktorju,
menjava dveh faktorjev spremeni predznak (
\begin_inset Formula $\left[u,v,w\right]=-\left[v,u,w\right]$
\end_inset
),
cikličen pomik ne spremeni vrednosti (
\begin_inset Formula $\left[u,v,w\right]=\left[v,w,u\right]=\left[w,u,v\right]$
\end_inset
).
\end_layout
\begin_layout Subsubsection
Premica v
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
\end_layout
\begin_layout Standard
Premico lahko podamo z
\end_layout
\begin_layout Itemize
dvema različnima točkama
\end_layout
\begin_layout Itemize
s točko
\begin_inset Formula $\vec{r_{0}}$
\end_inset
in neničelnim smernim vektorjem
\begin_inset Formula $\vec{p}$
\end_inset
.
Premica je tako množica točk
\begin_inset Formula $\left\{ \vec{r}=\vec{r_{0}}+t\vec{p};\forall t\in\mathbb{R}\right\} $
\end_inset
.
Taki enačbi premice rečemo parametrična.
\end_layout
\begin_layout Itemize
s točko in normalo (v
\begin_inset Formula $\mathbb{R}^{2}$
\end_inset
;
v
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
potrebujemo točko in
\begin_inset Formula $n-1$
\end_inset
normal)
\end_layout
\begin_layout Standard
Nadaljujmo s parametričnim zapisom
\begin_inset Formula $\vec{r}=\vec{r_{0}}+t\vec{p}$
\end_inset
.
Če točke zapišemo po komponentah,
dobimo parametrično enačbo premice po komponentah:
\begin_inset Formula $\left(x,y,z\right)=\left(x_{0},y_{0},z_{0}\right)+t\left(p_{1},p_{2},p_{3}\right)$
\end_inset
.
\begin_inset Formula
\[
x=x_{0}+tp_{1}
\]
\end_inset
\begin_inset Formula
\[
y=y_{0}+tp_{2}
\]
\end_inset
\begin_inset Formula
\[
z=z_{0}+tp_{3}
\]
\end_inset
\end_layout
\begin_layout Standard
Sedaj lahko iz vsake enačbe izrazimo
\begin_inset Formula $t$
\end_inset
in dobimo normalno enačbo premice v
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
:
\begin_inset Formula
\[
t=\frac{x-x_{0}}{p_{1}}=\frac{y-y_{0}}{p_{2}}=\frac{z-z_{0}}{p_{3}}\text{, oziroma v splošnem za premico v \ensuremath{\mathbb{R}^{n}}: }t=\frac{x_{1_{0}}-x_{1}}{p_{1}}=\cdots=\frac{x_{n_{0}}-x_{n}}{p_{n}}
\]
\end_inset
\end_layout
\begin_layout Standard
Osnovne naloge s premicami so projekcija točke na premico,
zrcaljenje točke čez premico in razdalja med točko in premico.
\end_layout
\begin_layout Paragraph*
Iskanje projekcije dane točke na dano premico
\end_layout
\begin_layout Standard
(skica prepuščena bralcu)
\begin_inset Formula $\vec{r_{1}}$
\end_inset
projiciramo na
\begin_inset Formula $\vec{r}=\vec{r_{0}}+t\vec{p}$
\end_inset
in dobimo
\begin_inset Formula $\vec{r_{1}'}$
\end_inset
.
Za
\begin_inset Formula $\vec{r_{1}'}$
\end_inset
vemo,
da leži na premici,
torej
\begin_inset Formula $\exists t\in\mathbb{R}\ni:\vec{r_{1}'}=\vec{r_{0}}+t\vec{p}$
\end_inset
.
Poleg tega vemo,
da je
\begin_inset Formula $\vec{r_{1}'}-\vec{r_{1}}$
\end_inset
pravokoten na premico oz.
njen smerni vektor
\begin_inset Formula $\vec{p}$
\end_inset
,
torej
\begin_inset Formula $\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{p}\right\rangle =0$
\end_inset
.
Ti dve enačbi združimo,
da dobimo
\begin_inset Formula $t$
\end_inset
,
ki ga nato vstavimo v prvo enačbo:
\begin_inset Formula
\[
\left\langle \vec{r_{0}}+t\vec{p}-\vec{r_{1},}\vec{p}\right\rangle =0\Longrightarrow\left\langle \vec{r_{0}},\vec{p}\right\rangle +t\left\langle \vec{p},\vec{p}\right\rangle -\left\langle \vec{r_{1}},\vec{p}\right\rangle =0\Longrightarrow t=\frac{\left\langle \vec{r_{1}},\vec{p}\right\rangle -\left\langle \vec{r_{0}},\vec{p}\right\rangle }{\left\langle \vec{p},\vec{p}\right\rangle }
\]
\end_inset
\begin_inset Formula
\[
\vec{r_{1}'}=\vec{r_{0}}+t\vec{p}=\vec{r_{0}}+\frac{\left\langle \vec{r_{1}},\vec{p}\right\rangle -\left\langle \vec{r_{0}},\vec{p}\right\rangle }{\left\langle \vec{p},\vec{p}\right\rangle }\vec{p}
\]
\end_inset
\end_layout
\begin_layout Standard
Spotoma si lahko izpišemo obrazec za oddaljenost točke od premice:
\begin_inset Formula $a=\left|\left|\vec{r_{1}'}-\vec{r_{1}}\right|\right|$
\end_inset
in obrazec za zrcalno sliko (
\begin_inset Formula $\vec{r_{1}''}$
\end_inset
):
\begin_inset Formula $\vec{r_{1}'}=\frac{\vec{r_{1}''}+\vec{r_{1}}}{2}\Longrightarrow\vec{r_{1}''}=2\vec{r_{1}'}-\vec{r_{1}}$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Ravnine v
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
\end_layout
\begin_layout Standard
Ravnino lahko podamo
\end_layout
\begin_layout Itemize
s tremi nekolinearnimi točkami
\end_layout
\begin_layout Itemize
s točko na ravnini in dvema neničelnima smernima vektorjema,
ki sta linarno neodvisna.
Ravnina je tako množica točk
\begin_inset Formula $\left\{ \vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q};\forall s,t\in\mathbb{R}\right\} $
\end_inset
.
Taki enačbi ravnine rečemo parametrična.
\end_layout
\begin_layout Itemize
s točko in na ravnini in normalo (v
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset
;
v
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
poleg točke potrebujemo
\begin_inset Formula $n-2$
\end_inset
normal)
\end_layout
\begin_layout Standard
Nadaljujmo s parametričnim zapisom
\begin_inset Formula $\vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q}$
\end_inset
.
Če točke zapišemo po komponentah,
dobimo parametrično enačbo ravnine po komponentah:
\begin_inset Formula $\left(x,y,z\right)=\left(x_{0},y_{0},z_{0}\right)+s\left(p_{1},p_{2},p_{3}\right)+t\left(q_{1},q_{2},q_{3}\right)$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
x=x_{0}+sp_{1}+tq_{1}
\]
\end_inset
\begin_inset Formula
\[
y=y_{0}+sp_{2}+tq_{2}
\]
\end_inset
\begin_inset Formula
\[
z=y_{0}+sp_{3}+tq_{3}
\]
\end_inset
\end_layout
\begin_layout Paragraph
Normalna enačba ravnine v
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset
\end_layout
\begin_layout Standard
(skica prepuščena bralcu) Vemo,
da je
\begin_inset Formula $\vec{n}$
\end_inset
(normala) pravokotna na vse vektorje v ravnini,
tudi na
\begin_inset Formula $\vec{r}-\vec{r_{0}}$
\end_inset
za poljuben
\begin_inset Formula $\vec{r}$
\end_inset
na ravnini.
Velja torej normalna enačba ravnine:
\begin_inset Formula $\left\langle \vec{r}-\vec{r_{0}},\vec{n}\right\rangle =0$
\end_inset
.
Razpišimo jo po komponentah,
da na koncu dobimo normalno enačbo ravnine po komponentah:
\begin_inset Formula
\[
\left\langle \left(x,y,z\right)-\left(x_{0},y_{0},z_{0}\right),\left(n_{1},n_{2},n_{3}\right)\right\rangle =0=\left\langle \left(x-x_{0},y-y_{0},z-z_{0}\right),\left(n_{1},n_{2},n_{3}\right)\right\rangle
\]
\end_inset
\begin_inset Formula
\[
n_{1}\left(x-x_{0}\right)+n_{2}\left(y-y_{0}\right)+n_{3}\left(z-z_{0}\right)=0=n_{1}x-n_{1}x_{0}+n_{2}y-n_{2}y_{0}+n_{3}z-n_{3}z_{0}=0
\]
\end_inset
\begin_inset Formula
\[
n_{1}x+n_{2}y+n_{3}z=n_{1}x_{0}+n_{2}y_{0}+n_{3}z_{0}=d
\]
\end_inset
\end_layout
\begin_layout Paragraph
Iskanje pravokotne projekcije dane točke na dano ravnino
\end_layout
\begin_layout Standard
(skica prepuščena bralcu) Projicirati želimo
\begin_inset Formula $\vec{r_{1}}$
\end_inset
v
\begin_inset Formula $\vec{r_{1}'}$
\end_inset
na ravnini
\begin_inset Formula $\vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q}$
\end_inset
.
Vemo,
da
\begin_inset Formula $\vec{r_{1}'}$
\end_inset
leži na ravnini,
zato
\begin_inset Formula $\exists s,t\in\mathbb{R}\ni:\vec{r_{1}'}=\vec{r_{0}}+s\vec{p}+t\vec{q}$
\end_inset
.
Poleg tega vemo,
da je
\begin_inset Formula $\vec{r_{1}'}-\vec{r_{1}}$
\end_inset
pravokoten na ravnino oz.
na
\begin_inset Formula $\vec{p}$
\end_inset
in na
\begin_inset Formula $\vec{q}$
\end_inset
hkrati,
torej
\begin_inset Formula $\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{q}\right\rangle $
\end_inset
.
Vstavimo
\begin_inset Formula $\vec{r_{1}'}$
\end_inset
iz prve enačbe v drugo in dobimo
\begin_inset Formula
\[
\left\langle \vec{r_{0}}+s\vec{p}+t\vec{q}-\vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{0}}+s\vec{p}+t\vec{q}-\vec{r_{1}},\vec{q}\right\rangle
\]
\end_inset
\begin_inset Formula
\[
\left\langle \vec{r_{0}},\vec{p}\right\rangle +s\left\langle \vec{p},\vec{p}\right\rangle +t\left\langle \vec{q},\vec{p}\right\rangle -\left\langle \vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{0}},\vec{q}\right\rangle +s\left\langle \vec{p},\vec{q}\right\rangle +t\left\langle \vec{q},\vec{q}\right\rangle -\left\langle \vec{r_{1}},\vec{q}\right\rangle
\]
\end_inset
dobimo sistem dveh enačb
\begin_inset Formula
\[
s\left\langle \vec{p},\vec{p}\right\rangle +t\left\langle \vec{q},\vec{p}\right\rangle =\left\langle \vec{r_{1}}-\vec{r_{0}},\vec{p}\right\rangle
\]
\end_inset
\begin_inset Formula
\[
s\left\langle \vec{p},\vec{q}\right\rangle +t\left\langle \vec{q},\vec{q}\right\rangle =\left\langle \vec{r_{1}}-\vec{r_{0}},\vec{q}\right\rangle
\]
\end_inset
sistem rešimo in dobljena
\begin_inset Formula $s,t$
\end_inset
vstavimo v prvo enačbo zgoraj,
da dobimo
\begin_inset Formula $\vec{r_{1}'}$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Regresijska premica
\end_layout
\begin_layout Standard
Regresijska premica je primer uporabe zgornje naloge.
V ravnini je danih
\begin_inset Formula $n$
\end_inset
točk
\begin_inset Formula $\left(x_{1},y_{1}\right),\dots,\left(x_{n},y_{n}\right)$
\end_inset
.
Iščemo tako premico
\begin_inset Formula $y=ax+b$
\end_inset
,
ki se najbolj prilega tem točkam.
Prileganje premice točkam merimo z metodo najmanjših kvadratov:
naj bo
\begin_inset Formula $d_{i}$
\end_inset
navpična razdalja med
\begin_inset Formula $\left(x_{i},y_{i}\right)$
\end_inset
in premico
\begin_inset Formula $y=ax+b$
\end_inset
,
torej razdalja med točkama
\begin_inset Formula $\left(x_{i},y_{i}\right)$
\end_inset
in
\begin_inset Formula $\left(x_{i},ax_{i}+b\right)$
\end_inset
,
kar je
\begin_inset Formula $\left|y_{i}-ax_{i}-b\right|$
\end_inset
.
Minimizirati želimo vsoto kvadratov navpičnih razdalj,
torej izraz
\begin_inset Formula $d_{1}^{2}+\cdots+d_{n}^{2}=\left(y_{1}-ax_{1}-b\right)^{2}+\cdots+\left(y_{n}-ax_{n}-b\right)^{2}=\left|\left|\left(y_{1}-ax_{1}-b,\dots,y_{n}-ax_{n}-b\right)\right|\right|^{2}=\left|\left|\left(y_{1},\dots,y_{n}\right)-a\left(x_{1},\dots,x_{n}\right)-b\left(1,\dots,1\right)\right|\right|^{2}$
\end_inset
.
\end_layout
\begin_layout Standard
Če je torej
\begin_inset Formula $\vec{r}=\vec{0}+a\left(x_{1},\dots,x_{n}\right)+b\left(1,\dots,1\right)$
\end_inset
hiperravnina v
\begin_inset Formula $n-$
\end_inset
dimenzionalnem prostoru,
bo norma,
ki jo želimo minimizirati,
najmanjša tedaj,
ko
\begin_inset Formula $a,b$
\end_inset
izberemo tako,
da najdemo projekcijo
\begin_inset Formula $\left(y_{1},\dots,y_{n}\right)$
\end_inset
na to hiperravnino (skica prepuščena bralcu).
Rešimo sedaj nalogo projekcije točke na ravnino:
\end_layout
\begin_layout Standard
Označimo
\begin_inset Formula $\vec{y}\coloneqq\left(y_{1},\dots,y_{n}\right)$
\end_inset
,
\begin_inset Formula $\vec{x}\coloneqq\left(x_{1},\dots,x_{n}\right)$
\end_inset
,
\begin_inset Formula $\vec{1}=\left(1,\dots,1\right)$
\end_inset
.
Vemo,
da
\begin_inset Formula $\vec{y}-a\vec{x}-b\vec{1}\perp\vec{x},\vec{1}$
\end_inset
,
torej
\begin_inset Formula $\left\langle \vec{y}-a\vec{x}-b\vec{1},\vec{x}\right\rangle =0=\left\langle \vec{y}-a\vec{x}-b\vec{1},\vec{1}\right\rangle $
\end_inset
in dobimo sistem enačb
\begin_inset Formula
\[
\left\langle \vec{y},\vec{x}\right\rangle =a\left\langle \vec{x},\vec{x}\right\rangle +b\left\langle \vec{1},\vec{x}\right\rangle
\]
\end_inset
\begin_inset Formula
\[
\left\langle \vec{y},\vec{1}\right\rangle =a\left\langle \vec{x},\vec{1}\right\rangle +b\left\langle \vec{1},\vec{1}\right\rangle .
\]
\end_inset
V sistem sedaj vstavimo definicije točk
\begin_inset Formula $\left(x_{i},y_{i}\right)$
\end_inset
in ga nato delimo s številom točk,
da dobimo sistem s povprečji,
ki ga nato rešimo (izluščimo
\begin_inset Formula $a,b$
\end_inset
):
\begin_inset Formula
\[
\sum_{i=1}^{n}y_{i}x_{i}=a\sum_{i=i}^{n}x_{i}^{2}+b\sum_{i=1}^{n}x_{i}\quad\quad\quad\quad/:n
\]
\end_inset
\begin_inset Formula
\[
\sum_{i=1}^{n}y_{i}=a\sum_{i=1}^{n}x_{i}+b\sum_{i=1}^{n}1=a\sum_{i=1}^{n}x_{i}+bn\quad\quad\quad\quad/:n
\]
\end_inset
\begin_inset Formula
\[
\overline{yx}=a\overline{x^{2}}+b\overline{x}
\]
\end_inset
\begin_inset Formula
\[
\overline{y}=a\overline{x}+b\Longrightarrow\overline{y}-a\overline{x}=b
\]
\end_inset
\begin_inset Formula
\[
\overline{yx}=a\overline{x^{2}}+\left(\overline{y}-a\overline{x}\right)\overline{x}=a\overline{x^{2}}+\overline{y}\cdot\overline{x}-a\overline{x}^{2}\Longrightarrow a\left(\overline{x^{2}}-\overline{x}^{2}\right)=\overline{yx}-\overline{y}\cdot\overline{x}\Longrightarrow a=\frac{\overline{yx}-\overline{y}\cdot\overline{x}}{\overline{x^{2}}-\overline{x}^{2}}
\]
\end_inset
\end_layout
\begin_layout Subsection
Sistemi linearnih enačb
\end_layout
\begin_layout Standard
Ta sekcija,
z izjemo prve podsekcije,
je precej dobesedno povzeta po profesorjevi beamer skripti.
\end_layout
\begin_layout Subsubsection
Linearna enačba
\end_layout
\begin_layout Definition*
\begin_inset Formula $\sim$
\end_inset
je enačba oblike
\begin_inset Formula $a_{1}x_{1}+\cdots+a_{n}x_{n}=b$
\end_inset
in vsebuje koeficiente,
spremenljivke in desno stran.
Množica rešitev so vse
\begin_inset Formula $n-$
\end_inset
terice realnih števil,
ki zadoščajo enačbi
\begin_inset Formula $R=\left\{ \left(x_{1},\dots,x_{n}\right)\in\mathbb{R}^{n};a_{1}x_{1}+\cdots+a_{n}x_{n}=b\right\} $
\end_inset
.
Če so vsi koeficienti 0,
pravimo,
da je enačba trivialna,
sicer (torej čim je en koeficient neničeln) je netrivialna.
\end_layout
\begin_layout Remark*
Za trivialno enačbo velja
\begin_inset Formula $R=\begin{cases}
\emptyset & ;b\not=0\\
\mathbb{R}^{n} & ;b=0
\end{cases}$
\end_inset
.
Za netrivialno enačbo pa velja
\begin_inset Formula $a_{i}\not=0$
\end_inset
,
torej:
\begin_inset Formula
\[
a_{1}x_{1}+\cdots+a_{i}x_{i}+\cdots+a_{n}x_{n}=b
\]
\end_inset
\begin_inset Formula
\[
a_{1}x_{1}+\cdots+a_{i-1}x_{i-1}+a_{i+1}x_{i+1}+\cdots+a_{n}x_{n}=b-a_{i}x_{i}=-a_{i}\left(x_{i}-\frac{b}{a_{i}}\right)
\]
\end_inset
\begin_inset Formula
\[
a_{1}x_{1}+\cdots+a_{i-1}x_{i-1}+a_{i}\left(x_{i}-\frac{b}{a_{i}}\right)+a_{i+1}x_{i+1}+\cdots+a_{n}x_{n}=0
\]
\end_inset
\begin_inset Formula
\[
\left\langle \left(a_{i},\dots,a_{n}\right),\left(x_{1},\dots,x_{i-1},x_{i}-\frac{b}{a_{i}},x_{i+1},\dots,x_{n}\right)\right\rangle =0=\left\langle \left(a_{i},\dots,a_{n}\right),\left(x_{1},\dots,x_{i},\dots,x_{n}\right)-\left(0,\dots,0,\frac{b}{a},0,\dots,0\right)\right\rangle
\]
\end_inset
Tu lahko označimo
\begin_inset Formula $\vec{n}\coloneqq\left(a_{i},\dots,a_{n}\right)$
\end_inset
,
\begin_inset Formula $\vec{r}=\left(x_{1},\dots,x_{i},\dots,x_{n}\right)$
\end_inset
,
\begin_inset Formula $\vec{r_{0}}=\left(0,\dots,0,\frac{b}{a},0,\dots,0\right)$
\end_inset
in dobimo
\begin_inset Formula $\left\langle \vec{n},\vec{r}-\vec{r_{0}}\right\rangle $
\end_inset
,
kar je normalna enačba premice v
\begin_inset Formula $\mathbb{R}^{2}$
\end_inset
,
normalna enačba ravnine v
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset
oziroma normalna enačba hiperravnine v
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Sistem linearnih enačb
\end_layout
\begin_layout Definition*
Sistem
\begin_inset Formula $m$
\end_inset
linearnih enačb z
\begin_inset Formula $n$
\end_inset
spremenljivkami je sistem enačb oblike
\begin_inset Formula
\[
\begin{array}{ccccccc}
a_{1,1}x_{1} & + & \cdots & + & a_{1,n}x_{n} & = & b_{1}\\
\vdots & & & & \vdots & & \vdots\\
a_{m,1}x_{1} & + & \cdots & + & a_{m,n}x_{n} & = & b_{m}
\end{array}.
\]
\end_inset
\end_layout
\begin_layout Fact*
Množica rešitev sistema je
\begin_inset Formula $\mathbb{R}^{n}\Leftrightarrow\forall i,j:a_{i,j}=b_{i}=0$
\end_inset
.
Sicer je množica rešitev presek hiperravnin v
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
—
rešitev posameznih enačb.
To vključuje tudi primer prazne množice rešitev,
saj je takšna na primer presek dveh vzporednih hiperravnin.
\end_layout
\begin_layout Example*
Množica rešitev
\begin_inset Formula $2\times2$
\end_inset
sistema je lahko
\end_layout
\begin_layout Itemize
cela ravnina
\end_layout
\begin_layout Itemize
premica v ravnini
\end_layout
\begin_layout Itemize
točka v ravnini
\end_layout
\begin_layout Itemize
prazna množica
\end_layout
\begin_layout Remark*
Enako velja za množico rešitev
\begin_inset Formula $3\times2$
\end_inset
sistema.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
Množica rešitev sistema
\begin_inset Formula $2\times3$
\end_inset
pa ne more biti točka v prostoru,
lahko pa je cel prostor,
ravnina v prostoru,
premica v prostoru ali prazna množica.
\end_layout
\begin_layout Paragraph*
Algebraičen pomen rešitev sistema
\end_layout
\begin_layout Standard
Rešitve sistema
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\begin{array}{ccccccc}
a_{1,1}x_{1} & + & \cdots & + & a_{1,n}x_{n} & = & b_{1}\\
\vdots & & & & \vdots & & \vdots\\
a_{m,1}x_{1} & + & \cdots & + & a_{m,n}x_{n} & = & b_{m}
\end{array}
\]
\end_inset
\end_layout
\begin_layout Standard
lahko zapišemo kot linearno kombinacijo stolpecv sistema in spremenljivk:
\begin_inset Formula
\[
\left(b_{1},\dots,b_{n}\right)=\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)=x_{1}\left(a_{1,1},\dots,a_{m,1}\right)+\cdots+x_{n}\left(a_{1,n},\dots,a_{m,n}\right)
\]
\end_inset
\begin_inset Formula
\[
\vec{b}=x_{1}\vec{a_{1}}+\cdots+x_{n}\vec{a_{n}}
\]
\end_inset
\end_layout
\begin_layout Subsubsection
Klasifikacija sistemov linearnih enačb
\end_layout
\begin_layout Standard
Sisteme linearnih enačb delimo glede na velikost na
\end_layout
\begin_layout Itemize
kvadratne (toliko enačb kot spremenljivk),
\end_layout
\begin_layout Itemize
poddoločene (več spremenljivk kot enačb),
\end_layout
\begin_layout Itemize
predoločene (več enačb kot spremenljivk);
\end_layout
\begin_layout Standard
glede na rešljivost na
\end_layout
\begin_layout Itemize
nerešljive (prazna množica rešitev),
\end_layout
\begin_layout Itemize
enolično rešljive (množica rešitev je singleton),
\end_layout
\begin_layout Itemize
neenolično rešljive (moč množice rešitev je več kot 1);
\end_layout
\begin_layout Standard
glede na obliko desnih strani na
\end_layout
\begin_layout Itemize
homogene (vektor desnih stani je ničeln)
\end_layout
\begin_layout Itemize
nehomogene (vektor desnih strani je neničen)
\end_layout
\begin_layout Remark*
Če sta
\begin_inset Formula $\vec{x}$
\end_inset
in
\begin_inset Formula $\vec{y}$
\end_inset
dve različni rešitvi sistema,
je rešitev sistema tudi
\begin_inset Formula $\left(1-t\right)\vec{x}+t\vec{y}$
\end_inset
za vsak realen
\begin_inset Formula $t$
\end_inset
,
torej ima vsak neenolično rešljiv sistem neskončno rešitev.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
Pogosto (a nikakor ne vedno) se zgodi,
da je kvadraten sistem enolično rešljiv,
predoločen sistem nerešljiv,
poddoločen sistem pa neenolično rešljiv.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
Homogen sistem je vedno rešljiv,
saj obstaja trivialna rešitev
\begin_inset Formula $\vec{0}$
\end_inset
.
Vprašanje pri homogenih sistemih je torej,
kdaj je enolično in kdaj neenolično rešljiv.
Dokazali bomo,
da je vsak poddoločen homogen sistem linearnih enačb neenolično rešljiv.
\end_layout
\begin_layout Subsubsection
Reševanje sistema
\end_layout
\begin_layout Standard
Sisteme lahko rešujemo z izločanjem spremenljivk.
Iz ene enačbe izrazimo spremenljivko in jo vstavimo v druge enačbe,
da izrazimo zopet nove spremenljivke,
ki jih spet vstavimo v nove enačbe,
iz katerih spremenljivk še nismo izražali in tako naprej,
vse dokler ne pridemo do zadnjega možnega izražanja (dodatno branje prepuščeno bralcu).
\end_layout
\begin_layout Standard
Sisteme pa lahko rešujemo tudi z Gaussovo metodo.
Trdimo,
da se rešitev sistema ne spremeni,
če na njem uporabimo naslednje elementarne vrstične transformacije:
\end_layout
\begin_layout Itemize
menjava vrstnega reda enačb,
\end_layout
\begin_layout Itemize
množenje enačbe z neničelno konstanto,
\end_layout
\begin_layout Itemize
prištevanje večkratnika ene enačbe k drugi.
\end_layout
\begin_layout Standard
Z Gaussovo metodo (dodatno branje prepuščeno bralcu) mrcvarimo razširjeno matriko sistema,
dokler ne dobimo reducirane kvadratne stopničaste forme (angl.
row echelon),
ki izgleda takole (
\begin_inset Formula $\times$
\end_inset
reprezentira poljubno realno številko,
\begin_inset Formula $0$
\end_inset
ničlo in
\begin_inset Formula $1$
\end_inset
enico):
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\left[\begin{array}{ccccccccccccccccc|c}
0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \cdots & \times\\
\vdots & & \vdots & 0 & 0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \cdots & \times\\
& & & \vdots & \vdots & & \vdots & 0 & 0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \cdots & \times\\
& & & & & & & \vdots & \vdots & & \vdots & 0 & 0 & \cdots & 0 & 1 & \cdots & \times\\
& & & & & & & & & & & \vdots & \vdots & & \vdots & 0 & \cdots & \vdots\\
\vdots & & \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots & & \vdots & \vdots & & \vdots\\
0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & \cdots & \times
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Subsubsection
Homogeni sistemi
\end_layout
\begin_layout Definition*
Sistem je homogen,
če je vektor desnih strani ničeln.
\end_layout
\begin_layout Standard
Vedno ima rešitev
\begin_inset Formula $\vec{0}$
\end_inset
.
Linearna kombinacija dveh rešitev homogenega sistema je spet njegova rešitev.
Splošna rešitev nehomogenega sistema je vsota partikularne rešitve tega nehomogenega sistema in splošne rešitve njemu prirejenega homogenega sistema.
\end_layout
\begin_layout Remark*
V tem razdelku nehomogen sistem pomeni nenujno homogen sistem (torej splošen sistem linearnih enačb),
torej je vsak homogen sistem nehomogen.
\end_layout
\begin_layout Claim
\begin_inset CommandInset label
LatexCommand label
name "claim:Vpoddol-hom-sist-ima-ne0-reš"
\end_inset
Vsak poddoločen homogen sistem ima vsaj eno netrivialno rešitev.
\end_layout
\begin_layout Proof
Dokaz z indukcijo po številu enačb.
\end_layout
\begin_deeper
\begin_layout Paragraph*
Baza
\end_layout
\begin_layout Standard
\begin_inset Formula $a_{1}x_{1}+\cdots+a_{n}x_{n}=0$
\end_inset
za
\begin_inset Formula $n\geq2$
\end_inset
.
Če je
\begin_inset Formula $a_{n}=0$
\end_inset
,
je netrivialna rešitev
\begin_inset Formula $\left(0,\dots,0,1\right)$
\end_inset
,
sicer pa
\begin_inset Formula $\left(0,\dots,0,-a_{n},a_{n-1}\right)$
\end_inset
.
\end_layout
\begin_layout Paragraph*
Korak
\end_layout
\begin_layout Standard
Denimo,
da velja za vse poddoločene homogene sisteme z
\begin_inset Formula $m-1$
\end_inset
vrsticami.
Vzemimo poljuben homogen sistem z
\begin_inset Formula $n>m$
\end_inset
stolpci (da je poddoločen).
Če je
\begin_inset Formula $a_{n}=0$
\end_inset
,
je netriviačna rešitev
\begin_inset Formula $\left(0,\dots,0,1\right)$
\end_inset
,
sicer pa iz ene od enačb izrazimo
\begin_inset Formula $x_{n}$
\end_inset
s preostalimi spremenljivkami.
Dobljen izraz vstavimo v preostalih
\begin_inset Formula $m-1$
\end_inset
enačb z
\begin_inset Formula $n-1$
\end_inset
spremenljivkami in dobljen sistem uredimo.
Po I.
P.
ima slednji netrivialno rešitev
\begin_inset Formula $\left(\alpha_{1},\dots,\alpha_{n-1}\right)$
\end_inset
.
To rešitev vstavimo v izraz za
\begin_inset Formula $x_{n}$
\end_inset
in dobimo
\begin_inset Formula $\alpha_{n}$
\end_inset
in s tem
\begin_inset Formula $\left(\alpha_{1},\dots,\alpha_{n-1},\alpha_{n}\right)$
\end_inset
kot netrivialno rešitev sistema z
\begin_inset Formula $m$
\end_inset
vrsticami.
\end_layout
\end_deeper
\begin_layout Claim*
Linearna kombinacija dveh rešitev homogenega sistema je spet njegova rešitev.
\end_layout
\begin_layout Proof
Če sta
\begin_inset Formula $\left(s_{1},\dots,s_{n}\right)$
\end_inset
in
\begin_inset Formula $\left(t_{1},\dots,t_{n}\right)$
\end_inset
dve rešitvi homogenega sistema,
velja za
\begin_inset Formula $\vec{s}$
\end_inset
\begin_inset Formula $\forall i:\left\langle \left(a_{i,1},\dots,a_{i,n}\right),\left(s_{1},\dots,s_{n}\right)\right\rangle =a_{i,1}s_{1}+\cdots+a_{i,n}s_{n}=0$
\end_inset
in enako za
\begin_inset Formula $\vec{t}$
\end_inset
.
Dokažimo
\begin_inset Formula $\forall\alpha,\beta\in\mathbb{R},i:\left\langle \left(a_{i,1},\dots,a_{i,n}\right),\alpha\left(s_{1},\dots,s_{n}\right)+\beta\left(t_{1},\dots,t_{n}\right)\right\rangle =0$
\end_inset
.
\begin_inset Formula
\[
\left\langle \left(a_{i,1},\dots,a_{i,n}\right),\alpha\left(s_{1},\dots,s_{n}\right)+\beta\left(t_{1},\dots,t_{n}\right)\right\rangle =\left\langle \alpha\left(s_{1},\dots,s_{n}\right)+\beta\left(t_{1},\dots,t_{n}\right),\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =
\]
\end_inset
\begin_inset Formula
\[
=\alpha\left\langle \vec{s},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle +\beta\left\langle \vec{t},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =\alpha0+\beta0
\]
\end_inset
\end_layout
\begin_layout Claim*
Splošna rešitev
\begin_inset Formula $\vec{x}$
\end_inset
rešljivega nehomogenega sistema s partikularno rešitvijo
\begin_inset Formula $\vec{p}$
\end_inset
je
\begin_inset Formula $\vec{x}=\vec{p}+\vec{h}$
\end_inset
,
kjer je
\begin_inset Formula $\vec{h}$
\end_inset
rešitev temu sistemu prirejenega homogenega sistema (desno stvar smo prepisali z ničlami).
\end_layout
\begin_layout Remark*
Trdimo,
da je množica rešitev nehomogenega sistema samo množica rešitev prirejenega homogenega sistema,
premaknjena za partikularno rešitev nehomogenega sistema.
\end_layout
\begin_layout Proof
Velja
\begin_inset Formula $\forall i:\left\langle \vec{p},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =b_{i}\wedge\left\langle \vec{h},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =0$
\end_inset
.
Dokažimo
\begin_inset Formula $\forall i:\left\langle \vec{p}+\vec{h},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =b_{i}$
\end_inset
.
\begin_inset Formula
\[
\left\langle \vec{p}+\vec{h},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =\left\langle \vec{p},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle +\left\langle \vec{h}\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =b_{i}+0=b_{i}
\]
\end_inset
\end_layout
\begin_layout Subsubsection
Predoločeni sistemi
\end_layout
\begin_layout Standard
Predoločen sistem,
torej tak z več enačbami kot spremenljivkami,
je običajno,
a ne nujno,
nerešljiv.
\end_layout
\begin_layout Definition*
Posplošena rešitev sistema linearnih enačb je taka
\begin_inset Formula $n-$
\end_inset
terica števil
\begin_inset Formula $\left(x_{1},\dots x_{n}\right)$
\end_inset
,
za katero je vektor levih strani
\begin_inset Formula $\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)$
\end_inset
najbližje vektorju desnih strani
\begin_inset Formula $\left(b_{1},\dots,b_{n}\right)$
\end_inset
.
\end_layout
\begin_layout Remark*
Če je sistem rešljiv,
se njegova rešitev ujema s posplošeno rešitvijo.
Po metodi najmanjših kvadratov želimo minimizirati izraz
\begin_inset Formula $\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n}-b_{1}\right)^{2}+\cdots+\left(a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}-b_{n}\right)^{2}$
\end_inset
oziroma kvadrat norme razlike
\begin_inset Formula $\left|\left|x_{1}\vec{a_{1}}+\cdots+x_{n}\vec{a_{n}}-\vec{b}\right|\right|^{2}$
\end_inset
.
\begin_inset Foot
status open
\begin_layout Plain Layout
Z
\begin_inset Formula $\vec{a_{i}}$
\end_inset
označujemo stolpične vektorje sistema,
torej
\begin_inset Formula $\vec{a_{i}}=\left(a_{1,i},\dots,a_{m,i}\right)$
\end_inset
.
\end_layout
\end_inset
Podobno kot pri regresijski premici želimo pravokotno projicirati
\begin_inset Formula $\vec{b}$
\end_inset
na
\begin_inset Formula $\Lin\left\{ \vec{a_{1}},\dots,\vec{a_{n}}\right\} $
\end_inset
.
Iščemo torej take skalarje
\begin_inset Formula $\left(x_{1},\dots,x_{n}\right)$
\end_inset
,
da je
\begin_inset Formula $\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}-\vec{b}\perp\vec{a_{1}},\dots,\vec{a_{n}}$
\end_inset
(hkrati pravokotna na vse vektorje,
ki določajo to linearno ogrinjačo).
Preuredimo skalarne produkte in zopet dobimo sistem enačb:
\begin_inset Formula
\[
\left\langle \vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}-\vec{b},\vec{a_{1}}\right\rangle =\cdots=\left\langle \vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}-\vec{b},\vec{a_{n}}\right\rangle =0
\]
\end_inset
\begin_inset Formula
\[
x_{1}\left\langle \vec{a_{1}},\vec{a_{1}}\right\rangle +\cdots+x_{n}\left\langle \vec{a_{n}},\vec{a_{1}}\right\rangle =\left\langle \vec{b},\vec{a_{1}}\right\rangle
\]
\end_inset
\begin_inset Formula
\[
\cdots
\]
\end_inset
\begin_inset Formula
\[
x_{1}\left\langle \vec{a_{1}},\vec{a_{n}}\right\rangle +\cdots+x_{n}\left\langle \vec{a_{n}},\vec{a_{n}}\right\rangle =\left\langle \vec{b},\vec{a_{n}}\right\rangle
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
Izkaže se,
da je zgornji sistem vedno rešljiv.
Enolično takrat,
ko so
\begin_inset Formula $\left\{ \vec{a_{1}},\dots,\vec{a_{n}}\right\} $
\end_inset
linearno neodvisni.
Če je neenolično rešljiv,
pa poiščemo njegovo najkrajšo rešitev.
\end_layout
\begin_layout Subsubsection
Poddoločeni sistemi
\end_layout
\begin_layout Claim*
Poddoločen sistem,
torej tak,
ki ima več spremenljivk kot enačb,
ima neskončno rešitev,
čim je rešljiv.
\end_layout
\begin_layout Proof
Sledi iz zgornjih dokazov,
da ima vsak poddoločen homogen sistem neskončno rešitev in da je
\begin_inset Formula $\vec{p}+\vec{h}$
\end_inset
splošna rešitev nehomogenega sistema,
če je
\begin_inset Formula $\vec{p}$
\end_inset
partikularna rešitev tega sistema in
\begin_inset Formula $\vec{h}$
\end_inset
splošna rešitev prirejenega homogenega sistema.
\end_layout
\begin_layout Remark*
Seveda je lahko poddoločen sistem nerešljiv.
Trivialen primer:
\begin_inset Formula $x+y+z=1$
\end_inset
,
\begin_inset Formula $x+y+z=2$
\end_inset
.
\end_layout
\begin_layout Standard
Kadar ima sistem neskončno rešitev,
nas često zanima najkrajša (recimo zadnja opomba v prejšnji sekciji).
Geometrijski gledano je najkrajša rešitev pravokotna projekcija izhodišča na presek hiperravnin,
ki so množica rešitve sistema.
Vsaka enačba določa eno hiperravnino v normalni obliki,
torej
\begin_inset Formula $\left\langle \vec{r},\vec{n_{i}}\right\rangle =b_{i}$
\end_inset
.
Projekcija izhodišča na hiperravnino v normalni obliki je presečišče premice,
ki gre skozi izhodišče in je pravokotna na ravnino,
torej
\begin_inset Formula $\vec{r}=t\vec{n_{i}}$
\end_inset
,
in ravnine same.
Vstavimo drugo enačbo v prvo in dobimo
\begin_inset Formula $\left\langle t\vec{n_{i}},\vec{n_{i}}\right\rangle =b_{i}$
\end_inset
in izrazimo
\begin_inset Formula $t=\frac{b}{\left\langle \vec{n_{i}},\vec{n_{i}}\right\rangle }$
\end_inset
,
s čimer dobimo
\begin_inset Formula $\vec{r}=\frac{b}{\left\langle \vec{n_{i}},\vec{n_{i}}\right\rangle }\vec{n_{i}}$
\end_inset
.
Doslej je to le projekcija na eno hiperravnino.
\end_layout
\begin_layout Standard
Za pravokotno projekcijo na presek hiperravnin pa najprej določimo ravnino,
ki je pravokotna na vse hiperravnine sistema,
torej
\begin_inset Formula $\vec{r}=t_{1}\vec{n_{1}}+\cdots+t_{m}\vec{n_{m}}$
\end_inset
,
in najdimo presek te ravnine z vsemi hiperravninami.
To storimo tako,
da enačbo ravnine vstavimo v enačbe hiperravnin in jih uredimo:
\begin_inset Formula $\left\langle \vec{r},\vec{n_{i}}\right\rangle =b_{i}\sim\left\langle t_{1}\vec{n_{1}}+\cdots+t_{m}\vec{n_{m}},\vec{n_{i}}\right\rangle =b_{i}\sim t_{1}\left\langle \vec{n_{1}},\vec{n_{i}}\right\rangle +\cdots+t_{m}\left\langle \vec{n_{m}},\vec{n_{i}}\right\rangle =b_{i}$
\end_inset
.
To nam da sistem enačb
\begin_inset Formula
\[
t_{1}\left\langle \vec{n_{1}},\vec{n_{1}}\right\rangle +\cdots+t_{m}\left\langle \vec{n_{m}},\vec{n_{1}}\right\rangle =b_{1}
\]
\end_inset
\begin_inset Formula
\[
\cdots
\]
\end_inset
\begin_inset Formula
\[
t_{1}\left\langle \vec{n_{1}},\vec{n_{m}}\right\rangle +\cdots+t_{m}\left\langle \vec{n_{m}},\vec{n_{m}}\right\rangle =b_{m}
\]
\end_inset
Rešimo sistem in dobimo
\begin_inset Formula $\left(t_{1},\dots,t_{m}\right)$
\end_inset
,
kar vstavimo v enačbo ravnine
\begin_inset Formula $\vec{r}=t_{1}\vec{n_{1}}+\cdots+t_{m}\vec{n_{m}}$
\end_inset
,
da dobimo najkrajšo rešitev.
\end_layout
\begin_layout Subsection
Matrike
\end_layout
\begin_layout Definition*
\begin_inset Formula $m\times n$
\end_inset
matrika je element
\begin_inset Formula $\left(\mathbb{R}^{n}\right)^{m}$
\end_inset
,
torej
\begin_inset Formula $A=\left(\left(a_{1,1},\dots,a_{1,n}\right),\dots,\left(a_{m,1},\dots,a_{m,n}\right)\right)$
\end_inset
.
Ima
\begin_inset Formula $m$
\end_inset
vrstic in
\begin_inset Formula $n$
\end_inset
stolpcev,
zato jo pišemo takole:
\begin_inset Formula
\[
A=\left[\begin{array}{ccc}
a_{1,1} & \cdots & a_{1,n}\\
\vdots & & \vdots\\
a_{m,1} & \cdots & a_{m,n}
\end{array}\right]
\]
\end_inset
Matrikam velikosti
\begin_inset Formula $1\times n$
\end_inset
pravimo vrstični vektor,
matrikam velikosti
\begin_inset Formula $m\times1$
\end_inset
pa stolpični vektor.
Obe vrsti običajno identificiramo z vektorji.
\begin_inset Formula $\left[1\right]$
\end_inset
identificiramo z 1.
Na preseku
\begin_inset Formula $i-$
\end_inset
te vrstice in
\begin_inset Formula $j-$
\end_inset
tega stolpca matrike se nahaja element
\begin_inset Formula $a_{i,j}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Seštevanje matrik je definirano le za matrike enakih dimenzij.
Vsota matrik
\begin_inset Formula $A+B$
\end_inset
je matrika
\begin_inset Formula
\[
A+B=\left[\begin{array}{ccc}
a_{1,1}+b_{1,1} & \cdots & a_{1,n}+b_{1,n}\\
\vdots & & \vdots\\
a_{m,1}+b_{m.1} & \cdots & a_{m,n}+b_{m,n}
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
Ničelna matrika 0 je aditivna enota.
\begin_inset Formula
\[
0=\left[\begin{array}{ccc}
0 & \cdots & 0\\
\vdots & & \vdots\\
0 & \cdots & 0
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Produkt matrike s skalarjem.
\begin_inset Formula
\[
A\cdot\alpha=\alpha\cdot A=\left[\begin{array}{ccc}
\alpha a_{1,1} & \cdots & \alpha a_{1,n}\\
\vdots & & \vdots\\
\alpha a_{m,1} & \cdots & \alpha a_{m,n}
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Produkt dveh matrik
\begin_inset Formula $A_{m\times n}\cdot B_{n\times p}=C_{m\times p}$
\end_inset
.
Velja
\begin_inset Formula $c_{i,j}=\sum_{k=1}^{n}a_{i,k}b_{j,k}$
\end_inset
.
(razmislek prepuščen bralcu)
\end_layout
\begin_layout Remark*
Kvadratna matrika identiteta
\begin_inset Formula $I$
\end_inset
je multiplikativna enota:
\begin_inset Formula $i_{ij}=\begin{cases}
0 & ;i\not=j\\
1 & ;i=j
\end{cases}$
\end_inset
.
\end_layout
\begin_layout Definition*
Transponiranje matrike
\begin_inset Formula $A_{m\times n}^{T}=B_{n\times m}$
\end_inset
.
\begin_inset Formula $b_{ij}=a_{ji}$
\end_inset
.
\end_layout
\begin_layout Remark*
Lastnosti transponiranja:
\begin_inset Formula $\left(A^{T}\right)^{T}=A$
\end_inset
,
\begin_inset Formula $\left(A+B\right)^{T}=A^{T}+B^{T}$
\end_inset
,
\begin_inset Formula $\left(\alpha A\right)^{T}=\alpha A^{T}$
\end_inset
,
\begin_inset Formula $\left(AB\right)^{T}=B^{T}A^{T}$
\end_inset
,
\begin_inset Formula $I^{T}=I$
\end_inset
,
\begin_inset Formula $0^{T}=0$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Matrični zapis sistema linearnih enačb
\end_layout
\begin_layout Standard
Matrika koeficientov vsebuje koeficiente,
imenujmo jo
\begin_inset Formula $A$
\end_inset
(ena vrstica matrike je ena enačba v sistemu).
Stolpični vektor spremenljivk vsebuje spremenljivke
\begin_inset Formula $\vec{x}=\left(x_{1},\dots,x_{n}\right)$
\end_inset
.
Vektor desne strani vsebuje desne strani
\begin_inset Formula $\vec{b}=\left(b_{1},\dots,b_{m}\right)$
\end_inset
.
Sistem torej zapišemo kot
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
.
\end_layout
\begin_layout Standard
Tudi Gaussovo metodo lahko zapišemo matrično.
Trem elementarnim preoblikovanjem,
ki ne spremenijo množice rešitev,
priredimo ustrezne t.
i.
elementarne matrike:
\end_layout
\begin_layout Itemize
\begin_inset Formula $E_{i,j}\left(\alpha\right)$
\end_inset
:
identiteta,
ki ji na
\begin_inset Formula $i,j-$
\end_inset
to mesto prištejemo
\begin_inset Formula $\alpha$
\end_inset
.
Ustreza prištevanju
\begin_inset Formula $\alpha-$
\end_inset
kratnika
\begin_inset Formula $j-$
\end_inset
te vrstice k
\begin_inset Formula $i-$
\end_inset
ti vrstici.
\end_layout
\begin_layout Itemize
\begin_inset Formula $P_{ij}$
\end_inset
:
v
\begin_inset Formula $I$
\end_inset
zamenjamo
\begin_inset Formula $i-$
\end_inset
to in
\begin_inset Formula $j-$
\end_inset
to vrstico.
Ustreza zamenjavi
\begin_inset Formula $i-$
\end_inset
te in
\begin_inset Formula $j-$
\end_inset
te vrstice.
\end_layout
\begin_layout Itemize
\begin_inset Formula $E_{i}\left(\alpha\right)$
\end_inset
:
v
\begin_inset Formula $I$
\end_inset
pomnožiš
\begin_inset Formula $i-$
\end_inset
to vrstico z
\begin_inset Formula $\alpha$
\end_inset
.
Ustreza množenju
\begin_inset Formula $i-$
\end_inset
te vrstice s skalarjem
\begin_inset Formula $\alpha$
\end_inset
.
\end_layout
\begin_layout Fact*
Vsako matriko je moč z levim množenjem z elementarnimi matrikami (Gaussova metoda) prevesti na reducirano vrstično stopničasto formo/obliko.
ZDB
\begin_inset Formula $\forall A\in M\left(\mathbb{R}\right)\exists E_{1},\dots,E_{k}\ni:R=E_{1}\cdot\cdots\cdot E_{k}\cdot A$
\end_inset
je r.
v.
s.
f.
Ko rešujemo sistem s temi matrikami množimo levo in desno stran sistema.
\end_layout
\begin_layout Subsubsection
Postopek iskanja posplošene rešitve predoločenega sistema
\end_layout
\begin_layout Enumerate
Sistem
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
z leve pomnožimo z
\begin_inset Formula $A^{T}$
\end_inset
in dobimo sistem
\begin_inset Formula $A^{T}A\vec{x}=A^{T}\vec{b}$
\end_inset
.
\end_layout
\begin_layout Enumerate
Poiščemo običajno rešitev dobljenega sistema,
za katero se izkaže,
da vselej obstaja (dokaz v 2.
semestru).
\end_layout
\begin_layout Enumerate
Dokažemo,
da je običajna rešitev
\begin_inset Formula $A^{T}A\vec{x}=A^{T}\vec{b}$
\end_inset
enaka posplošeni rešitvi
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
.
\end_layout
\begin_layout Proof
\begin_inset Formula $\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}$
\end_inset
bi radi minimizirali.
Naj bo
\begin_inset Formula $\vec{x_{0}}$
\end_inset
običajna rešitev sistema
\begin_inset Formula $A^{T}A\vec{x}=A^{T}\vec{b}$
\end_inset
.
\begin_inset Formula
\[
\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}=\left|\left|A\vec{x}-A\vec{x_{0}}+A\vec{x_{0}}-\vec{b}\right|\right|^{2}
\]
\end_inset
\end_layout
\begin_layout Proof
Naj bosta
\begin_inset Formula $\vec{u}=A\vec{x}-A\vec{x_{0}}$
\end_inset
in
\begin_inset Formula $\vec{v}=A\vec{x_{0}}-\vec{b}$
\end_inset
.
Trdimo,
da
\begin_inset Formula $\vec{u}\perp\vec{v}$
\end_inset
,
torej
\begin_inset Formula $\left\langle \vec{u},\vec{v}\right\rangle =0$
\end_inset
.
Dokažimo:
\begin_inset Formula
\[
\left\langle A\vec{x}-A\vec{x_{0}},A\vec{x_{0}}-\vec{b}\right\rangle =\left\langle A\left(\vec{x}-\vec{x_{0}}\right),A\vec{x_{0}}-\vec{b}\right\rangle =\left(A\vec{x_{0}}-\vec{b}\right)^{T}A\left(\vec{x}-\vec{x_{0}}\right)=\left(A\vec{x_{0}}-\vec{b}\right)^{T}\left(A^{T}\right)^{T}\left(\vec{x}-\vec{x_{0}}\right)=
\]
\end_inset
\begin_inset Formula
\[
=\left(A^{T}\left(A\vec{x_{0}}-\vec{b}\right)\right)^{T}\left(\vec{x}-\vec{x_{0}}\right)=\left(A^{T}A\vec{x_{0}}-A^{T}\vec{b}\right)\left(\vec{x}-\vec{x_{0}}\right)\overset{\text{predpostavka o }\vec{x_{0}}}{=}0\left(\vec{x}-\vec{x_{0}}\right)
\]
\end_inset
\end_layout
\begin_layout Proof
Ker sedaj vemo,
da sta
\begin_inset Formula $\vec{u}$
\end_inset
in
\begin_inset Formula $\vec{v}$
\end_inset
pravokotna,
lahko uporabimo Pitagorov izrek,
ki za njiju pravi
\begin_inset Formula $\left|\left|\vec{u}+\vec{v}\right|\right|^{2}=\left|\left|\vec{u}\right|\right|^{2}+\left|\left|\vec{v}\right|\right|^{2}$
\end_inset
.
V naslednjih izpeljavah je
\begin_inset Formula $\vec{x}$
\end_inset
poljuben,
\begin_inset Formula $\vec{x_{0}}$
\end_inset
pa kot prej.
\begin_inset Formula
\[
\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}=\left|\left|A\vec{x}-A\vec{x_{0}}+A\vec{x_{0}}-\vec{b}\right|\right|^{2}=\left|\left|A\vec{x}-A\vec{x_{0}}\right|\right|^{2}+\left|\left|A\vec{x_{0}}-\vec{b}\right|\right|^{2}\geq\left|\left|A\vec{x_{0}}-\vec{b}\right|\right|^{2}
\]
\end_inset
\begin_inset Formula
\[
\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}\geq\left|\left|A\vec{x_{0}}-\vec{b}\right|\right|^{2},
\]
\end_inset
kar pomeni ,
da je
\begin_inset Formula $\vec{x_{0}}$
\end_inset
manjši ali enak kot vsi ostale
\begin_inset Formula $n-$
\end_inset
terice spremenljivk.
\end_layout
\begin_layout Subsubsection
Najkrajša rešitev sistema
\end_layout
\begin_layout Standard
Ta sekcija je precej dobesedno povzeta po profesorjevi beamer skripti.
\end_layout
\begin_layout Standard
Sistem
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
je lahko neenolično rešljiv.
Tedaj nas često zanima po normi najkrajša rešitev sistema.
\end_layout
\begin_layout Claim*
Najkrajša rešitev sistema
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
je
\begin_inset Formula $A^{T}\vec{y_{0}}$
\end_inset
,
kjer je
\begin_inset Formula $\vec{y_{0}}$
\end_inset
poljubna rešitev sistema
\begin_inset Formula $AA^{T}\vec{y}=\vec{b}$
\end_inset
.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $\vec{x_{0}}$
\end_inset
poljubna rešitev sistema
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
in
\begin_inset Formula $\vec{y_{0}}$
\end_inset
poljubna rešitev sistema
\begin_inset Formula $AA^{T}\vec{y}=\vec{b}$
\end_inset
.
Dokazali bi radi,
da velja
\begin_inset Formula $\left|\left|A^{T}\vec{y_{0}}\right|\right|^{2}\leq\left|\left|\vec{x_{0}}\right|\right|^{2}$
\end_inset
.
Podobno,
kot v prejšnji sekciji:
\end_layout
\begin_layout Proof
\begin_inset Formula
\[
\left|\left|\vec{x_{0}}\right|\right|^{2}=\left|\left|\vec{x_{0}}-A^{T}\vec{y_{0}}+A^{T}\vec{y_{0}}\right|\right|^{2}=\left|\left|u+v\right|\right|^{2}
\]
\end_inset
\end_layout
\begin_layout Proof
Dokažimo,
da sta
\begin_inset Formula $\vec{u}=\vec{x_{0}}-A^{T}\vec{y_{0}}$
\end_inset
in
\begin_inset Formula $\vec{v}=A^{T}\vec{y_{0}}$
\end_inset
pravokotna,
da lahko uporabimo pitagorov izrek v drugi vrstici:
\begin_inset Formula
\[
\left\langle \vec{x_{0}}-A^{T}\vec{y_{0}},A^{T}\vec{y_{0}}\right\rangle =\left(\vec{x_{0}}-A^{T}\vec{y_{0}}\right)^{T}A^{T}\vec{y_{0}}=\left(A\left(\vec{x_{0}}-A^{T}\vec{y_{0}}\right)\right)^{T}\vec{y_{0}}=\left(A\vec{x_{0}}-AA^{T}\vec{y_{0}}\right)^{T}\vec{y_{0}}=\left(\vec{b}-\vec{b}\right)^{T}\vec{y_{0}}=0
\]
\end_inset
\begin_inset Formula
\[
\left|\left|u+v\right|\right|^{2}=\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}=\left|\left|\vec{x_{0}}-A^{T}\vec{y_{0}}+A^{T}\vec{y_{0}}\right|\right|^{2}=\left|\left|\vec{x_{0}}-A^{T}\vec{y_{0}}\right|\right|^{2}+\left|\left|A^{T}\vec{y_{0}}\right|\right|^{2}\geq\left|\left|A^{T}\vec{y_{0}}\right|\right|
\]
\end_inset
\begin_inset Formula
\[
\left|\left|\vec{x_{0}}\right|\right|^{2}\geq\left|\left|A^{T}\vec{y_{0}}\right|\right|
\]
\end_inset
\end_layout
\begin_layout Remark*
Iz rešljivosti
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
sledi rešljivost
\begin_inset Formula $AA^{T}\vec{y}=\vec{b}$
\end_inset
,
toda to znamo dokazati šele v drugem semestru.
\end_layout
\begin_layout Subsubsection
Inverzi matrik
\end_layout
\begin_layout Definition*
Matrika
\begin_inset Formula $B$
\end_inset
je inverz matrike
\begin_inset Formula $A$
\end_inset
,
če velja
\begin_inset Formula $AB=I$
\end_inset
in
\begin_inset Formula $BA=I$
\end_inset
.
Matrika
\begin_inset Formula $A$
\end_inset
je obrnljiva,
če ima inverz,
sicer je neobrnljiva.
\end_layout
\begin_layout Claim*
Če inverz obstaja,
je enoličen.
\end_layout
\begin_layout Proof
Naj bosta
\begin_inset Formula $B_{1}$
\end_inset
in
\begin_inset Formula $B_{2}$
\end_inset
inverza
\begin_inset Formula $A$
\end_inset
.
Velja
\begin_inset Formula $AB_{1}=B_{1}A=AB_{2}=B_{2}A=I$
\end_inset
.
\begin_inset Formula $B_{1}=B_{1}I=B_{1}\left(AB_{2}\right)=\left(B_{1}A\right)B_{2}=IB_{2}=B_{2}$
\end_inset
.
\end_layout
\begin_layout Definition*
Če inverz
\begin_inset Formula $A$
\end_inset
obstaja,
ga označimo z
\begin_inset Formula $A^{-1}$
\end_inset
.
\end_layout
\begin_layout Example*
Primeri obrnljivih matrik:
\end_layout
\begin_deeper
\begin_layout Itemize
Identična matrika
\begin_inset Formula $I$
\end_inset
:
\begin_inset Formula $I\cdot I=I$
\end_inset
,
\begin_inset Formula $I^{-1}=I$
\end_inset
\end_layout
\begin_layout Itemize
Elementarne matrike:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $E_{ij}\left(\alpha\right)\cdot E_{ij}\left(-\alpha\right)=I$
\end_inset
,
torej
\begin_inset Formula $E_{ij}\left(\alpha\right)^{-1}=E_{ij}\left(-\alpha\right)$
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $P_{ij}\cdot P_{ij}=I$
\end_inset
,
torej
\begin_inset Formula $P_{ij}^{-1}=P_{ij}$
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $E_{i}\left(\alpha\right)\cdot E_{i}\left(\alpha^{-1}\right)=I$
\end_inset
,
torej
\begin_inset Formula $E_{i}\left(\alpha\right)^{-1}=E_{i}\left(\alpha^{-1}\right)$
\end_inset
\end_layout
\end_deeper
\end_deeper
\begin_layout Claim*
Produkt obrnljivih matrik je obrnljiva matrika.
\end_layout
\begin_layout Proof
Naj bodo
\begin_inset Formula $A_{1},\dots,A_{n}$
\end_inset
obrnljive matrike,
torej po definiciji velja
\begin_inset Formula $A_{1}\cdot\cdots\cdot A_{n}\cdot A_{n}^{-1}\cdot\cdots\cdot A_{1}^{-1}=A_{n}\cdot\cdots\cdot A_{1}\cdot A_{1}^{-1}\cdot\cdots\cdot A_{n}^{-1}=I$
\end_inset
.
Opazimo,
da velja
\begin_inset Formula $\left(A_{1}\cdot\cdots\cdot A_{n}\right)^{-1}=A_{1}^{-1}\cdot\cdots\cdot A_{n}^{-1}$
\end_inset
.
\end_layout
\begin_layout Remark*
Vsaka obrnljiva matrika je produkt elementarnih matrik.
Dokaz sledi kasneje.
\end_layout
\begin_layout Example*
Primeri neobrnljivih matrik:
\end_layout
\begin_deeper
\begin_layout Itemize
Ničelna matrika,
saj pri množenju s katerokoli matriko pridela ničelno matriko in velja
\begin_inset Formula $I\not=0$
\end_inset
.
\end_layout
\begin_layout Itemize
Matrike z ničelnim stolpcem/vrstico.
\end_layout
\begin_deeper
\begin_layout Proof
Naj ima
\begin_inset Formula $A$
\end_inset
vrstico samih ničel.
Tedaj za vsako
\begin_inset Formula $B$
\end_inset
velja,
da ima
\begin_inset Formula $AB$
\end_inset
vrstico samih ničel (očitno po definiciji množenja).
\begin_inset Formula $AB$
\end_inset
zato ne more biti
\begin_inset Formula $I$
\end_inset
,
saj
\begin_inset Formula $I$
\end_inset
ne vsebuje nobene vrstice samih ničel.
Podobno za ničelni stolpec.
\end_layout
\end_deeper
\begin_layout Itemize
Nekvadratne matrike
\end_layout
\begin_deeper
\begin_layout Proof
Naj ima
\begin_inset Formula $A_{m\times n}$
\end_inset
več vrstic kot stolpcev (
\begin_inset Formula $m>n$
\end_inset
).
PDDRAA obstaja
\begin_inset Formula $B$
\end_inset
,
da
\begin_inset Formula $AB=I$
\end_inset
.
Uporabimo Gaussovo metodo na
\begin_inset Formula $A$
\end_inset
.
Z levim množenjem
\begin_inset Formula $A$
\end_inset
z nekimi elementarnimi matrikami lahko pridelamo RVSO.
\begin_inset Formula $E_{1}\cdots E_{n}A=R$
\end_inset
.
\begin_inset Formula $E_{1}\cdots E_{n}AB=E_{1}\cdots E_{n}I=E_{1}\cdots E_{n}=RB$
\end_inset
.
Toda
\begin_inset Formula $R$
\end_inset
ima po konstrukciji ničelno vrstico (je namreč
\begin_inset Formula $A$
\end_inset
podobna RVSO in a ima več vrstic kot stolpcev).
Potemtakem ima tudi
\begin_inset Formula $RB$
\end_inset
ničelno vrstico,
torej je neobrnljiva,
toda
\begin_inset Formula $RB$
\end_inset
je enak produktu elementarnih matrik,
torej bi morala biti obrnljiva.
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
\end_layout
\end_deeper
\end_deeper
\begin_layout Remark*
Iz
\begin_inset Formula $AB=I$
\end_inset
ne sledi nujno
\begin_inset Formula $BA=I$
\end_inset
.
Primer:
\begin_inset Formula $A=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0
\end{array}\right]$
\end_inset
,
\begin_inset Formula $B=\left[\begin{array}{cc}
1 & 0\\
0 & 1\\
0 & 0
\end{array}\right]$
\end_inset
,
\begin_inset Formula $AB=\left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right]$
\end_inset
,
\begin_inset Formula $BA=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{array}\right]$
\end_inset
.
Velja pa to za kvadratne matrike.
Dokaz kasneje.
\end_layout
\begin_layout Subsubsection
Karakterizacija obrnljivih matrik
\end_layout
\begin_layout Theorem*
Za vsako kvadratno matriko
\begin_inset Formula $A$
\end_inset
so naslednje trditve ekvivalentne:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $A$
\end_inset
je obrnljiva
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A$
\end_inset
ima levi inverz (
\begin_inset Formula $\exists B\ni:BA=I$
\end_inset
)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A$
\end_inset
ima desni inverz (
\begin_inset Formula $\exists B\ni:AB=I$
\end_inset
)
\end_layout
\begin_layout Enumerate
stolpci
\begin_inset Formula $A$
\end_inset
so linearno neodvisni
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall\vec{x}:A\vec{x}=0\Rightarrow\vec{x}=0$
\end_inset
\end_layout
\begin_layout Enumerate
stolpci
\begin_inset Formula $A$
\end_inset
so ogrodje
\end_layout
\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:VbEx:Ax=b"
\end_inset
\begin_inset Formula $\forall\vec{b}\exists\vec{x}\ni:A\vec{x}=\vec{b}$
\end_inset
\end_layout
\begin_layout Enumerate
RVSO
\begin_inset Formula $A$
\end_inset
je
\begin_inset Formula $I$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A$
\end_inset
je produkt elementarnih matrik
\end_layout
\end_deeper
\begin_layout Standard
Shema dokaza teh ekvivalenc je zanimiv graf.
Bralcu je prepuščena njegova skica.
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco.
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(1\Rightarrow2\right)$
\end_inset
Sledi iz definicije.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(1\Rightarrow3\right)$
\end_inset
Sledi iz definicije.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(2\Rightarrow5\right)$
\end_inset
Naj
\begin_inset Formula $\exists B\ni:BA=I$
\end_inset
.
Dokažimo,
da
\begin_inset Formula $\forall\vec{x}:A\vec{x}=0\Rightarrow\vec{x}=0$
\end_inset
.
Pa dajmo:
\begin_inset Formula $A\vec{x}=0\Rightarrow B\left(A\vec{x}\right)=B0=0=\left(BA\right)\vec{x}=I\vec{x}=\vec{x}$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(3\Rightarrow7\right)$
\end_inset
Naj
\begin_inset Formula $\exists B\ni:AB=I$
\end_inset
.
Dokažimo,
da
\begin_inset Formula $\forall\vec{b}\exists\vec{x}\ni:A\vec{x}=\vec{b}$
\end_inset
.
Vzemimo
\begin_inset Formula $\vec{x}=B\vec{b}$
\end_inset
.
Tedaj
\begin_inset Formula $A\vec{x}=AB\vec{b}=I\vec{b}=\vec{b}$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(5\Rightarrow4\right)$
\end_inset
Naj
\begin_inset Formula $\forall\vec{x}:A\vec{x}=0\Rightarrow\vec{x}=0$
\end_inset
.
Dokažimo,
da so stolpci
\begin_inset Formula $A$
\end_inset
linearno neodvisni.
Naj bo
\begin_inset Formula $A=\left[\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots & & \vdots\\
a_{n1} & \cdots & a_{mn}
\end{array}\right]$
\end_inset
,
\begin_inset Formula $\vec{x}=\left[\begin{array}{c}
x_{1}\\
\vdots\\
x_{n}
\end{array}\right]$
\end_inset
.
Tedaj
\begin_inset Formula $A\vec{x}=\left[\begin{array}{ccc}
a_{11}x_{1} & \cdots & a_{1n}x_{n}\\
\vdots & & \vdots\\
a_{n1}x_{1} & \cdots & a_{mn}x_{n}
\end{array}\right]=\left[\begin{array}{c}
a_{11}\\
\vdots\\
a_{m1}
\end{array}\right]x_{1}+\cdots+\left[\begin{array}{c}
a_{1n}\\
\vdots\\
a_{mn}
\end{array}\right]x_{n}=\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}$
\end_inset
.
Po definiciji
\begin_inset CommandInset ref
LatexCommand ref
reference "def:vsi0"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
za linearno neodvisnost mora veljati
\begin_inset Formula $\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}=0\Rightarrow x_{1}=\cdots=x_{n}=0$
\end_inset
.
Ravno to pa smo predpostavili.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(7\Rightarrow6\right)$
\end_inset
Uporabimo iste oznake kot zgoraj.
Za poljuben
\begin_inset Formula $\vec{b}$
\end_inset
iščemo tak
\begin_inset Formula $\vec{x}$
\end_inset
,
da je
\begin_inset Formula $\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}=A\vec{x}=\vec{b}$
\end_inset
(definicija ogrodja).
Po predpostavki
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:VbEx:Ax=b"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
velja,
da
\begin_inset Formula $\forall\vec{b}\exists\vec{x}\ni:A\vec{x}=\vec{b}$
\end_inset
.
Torej po predpostavki najdemo ustrezen
\begin_inset Formula $\vec{x}$
\end_inset
za poljuben
\begin_inset Formula $\vec{b}$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(4\Rightarrow8\right)$
\end_inset
Za dokaz uvedimo nekaj lem,
ki dokažejo trditev.
\begin_inset CommandInset counter
LatexCommand set
counter "theorem"
value "0"
lyxonly "false"
\end_inset
\end_layout
\begin_deeper
\begin_layout Lemma
\begin_inset CommandInset label
LatexCommand label
name "lem:kom1"
\end_inset
Če ima
\begin_inset Formula $A_{n\times n}$
\end_inset
LN stolpce in če je
\begin_inset Formula $C_{n\times n}$
\end_inset
obrnljiva,
ima tudi
\begin_inset Formula $CA$
\end_inset
LN stolpce.
\end_layout
\begin_layout Proof
Naj bodo
\begin_inset Formula $a_{1},\dots,a_{n}$
\end_inset
stolpci
\begin_inset Formula $A$
\end_inset
.
Velja
\begin_inset Formula $Ax=0\Rightarrow x=0$
\end_inset
.
Dokazati želimo,
da
\begin_inset Formula $CAx=0\Rightarrow x=0$
\end_inset
.
Predpostavimo
\begin_inset Formula $CAx=0$
\end_inset
.
Množimo obe strani z
\begin_inset Formula $C^{-1}$
\end_inset
.
\begin_inset Formula $C^{-1}CAx=C^{-1}0\sim IAx=0\sim Ax=0\Rightarrow x=0$
\end_inset
.
\end_layout
\begin_layout Lemma
Če ima
\begin_inset Formula $A$
\end_inset
LN stolpce,
ima njena RVSO LN stolpce.
\end_layout
\begin_layout Proof
Po Gaussu obstajajo take elementarne
\begin_inset Formula $E_{1},\dots,E_{n}$
\end_inset
,
da je
\begin_inset Formula $E_{n}\cdots E_{1}A=R$
\end_inset
RVSO.
Po lemi
\begin_inset CommandInset ref
LatexCommand ref
reference "lem:kom1"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
ima
\begin_inset Formula $E_{1}A$
\end_inset
LN stolpce,
prav tako
\begin_inset Formula $E_{2}E_{1}A$
\end_inset
in tako dalje,
vse do
\begin_inset Formula $E_{n}\cdots E_{1}A=R$
\end_inset
.
\end_layout
\begin_layout Lemma
Če ima RVSO
\begin_inset Formula $R$
\end_inset
LN stolpce,
je enaka identiteti.
\end_layout
\begin_layout Proof
PDDRAA
\begin_inset Formula $R\not=I$
\end_inset
.
Tedaj ima bodisi ničelni stolpec bodisi stopnico,
daljšo od 1.
Če ima ničelni stolpec,
ni LN.
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
.
Če ima stopnico,
daljšo od 1,
kar pomeni,
da v vrstici takoj za prvo enico obstajajo neki neničelni
\begin_inset Formula $\times-$
\end_inset
i,
pa je stolpec z nekim neničelnim
\begin_inset Formula $\times-$
\end_inset
om linearna kombinacija ostalih stolpcev,
torej stolpci
\begin_inset Formula $R$
\end_inset
niso LN
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(6\Rightarrow8\right)$
\end_inset
Predpostavimo,
da so stolpci
\begin_inset Formula $A$
\end_inset
ogrodje in dokazujemo,
da RVSO
\begin_inset Formula $A$
\end_inset
je
\begin_inset Formula $I$
\end_inset
.
\begin_inset CommandInset counter
LatexCommand set
counter "theorem"
value "0"
lyxonly "false"
\end_inset
\end_layout
\begin_deeper
\begin_layout Lemma
\begin_inset CommandInset label
LatexCommand label
name "lem:68kom1"
\end_inset
Če so stolpci
\begin_inset Formula $A_{n\times n}$
\end_inset
ogrodje in če je
\begin_inset Formula $C_{n\times n}$
\end_inset
obrnljica,
so tudi stolpci
\begin_inset Formula $CA$
\end_inset
ogrodje.
\end_layout
\begin_layout Proof
Naj bodo stolpci
\begin_inset Formula $A$
\end_inset
ogrodje.
Torej
\begin_inset Formula $\forall b\exists x\ni:Ax=C^{-1}b$
\end_inset
.
Množimo obe strani z
\begin_inset Formula $C^{-1}$
\end_inset
.
\begin_inset Formula $\forall b\exists x\ni:CAx=b$
\end_inset
—
stolpci
\begin_inset Formula $CA$
\end_inset
so ogrodje.
\end_layout
\begin_layout Lemma
Če so stolpci
\begin_inset Formula $A$
\end_inset
ogrodje,
so stolpci njene RVSO ogrodje.
\end_layout
\begin_layout Proof
Po Gaussu obstajajo take elementarne
\begin_inset Formula $E_{1},\dots,E_{n}$
\end_inset
,
da je
\begin_inset Formula $E_{n}\cdots E_{1}A=R$
\end_inset
RVSO.
Po lemi
\begin_inset CommandInset ref
LatexCommand ref
reference "lem:68kom1"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
so stolpci
\begin_inset Formula $E_{1}A$
\end_inset
ogrodje in tudi stolpci
\begin_inset Formula $E_{2}E_{1}A$
\end_inset
so ogrodje in tako dalje vse do
\begin_inset Formula $R$
\end_inset
.
\end_layout
\begin_layout Lemma
Če so stolpci RVSO
\begin_inset Formula $R$
\end_inset
ogrodje,
je
\begin_inset Formula $R=I$
\end_inset
.
\end_layout
\begin_layout Proof
PDDRAA
\begin_inset Formula $R\not=I$
\end_inset
.
Tedaj ima bodisi ničelni stolpec bodisi stopnico,
daljšo od 1.
Če ima ničelni stolpec,
stolpci niso ogrodje zaradi enoličnosti moči baze (dimenzije prostora).
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
Če ima stopnico,
daljšo od 1,
pa je stolpec z nekim neničelnim
\begin_inset Formula $\times-$
\end_inset
om linearna kombinacija ostalih stolpcev,
torej stolpci
\begin_inset Formula $R$
\end_inset
niso ogrodje zaradi enoličnosti moči baze (dimenzije prostora)
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
.
\begin_inset Note Note
status open
\begin_layout Plain Layout
(tegale ne razumem zares dobro,
niti med predavanji nismo dokazali) mogoče čim ima stopnico,
daljšo od 1,
ima ničelno vrstico?
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(8\Rightarrow9\right)$
\end_inset
Predpostavimo,
da je
\begin_inset Formula $R\coloneqq\text{RVSO}\left(A\right)=I$
\end_inset
.
Dokažimo,
da je
\begin_inset Formula $A$
\end_inset
produkt elementarnih matrik.
Po Gaussu obstajajo take elementarne matrike
\begin_inset Formula $E_{1},\dots,E_{n}$
\end_inset
,
da
\begin_inset Formula $E_{n}\cdots E_{1}A=R$
\end_inset
.
Elementarne matrike so obrnljive,
zato množimo z leve najprej z
\begin_inset Formula $E_{n}^{-1}$
\end_inset
,
nato z
\begin_inset Formula $E_{n-1}^{-1}$
\end_inset
,
vse do
\begin_inset Formula $E_{1}^{-1}$
\end_inset
in dobimo
\begin_inset Formula $A=E_{1}^{-1}\cdots E_{n}^{-1}R$
\end_inset
.
Upoštevamo,
da je inverz elementarne matrike elementarna matrika in da je
\begin_inset Formula $R=I$
\end_inset
.
Tedaj
\begin_inset Formula $A=E_{1}^{-1}\cdots E_{n}^{-1}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Claim*
\begin_inset Formula $A$
\end_inset
je obrnljiva
\begin_inset Formula $\Leftrightarrow A^{T}$
\end_inset
obrnljiva.
\end_layout
\begin_layout Proof
Velja
\begin_inset Formula $AB=I\Leftrightarrow\left(AB\right)^{T}=I^{T}\Leftrightarrow B^{T}A^{T}=I$
\end_inset
in
\begin_inset Formula $BA=I\Leftrightarrow\left(BA\right)^{T}=I^{T}\Leftrightarrow A^{T}B^{T}=I$
\end_inset
.
\end_layout
\begin_layout Corollary*
\begin_inset Formula $\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}$
\end_inset
in vrstice so LN in ogrodje.
\end_layout
\begin_layout Remark*
Inverz
\begin_inset Formula $A$
\end_inset
lahko izračunamo po Gaussu.
Zapišemo razširjeno matriko
\begin_inset Formula $\left[A,I\right]$
\end_inset
in na obeh applyamo iste elementarne transformacije,
da
\begin_inset Formula $A$
\end_inset
pretvorimo v RVSO.
Če je
\begin_inset Formula $A$
\end_inset
obrnljiva,
dobimo na levi identiteto,
na desni pa
\begin_inset Formula $A^{-1}$
\end_inset
.
\end_layout
\begin_layout Subsection
Determinante
\end_layout
\begin_layout Definition*
Vsaki kvadratni matriki
\begin_inset Formula $A$
\end_inset
priredimo število
\begin_inset Formula $\det A$
\end_inset
.
Definicija za
\begin_inset Formula $1\times1$
\end_inset
matrike:
\begin_inset Formula $\det\left[a\right]\coloneqq a$
\end_inset
.
Rekurzivna definicija za
\begin_inset Formula $n\times n$
\end_inset
matrike:
\begin_inset Formula
\[
\det\left[\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots & & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{array}\right]=\sum_{k=1}^{n}\left(-1\right)^{k+1}a_{1k}\det A_{1k},
\]
\end_inset
kjer
\begin_inset Formula $A_{ij}$
\end_inset
predstavja
\begin_inset Formula $A$
\end_inset
brez
\begin_inset Formula $i-$
\end_inset
te vrstice in
\begin_inset Formula $j-$
\end_inset
tega stolpca.
Tej formuli razvoja se reče
\begin_inset Quotes gld
\end_inset
razvoj determinante po prvi vrstici
\begin_inset Quotes grd
\end_inset
.
\end_layout
\begin_layout Example*
\begin_inset Formula $2\times2$
\end_inset
determinanta.
\begin_inset Formula $\det\left[\begin{array}{cc}
a & b\\
c & d
\end{array}\right]=ad-bc$
\end_inset
.
Geometrijski pomen je ploščina paralelograma,
ki ga razpenjata
\begin_inset Formula $\left(c,d\right)$
\end_inset
in
\begin_inset Formula $\left(a,b\right)$
\end_inset
,
kajti ploščina bi bila
\begin_inset Formula $\left(a+c\right)\left(b+d\right)-2bc-2\frac{cd}{2}-2\frac{ab}{2}=\cancel{ab}+\cancel{cb}+ad+\cancel{cd}-\cancel{2}bc-\cancel{cd}-\cancel{ab}=ad-bc$
\end_inset
.
Če zamenjamo vrstni red vektorjev,
pa dobimo za predznak napačen rezultat,
torej je ploščina enaka
\begin_inset Formula $\left|\det\left[\begin{array}{cc}
a & b\\
c & d
\end{array}\right]\right|$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula $3\times3$
\end_inset
determinanta.
\begin_inset Formula
\[
\det\left[\begin{array}{ccc}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{array}\right]=a_{11}\det\left[\begin{array}{cc}
a_{22} & a_{23}\\
a_{32} & a_{33}
\end{array}\right]-a_{12}\det\left[\begin{array}{cc}
a_{21} & a_{23}\\
a_{31} & a_{33}
\end{array}\right]+a_{13}\left[\begin{array}{cc}
a_{21} & a_{22}\\
a_{31} & a_{32}
\end{array}\right]=
\]
\end_inset
\begin_inset Formula
\[
a_{11}\left(a_{22}a_{33}-a_{23}a_{32}\right)-a_{12}\left(a_{21}a_{33}-a_{23}a_{31}\right)+a_{13}\left(a_{21}a_{32}-a_{22}a_{31}\right)
\]
\end_inset
To si lahko zapomnimo s Saurusovim pravilom.
Pripišemo na desno stran prva dva stolpca in seštejemo produkte po šestih diagonalah.
Naraščajoče diagonale (tiste s pozitivnim koeficientom,
če bi jih risali kot premice v ravnini) prej negiramo.
Geometrijski
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Vektorski produkt.
Velja:
\begin_inset Formula
\[
\left\langle \left(x,y,z\right),\left(a_{21},a_{22},a_{23}\right)\times\left(a_{31},a_{32},a_{33}\right)\right\rangle =\det\left[\begin{array}{ccc}
x & y & z\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{array}\right],
\]
\end_inset
torej je
\begin_inset Formula $\left[\left(a_{11},a_{12},a_{13}\right),\left(a_{21},a_{22},a_{23}\right),\left(a_{31},a_{32},a_{33}\right)\right]$
\end_inset
(mešani produkt) determinanta matrike
\begin_inset Formula $A$
\end_inset
,
torej je
\begin_inset Formula $\left|\det A\right|$
\end_inset
ploščina paralelpipeda,
ki ga razpenjajo trije vrstični vektorji
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Računanje determinant
\end_layout
\begin_layout Standard
Determinante računati po definiciji je precej zahtevno (bojda
\begin_inset Formula $O\left(n!\right)$
\end_inset
) za
\begin_inset Formula $n\times n$
\end_inset
determinanto.
Boljšo računsko zahtevnost dobimo z Gaussovo metodo.
Oglejmo si najprej posplošeno definicijo determinante:
\begin_inset Quotes gld
\end_inset
razvoj po poljubni
\begin_inset Formula $i-$
\end_inset
ti vrstici
\begin_inset Quotes grd
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\det A=\sum_{j=1}^{n}\left(-1\right)^{i+j}a_{ij}\det A_{ij}
\]
\end_inset
\begin_inset Quotes grd
\end_inset
razvoj po poljubnem
\begin_inset Formula $j-$
\end_inset
tem stolpcu
\begin_inset Quotes grd
\end_inset
\begin_inset Formula
\[
\det A=\sum_{i=1}^{n}\left(-1\right)^{i+j}a_{ij}\det A_{ij}
\]
\end_inset
Ti dve formuli sta še vedno nepolinomske zahtevnosti,
uporabni pa sta v primerih,
ko imamo veliko ničel na kaki vrstici/stolpcu.
Determinanta zgornjetrikotne matrike je po tej formuli produkt diagonalcev.
\end_layout
\begin_layout Standard
Kako pa se determinanta obnaša pri elementarnih vrstičnih transformacijah iz Gaussove metode?
\end_layout
\begin_layout Itemize
menjava vrstic
\begin_inset Formula $\Longrightarrow$
\end_inset
determinanti se spremeni predznak
\end_layout
\begin_layout Itemize
množenje vrstice z
\begin_inset Formula $\alpha\Longrightarrow$
\end_inset
determinanta se pomnoži z
\begin_inset Formula $\alpha$
\end_inset
\end_layout
\begin_layout Itemize
prištevanje večkratnika ene vrstice k drugi
\begin_inset Formula $\Longrightarrow$
\end_inset
determinanta se ne spremeni
\end_layout
\begin_layout Standard
Časovna zahtevnost Gaussove metode je bojda polinomska
\begin_inset Formula $O\left(n^{3}\right)$
\end_inset
.
\end_layout
\begin_layout Standard
Ideja dokaza veljavnosti Gaussove metode:
Indukcija po velikosti matrike.
\end_layout
\begin_layout Standard
Baza:
\begin_inset Formula $2\times2$
\end_inset
matrike
\end_layout
\begin_layout Standard
Korak:
Razvoj po vrstici,
ki je elementarna transformacija ne spremeni,
dobiš
\begin_inset Formula $n$
\end_inset
\begin_inset Formula $\left(n-1\right)\times\left(n-1\right)$
\end_inset
determimant,
ki so veljavne po I.
P.
\end_layout
\begin_layout Subsubsection
Lastnosti determinante
\end_layout
\begin_layout Claim*
Velja
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\det\left(AB\right)=\det A\det B$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\det A^{T}=\det A$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\det\left[\begin{array}{cc}
A & B\\
0 & C
\end{array}\right]=\det A\det C$
\end_inset
\end_layout
\end_deeper
\begin_layout Proof
Dokazujemo tri trditve
\end_layout
\begin_deeper
\begin_layout Enumerate
Dokazujemo
\begin_inset Formula $\det\left(AB\right)=\det A\det B$
\end_inset
.
Obravnavajmo dva posebna primera:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $A$
\end_inset
je elementarna:
obrat pomeni množenje determinante z
\begin_inset Formula $-1$
\end_inset
,
množenje vrstice z
\begin_inset Formula $\alpha$
\end_inset
množi determinanto z
\begin_inset Formula $\alpha$
\end_inset
,
prištevanje večkratnika vrstice k drugi vrstici množi determinanto z
\begin_inset Formula $1$
\end_inset
.
Očitno torej trditev velja,
če je
\begin_inset Formula $A$
\end_inset
elementarna.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A$
\end_inset
ima ničelno vrstico:
tedaj ima tudi
\begin_inset Formula $AB$
\end_inset
ničelno vrstico in je
\begin_inset Formula $\det A=0$
\end_inset
in
\begin_inset Formula $\det AB=0$
\end_inset
,
torej očitno trditev velja,
če ima
\begin_inset Formula $A$
\end_inset
ničelno vrstico.
\end_layout
\begin_layout Standard
Obravnavajmo še splošen primer:
Po Gaussovi metodi obstajajo take elementarne
\begin_inset Formula $E_{1},\dots,E_{n}$
\end_inset
,
da je
\begin_inset Formula $E_{n}\cdots E_{1}A=R$
\end_inset
RVSO.
Ker je
\begin_inset Formula $A$
\end_inset
kvadratna,
je tudi
\begin_inset Formula $R$
\end_inset
kvadratna.
Ločimo dva primera:
\end_layout
\begin_layout Enumerate
\begin_inset Formula $R=I$
\end_inset
.
Tedaj
\begin_inset Formula $\det\left(E_{n}\cdots E_{1}AB\right)=\det E_{n}\cdots\det E_{1}\det AB=\det\left(RB\right)=\det\left(IB\right)=\det B$
\end_inset
\begin_inset Formula
\[
\det I=\det R=\det E_{n}\cdots\det E_{1}\det A\quad\quad\quad\quad/\cdot\det B
\]
\end_inset
\begin_inset Formula
\[
\det I\det B=\det B=\det E_{n}\cdots\det E_{1}\det A\det B
\]
\end_inset
\begin_inset Formula $\det B$
\end_inset
zapišimo na dva načina ne levo in desno stran enačbe.
\begin_inset Formula
\[
\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det AB=\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A\det B
\]
\end_inset
\begin_inset Formula
\[
\det AB=\det A\det B
\]
\end_inset
\end_layout
\begin_deeper
\begin_layout Remark*
\begin_inset Formula $\exists A^{-1}\Leftrightarrow\det A\not=0$
\end_inset
\end_layout
\begin_deeper
\begin_layout Proof
Predpostavimo
\begin_inset Formula $A$
\end_inset
je obrnljiva.
Tedaj
\begin_inset Formula $\exists A^{-1}=B\ni:AB=I\overset{\circ\det}{\Longrightarrow}\det\left(AB\right)=\det I$
\end_inset
.
PDDRAA
\begin_inset Formula $\det A\not=0$
\end_inset
,
tedaj
\begin_inset Formula $\det AB=\det B\det A=0\not=\det I=1$
\end_inset
.
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
\end_layout
\begin_layout Proof
Predpostavimo sedaj
\begin_inset Formula $A$
\end_inset
ni obrnljiva.
Tedaj
\begin_inset Formula $\nexists A^{-1}\Rightarrow\exists E_{n},\dots,E_{1}\ni:E_{n}\cdots E_{1}A=R$
\end_inset
ima ničelno vrstico.
Uporabimo isti razmislek kot spodaj,
torej
\begin_inset Formula $\det R=0\Rightarrow0=\det R=\det E_{n}\cdots\det E_{1}\det A$
\end_inset
.
Ker so determinante elementarnih matrik vse neničelne,
mora biti
\begin_inset Formula $\det A$
\end_inset
ničeln,
da je produkt ničeln.
\end_layout
\end_deeper
\end_deeper
\begin_layout Enumerate
\begin_inset Formula $R$
\end_inset
ima ničelno vrstico.
Tedaj
\begin_inset Formula $\det\left(R\right)=0\Rightarrow0=\det R=\det E_{n}\cdots\det E_{1}\det A$
\end_inset
.
Ker so determinante elementarnih matrik vse neničelne,
mora biti
\begin_inset Formula $\det A$
\end_inset
ničeln,
da je produkt ničeln.
\end_layout
\end_deeper
\begin_layout Enumerate
Dokazujemo
\begin_inset Formula $\det A^{T}=\det A$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Enumerate
Če je
\begin_inset Formula $A$
\end_inset
elementarna matrika,
to drži:
\begin_inset Formula $\det P_{ij}=-1=\det P_{ij}^{T}=\det P_{ij}$
\end_inset
,
\begin_inset Formula $\det E_{i}\left(\alpha\right)=\alpha=\det E_{i}\left(\alpha\right)^{T}=\det E_{i}\left(\alpha\right)$
\end_inset
,
\begin_inset Formula $\det E_{ij}\left(\alpha\right)=1=\det E_{ji}\left(\alpha\right)^{T}=\det E_{ij}\left(\alpha\right)^{T}$
\end_inset
.
\end_layout
\begin_layout Enumerate
Če ima
\begin_inset Formula $A$
\end_inset
ničelno vrstico,
to drži,
saj ima tedaj
\begin_inset Formula $A^{T}$
\end_inset
ničeln stolpec in
\begin_inset Formula $\det A=0=\det A^{T}$
\end_inset
.
\end_layout
\begin_layout Enumerate
Splošen primer:
Po Gaussovi metodi
\begin_inset Formula $\exists E_{n},\dots,E_{1}\ni:E_{n}\cdots E_{1}A=\text{RVSO}\left(A\right)=R$
\end_inset
.
Zopet ločimo dva primera:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $R=I$
\end_inset
.
\begin_inset Formula $\det R=\det R^{T}=1$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $R$
\end_inset
ima ničelno vrstico.
\begin_inset Formula $\det R=\det R^{T}=0$
\end_inset
\end_layout
\begin_layout Standard
Sedaj vemo,
da
\begin_inset Formula $\det R=\det R^{T}$
\end_inset
.
Računajmo:
\begin_inset Formula
\[
\det R=\det R^{T}
\]
\end_inset
\begin_inset Formula
\[
\det\left(E_{n}\cdots E_{1}A\right)=\det\left(E_{n}\cdots E_{1}A\right)^{T}
\]
\end_inset
\begin_inset Formula
\[
\det\left(E_{n}\cdots E_{1}A\right)=\det\left(A^{T}E_{1}^{T}\cdots E_{n}^{T}\right)
\]
\end_inset
\begin_inset Formula
\[
\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A=\det A^{T}\cancel{\det E_{1}^{T}}\cdots\cancel{\det E_{n}^{T}}
\]
\end_inset
\begin_inset Formula
\[
\det A=\det A^{T}
\]
\end_inset
\end_layout
\end_deeper
\end_deeper
\begin_layout Enumerate
Dokazujemo
\begin_inset Formula $\det\left[\begin{array}{cc}
A & B\\
0 & C
\end{array}\right]=\det A\det C$
\end_inset
.
Levi izraz v enačbi vsebuje t.
i.
bločno matriko.
Upoštevamo poprej dokazano multiplikativnost determinante in opazimo,
da pri bločnem množenju matrik velja
\begin_inset Formula
\[
\left[\begin{array}{cc}
A & B\\
0 & C
\end{array}\right]=\left[\begin{array}{cc}
I & 0\\
0 & C
\end{array}\right]\left[\begin{array}{cc}
A & B\\
0 & I
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
\det\left[\begin{array}{cc}
A & B\\
0 & C
\end{array}\right]=\det\left[\begin{array}{cc}
I & 0\\
0 & C
\end{array}\right]\det\left[\begin{array}{cc}
A & B\\
0 & I
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
\det\left[\begin{array}{cc}
A & B\\
0 & C
\end{array}\right]=\det C\det A
\]
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
Pojasnilo:
Za
\begin_inset Formula $\det C$
\end_inset
si razpišemo bločno matriko,
za
\begin_inset Formula $\det A$
\end_inset
si zopet razpišemo bločno matriko in nato z Gaussovimi transformacijami z enicami iz spodnjega desnega bloka izničimo zgornji desni blok (
\begin_inset Formula $B$
\end_inset
).
\end_layout
\end_deeper
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsubsection
Cramerjevo pravilo —
eksplicitna formula za rešitve kvadratnega sistema linearnih enačb
\end_layout
\begin_layout Standard
Radi bi dobili eksplicitne formule za komponente rešitve
\begin_inset Formula $x_{i}$
\end_inset
kvadratnega sistema linearnih enačb
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
.
Izpeljimo torej eksplicitno formulo .
Druga/srednja matrika je identična,
v kateri smo
\begin_inset Formula $i-$
\end_inset
ti stolpec zamenjali z vektorjem spremenljivk
\begin_inset Formula $\vec{x}$
\end_inset
(to označimo z
\begin_inset Formula $I_{i}\left(\vec{x}\right)$
\end_inset
),
tretja/desna matrika pa je matrika koeficientov v kateri smo
\begin_inset Formula $i-$
\end_inset
ti stolpec zamenjali z vektorjem desnih strani
\begin_inset Formula $b$
\end_inset
(to označimo z
\begin_inset Formula $A_{i}\left(\vec{x}\right)$
\end_inset
).
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\left[\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots & & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{array}\right]\left[\begin{array}{ccccccc}
1 & & 0 & x_{1} & & & 0\\
& \ddots & & \vdots\\
& & 1 & x_{i-1}\\
& & & x_{i}\\
& & & x_{i+1} & 1\\
& & & \vdots & & \ddots\\
0 & & & x_{n} & 0 & & 1
\end{array}\right]=\left[\begin{array}{ccccc}
a_{11} & \cdots & b_{1} & \cdots & a_{1n}\\
\vdots & \ddots & \vdots & \iddots & \vdots\\
a_{i1} & & b_{i} & & a_{in}\\
\vdots & \iddots & \vdots & \ddots & \vdots\\
a_{n1} & \cdots & b_{n} & \cdots & a_{nn}
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\left[\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots & & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{array}\right]\left[\begin{array}{ccccccc}
1 & & 0 & x_{1} & & & 0\\
& \ddots & & \vdots\\
& & 1 & x_{i-1}\\
& & & x_{i}\\
& & & x_{i+1} & 1\\
& & & \vdots & & \ddots\\
0 & & & x_{n} & 0 & & 1
\end{array}\right]=\left[\begin{array}{ccccc}
a_{11} & \cdots & a_{11}x_{1}+\cdots+a_{1n}x_{n} & \cdots & a_{1n}\\
\vdots & \ddots & \vdots & \iddots & \vdots\\
a_{i1} & & a_{i1}x_{1}+\cdots+a_{in}x_{n} & & a_{in}\\
\vdots & \iddots & \vdots & \ddots & \vdots\\
a_{n1} & \cdots & a_{n1}x_{1}+\cdots+a_{nn}x_{n} & \cdots & a_{nn}
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
AI_{i}\left(\vec{x}\right)=A_{i}\left(\vec{b}\right)\quad\quad\quad\quad/\det
\]
\end_inset
\begin_inset Formula
\[
\det\left(AI_{i}\left(\vec{x}\right)\right)=\det A_{i}\left(\vec{b}\right)
\]
\end_inset
\begin_inset Formula
\[
\det A\det I_{i}\left(\vec{x}\right)=\det A_{i}\left(\vec{b}\right)
\]
\end_inset
Izračunamo
\begin_inset Formula $\det I_{i}\left(\vec{x}\right)$
\end_inset
z razvojem po
\begin_inset Formula $i-$
\end_inset
ti vrstici.
\begin_inset Formula
\[
\det A\cdot x_{i}=\det A_{i}\left(\vec{b}\right)
\]
\end_inset
\begin_inset Formula
\[
x_{i}=\frac{\det A_{i}\left(\vec{b}\right)}{\det A}
\]
\end_inset
\end_layout
\begin_layout Subsubsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Formula-za-inverz-matrike"
\end_inset
Formula za inverz matrike
\end_layout
\begin_layout Standard
Za dano obrnljivo
\begin_inset Formula $A_{n\times n}$
\end_inset
iščemo eksplicitno formulo za celice
\begin_inset Formula $X$
\end_inset
,
da velja
\begin_inset Formula $AX=I$
\end_inset
.
Ideja:
najprej bomo problem prevedli na reševanje sistemov linearnih enačb in uporabili Cramerjevo pravilo ter končno poenostavili formule.
Naj bodo
\begin_inset Formula $\vec{x_{1}},\dots,\vec{x_{n}}$
\end_inset
stolpci
\begin_inset Formula $X$
\end_inset
in
\begin_inset Formula $\vec{i_{1}},\dots,\vec{i_{n}}$
\end_inset
.
Potemtakem je
\begin_inset Formula $\left[\begin{array}{ccc}
A\vec{x_{1}} & \cdots & A\vec{x_{n}}\end{array}\right]=A\left[\begin{array}{ccc}
\vec{x_{1}} & \cdots & \vec{x_{n}}\end{array}\right]=AX=I=\left[\begin{array}{ccc}
\vec{i_{1}} & \cdots & \vec{i_{n}}\end{array}\right]$
\end_inset
.
Primerjajmo sedaj stolpce na obeh straneh:
\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :A\vec{x_{i}}=\vec{i_{i}}$
\end_inset
.
ZDB za vsak stolpec
\begin_inset Formula $X$
\end_inset
smo dobili sistem
\begin_inset Formula $n\times n$
\end_inset
linearnih enačb.
Te sisteme
\begin_inset Formula $A\vec{x_{j}}=\vec{i_{j}}$
\end_inset
\begin_inset Foot
status open
\begin_layout Plain Layout
Tokrat uporabimo indeks
\begin_inset Formula $j$
\end_inset
,
ker z njim reprezentiramo stolpec in ponavadi,
ko govorimo o elementu
\begin_inset Formula $x_{ij}$
\end_inset
matrike
\begin_inset Formula $X$
\end_inset
,
z
\begin_inset Formula $i$
\end_inset
označimo vrstico.
\end_layout
\end_inset
bomo rešili s Cramerjevim pravilom.
\begin_inset Formula
\[
x_{ij}=\left(\vec{x}_{j}\right)_{i}=\frac{\det A_{i}\left(\vec{i_{j}}\right)}{\det A}\overset{\text{razvoj po \ensuremath{j-}ti vrstici}}{=}\frac{\det A_{ji}\cdot\left(-1\right)^{j+i}}{\det A}
\]
\end_inset
\begin_inset Formula
\[
X=A^{-1}=\left[\begin{array}{ccc}
\frac{\det A_{11}\cdot\left(-1\right)^{1+1}}{\det A} & \cdots & \frac{\det A_{n1}\cdot\left(-1\right)^{n+1}}{\det A}\\
\vdots & & \vdots\\
\frac{\det A_{1n}\cdot\left(-1\right)^{1+n}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A}
\end{array}\right]=\frac{1}{\det A}\left[\begin{array}{ccc}
\det A_{11}\cdot\left(-1\right)^{1+1} & \cdots & \det A_{1n}\cdot\left(-1\right)^{1+1}\\
\vdots & & \vdots\\
\det A_{n1}\cdot\left(-1\right)^{n+1} & \cdots & \det A_{nn}\cdot\left(-1\right)^{n+n}
\end{array}\right]^{T}=\frac{1}{\det A}\tilde{A}^{T},
\]
\end_inset
\end_layout
\begin_layout Standard
kjer
\begin_inset Formula $\tilde{A}$
\end_inset
pravimo kofaktorska matrika.
\end_layout
\begin_layout Subsection
Algebrske strukture
\end_layout
\begin_layout Subsubsection
Uvod
\end_layout
\begin_layout Standard
Naj bo
\begin_inset Formula $M$
\end_inset
neprazna množica.
Operacija na
\begin_inset Formula $M$
\end_inset
pove,
kako iz dveh elementov
\begin_inset Formula $M$
\end_inset
dobimo nov element
\begin_inset Formula $M$
\end_inset
.
Na primer,
če
\begin_inset Formula $a,b\in M$
\end_inset
,
je
\begin_inset Formula $a\circ b$
\end_inset
nov element
\begin_inset Formula $M$
\end_inset
.
\end_layout
\begin_layout Definition*
Operacija na
\begin_inset Formula $M$
\end_inset
je funkcija
\begin_inset Formula $\circ:M\times M\to M$
\end_inset
,
kjer je
\begin_inset Formula $M\times M$
\end_inset
kartezični produkt (urejeni pari).
\begin_inset Formula $\left(a,b\right)\mapsto\circ\left(a,b\right)$
\end_inset
,
slednje pa označimo z
\begin_inset Formula $\circ\left(a,b\right)=a\circ b$
\end_inset
.
\end_layout
\begin_layout Standard
Na isti množici imamo lahko več različno definiranih operacij.
Ločimo jih tako,
da uvedemo pojem grupoida.
\end_layout
\begin_layout Definition*
Grupoid je
\begin_inset Formula $\left(\text{neprazna množica},\text{izbrana operacija }\circ:M\times M\to M\right)$
\end_inset
.
Na primer
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
.
\end_layout
\begin_layout Standard
Še posebej nas zanimajo operacije z lepimi lastnostmi,
denimo asociativnost,
komutativnost,
obstoj enot,
inverzov.
\end_layout
\begin_layout Definition*
Grupoid,
katerega
\begin_inset Formula $\circ$
\end_inset
je asociativna
\begin_inset Formula $\Leftrightarrow\forall a,b,c\in M:\left(a\circ b\right)\circ c=a\circ\left(b\circ c\right)$
\end_inset
,
je polgrupa.
Tedaj skladnja dopušča pisanje brez oklepajev:
\begin_inset Formula $a\circ b\circ c\circ d$
\end_inset
je nedvoumen/veljaven izraz,
ko je
\begin_inset Formula $\circ$
\end_inset
asociativna.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Komutativnost:
\begin_inset Formula $\circ$
\end_inset
je komutativna
\begin_inset Formula $\Leftrightarrow\forall a,b\in M:a\circ b=b\circ a$
\end_inset
.
Grupoidom s komutativno operacijo pravimo,
da so komutativni.
\end_layout
\begin_layout Example*
Asociativni in komutativni grupoidi (komutativne polgrupe):
\begin_inset Formula $\left(\mathbb{N},\cdot\right)$
\end_inset
,
\begin_inset Formula $\left(\mathbb{Q},\cdot\right)$
\end_inset
,
\begin_inset Formula $\left(\mathbb{N},+\right)$
\end_inset
—
številske operacije.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Asociativni,
a ne komutativni grupoidi (nekomutativne polgrupe):
\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
\end_inset
—
množenje matrik.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Komutativni,
a ne asociativni grupoidi:
Jordanski produkt matrik:
\begin_inset Formula $A\circ B=\frac{1}{2}\left(AB+BA\right)$
\end_inset
\SpecialChar endofsentence
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Niti komutativni niti asociativni grupoidi:
Vektorski produkt v
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset
:
\begin_inset Formula $\left(\mathbb{R}^{3},\times\right)$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula $M\not=\emptyset$
\end_inset
.
\begin_inset Formula $F$
\end_inset
naj bodo vse funkcije
\begin_inset Formula $M\to M$
\end_inset
,
\begin_inset Formula $\circ$
\end_inset
pa kompozitum dveh funkcij.
Izkaže se,
da:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\left(F,\circ\right)$
\end_inset
je vedno polgrupa.
\end_layout
\begin_deeper
\begin_layout Proof
Definicija kompozituma:
\begin_inset Formula $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
\end_inset
.
\begin_inset Formula
\[
\left(f\circ g\right)\circ h\overset{?}{=}f\circ\left(g\circ h\right)
\]
\end_inset
\begin_inset Formula
\[
\forall x:\left(\left(f\circ g\right)\circ h\right)\left(x\right)=\left(f\circ g\right)\left(h\left(x\right)\right)=f\left(g\left(h\left(x\right)\right)\right)
\]
\end_inset
\begin_inset Formula
\[
\forall x:\left(f\circ\left(g\circ h\right)\right)\left(x\right)=f\left(\left(g\circ h\right)\left(x\right)\right)=f\left(g\left(h\left(x\right)\right)\right)
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Itemize
Čim ima
\begin_inset Formula $M$
\end_inset
vsaj tri elemente,
\begin_inset Formula $\left(F,\circ\right)$
\end_inset
ni komutativna.
\end_layout
\end_deeper
\begin_layout Definition*
Naj bo
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
grupoid.
Element
\begin_inset Formula $e\in M$
\end_inset
je enota,
če
\begin_inset Formula $\forall a\in M:e\circ a=a\wedge a\circ e=a$
\end_inset
.
Če velja le eno v konjunkciji,
je
\begin_inset Formula $e$
\end_inset
bodisi leva bodisi desna enota (respectively) in v takem primeru
\begin_inset Formula $e$
\end_inset
ni enota.
\end_layout
\begin_layout Example*
Ali spodnji grupoidi imajo enoto in kakšna je?
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{R},+\right)$
\end_inset
:
enota je 0.
\end_layout
\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{N},\cdot\right)$
\end_inset
:
enota je 1.
\end_layout
\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{N},+\right)$
\end_inset
:
ni enote,
kajti
\begin_inset Formula $0\not\in\mathbb{N}$
\end_inset
.
\end_layout
\begin_layout Itemize
\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
\end_inset
:
enota je
\begin_inset Formula $I_{n\times n}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Claim*
Vsak grupoid ima kvečjemu eno enoto.
Dve enoti v istem grupoidu sta enaki.
Še več:
vsaka leva enota je enaka vsaki desni enoti.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $e$
\end_inset
leva enota in
\begin_inset Formula $f$
\end_inset
desna enota,
torej
\begin_inset Formula $\forall a:e\circ a=a\wedge a\circ f=a$
\end_inset
.
Tedaj
\begin_inset Formula $e\circ f=f$
\end_inset
in
\begin_inset Formula $e\circ f=e$
\end_inset
.
Ker je vsaka leva enota vsaki desni,
sta poljubni enoti enaki.
Enota je,
če obstaja,
ena sama in je obenem edina leva in edina desna enota.
\end_layout
\begin_layout Example*
Lahko se zgodi,
da obstaja poljubno različnih levih,
a nobene desne enote.
Primer so vse matrike oblike
\begin_inset Formula $\left[\begin{array}{cc}
a & b\\
0 & 0
\end{array}\right]$
\end_inset
.
Račun
\begin_inset Formula $\left[\begin{array}{cc}
a & b\\
0 & 0
\end{array}\right]\cdot\left[\begin{array}{cc}
c & d\\
0 & 0
\end{array}\right]=\left[\begin{array}{cc}
ac & ad\\
0 & 0
\end{array}\right]$
\end_inset
pokaže,
da so vsi elementi
\begin_inset Formula $\left[\begin{array}{cc}
1 & \times\\
0 & 0
\end{array}\right]$
\end_inset
leve enote.
Iz dejstva,
da je več (tu celo neskončno) levih enot,
sledi dejstvo,
da ni desnih.
\end_layout
\begin_layout Definition*
Polgrupi z enoto pravimo monoid.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
monoid z enoto
\begin_inset Formula $e$
\end_inset
.
Inverz elementa
\begin_inset Formula $a\in M$
\end_inset
je tak
\begin_inset Formula $b\in M\ni:b\circ a=e\wedge a\circ b=e$
\end_inset
.
Elementu,
ki zadošča levi strani konjunkcije,
pravimo levi inverz
\begin_inset Formula $a$
\end_inset
,
elemetu,
ki zadošča desni strani konjunkcije,
pa desni inverz
\begin_inset Formula $a$
\end_inset
.
Inverz
\begin_inset Formula $a$
\end_inset
je torej tak element,
ki je hkrati levi in desni inverz
\begin_inset Formula $a$
\end_inset
.
\end_layout
\begin_layout Remark*
Ni nujno,
da ima vsak element monoida inverz.
Primer je
\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
\end_inset
;
niso vse matrike obrnljive.
\end_layout
\begin_layout Claim*
Vsak element monoida ima kvečjemu en inverz.
Vsak levi inverz je enak vsakemu desnemu.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $b$
\end_inset
levi in
\begin_inset Formula $c$
\end_inset
desni inverz
\begin_inset Formula $a$
\end_inset
,
torej
\begin_inset Formula $b\circ a=e=a\circ c$
\end_inset
.
Računajmo:
\begin_inset Formula $b=b\circ e=b\circ\left(a\circ c\right)=\left(b\circ a\right)\circ c=e\circ c=c$
\end_inset
.
Če obstaja,
je torej inverz en sam,
in ta je edini levi in edini desni inverz.
\end_layout
\begin_layout Definition*
Ker vemo,
da je inverz enoličen,
lahko vpeljemo oznako
\begin_inset Formula $a^{-1}$
\end_inset
za inverz elementa
\begin_inset Formula $a$
\end_inset
.
\end_layout
\begin_layout Example*
Ali v spodnjih monoidih obstajajo inverzi in kakšni so?
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{Z},+\right)$
\end_inset
:
inverz
\begin_inset Formula $a$
\end_inset
je
\begin_inset Formula $-a$
\end_inset
.
\end_layout
\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{Z},\cdot\right)$
\end_inset
:
inverz
\begin_inset Formula $1$
\end_inset
je
\begin_inset Formula $1$
\end_inset
,
inverz
\begin_inset Formula $-1$
\end_inset
je
\begin_inset Formula $-1$
\end_inset
,
ostali elementi pa inverza nimajo.
\end_layout
\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{Q}\setminus\left\{ 0\right\} ,\cdot\right)$
\end_inset
:
inverz
\begin_inset Formula $a$
\end_inset
je
\begin_inset Formula $\frac{1}{a}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Remark*
Če desnega inverza ni,
je lahko levih inverzov več.
Primer:
Naj bodo
\begin_inset Formula $M$
\end_inset
vse funkcije
\begin_inset Formula $\mathbb{N}\to\mathbb{N}$
\end_inset
in naj bo
\begin_inset Formula $\circ$
\end_inset
kompozitum funkcij.
Tedaj velja:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $f\in M$
\end_inset
ima levi inverz
\begin_inset Formula $\Leftrightarrow f$
\end_inset
injektivna.
\end_layout
\begin_layout Itemize
\begin_inset Formula $f\in M$
\end_inset
ima desni inverz
\begin_inset Formula $\Leftrightarrow f$
\end_inset
surjektivna.
\end_layout
\begin_layout Itemize
\begin_inset Formula $f\in M$
\end_inset
ima inverz
\begin_inset Formula $\Leftrightarrow f$
\end_inset
bijektivna.
\end_layout
\end_deeper
\begin_layout Example*
\begin_inset Formula $f\left(n\right)=n+1$
\end_inset
je injektivna,
a ne surjektivna.
Vsi za komponiranje levi inverzi
\begin_inset Formula $f$
\end_inset
so funkcije oblike
\begin_inset Formula $g\left(x\right)=\begin{cases}
x-1 & ;x>1\\
\times & ;x=1
\end{cases}$
\end_inset
ZDB
\begin_inset Formula $x$
\end_inset
lahko slikajo v karkoli,
pa bo
\begin_inset Formula $\left(g\circ f\right)$
\end_inset
še vedno funkcija identiteta.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
V
\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
\end_inset
je vsak levi inverz tudi desni inverz.
To je res tudi za funkcije na končni množici,
toda ni res v splošnem.
\end_layout
\begin_layout Definition*
Grupa je tak monoid,
v katerem ima vsak element inverz.
Daljše:
grupa je taka neprazna množica
\begin_inset Formula $G$
\end_inset
z operacijo
\begin_inset Formula $\circ$
\end_inset
,
ki zadošča asociativnosti,
obstaja enota in za vsak element obstaja njegov inverz.
Grupi s komutativno operacijo pravimo Abelova grupa.
\end_layout
\begin_layout Example*
Nekaj abelovih grup:
\begin_inset Formula $\left(\mathbb{Z},+\right)$
\end_inset
,
\begin_inset Formula $\left(\mathbb{Q}\setminus\left\{ 0\right\} ,\cdot\right)$
\end_inset
,
\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),+\right)$
\end_inset
,
\begin_inset Formula $\left(\mathbb{R}^{n},+\right)$
\end_inset
.
Nekaj neabelovih grup:
\family roman
\series medium
\shape up
\size normal
\emph off
\nospellcheck off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
vse obrnljive matrike fiksne dimenzije
\family default
\series default
\shape default
\size default
\emph default
\nospellcheck default
\bar default
\strikeout default
\xout default
\uuline default
\uwave default
\noun default
\color inherit
,
\family roman
\series medium
\shape up
\size normal
\emph off
\nospellcheck off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
vse permutacije neprazne končne množice
\family default
\series default
\shape default
\size default
\emph default
\nospellcheck default
\bar default
\strikeout default
\xout default
\uuline default
\uwave default
\noun default
\color inherit
.
\end_layout
\begin_layout Subsubsection
Podstrukture
\end_layout
\begin_layout Standard
Naj bo
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
grupoid.
Reciumi,
da je
\begin_inset Formula $N$
\end_inset
neprazna podmnožica
\begin_inset Formula $M$
\end_inset
.
Pod temi pogoji se lahko zgodi,
da
\begin_inset Formula $\exists a,b\in N\ni:a\circ b\not\in N$
\end_inset
.
\end_layout
\begin_layout Example*
Oglejmo si grupoid
\begin_inset Formula $\left(\mathbb{Z},+\right)$
\end_inset
.
\begin_inset Formula $N\subseteq\mathbb{Z}$
\end_inset
naj bodo liha cela števila.
\begin_inset Formula $\forall a,b\in N:a+b\not\in N\Rightarrow\exists a,b\in N\ni:a+b\not\in N$
\end_inset
,
kajti vsota lihih števil je soda.
\end_layout
\begin_layout Definition*
Pravimo,
da je podmnožica
\begin_inset Formula $N\subseteq M$
\end_inset
zaprta za
\begin_inset Formula $\circ$
\end_inset
,
če
\begin_inset Formula $\forall a,b\in N:a\circ b\in N$
\end_inset
.
\end_layout
\begin_layout Example*
Oglejmo si spet grupoid
\begin_inset Formula $\left(\mathbb{Z},+\right)$
\end_inset
.
\begin_inset Formula $N\subseteq\mathbb{Z}$
\end_inset
naj bodo soda cela števila.
\begin_inset Formula $N$
\end_inset
je zaprta za
\begin_inset Formula $+$
\end_inset
.
\end_layout
\begin_layout Definition*
Takemu
\begin_inset Formula $N$
\end_inset
,
kjer je
\begin_inset Formula $N\subseteq M$
\end_inset
,
z implicitno podedovano operacijo (
\begin_inset Formula $a\circ_{N}b=a\circ b$
\end_inset
) pravimo podgrupoid
\begin_inset Formula $\left(N,\circ_{N}\right)$
\end_inset
.
\end_layout
\begin_layout Exercise*
Pokaži,
da je
\begin_inset Quotes gld
\end_inset
general linear
\begin_inset Quotes grd
\end_inset
\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)\coloneqq\left\{ A\in M_{n\times n}\left(\mathbb{R}\right);\det A\not=0\right\} $
\end_inset
grupa za matrično množenje.
\end_layout
\begin_deeper
\begin_layout Standard
Asociativnost je dokazana zgoraj.
Enota je
\begin_inset Formula $I_{n}$
\end_inset
.
Inverzi obstajajo,
ker so determinante neničelne in tudi inverzi imajo neničelne determinante.
Preveriti je treba še vsebovanost,
torej
\begin_inset Formula $\forall A,B\in GL_{n}\left(\mathbb{R}\right):A\cdot B\in GL_{n}\left(\mathbb{R}\right)$
\end_inset
.
Vzemimo poljubni
\begin_inset Formula $A,B\in GL_{n}\left(\mathbb{R}\right)$
\end_inset
,
torej
\begin_inset Formula $\det A\not=0\wedge\det B\not=0$
\end_inset
.
\begin_inset Formula $\det\left(AB\right)=\det A\det B=0\Leftrightarrow\det A=0\vee\det B=0$
\end_inset
,
toda ker noben izmed izrazov disjunkcije ne drži,
determinanta
\begin_inset Formula $AB$
\end_inset
nikdar ni 0.
Enota
\begin_inset Formula $I$
\end_inset
je vsebovana v
\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$
\end_inset
,
saj
\begin_inset Formula $\det I=1\not=0$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
Ali je
\begin_inset Quotes gld
\end_inset
special linear
\begin_inset Quotes grd
\end_inset
\begin_inset Formula $SL_{n}\left(\mathbb{R}\right)\coloneqq\left\{ A\in M_{n\times n}\left(\mathbb{R}\right);\det A=1\right\} $
\end_inset
grupa za matrično množenje?
\end_layout
\begin_deeper
\begin_layout Standard
Vse lastnosti (razen vsebovanosti) smo preverili zgoraj.
Preveriti je treba vsebovanost,
torej ali
\begin_inset Formula $\forall A,B\in SL_{n}\left(\mathbb{R}\right):A\cdot B\in SL_{n}\left(\mathbb{R}\right)$
\end_inset
.
Vzemimo poljubni
\begin_inset Formula $A,B\in SL_{n}\left(\mathbb{R}\right)$
\end_inset
,
torej
\begin_inset Formula $\det A=1\wedge\det B=1$
\end_inset
.
\begin_inset Formula $\det\left(AB\right)=\det A\det B=1\cdot1=1$
\end_inset
.
Preveriti je treba še,
da so inverzi vsebovani.
Za poljubno
\begin_inset Formula $A\in SL_{n}\left(\mathbb{R}\right)$
\end_inset
je
\begin_inset Formula $\det A^{-1}=\frac{1}{\det A}=1$
\end_inset
,
ker je
\begin_inset Formula $\det A=1$
\end_inset
.
Enota
\begin_inset Formula $I$
\end_inset
je vsebovana v
\begin_inset Formula $SL_{n}\left(\mathbb{R}\right)$
\end_inset
,
saj
\begin_inset Formula $\det I=1$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Fact*
Za podedovano operacijo
\begin_inset Formula $\circ_{N}$
\end_inset
v podstrukturi se asociativnost in komutativnost podedujeta,
ni pa nujno,
da če obstaja enota v
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
,
obstaja enota tudi v
\begin_inset Formula $\left(N,\circ_{N}\right)$
\end_inset
.
Prav tako ni rečeno,
da se podeduje obstoj inverzov.
\end_layout
\begin_layout Definition*
Če je
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
polgrupa (asociativen grupoid) in
\begin_inset Formula $N\subseteq M$
\end_inset
,
pravimo,
da je
\begin_inset Formula $N$
\end_inset
podpolgrupa,
če je zaprta za
\begin_inset Formula $\circ$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:podmonoid"
\end_inset
Če je
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
monoid (polgrupa z enoto) in
\begin_inset Formula $N\subseteq M$
\end_inset
,
je
\begin_inset Formula $N$
\end_inset
podmonoid,
če je zaprt za
\begin_inset Formula $\circ$
\end_inset
in vsebuje enoto iz
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
(prav tisto enoto,
glej primer
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:nxnmonoid"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
spodaj).
\end_layout
\begin_layout Example*
\begin_inset Formula $\left(\mathbb{N},\cdot\right)$
\end_inset
je monoid.
Soda števila so podpolgrupa (zaprta so za množenje),
niso pa podmonoid,
saj ne vsebujejo enice (enote).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:nxnmonoid"
\end_inset
\begin_inset Formula $\left(\mathbb{N}\times\mathbb{N},\circ\right)$
\end_inset
je monoid za operacijo
\begin_inset Formula $\left(a,b\right)\circ\left(c,d\right)=\left(ac,bd\right)$
\end_inset
,
saj je enota
\begin_inset Formula $\left(1,1\right)$
\end_inset
.
\begin_inset Formula $\left(\mathbb{N}\times\left\{ 0\right\} ,\circ\right)$
\end_inset
pa za
\begin_inset Formula $\circ$
\end_inset
kot prej je sicer podpolgrupa v
\begin_inset Formula $\left(\mathbb{N}\times\mathbb{N},\circ\right)$
\end_inset
in ima enoto
\begin_inset Formula $\left(1,0\right)$
\end_inset
,
vendar,
ker
\begin_inset Formula $\left(1,0\right)\not=\left(1,1\right)$
\end_inset
,
to ni podmonoid.
Enota mora torej biti,
kot pravi definicija
\begin_inset CommandInset ref
LatexCommand ref
reference "def:podmonoid"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
,
ista kot enota v
\begin_inset Quotes gld
\end_inset
starševski
\begin_inset Quotes grd
\end_inset
strukturi.
\end_layout
\begin_layout Definition*
Če je
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
grupa in
\begin_inset Formula $N\subseteq M$
\end_inset
,
pravimo,
da je
\begin_inset Formula $N$
\end_inset
podgrupa
\begin_inset Formula $\Longleftrightarrow$
\end_inset
hkrati velja
\end_layout
\begin_deeper
\begin_layout Itemize
je zaprta za
\begin_inset Formula $\circ$
\end_inset
,
\end_layout
\begin_layout Itemize
vsebuje isto enoto kot
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
in
\end_layout
\begin_layout Itemize
vsebuje inverz vsakega svojega elementa;
ti inverzi pa so itak po enoličnosti enaki inverzom iz
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Example*
special linear,
\begin_inset Formula $SL_{n}$
\end_inset
,
grupa vseh matrik z determinanto enako 1,
je podgrupa
\begin_inset Quotes gld
\end_inset
general linear
\begin_inset Quotes grd
\end_inset
,
\begin_inset Formula $GL_{n}$
\end_inset
,
grupe vseh obrnljivih
\begin_inset Formula $n\times n$
\end_inset
matrik,
kajti
\begin_inset Formula $\det I=1$
\end_inset
,
\begin_inset Formula $\det$
\end_inset
je multiplikativna (glej vajo zgoraj) in
\begin_inset Formula $\det A=1\Leftrightarrow\det A^{-1}=1$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
ortogonalne matrike,
\begin_inset Formula $O_{n}$
\end_inset
,
vse
\begin_inset Formula $n\times n$
\end_inset
matrike
\begin_inset Formula $A$
\end_inset
,
ki zadoščajo
\begin_inset Formula $A^{T}A=I$
\end_inset
,
je podgrupa
\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$
\end_inset
,
kajti:
\end_layout
\begin_deeper
\begin_layout Itemize
Je zaprta:
\begin_inset Formula
\[
A,B\in O_{n}\overset{?}{\Longrightarrow}AB\in O_{n}
\]
\end_inset
\begin_inset Formula
\[
\left(AB\right)^{T}\left(AB\right)\overset{?}{=}I
\]
\end_inset
\begin_inset Formula
\[
B^{T}\left(A^{T}A\right)B\overset{?}{=}I
\]
\end_inset
\begin_inset Formula
\[
I=I
\]
\end_inset
\end_layout
\begin_layout Itemize
Vsebuje enoto
\begin_inset Formula $I$
\end_inset
:
\begin_inset Formula
\[
I^{T}I=I
\]
\end_inset
\end_layout
\begin_layout Itemize
Vsebuje inverze vseh svojih elementov:
Uporabimo
\begin_inset Formula $A^{T}A=I\Rightarrow A^{T}=A^{-1}$
\end_inset
\begin_inset Formula
\[
A\in O_{n}\overset{?}{\Longrightarrow}A^{-1}\in O_{n}
\]
\end_inset
\begin_inset Formula
\[
\left(A^{-1}\right)^{T}A^{-1}\overset{?}{=}I
\]
\end_inset
\begin_inset Formula
\[
\left(A^{T}\right)^{T}A^{T}=AA^{T}=I
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Fact*
specialna ortogonalna grupa,
\begin_inset Formula $SO_{n}\coloneqq O_{n}\cap SL_{n}$
\end_inset
je podgrupa
\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$
\end_inset
.
Dokazati je moč še bolj splošno,
namreč,
da je presek dveh podgrup spet podgrupa.
\begin_inset Note Note
status open
\begin_layout Plain Layout
DOKAŽI?????
\end_layout
\end_inset
\end_layout
\begin_layout Claim*
Naj bo
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
grupa in
\begin_inset Formula $N\subseteq M$
\end_inset
neprazna.
Tedaj velja
\begin_inset Formula $N$
\end_inset
podgrupa
\begin_inset Formula $\Leftrightarrow\forall a,b\in N:a\circ b^{-1}\in N$
\end_inset
(zaprtost za odštevanje —
v abelovih grupah namreč običajno operacijo označimo s
\begin_inset Formula $+$
\end_inset
in označimo
\begin_inset Formula $a+b^{-1}=a-b$
\end_inset
).
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Naj bo
\begin_inset Formula $N$
\end_inset
podgrupa v
\begin_inset Formula $\left(M,\circ\right)$
\end_inset
.
Vzemimo
\begin_inset Formula $a,b\in N$
\end_inset
.
Upoštevamo
\begin_inset Formula $b\in N\Rightarrow b^{-1}\in N$
\end_inset
iz definicije podgrupe.
Torej velja
\begin_inset Formula $a,b^{-1}\in N\Rightarrow a\circ b^{-1}\in N$
\end_inset
,
zopet iz definicije podgrupe.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
Naj
\begin_inset Formula $\forall a,b\in N:a\circ b^{-1}\in N$
\end_inset
.
Preverimo lastnosti iz definicije podgrupe:
\end_layout
\begin_deeper
\begin_layout Itemize
Vsebovanost enote:
Ker je
\begin_inset Formula $N$
\end_inset
neprazna,
vsebuje nek
\begin_inset Formula $a$
\end_inset
.
Po predpostavki je
\begin_inset Formula $a\circ a^{-1}\in N$
\end_inset
,
\begin_inset Formula $a\circ a^{-1}$
\end_inset
pa je po definiciji inverza enota.
\end_layout
\begin_layout Itemize
Vsebovanost inverzov:
Naj bo
\begin_inset Formula $a\in N$
\end_inset
poljuben.
Od prej vemo,
da
\begin_inset Formula $e\in N$
\end_inset
.
Po predpostavki,
ker
\begin_inset Formula $e,a\in N\Rightarrow e\circ a^{-1}\in N$
\end_inset
,
\begin_inset Formula $e\circ a^{-1}$
\end_inset
pa je po definiciji enote
\begin_inset Formula $a^{-1}$
\end_inset
.
\end_layout
\begin_layout Itemize
Zaprtost:
Naj bosta
\begin_inset Formula $a,b\in N$
\end_inset
poljubna.
Od prej vemo,
da
\begin_inset Formula $b^{-1}\in N$
\end_inset
.
Po predpostavki,
ker
\begin_inset Formula $a,b^{-1}\in N\Rightarrow a\circ\left(b^{-1}\right)^{-1}\in N$
\end_inset
,
\begin_inset Formula $a\circ\left(b^{-1}\right)^{-1}$
\end_inset
pa je po definiciji inverza
\begin_inset Formula $a\circ b$
\end_inset
.
\end_layout
\end_deeper
\end_deeper
\begin_layout Subsubsection
Homomorfizmi
\end_layout
\begin_layout Standard
\begin_inset Formula $\sim$
\end_inset
so operacije,
ki
\begin_inset Quotes gld
\end_inset
ohranjajo strukturo
\begin_inset Quotes grd
\end_inset
.
\end_layout
\begin_layout Definition*
Naj bosta
\begin_inset Formula $\left(M_{1},\circ_{1}\right)$
\end_inset
in
\begin_inset Formula $\left(M_{2},\circ_{2}\right)$
\end_inset
dva grupoida.
Preslikava
\begin_inset Formula $f:M_{1}\to M_{2}$
\end_inset
je homomorfizem grupoidov,
če
\begin_inset Formula $\forall a,b\in M_{1}:f\left(a\circ_{1}b\right)=f\left(a\right)\circ_{2}f\left(b\right)$
\end_inset
.
Enaka definicija v polgrupah.
Za homomorfizem monoidov zahtevamo še,
da
\begin_inset Formula $f\left(e_{1}\right)=e_{2}$
\end_inset
,
kjer je
\begin_inset Formula $e_{1}$
\end_inset
enota
\begin_inset Formula $M_{1}$
\end_inset
in
\begin_inset Formula $e_{2}$
\end_inset
enota
\begin_inset Formula $M_{2}$
\end_inset
.
\end_layout
\begin_layout Example*
\begin_inset Formula $f:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$
\end_inset
,
ki slika
\begin_inset Formula $a\mapsto\left(a,0\right)$
\end_inset
.
\begin_inset Formula $\circ_{1}$
\end_inset
naj bo množenje,
\begin_inset Formula $\circ_{2}$
\end_inset
pa
\begin_inset Formula $\left(a,b\right)\circ_{2}\left(c,d\right)=\left(ac,bd\right)$
\end_inset
(množenje po komponentah).
\begin_inset Formula $\left(1,1\right)$
\end_inset
je enota v
\begin_inset Formula $\mathbb{N\times\mathbb{N}}$
\end_inset
,
\begin_inset Formula $1$
\end_inset
pa je enota v
\begin_inset Formula $\mathbb{N}$
\end_inset
.
\begin_inset Formula $f$
\end_inset
je homomorfizem,
ker
\begin_inset Formula $f\left(a\circ_{1}b\right)=\left(a\cdot b,0\right)=\left(a,0\right)\circ_{2}\left(b,0\right)=f\left(a\right)\circ_{2}f\left(b\right)$
\end_inset
,
ni pa homomorfizem monoidov,
saj
\begin_inset Formula $f\left(1\right)=\left(1,0\right)\not=\left(1,1\right)$
\end_inset
.
\end_layout
\begin_layout Definition*
Za homomorfizem grup zahtevamo še,
da
\begin_inset Formula $f\left(a^{-1}\right)=f\left(a\right)^{-1}$
\end_inset
.
\end_layout
\begin_layout Remark*
Izkaže se,
da ohranjanje enote in inverzov pri homomorfizmih grup sledi že iz definicije homomorfizmov grupoidov.
\end_layout
\begin_layout Claim*
Naj bosta
\begin_inset Formula $\left(M_{1},\circ_{1}\right)$
\end_inset
in
\begin_inset Formula $\left(M_{2},\circ_{2}\right)$
\end_inset
grupi.
Naj bo
\begin_inset Formula $f:M_{1}\to M_{2}$
\end_inset
preslikava,
ki je homomorfizem grupoidov.
Trdimo,
da slika enoto v enoto in inverze v inverze.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $e_{1}$
\end_inset
enota za
\begin_inset Formula $\left(M_{1},\circ_{1}\right)$
\end_inset
in
\begin_inset Formula $e_{2}$
\end_inset
enota za
\begin_inset Formula $\left(M_{2},\circ_{2}\right)$
\end_inset
.
Dokažimo,
da
\begin_inset Formula $f\left(e_{1}\right)\overset{?}{=}e_{2}$
\end_inset
.
\begin_inset Formula
\[
f\left(e_{1}\right)=f\left(e_{1}\circ_{1}e_{1}\right)=f\left(e_{1}\right)\circ_{2}f\left(e_{1}\right)=f\left(e_{1}\right)^{-1}\circ f\left(e_{1}\right)\circ e_{2}=e_{2}\circ e_{2}=e_{2}
\]
\end_inset
Dokažimo še ohranjanje inverzov,
se pravi
\begin_inset Formula $b$
\end_inset
je inverz
\begin_inset Formula $a\overset{?}{\Longrightarrow}f\left(b\right)$
\end_inset
je inverz
\begin_inset Formula $f\left(a\right)$
\end_inset
.
\begin_inset Formula
\[
a\circ_{1}b=e_{1}\overset{?}{\Longrightarrow}f\left(a\right)\circ_{2}f\left(b\right)=f\left(a\circ_{1}b\right)=f\left(e_{1}\right)=e_{2}
\]
\end_inset
\begin_inset Formula
\[
b\circ_{1}a=e_{1}\overset{?}{\Longrightarrow}f\left(b\right)\circ_{2}f\left(a\right)=f\left(b\circ_{1}a\right)=f\left(e_{1}\right)=e_{2}
\]
\end_inset
\end_layout
\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:primeri-homomorfizmov"
\end_inset
Primeri homomorfizmov.
\end_layout
\begin_deeper
\begin_layout Enumerate
Determinanta:
\begin_inset Formula $M_{n}\left(\mathbb{R}\right)\to\mathbb{R}$
\end_inset
je homomorfizem,
ker ima multiplikativno lastnost:
\begin_inset Formula $\det\left(AB\right)=\det A\det B$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:permutacijska-matrika"
\end_inset
\begin_inset Formula $S_{n}$
\end_inset
so vse permutacije množice
\begin_inset Formula $\left\{ 1..n\right\} $
\end_inset
.
Vsaki permutaciji
\begin_inset Formula $\sigma\in S_{n}$
\end_inset
priredimo permutacijsko matriko
\begin_inset Formula $P_{\sigma}\in M_{n}\left(\mathbb{R}\right)$
\end_inset
tako,
da vsebuje vektorje standardne baze
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
kot stolpce:
\begin_inset Formula
\[
P_{\sigma}\coloneqq\left[\begin{array}{ccc}
\vec{e_{\sigma\left(1\right)}} & \cdots & \vec{e_{\sigma\left(n\right)}}\end{array}\right]
\]
\end_inset
Imamo preslikavo
\begin_inset Formula $S\to M_{n}\left(\mathbb{R}\right)$
\end_inset
,
ki slika
\begin_inset Formula $\sigma\mapsto P_{\sigma}$
\end_inset
in trdimo,
da je homomorfizem.
Dokažimo,
da je
\begin_inset Formula $\forall\sigma,\tau\in S_{n}:P_{\sigma\circ\tau}=P_{\sigma}\cdot P_{\tau}$
\end_inset
.
Opazimo,
da je
\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :P_{\sigma}\vec{e_{i}}=\vec{e_{\sigma\left(i\right)}}$
\end_inset
(tu množimo matriko z vektorjem).
Če namesto
\begin_inset Formula $i$
\end_inset
pišemo
\begin_inset Formula $\tau\left(i\right)$
\end_inset
,
dobimo
\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :P_{\sigma}\vec{e_{\tau\left(i\right)}}=\vec{e_{\left(\sigma\circ\tau\right)\left(i\right)}}$
\end_inset
.
Preverimo sedaj množenje
\begin_inset Formula $P_{\sigma}P_{\tau}=P_{\sigma}\left[\begin{array}{ccc}
\vec{e_{\tau\left(1\right)}} & \cdots & \vec{e_{\tau\left(n\right)}}\end{array}\right]=\left[\begin{array}{ccc}
P_{\sigma}\vec{e_{\tau\left(1\right)}} & \cdots & P_{\sigma}\vec{e_{\tau\left(n\right)}}\end{array}\right]=\left[\begin{array}{ccc}
\vec{e_{\left(\sigma\circ\tau\right)\left(1\right)}} & \cdots & \vec{e_{\left(\sigma\circ\tau\right)\left(n\right)}}\end{array}\right]=P_{\sigma\circ\tau}$
\end_inset
.
Preslikava je res homomorfizem.
\end_layout
\end_deeper
\begin_layout Claim*
Kompozitum dveh homomorfizmov je tudi sam zopet homomorfizem.
\end_layout
\begin_layout Proof
Imejmo tri grupoide in homomorfizma,
ki slikata med njimi takole:
\begin_inset Formula $\left(M_{1},\circ_{1}\right)\overset{f}{\longrightarrow}\left(M_{2},\circ_{2}\right)\overset{g}{\longrightarrow}\left(M_{3},\circ_{3}\right)$
\end_inset
.
Dokažimo,
da je
\begin_inset Formula $g\circ f$
\end_inset
spet homorfizem.
\begin_inset Formula
\[
\left(g\circ f\right)\left(a\circ_{1}b\right)=g\left(f\left(a\circ_{1}b\right)\right)=g\left(f\left(a\right)\circ_{2}f\left(b\right)\right)=g\left(f\left(a\right)\right)\circ_{3}g\left(f\left(b\right)\right)=\left(g\circ f\right)\left(a\right)\circ_{3}\left(g\circ f\right)\left(b\right)
\]
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula $S_{n}\overset{\sigma}{\longrightarrow}M_{n}\left(\mathbb{R}\right)\overset{\det}{\rightarrow}\mathbb{R}$
\end_inset
,
kjer je
\begin_inset Formula $\sigma$
\end_inset
preslikava iz točke
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:permutacijska-matrika"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
zgleda
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:primeri-homomorfizmov"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
zgoraj.
\begin_inset Formula $\sgn=\det\circ\sigma$
\end_inset
,
kjer je
\begin_inset Formula $\sgn$
\end_inset
parnost permutacije.
Preslikava
\begin_inset Formula $\sgn$
\end_inset
je homomorfizem,
ker je kompozitum dveh homomorfizmov.
\end_layout
\begin_layout Definition*
Izomorfizem je preslikava,
ki je bijektivna in je homomorfizem.
Dve grupi sta izomorfni,
kadar med njima obstaja izomorfizem.
\end_layout
\begin_layout Remark*
S stališča algebre sta dve izomorfni grupi v abstraktnem smislu enaki,
saj je izomorfizem zgolj reverzibilno preimenovanje elementov.
\end_layout
\begin_layout Subsubsection
Bigrupoidi,
polkolobarji,
kolobarji
\end_layout
\begin_layout Definition*
Neprazni množici
\begin_inset Formula $M$
\end_inset
z dvema operacijama
\begin_inset Formula $\circ_{1}$
\end_inset
in
\begin_inset Formula $\circ_{2}$
\end_inset
pravimo bigrupoid in ga označimo z
\begin_inset Formula $\left(M,\circ_{1},\circ_{2}\right)$
\end_inset
.
Običajno operaciji označimo z
\begin_inset Formula $+,\cdot$
\end_inset
,
tedaj bigrupoid pišemo kot
\begin_inset Formula $\left(M,+,\cdot\right)$
\end_inset
.
\end_layout
\begin_layout Quotation
\begin_inset Quotes gld
\end_inset
Če
\begin_inset Formula $+$
\end_inset
in
\begin_inset Formula $\cdot$
\end_inset
ena z drugo nimata nobene zveze,
je vseeno,
če ju študiramo skupaj ali posebej.
\begin_inset Quotes grd
\end_inset
\end_layout
\begin_layout Definition*
Distributivnost je značilnost bigrupoida
\begin_inset Formula $\left(M,+,\cdot\right)$
\end_inset
.
Ločimo levo distributivnost:
\begin_inset Formula $\forall a,b,c\in M:a\cdot\left(b+c\right)=a\cdot b+a\cdot c$
\end_inset
in desno distributivnost:
\begin_inset Formula $\forall a,b,c\in M:\left(a+b\right)\cdot c=a\cdot c+b\cdot c$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Bigrupoid,
ki zadošča levi in desni distributivnosti,
je distributiven.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Distributiven bigrupoid,
je polkolobar,
če je
\begin_inset Formula $\left(M,+\right)$
\end_inset
komutativna polgrupa.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Distributiven grupoid je kolobar,
če je
\begin_inset Formula $\left(M,+\right)$
\end_inset
komutativna grupa.
\end_layout
\begin_layout Example*
Primer polkolobarja,
ki ni kolobar,
je
\begin_inset Formula $\left(\mathbb{N},+,\cdot\right)$
\end_inset
.
Ni enote niti inverza za
\begin_inset Formula $+$
\end_inset
,
\begin_inset Formula $\left(\mathbb{N},+\right)$
\end_inset
pa je polgrupa.
\end_layout
\begin_layout Standard
Kolobarje delimo glede na lastnosti operacije
\begin_inset Formula $\cdot$
\end_inset
:
\end_layout
\begin_layout Definition*
Asociativen kolobar je tak,
kjer je
\begin_inset Formula $\cdot$
\end_inset
asociativna operacija
\begin_inset Formula $\sim\left(M,\cdot\right)$
\end_inset
je polgrupa.
\end_layout
\begin_layout Example*
Primer kolobarja,
ki ni asociativen,
je
\begin_inset Formula $\left(\mathbb{R}^{3},+,\times\right)$
\end_inset
,
kjer je
\begin_inset Formula $\times$
\end_inset
vektorski produkt.
Primer kolobarja,
ki je asociativen,
je
\begin_inset Formula $\left(M_{n}\left(\mathbb{R}\right),+,\cdot\right)$
\end_inset
,
kjer je
\begin_inset Formula $\cdot$
\end_inset
matrično množenje.
\end_layout
\begin_layout Definition*
Asociativen kolobar z enoto je tak,
ki ima multiplikativno enoto,
torej enoto za drugo operacijo
\begin_inset Formula $\sim\left(M,\cdot\right)$
\end_inset
je monoid.
Tipično se enoto za
\begin_inset Formula $\cdot$
\end_inset
označi z 1,
enoto za
\begin_inset Formula $+$
\end_inset
pa z 0.
\end_layout
\begin_layout Example*
Primer asociativnega kolobarja brez enote je
\begin_inset Formula $\left(\text{soda }\mathbb{N},+,\cdot\right)$
\end_inset
.
Primer asociativnega kolobarja z enoto je
\begin_inset Formula $\left(\mathbb{N},+,\cdot\right)$
\end_inset
.
\end_layout
\begin_layout Definition*
\begin_inset Formula $b$
\end_inset
je inverz
\begin_inset Formula $a$
\end_inset
,
če
\begin_inset Formula $b\cdot a=e$
\end_inset
in
\begin_inset Formula $a\cdot b=e$
\end_inset
,
kjer je
\begin_inset Formula $e$
\end_inset
multiplikativna enota kolobarja.
\end_layout
\begin_layout Remark*
Element 0 nima nikoli inverza,
ker
\begin_inset Formula $\forall a\in M:0\cdot a=0$
\end_inset
.
\end_layout
\begin_layout Proof
\begin_inset Formula $\cancel{0\cdot a}=\left(0+0\right)\cdot a=0\cdot a+\cancel{0\cdot a}$
\end_inset
(dokaz velja za kolobarje,
ne pa polkolobarje,
ker imamo pravilo krajšanja
\begin_inset Foot
status open
\begin_layout Plain Layout
Dokaz v mojih Odgovorih na vprašanja za ustni izpit Diskretnih struktur 2 IŠRM
\end_layout
\end_inset
le,
kadar je
\begin_inset Formula $\left(M,+\right)$
\end_inset
grupa).
\end_layout
\begin_layout Definition*
Asociativen kolobar z enoto,
v katerem ima vsak neničen element inverz,
je obseg.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Kolobar je komutativen,
če je
\begin_inset Formula $\cdot$
\end_inset
komutativna operacija (
\begin_inset Formula $+$
\end_inset
je itak po definiciji že komutativna).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Komutativen obseg je polje.
\end_layout
\begin_layout Example*
Primeri polj:
\begin_inset Formula $\left(\mathbb{Q},+,\cdot\right)$
\end_inset
,
\begin_inset Formula $\left(\mathbb{R},+,\cdot\right)$
\end_inset
,
\begin_inset Formula $\left(\mathbb{C},+,\cdot\right)$
\end_inset
,
\begin_inset Formula $\left(F\left[\mathbb{R}\right],+,\cdot\right)$
\end_inset
,
kjer je
\begin_inset Formula $F\left[\mathbb{R}\right]$
\end_inset
polje racionalnih funkcij.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Primer obsega,
ki ni polje:
\begin_inset Formula $\left(\mathbb{H},+,\cdot\right)$
\end_inset
.
\end_layout
\begin_layout Definition*
Kvaternioni so
\begin_inset Formula $M_{2\times2}\left(\mathbb{R}\right)$
\end_inset
take oblike:
za
\begin_inset Formula $\alpha,\beta\in\mathbb{C}$
\end_inset
je
\begin_inset Formula $\mathbb{H}\coloneqq\left[\begin{array}{cc}
\alpha & \beta\\
-\overline{\beta} & \overline{\alpha}
\end{array}\right]=\left[\begin{array}{cc}
a+bi & c+di\\
-c+di & a-bi
\end{array}\right]=\left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right]a+\left[\begin{array}{cc}
i & 0\\
0 & -i
\end{array}\right]b+\left[\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right]c+\left[\begin{array}{cc}
0 & i\\
i & 0
\end{array}\right]d=1a+bi+cj+dk$
\end_inset
za
\begin_inset Formula $a,b,c,d\in\mathbb{R}$
\end_inset
in dimenzije
\begin_inset Formula $1,i,j,k$
\end_inset
.
\end_layout
\begin_layout Example*
Primer kolobarja:
Naj bo
\begin_inset Formula $X$
\end_inset
neprazna množica in
\begin_inset Formula $R$
\end_inset
kolobar.
\begin_inset Formula $R^{X}$
\end_inset
so vse funkcije
\begin_inset Formula $X\to R$
\end_inset
.
Naj bosta
\begin_inset Formula $f,g\in R^{X}$
\end_inset
.
Definirajmo operaciji:
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $+$
\end_inset
\begin_inset Formula $f+g\coloneqq\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\cdot$
\end_inset
\begin_inset Formula $f\cdot g\coloneqq\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$
\end_inset
\end_layout
\end_deeper
\begin_layout Subsubsection
Podkolobarji
\end_layout
\begin_layout Definition*
Podbigrupoid od
\begin_inset Formula $\left(M,+,\cdot\right)$
\end_inset
je taka podmnožica
\begin_inset Formula $N\subseteq M$
\end_inset
,
ki je zaprta za
\begin_inset Formula $+$
\end_inset
in
\begin_inset Formula $\cdot$
\end_inset
ZDB
\begin_inset Formula $N\subseteq M$
\end_inset
je podgrupoid v
\begin_inset Formula $\left(M,+\right)$
\end_inset
in
\begin_inset Formula $\left(M,\cdot\right)$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Podkolobar kolobarja
\begin_inset Formula $\left(M,+,\cdot\right)$
\end_inset
je taka podmnožica
\begin_inset Formula $N\subseteq M$
\end_inset
,
da je
\begin_inset Formula $N$
\end_inset
podgrupa v
\begin_inset Formula $\left(M,+\right)$
\end_inset
in
\begin_inset Formula $N$
\end_inset
podgrupoid v
\begin_inset Formula $\left(N,\cdot\right)\Leftrightarrow N$
\end_inset
zaprta za
\begin_inset Formula $\cdot$
\end_inset
.
Skrajšana definicija je torej,
da je
\begin_inset Formula $\forall a,b\in N:a+b^{-1}\in N\wedge a\cdot b\in N$
\end_inset
,
torej zaprtost za odštevanje in množenje.
\end_layout
\begin_layout Example*
Primeri podkolobarjev
\end_layout
\begin_deeper
\begin_layout Itemize
v
\begin_inset Formula $\left(M_{n}\left(\mathbb{R}\right),+,\cdot\right)$
\end_inset
\end_layout
\begin_deeper
\begin_layout Itemize
zgornjetrikotne matrike
\end_layout
\begin_layout Itemize
diagonalne matrike
\end_layout
\begin_layout Itemize
matrike s spodnjo vrstico ničelno
\end_layout
\begin_layout Itemize
matrike z ničelnim
\begin_inset Formula $i-$
\end_inset
tim stolpcem
\end_layout
\begin_layout Itemize
\begin_inset Formula $M_{n}\left(\mathbb{Z}\right)$
\end_inset
,
\begin_inset Formula $M_{n}\left(\mathbb{Q}\right)$
\end_inset
\end_layout
\begin_layout Itemize
matrike oblike
\begin_inset Formula $\left[\begin{array}{cc}
a & b\\
b & a
\end{array}\right]$
\end_inset
\end_layout
\end_deeper
\begin_layout Itemize
v
\begin_inset Formula $\left(\mathbb{R}^{[a,b]},+,\cdot\right)$
\end_inset
(vse funkcije
\begin_inset Formula $\left[a,b\right]\to\mathbb{R}$
\end_inset
za seštevanje in množenje)
\end_layout
\begin_deeper
\begin_layout Itemize
vse omejene funkcije
\end_layout
\begin_layout Itemize
vse zvezne funkcije
\end_layout
\begin_layout Itemize
vse odvedljive funkcije
\end_layout
\end_deeper
\end_deeper
\begin_layout Definition*
Podobseg obsega
\begin_inset Formula $\left(M,+,\cdot\right)$
\end_inset
je taka
\begin_inset Formula $N\subseteq M$
\end_inset
,
da velja:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $N$
\end_inset
podgrupa v
\begin_inset Formula $\left(M,+\right)$
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $N\setminus\left\{ 0\right\} $
\end_inset
podgrupa v
\begin_inset Formula $\left(M\setminus\left\{ 0\right\} ,\cdot\right)$
\end_inset
\end_layout
\begin_layout Standard
ZDB:
\begin_inset Formula $N$
\end_inset
je zaprta za odštevanje (seštevanje z aditivnim inverzom) in za deljenje (množenje z multiplikativnim inverzom) z neničelnimi elementi.
\end_layout
\end_deeper
\begin_layout Example*
Primeri podobsegov:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\mathbb{R}$
\end_inset
je podobseg v
\begin_inset Formula $\left(\mathbb{C},+,\cdot\right)$
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $\mathbb{Q}$
\end_inset
je podobseg v
\begin_inset Formula $\left(\mathbb{R},+,\cdot\right)$
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Izkaže se,
da je najmanjše podpolje v
\begin_inset Formula $\mathbb{R}$
\end_inset
,
ki vsebuje
\begin_inset Formula $\mathbb{Q}$
\end_inset
in
\begin_inset Formula $\sqrt{3}$
\end_inset
množica
\begin_inset Formula $\left\{ a+b\sqrt{3};\forall a,b\in\mathbb{Q}\right\} $
\end_inset
.
Očitno je zaprt za odštevanje.
Za deljenje?
\end_layout
\begin_deeper
\begin_layout Standard
\begin_inset Formula
\[
\frac{a+b\sqrt{3}}{c+d\sqrt{3}}=\frac{\left(a+b\sqrt{3}\right)\left(c-d\sqrt{3}\right)}{\left(c+d\sqrt{3}\right)\left(c-d\sqrt{3}\right)}=\frac{ac-ad\sqrt{3}+bc\sqrt{3}-3bd}{c^{2}-3d^{2}}=\frac{ac-3bd+\left(bc-ad\right)\sqrt{3}}{c^{2}-3d^{2}}=
\]
\end_inset
\begin_inset Formula
\[
=\frac{ac-3bd}{c^{2}-3d^{2}}+\frac{bc-ad}{c^{2}-3d^{2}}\sqrt{3}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Subsubsection
Homomorfizmi kolobarjev
\end_layout
\begin_layout Definition*
Naj bosta
\begin_inset Formula $\left(M_{1},+_{1},\cdot_{1}\right)$
\end_inset
in
\begin_inset Formula $\left(M_{2},+_{2},\cdot_{2}\right)$
\end_inset
kolobarja.
\begin_inset Formula $f:M_{1}\to M_{2}$
\end_inset
je homomorfizem kolobarjev
\begin_inset Formula $\Leftrightarrow\forall a,b\in M_{1}:f\left(a+_{1}b\right)=f\left(a\right)+_{2}f\left(b\right)\wedge f\left(a\cdot_{1}b\right)=f\left(a\right)\cdot_{2}f\left(b\right)$
\end_inset
.
ZDB
\begin_inset Formula $f$
\end_inset
mora biti homomorfizem grupoidov
\begin_inset Formula $\left(M_{1},+_{1}\right)\to\left(M_{2},+_{2}\right)$
\end_inset
in
\begin_inset Formula $\left(M_{1},\cdot_{1}\right)\to\left(M_{2},\cdot_{2}\right)$
\end_inset
.
Za homomorfizem kolobarjev z enoto zahtevamo še
\begin_inset Formula $f\left(1_{1}\right)=1_{2}$
\end_inset
.
\end_layout
\begin_layout Example*
\begin_inset Formula $f:M_{2}\left(\mathbb{R}\right)\to M_{3}\left(\mathbb{R}\right)$
\end_inset
s predpisom
\begin_inset Formula $\left[\begin{array}{cc}
a & b\\
c & d
\end{array}\right]\mapsto\left[\begin{array}{ccc}
a & b & 0\\
c & d & 0\\
0 & 0 & 0
\end{array}\right]$
\end_inset
je homomorfizem kolobarjev,
ni pa homomorfizem kolobarjev z enoto,
kajti
\begin_inset Formula $f\left(\left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right]\right)=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{array}\right]$
\end_inset
,
kar ni enota v
\begin_inset Formula $M_{3}\left(\mathbb{R}\right)$
\end_inset
(
\begin_inset Formula $I_{3}$
\end_inset
) za implicitni operaciji
\begin_inset Formula $+$
\end_inset
in
\begin_inset Formula $\cdot$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula $g:M_{n}\left(\mathbb{R}\right)\to M_{n}\left(\mathbb{R}\right)$
\end_inset
ki slika
\begin_inset Formula $A\mapsto S^{-1}AS$
\end_inset
,
kjer je
\begin_inset Formula $S$
\end_inset
neka fiksna obrnljiva matrika v
\begin_inset Formula $M_{n}\left(\mathbb{R}\right)$
\end_inset
.
Uporabimo implicitni operaciji
\begin_inset Formula $+$
\end_inset
in
\begin_inset Formula $\cdot$
\end_inset
za matrike.
Računa
\begin_inset Formula $g\left(A+B\right)=S^{-1}\left(A+B\right)S=S^{-1}AS+S^{-1}BS=g\left(A\right)+g\left(B\right)$
\end_inset
in
\begin_inset Formula $g\left(AB\right)=S^{-1}ABS=S^{-1}AIBS=S^{-1}ASS^{-1}BS=g\left(A\right)g\left(B\right)$
\end_inset
pokažeta,
da je
\begin_inset Formula $g$
\end_inset
homomorfizem kolobarjev,
celo z enoto,
kajti
\begin_inset Formula $g\left(I\right)=S^{-1}IS=S^{-1}S=I$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula $h:\mathbb{C}\to M_{n}\left(\mathbb{R}\right)$
\end_inset
s predpisom
\begin_inset Formula $\alpha+\beta i\to\left[\begin{array}{cc}
\alpha & \beta\\
-\beta & \alpha
\end{array}\right]$
\end_inset
je homomorfizem kolobarjev z enoto.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Kolobar ostankov
\begin_inset Formula $\mathbb{Z}_{n}\coloneqq\left\{ 0..\left(n-1\right)\right\} $
\end_inset
je asociativni kolobar z enoto.
Če je
\begin_inset Formula $p$
\end_inset
praštevilo,
pa je celo
\begin_inset Formula $\mathbb{Z}_{p}$
\end_inset
polje za implicitni operaciji seštevanje in množenja po modulu.
\end_layout
\begin_layout Subsection
Vektorski prostori
\end_layout
\begin_layout Standard
Ideja:
Vektorski prostor je Abelova grupa z dodatno strukturo —
množenje s skalarjem.
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $\left(F,+,\cdot\right)$
\end_inset
polje.
Vektorski prostor z operacijama
\begin_inset Formula $V+V\to V$
\end_inset
in
\begin_inset Formula $F\cdot V\to V$
\end_inset
nad
\begin_inset Formula $F$
\end_inset
je taka
\begin_inset Formula $\left(V,+,\cdot\right)$
\end_inset
,
da velja:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\left(V,+\right)$
\end_inset
je abelova grupa:
komutativnost,
asociativnost,
enota,
aditivni inverzi
\end_layout
\begin_layout Enumerate
Lastnosti množenja s skalarjem.
\begin_inset Formula $\forall\alpha,\beta\in F,a,b\in V:$
\end_inset
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\alpha\left(a+b\right)=\alpha a+\alpha b$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\left(\alpha+\beta\right)a=\alpha a+\beta a$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\left(\alpha\cdot\beta\right)\cdot a=\alpha\cdot\left(\beta\cdot a\right)$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $1\cdot a=a$
\end_inset
\end_layout
\begin_layout Standard
Alternativna abstraktna formulacija aksiomov množenja s skalarjem se glasi:
\end_layout
\begin_layout Standard
\begin_inset Formula $\forall\alpha\in F$
\end_inset
priredimo preslikavo
\begin_inset Formula $\varphi_{\alpha}:V\to V$
\end_inset
,
ki pošlje
\begin_inset Formula $v\mapsto\alpha v$
\end_inset
.
Štiri zgornje aksiome množenja s skalarjem sedaj označimo z abstraktnimi formulacijami:
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\varphi_{\alpha}\left(a+b\right)\overset{\text{def.}}{=}\alpha\left(a+b\right)=\alpha b+\alpha b\overset{\text{def.}}{=}\varphi_{\alpha}\left(a\right)+\varphi_{\alpha}\left(b\right)$
\end_inset
—
vidimo,
da je
\begin_inset Formula $\varphi_{\alpha}$
\end_inset
homomorfizem iz
\begin_inset Formula $\left(V,+\right)$
\end_inset
v
\begin_inset Formula $\left(V,+\right)$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\varphi_{\alpha+\beta}\left(a\right)\overset{\text{def.}}{=}\left(\alpha+\beta\right)a=\alpha a+\beta a\overset{\text{def.}}{=}\varphi_{\alpha}a+\varphi_{\beta}a$
\end_inset
—
torej
\begin_inset Formula $\varphi_{\alpha+\beta}=\varphi_{\alpha}+\varphi_{\beta}$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\varphi_{\alpha\beta}a\overset{\text{def.}}{=}\left(\alpha\beta\right)a=\alpha\left(\beta a\right)\overset{\text{def.}}{=}\varphi_{\alpha}\left(\varphi_{\beta}\left(a\right)\right)=\left(\varphi_{\alpha}\circ\varphi_{\beta}\right)\left(a\right)$
\end_inset
—
torej
\begin_inset Formula $\varphi_{\alpha\beta}=\varphi_{\alpha}\circ\varphi_{\beta}$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\varphi_{1}a\overset{\text{def.}}{=}1a=a$
\end_inset
—
torej
\begin_inset Formula $\varphi_{1}=id$
\end_inset
.
\end_layout
\begin_layout Paragraph
\begin_inset Note Note
status open
\begin_layout Plain Layout
TODO ALTERNATIVNA DEFINICIJA VEKTORSKEGA PROSTORA Z GRUPO ENDOMORFIZMOV
\end_layout
\end_inset
\end_layout
\end_deeper
\end_deeper
\begin_layout Remark*
Če v definiciji vektorskega prostora zamenjamo polje
\begin_inset Formula $F$
\end_inset
s kolobarjem
\begin_inset Formula $F$
\end_inset
,
dobimo definicijo
\series bold
modula
\series default
nad
\begin_inset Formula $F$
\end_inset
.
\end_layout
\begin_layout Example*
Primeri vektorskih prostorov:
\end_layout
\begin_deeper
\begin_layout Itemize
standarden primer:
naj bo
\begin_inset Formula $F$
\end_inset
pojle in
\begin_inset Formula $n\in\mathbb{N}$
\end_inset
.
Naj bo
\begin_inset Formula $V=F^{n}$
\end_inset
,
\begin_inset Formula $+$
\end_inset
seštevanje po komponentah in
\begin_inset Formula $\cdot$
\end_inset
množenje s skalarjem po komponentah.
Pod temi pogoji je
\begin_inset Formula $\left(V,+,\cdot\right)$
\end_inset
vektorski prostor —
ustreza vsem osmim aksiomom.
\end_layout
\begin_layout Itemize
Naj bo
\begin_inset Formula $F$
\end_inset
polje in
\begin_inset Formula $n,m\in\mathbb{N}$
\end_inset
.
Naj bo
\begin_inset Formula $V\coloneqq M_{m,n}\left(\mathbb{F}\right)=m\times n$
\end_inset
matrike nad
\begin_inset Formula $F$
\end_inset
.
\begin_inset Formula $+$
\end_inset
in
\begin_inset Formula $\cdot$
\end_inset
definiramo kot pri matrikah.
\end_layout
\begin_layout Itemize
Naj bo
\begin_inset Formula $F$
\end_inset
polje,
\begin_inset Formula $S\not=\emptyset$
\end_inset
množica.
Naj bo
\begin_inset Formula $V\coloneqq F^{S}$
\end_inset
(vse funkcije
\begin_inset Formula $S\to F$
\end_inset
).
Naj bosta
\begin_inset Formula $\varphi,\tau:S\to F$
\end_inset
.
Definirajmo
\begin_inset Formula $\forall s\in S$
\end_inset
operaciji
\begin_inset Formula $\left(\varphi+\tau\right)\left(s\right)=\varphi\left(s\right)+\tau\left(s\right)$
\end_inset
in
\begin_inset Formula $\left(\varphi\cdot\tau\right)\left(s\right)=\varphi\left(s\right)\cdot\tau\left(s\right)$
\end_inset
.
Tedaj je
\begin_inset Formula $V$
\end_inset
vektorski prostor.
Ta definicija je podobna kot definiciji z
\begin_inset Formula $n-$
\end_inset
terico elementov polja,
saj lahko
\begin_inset Formula $n-$
\end_inset
terico identificiramo s funkcijo
\begin_inset Formula $\left\{ \alpha_{1},\dots,\alpha_{n}\right\} \to F$
\end_inset
,
toda ta primer dovoli neskončno razsežne vektorske prostore,
saj
\begin_inset Formula $S$
\end_inset
ni nujno končna,
\begin_inset Formula $n-$
\end_inset
terica pa nekako implicitno je,
saj
\begin_inset Formula $n\in\mathbb{N}$
\end_inset
.
\end_layout
\begin_layout Itemize
Polinomi.
Naj bo
\begin_inset Formula $V\coloneqq F\left[x\right]$
\end_inset
(polinomi v spremenljivki
\begin_inset Formula $x$
\end_inset
s koeficienti v
\begin_inset Formula $F$
\end_inset
).
Seštevanje definirajmo po komponentah:
\begin_inset Formula $\left(\alpha+\beta x+\gamma x^{2}\right)+\left(\pi+\tau x\right)=\left(\alpha+\pi+\left(\beta+\tau\right)x+\gamma x^{2}\right)$
\end_inset
,
množenje s skalarjem pa takole:
\begin_inset Formula $\alpha\left(a+bx+cx^{2}\right)=\alpha a+\alpha bx+\alpha cx^{2}$
\end_inset
.
\end_layout
\begin_layout Itemize
Naj bosta
\begin_inset Formula $V_{1}$
\end_inset
in
\begin_inset Formula $V_{2}$
\end_inset
dva vektorska prostora nad istim poljem
\begin_inset Formula $F$
\end_inset
.
Tvorimo nov vektorski prostor nad
\begin_inset Formula $F$
\end_inset
,
ki mu pravimo
\begin_inset Quotes gld
\end_inset
direktna vsota
\begin_inset Quotes grd
\end_inset
\begin_inset Formula $V_{1}$
\end_inset
in
\begin_inset Formula $V_{2}$
\end_inset
in ga označimo z
\begin_inset Formula $V_{1}\oplus V_{2}\coloneqq$
\end_inset
\begin_inset Formula $\left\{ \left(v_{1},v_{2}\right);\forall v_{1}\in V_{1},v_{2}\in V:2\right\} $
\end_inset
.
Seštevamo po komponentah:
\begin_inset Formula $\left(v_{1},v_{2}\right)+\left(v_{1}',v_{2}'\right)=\left(v_{1}+v_{1}',v_{2}+v_{2}'\right)$
\end_inset
,
s skalarjem pa množimo prvi komponento:
\begin_inset Formula $\forall\alpha\in F:\alpha\left(v_{1},v_{2}\right)=\left(\alpha v_{1},v_{2}\right)$
\end_inset
.
Definicijo lahko posplošimo na
\begin_inset Formula $n$
\end_inset
vektorskih prostorov.
Tedaj so elementi prostora urejene
\begin_inset Formula $n-$
\end_inset
terice.
\end_layout
\end_deeper
\begin_layout Subsubsection
Podprostori vekrorskih prostorov —
vektorski podprostori
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $\left(V,+,\cdot\right)$
\end_inset
vektorski prostor nad
\begin_inset Formula $F$
\end_inset
.
Vektorski podprostor je taka neprazna podmnožica
\begin_inset Formula $V$
\end_inset
,
ki je zaprta za seštevanje in množenje s skalarjem.
Natančneje:
\begin_inset Formula $\left(W,+,\cdot\right)$
\end_inset
je vektorski podprostor
\begin_inset Formula $\left(V,+,\cdot\right)\Longleftrightarrow$
\end_inset
velja hkrati:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $W\subseteq V$
\end_inset
in
\begin_inset Formula $W\not=\emptyset$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:zaprtost+"
\end_inset
\begin_inset Formula $\forall a,b\in W:a+b\in W$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:zaprtostskalar"
\end_inset
\begin_inset Formula $\forall a\in W,\alpha\in F:\alpha a\in W$
\end_inset
\end_layout
\begin_layout Standard
Lastnosti
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:zaprtost+"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
in
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:zaprtostskalar"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
je moč združiti v eno:
\begin_inset Formula $\forall a_{i},a_{2}\in W,\alpha_{1},\alpha_{2}\in F:\alpha_{1}a_{1}+\alpha_{2}a_{2}\in W$
\end_inset
.
\end_layout
\begin_layout Standard
Z drugimi besedami je vektorski podprostor taka podmnožica,
ki vsebuje vse linearne kombinacije svojih elementov.
Odštevanje
\begin_inset Formula $a-b$
\end_inset
je poseben primer linearne kombinacije,
kajti
\begin_inset Formula $a_{1}-a_{2}=1a_{1}+\left(-1\right)a_{2}$
\end_inset
.
Sledi,
da mora biti
\begin_inset Formula $\left(W,+\right)$
\end_inset
podgrupa
\begin_inset Formula $\left(V,+\right)$
\end_inset
,
torej taka podmnožica
\begin_inset Formula $V$
\end_inset
,
ki je zaprta za odštevanje.
\end_layout
\end_deeper
\begin_layout Example*
Primeri vektorskih podprostorov:
\end_layout
\begin_deeper
\begin_layout Itemize
Naj bo
\begin_inset Formula $V=\mathbb{R}^{2}$
\end_inset
(ravnina).
Vsi vektorski podprostori
\begin_inset Formula $V$
\end_inset
so premice,
ki gredo skozi izhodišče,
izhodišče samo in cela ravnina.
Slednja sta t.
i.
trivialna podprostora.
\end_layout
\end_deeper
\begin_layout Remark*
\begin_inset Formula $\forall\left(V,+,\cdot\right)$
\end_inset
vektorski prostor
\begin_inset Formula $:\left\{ 0\right\} ,V$
\end_inset
sta vektorska podprostora.
Imenujemo ju trivialna vektorska podprostora.
\end_layout
\begin_layout Claim*
Vsak podprostor vsebuje aditivno enoto 0.
\end_layout
\begin_layout Proof
Po definiciji je vsak vektorski podprostor neprazen,
torej
\begin_inset Formula $\exists w\in W$
\end_inset
.
Polje gotovo vsebuje aditivno enoto 0,
torej po aksiomu
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:zaprtostskalar"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
za podprostore sledi
\begin_inset Formula $0\cdot w\in W$
\end_inset
.
Dokažimo
\begin_inset Formula $0\cdot w\overset{?}{=}0$
\end_inset
:
\begin_inset Formula $\cancel{0\cdot w}=\left(0+0\right)\cdot w=0\cdot w+\cancel{0\cdot w}$
\end_inset
(pravilo krajšanja v grupi),
torej
\begin_inset Formula $0=0\cdot w$
\end_inset
.
\end_layout
\begin_layout Claim*
Množica rešitev homogene (desna stran je 0) linearne enačbe je vselej vektorski podprostor.
\end_layout
\begin_layout Proof
Imamo
\begin_inset Formula $\alpha_{1}x_{1}+\cdots+\alpha_{n}x_{n}=0$
\end_inset
.
Če sta
\begin_inset Formula $\vec{a}=\left(a_{1},\dots,a_{n}\right)$
\end_inset
in
\begin_inset Formula $\vec{b}=\left(b_{1},\dots,b_{n}\right)$
\end_inset
rešitvi,
velja
\begin_inset Formula $\alpha_{1}a_{1}+\cdots+\alpha_{n}a_{n}=0$
\end_inset
in
\begin_inset Formula $\alpha_{1}b_{1}+\cdots+\alpha_{n}b_{n}=0$
\end_inset
.
Vzemimo poljubna
\begin_inset Formula $\alpha,\beta\in F$
\end_inset
in si oglejmo
\begin_inset Formula $\alpha\vec{a}+\beta\vec{b}$
\end_inset
:
\begin_inset Formula
\[
\alpha\left(\alpha_{1}a_{1}+\cdots+\alpha_{n}a_{n}\right)+\beta\left(\alpha_{1}b_{1}+\cdots+\alpha_{n}b_{n}\right)=0
\]
\end_inset
\begin_inset Formula
\[
\alpha_{1}\left(\alpha a_{1}+\beta b_{1}\right)+\cdots+\alpha_{n}\left(\alpha a_{n}+\beta b_{n}\right)=0
\]
\end_inset
Vzemimo koeficiente v oklepajih pred
\begin_inset Formula $\alpha_{i}$
\end_inset
v enačbi pred to vrstico in jih zložimo v vektor.
Tedaj je
\begin_inset Formula $\alpha\vec{a}+\beta\vec{b}=\left(\alpha a_{1}+\beta b_{1},\dots,\alpha a_{n}+\beta b_{n}\right)$
\end_inset
spet rešitev homogene linearne enačbe.
Ker je linearna kombinacija elementov vektorskega podprostora spet element vektorskega podprostora,
je po definiciji množica rešitev homogene linearne enačbe res vselej vektorski podprostor.
\end_layout
\begin_layout Remark*
Podoben računa velja tudi za množico rešitev sistema linearnih enačb,
kar sicer sledi tudi iz naslednje trditve.
\end_layout
\begin_layout Claim*
Presek dveh podprostorov je tudi sam spet podprostor.
\end_layout
\begin_layout Proof
Naj bosta
\begin_inset Formula $W_{1},W_{2}$
\end_inset
podprostora v
\begin_inset Formula $V$
\end_inset
.
Dokažimo,
da je
\begin_inset Formula $W_{1}\cap V_{2}$
\end_inset
spet podprostor.
Vzemimo poljubna
\begin_inset Formula $a,b\in W_{1}\cap W_{2}$
\end_inset
in poljubna
\begin_inset Formula $\alpha,\beta\in F$
\end_inset
.
Dokažimo,
da je
\begin_inset Formula $\alpha a+\beta b\in W_{1}\cap W_{2}$
\end_inset
.
Vemo,
da
\begin_inset Formula $a,b\in W_{1}$
\end_inset
in
\begin_inset Formula $a,b\in W_{2}$
\end_inset
.
Ker je podprostor po definiciji zaprt za linearne kombinacije svojih elementov,
je
\begin_inset Formula $\alpha a+\beta b\in W_{1}$
\end_inset
in
\begin_inset Formula $\alpha a+\beta b\in W_{2}$
\end_inset
,
torej
\begin_inset Formula $\alpha a+\beta b\in W_{1}\cap W_{2}$
\end_inset
,
torej je presek podprostorov res zaprt za LK svojih elementov in je s tem tudi sam podprostor.
\end_layout
\begin_deeper
\begin_layout Remark*
Slednji dokaz lahko očitno posplošimo na več podprostorov.
Presek nikdar ni prazen,
saj vsi podprostori vsebujejo aditivno enoto 0 (dokaz za to je malce višje).
\end_layout
\end_deeper
\begin_layout Subsubsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Vsota-podprostorov"
\end_inset
Vsota podprostorov
\end_layout
\begin_layout Definition*
Naj bosta
\begin_inset Formula $W_{1}$
\end_inset
in
\begin_inset Formula $W_{2}$
\end_inset
podprostora v
\begin_inset Formula $V$
\end_inset
.
Vsoto podprostorov
\begin_inset Formula $W_{1}$
\end_inset
in
\begin_inset Formula $W_{2}$
\end_inset
označimo z
\begin_inset Formula $W_{1}+W_{2}=\left\{ w_{1}+w_{w};\forall w_{1}\in W_{1},w_{2}\in W_{2}\right\} $
\end_inset
.
\end_layout
\begin_layout Claim*
Vsota podprostorov je tudi sama spet podprostor.
\end_layout
\begin_layout Proof
Naj bosta
\begin_inset Formula $a,b\in W_{1}+W_{2}$
\end_inset
poljubna.
Tedaj po definiciji
\begin_inset Formula $a=a_{1}+a_{2}$
\end_inset
,
kjer
\begin_inset Formula $a_{1}\in W_{1}$
\end_inset
in
\begin_inset Formula $a_{2}\in W_{2}$
\end_inset
,
in
\begin_inset Formula $b=b_{1}+b_{2}$
\end_inset
,
kjer
\begin_inset Formula $b_{1}\in W_{1}$
\end_inset
in
\begin_inset Formula $b_{2}\in W_{2}$
\end_inset
.
\begin_inset Formula $\forall\alpha,\beta\in F$
\end_inset
:
\end_layout
\begin_layout Proof
\begin_inset Formula
\[
\alpha a+\beta b=\alpha\left(a_{1}+a_{2}\right)+\beta\left(b_{1}+b_{2}\right)=\alpha a_{1}+\alpha a_{2}+\beta b_{1}+\beta b_{2}=\left(\alpha a_{1}+\beta b_{1}\right)+\left(\alpha a_{2}+\beta b_{2}\right)\in W_{1}+W_{2},
\]
\end_inset
kajti
\begin_inset Formula $\left(\alpha a_{1}+\beta b_{1}\right)\in W_{1}$
\end_inset
in
\begin_inset Formula $\left(\alpha a_{2}+\beta b_{2}\right)\in W_{2}$
\end_inset
,
saj sta to linearni kombinaciji elementov prostorov.
Njuna vsota pa je element
\begin_inset Formula $W_{1}+W_{2}$
\end_inset
po definiciji vsote podprostorov.
\end_layout
\begin_layout Subsubsection
Baze
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $V$
\end_inset
vektorski prostor nad poljem
\begin_inset Formula $F$
\end_inset
.
Množica
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
je baza,
če je LN in če je ogrodje.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Množica
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
je LN,
če za vsake
\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in F$
\end_inset
,
ki zadoščajo
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
\end_inset
velja
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$
\end_inset
.
Ekvivalentni definiciji LN:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
je LN
\begin_inset Formula $\Leftrightarrow\forall v\in V$
\end_inset
se da kvečjemu na en način izraziti kot linearno kombinacijo
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
.
\end_layout
\begin_layout Itemize
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
je LN
\begin_inset Formula $\Leftrightarrow\nexists v\in\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
,
da bi se ga dalo izraziti kot LK preostalih elementov.
\end_layout
\begin_layout Standard
Dokaz ekvivalentnosti teh definicij je enak tistemu za
\begin_inset Formula $V=\mathbb{R}^{n}$
\end_inset
višje.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Množica
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
je ogrodje
\begin_inset Formula $\Leftrightarrow\forall v\in V$
\end_inset
se da na vsaj en način izraziti kot LK te množice
\begin_inset Formula $\Leftrightarrow\Lin\left\{ v_{1},\dots,v_{n}\right\} =V$
\end_inset
.
\end_layout
\begin_layout Example*
Primeri baz:
\end_layout
\begin_deeper
\begin_layout Itemize
standardna baza:
Naj bo
\begin_inset Formula $V=F^{n}$
\end_inset
.
\begin_inset Formula $v_{1}=\left(1,0,0,\dots,0,0\right)$
\end_inset
,
\begin_inset Formula $v_{2}=\left(0,1,0,\dots,0,0\right)$
\end_inset
,
...,
\begin_inset Formula $v_{n}=\left(0,0,0,\dots,0,1\right)$
\end_inset
.
Da je
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} \subseteq F^{n}$
\end_inset
res baza,
preverimo z determinanto (
\begin_inset Formula $\det A\not=0\Leftrightarrow\exists A^{-1}\Leftrightarrow$
\end_inset
stolpci so baza prostora):
\begin_inset Formula
\[
\det\left[\begin{array}{ccc}
v_{1} & \cdots & v_{n}\end{array}\right]=0\Leftrightarrow\left\{ v_{1},\dots,v_{n}\right\} \text{ \textbf{ni} baza}
\]
\end_inset
\end_layout
\begin_layout Itemize
baze v
\begin_inset Formula $F\left[x\right]_{<n}$
\end_inset
(polinomi stopnje,
manjše od
\begin_inset Formula $n$
\end_inset
)
\end_layout
\begin_deeper
\begin_layout Itemize
standardna baza:
\begin_inset Formula $\left\{ 1,x,x^{2},x^{3},\dots,x^{n-1}\right\} $
\end_inset
\end_layout
\begin_layout Itemize
vzemimo paroma različne
\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in F$
\end_inset
in definirajmo
\begin_inset Formula $p_{i}\left(x\right)=\left(x-\alpha_{1}\right)\cdots\left(x-\alpha_{i-1}\right)\left(x-\alpha_{i+1}\right)\cdots\left(x-\alpha_{n}\right)$
\end_inset
za vsak
\begin_inset Formula $i\in\left\{ 1..n\right\} $
\end_inset
,
kar je polimom stopnje
\begin_inset Formula $n-1$
\end_inset
.
\begin_inset Formula $\left\{ \alpha_{1}p_{1}\left(x\right),\dots,\alpha_{n}p_{n}\left(x\right)\right\} $
\end_inset
je baza za
\begin_inset Formula $F\left[x\right]_{<n}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Proof
Dokazujemo,
da so LN in ogrodje:
\end_layout
\begin_layout Itemize
LN:
\begin_inset Formula $\beta_{1}p_{i}\left(x\right)+\cdots+\beta_{n}p_{n}\left(x\right)=0\overset{?}{\Longrightarrow}\beta_{1}=\cdots=\beta_{n}=0$
\end_inset
.
Opazimo,
da
\begin_inset Formula $p_{i}\left(\alpha_{j}\right)=0\Leftrightarrow i=j$
\end_inset
.
Torej če za
\begin_inset Formula $x$
\end_inset
vstavimo katerikoli
\begin_inset Formula $\alpha_{i}$
\end_inset
,
bodo vsi členi 0,
razen
\begin_inset Formula $\beta_{i}p_{i}\left(x\right)$
\end_inset
.
Ker pa
\begin_inset Formula $\alpha_{i}$
\end_inset
ni ničla
\begin_inset Formula $p_{i}\left(x\right)$
\end_inset
,
je
\begin_inset Formula $\beta_{i}=0$
\end_inset
,
čim je
\begin_inset Formula $\beta_{i}p_{i}\left(x\right)=0$
\end_inset
.
Preverjati je treba le
\begin_inset Formula $\alpha_{i}$
\end_inset
,
ker je dovolj najti eno vrednost spremenljivke,
v kateri se vrednosti polinomov ne ujemajo,
da lahko rečemo,
da polinomi niso isti.
\end_layout
\begin_layout Itemize
ogrodje:
Trdimo,
da za vsak polimom velja formula
\begin_inset Formula $f\left(x\right)=\frac{f\left(\alpha_{1}\right)}{p_{1}\left(\alpha_{1}\right)}p_{1}\left(x\right)+\cdots+\frac{f\left(\alpha_{n}\right)}{p_{n}\left(\alpha_{n}\right)}p_{n}\left(x\right)$
\end_inset
.
Obe strani enačbe imata stopnjo največ
\begin_inset Formula $n-1$
\end_inset
in se ujemata v
\begin_inset Formula $n$
\end_inset
različnih točkah.
Če za
\begin_inset Formula $x$
\end_inset
vstavimo
\begin_inset Formula $\alpha_{i}$
\end_inset
za vsak
\begin_inset Formula $i$
\end_inset
,
dobimo 0 v vseh členih,
razen v
\begin_inset Formula $i-$
\end_inset
tem,
kjer se vrednost
\begin_inset Formula $f\left(x\right)$
\end_inset
ujema z vrednostjo
\begin_inset Formula $f\left(\alpha_{1}\right)$
\end_inset
.
\end_layout
\end_deeper
\end_deeper
\end_deeper
\begin_layout Subsubsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Obstoj-baze"
\end_inset
Obstoj baze
\end_layout
\begin_layout Standard
Omejimo se na končno razsežne vektorske prostore.
\end_layout
\begin_layout Definition*
Vektorski prostor je končno razsežen,
če ima končno ogrodje:
\begin_inset Formula $\exists n\in\mathbb{N}\exists v_{1},\dots,v_{n}\ni:V=\Lin\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
.
\end_layout
\begin_layout Theorem*
obstoj baze.
Vsak končno razsežen vektorski prostor ima vsaj eno bazo.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $V$
\end_inset
KRVP in naj bo
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
njegovo ogrodje.
Ker ogrodje ni nujno LN,
naj bo
\begin_inset Formula $S$
\end_inset
minimalna/najmanjša podmnožica
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
,
ki je še ogrodje za
\begin_inset Formula $V$
\end_inset
.
Trdimo,
da je
\begin_inset Formula $S$
\end_inset
baza za
\begin_inset Formula $V$
\end_inset
.
Po konstrukciji je ogrodje,
dokažimo še,
da je LN:
PDDRAA
\begin_inset Formula $S$
\end_inset
je linearno odvisna.
Tedaj
\begin_inset Formula $\exists v_{i}\in S\ni:v_{i}$
\end_inset
je LK
\begin_inset Formula $S\setminus\left\{ v_{i}\right\} $
\end_inset
.
Dokažimo,
da je
\begin_inset Formula $S\setminus\left\{ v_{i}\right\} $
\end_inset
ogrodje manjše moči,
kar bi bilo v protislovju s predpostavko.
Tedaj obstajajo koeficienti,
da velja
\begin_inset Formula $v_{i}=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$
\end_inset
.
Vzemimo poljuben
\begin_inset Formula $v\in V$
\end_inset
.
Ker je
\begin_inset Formula $S$
\end_inset
ogrodje
\begin_inset Formula $V$
\end_inset
,
obstajajo neki koeficienti
\begin_inset Formula $\beta_{1},\dots,\beta_{n}$
\end_inset
,
da velja
\begin_inset Formula
\[
v=\beta_{1}v_{1}+\cdots+\beta_{i}v_{i}+\cdots+\beta_{n}v_{n}=\beta_{1}v_{1}+\cdots+\beta_{i}\left(\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}\right)+\cdots+\beta_{n}v_{n}=
\]
\end_inset
\begin_inset Formula
\[
=\left(\beta_{1}+\beta_{i}\alpha_{1}\right)v_{1}+\cdots+\left(\beta_{i-1}+\beta_{i}\alpha_{i-1}\right)v_{i-1}+\left(\beta_{i+1}+\beta_{i}\alpha_{i+1}\right)v_{i+1}+\cdots+\left(\beta_{n}+\beta_{i}\alpha_{n}\right)v_{n}
\]
\end_inset
To pa je
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
,
saj je bilo rečeno,
da je
\begin_inset Formula $S$
\end_inset
najmanjše ogrodje,
mi pa smo razvili poljuben
\begin_inset Formula $v$
\end_inset
po manjšem ogrodju.
Torej ima vsak KRVP bazo in vsako ogrodje ima podmnožico,
ki je baza.
\end_layout
\begin_layout Claim
\begin_inset CommandInset label
LatexCommand label
name "enoličnost-moči-baze."
\end_inset
enoličnost moči baze.
Naj bo
\begin_inset Formula $V$
\end_inset
KRVP z
\begin_inset Formula $n-$
\end_inset
elementno bazo.
Tedaj velja vse to:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset
LN množica
\begin_inset Formula $A$
\end_inset
v
\begin_inset Formula $V$
\end_inset
ima
\begin_inset Formula $\leq n$
\end_inset
elementov
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset
ogrodje v
\begin_inset Formula $V$
\end_inset
ima
\begin_inset Formula $\geq n$
\end_inset
elementov
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset
baza v
\begin_inset Formula $V$
\end_inset
ima
\begin_inset Formula $n$
\end_inset
elementov
\end_layout
\end_deeper
\begin_layout Proof
Dokaz je dolg.
\begin_inset CommandInset counter
LatexCommand set
counter "theorem"
value "0"
lyxonly "false"
\end_inset
\end_layout
\begin_deeper
\begin_layout Lemma
\begin_inset CommandInset label
LatexCommand label
name "lem:Vsak-poddoločen-homogen"
\end_inset
Vsak poddoločen homogen sistem linearnih enačb ima netrivialno rešitev.
\end_layout
\begin_deeper
\begin_layout Proof
Dokaz se nahaja pod identično trditvijo
\begin_inset CommandInset ref
LatexCommand vref
reference "claim:Vpoddol-hom-sist-ima-ne0-reš"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Lemma
\begin_inset CommandInset label
LatexCommand label
name "lem:ln<=ogr"
\end_inset
Če je
\begin_inset Formula $u_{1},\dots,u_{m}$
\end_inset
LN množica v
\begin_inset Formula $V$
\end_inset
in
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
ogrodje za
\begin_inset Formula $V$
\end_inset
,
je
\begin_inset Formula $m\leq n$
\end_inset
.
ZDB moč katerekoli LN množice je manjša ali enaka od kateregakoli ogrodja v
\begin_inset Formula $V$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Proof
RAAPDD
\begin_inset Formula $u_{1},\dots,u_{m}$
\end_inset
je LN,
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
je ogrodje in
\begin_inset Formula $m>n$
\end_inset
.
Iščemo protislovje.
Vsakega od
\begin_inset Formula $u_{i}$
\end_inset
lahko razvijemo po
\begin_inset Formula $v$
\end_inset
.
\begin_inset Formula
\[
\begin{array}{ccccccc}
u_{1} & = & \alpha_{11}v_{1} & + & \cdots & + & \alpha_{1n}v_{n}\\
\vdots & & \vdots & & & & \vdots\\
u_{m} & = & \alpha_{m1}v_{1} & + & \cdots & + & \alpha_{mn}v_{n}
\end{array}
\]
\end_inset
\begin_inset Formula $\forall i\in\left\{ 1..m\right\} $
\end_inset
pomnožimo
\begin_inset Formula $i-$
\end_inset
to enačbo s skalarjem
\begin_inset Formula $x_{i}$
\end_inset
in jih seštejmo.
\begin_inset Formula $\vec{x}$
\end_inset
so abstraktne spremenljivke.
Tedaj:
\begin_inset Formula
\[
x_{1}u_{1}+\cdots+x_{m}u_{m}=x_{1}\left(\alpha_{11}v_{1}+\cdots+\alpha_{1n}v_{n}\right)+\cdots+x_{m}\left(\alpha_{m1}v_{1}+\cdots+\alpha_{mn}v_{n}\right)=
\]
\end_inset
\begin_inset Formula
\[
=v_{1}\left(\alpha_{11}x_{1}+\cdots+\alpha_{m1}x_{m}\right)+\cdots+v_{n}\left(\alpha_{1n}x_{1}+\cdots+\alpha_{mn}x_{m}\right)
\]
\end_inset
\end_layout
\begin_layout Proof
Izenačimo koeficiente za
\begin_inset Formula $v_{i}$
\end_inset
z 0 in dobimo poddoločen homogen sistem enačb (ima
\begin_inset Formula $n$
\end_inset
enačb in
\begin_inset Formula $m$
\end_inset
spremenljivk,
po predpostavki pa velja
\begin_inset Formula $m>n$
\end_inset
):
\begin_inset Formula
\[
\begin{array}{ccccccc}
\alpha_{11}x_{1} & + & \cdots & + & \alpha_{m1}x_{m} & = & 0\\
\vdots & & & & \vdots & & \vdots\\
\alpha_{1n}x_{1} & + & \cdots & + & \alpha_{mn}x_{m} & = & 0
\end{array}
\]
\end_inset
Po lemi
\begin_inset CommandInset ref
LatexCommand vref
reference "lem:Vsak-poddoločen-homogen"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
ima ta sistem netrivialno rešitev,
recimo
\begin_inset Formula $\left(\mu_{1},\dots,\mu_{m}\right)$
\end_inset
.
Če to rešitev vstavimo v
\begin_inset Formula $u_{1}x_{1}+\cdots+u_{m}x_{m}$
\end_inset
,
dobimo
\begin_inset Formula $u_{1}\mu_{1}+\cdots+u_{m}\mu_{m}=0$
\end_inset
.
Ker so
\begin_inset Formula $u_{1},\dots,u_{m}$
\end_inset
LN,
so
\begin_inset Formula $\mu_{1}=\cdots=\mu_{m}=0$
\end_inset
,
kar je v
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
s predpostavko.
\end_layout
\end_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset
baza je ogrodje
\begin_inset Formula $\Rightarrow$
\end_inset
po lemi
\begin_inset CommandInset ref
LatexCommand vref
reference "lem:ln<=ogr"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
velja,
da ima vsaka LN množica manj ali enako elementov kot vsako ogrodje,
torej tudi manj ali enako kot
\begin_inset Formula $n$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset
baza je LN
\begin_inset Formula $\Rightarrow$
\end_inset
po lemi
\begin_inset CommandInset ref
LatexCommand vref
reference "lem:ln<=ogr"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
velja,
da ima vsako ogrodje več ali enako elementov kot vsaka LN,
torej tudi več ali enako kot
\begin_inset Formula $n$
\end_inset
.
\end_layout
\begin_layout Enumerate
Sledi iz zgornjih dveh točk,
saj je baza tako ogrodje kot LN hkrati.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $V$
\end_inset
KRVP.
Njegova dimenzija,
\begin_inset Formula $\dim V$
\end_inset
,
je moč baze v
\begin_inset Formula $V$
\end_inset
.
\end_layout
\begin_layout Example*
\begin_inset Formula $\dim F^{n}=n$
\end_inset
,
\begin_inset Formula $\dim M_{m\times n}\left(\mathbb{F}\right)=m\cdot n$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Dopolnitev LN množice do baze
\end_layout
\begin_layout Claim*
Naj bo
\begin_inset Formula $V$
\end_inset
vektorski prostor z dimenzijo
\begin_inset Formula $n$
\end_inset
.
Trdimo,
da
\end_layout
\begin_deeper
\begin_layout Enumerate
ima vsaka LN množica
\begin_inset Formula $\leq n$
\end_inset
elementov,
\end_layout
\begin_layout Enumerate
je vsaka LN množica v
\begin_inset Formula $V$
\end_inset
z
\begin_inset Formula $n$
\end_inset
elementi baza,
\end_layout
\begin_layout Enumerate
lahko vsako LN množico v
\begin_inset Formula $V$
\end_inset
dopolnimo do baze.
\end_layout
\end_deeper
\begin_layout Proof
Dokaz je dolg
\begin_inset CommandInset counter
LatexCommand set
counter "theorem"
value "0"
lyxonly "false"
\end_inset
\end_layout
\begin_deeper
\begin_layout Lemma
\begin_inset CommandInset label
LatexCommand label
name "lem:večja-ln"
\end_inset
Če so
\begin_inset Formula $v_{1},\dots,v_{m}\in V$
\end_inset
LN in če
\begin_inset Formula $v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset
,
potem so tudi
\begin_inset Formula $v_{1},\dots,v_{m},v_{m+1}$
\end_inset
LN.
\end_layout
\begin_deeper
\begin_layout Proof
Naj velja
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{m+1}v_{m+1}=0$
\end_inset
za nek
\begin_inset Formula $\vec{\alpha}\in F^{m+1}$
\end_inset
.
Dokažimo
\begin_inset Formula $\vec{a}=\vec{0}$
\end_inset
.
Če
\begin_inset Formula $\alpha_{m+1}=0$
\end_inset
,
sledi
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{m}v_{m}=0$
\end_inset
,
ker pa so po predpostavki
\begin_inset Formula $v_{1},\dots,v_{m}$
\end_inset
LN,
je
\begin_inset Formula $\vec{\alpha}=\vec{0}$
\end_inset
.
Sicer pa,
če PDDRAA
\begin_inset Formula $\alpha_{m+1}\not=0$
\end_inset
,
lahko z
\begin_inset Formula $a_{m+1}$
\end_inset
delimo:
\begin_inset Formula
\[
\alpha_{1}v_{1}+\cdots+\alpha_{m+1}v_{m+1}=0
\]
\end_inset
\begin_inset Formula
\[
\alpha_{m+1}v_{m+1}=-\alpha_{1}v_{1}-\cdots-\alpha_{m}v_{m}
\]
\end_inset
\begin_inset Formula
\[
v_{m+1}=\frac{-\alpha_{1}}{\alpha_{m+1}}v_{1}+\cdots+\frac{-\alpha_{m}}{\alpha_{m+1}}v_{m}
\]
\end_inset
Tedaj pridemo do
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
,
saj smo
\begin_inset Formula $v_{m+1}$
\end_inset
izrazili kot LK
\begin_inset Formula $\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset
,
po predpostavki pa je vendar
\begin_inset Formula $v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset
.
\end_layout
\end_deeper
\begin_layout Enumerate
že dokazano z dokazom trditve
\begin_inset CommandInset ref
LatexCommand vref
reference "enoličnost-moči-baze."
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
v razdelku
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:Obstoj-baze"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
udensdash{Vsaka LN množica v
\backslash
ensuremath{V} z
\backslash
ensuremath{n} elementi je baza.}
\end_layout
\end_inset
PDDRAA
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
je LN,
ki ni baza.
Tedaj
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
ni ogrodje.
Tedaj
\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n}\right\} \not=V$
\end_inset
.
Zatorej
\begin_inset Formula $\exists v_{n+1}\in V\ni:\left\{ v_{1},\dots,v_{n},v_{n+1}\right\} $
\end_inset
je LN,
kar je v
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
s trditvijo,
da ima vsaka
\begin_inset Formula $LN$
\end_inset
množica v
\begin_inset Formula $V$
\end_inset
kvečjemu
\begin_inset Formula $n$
\end_inset
elementov.
\end_layout
\begin_layout Enumerate
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
udensdash{Vsako LN množico v $V$ z $n$ elementi lahko dopolnimo do baze.}
\end_layout
\end_inset
Naj bo
\begin_inset Formula $v_{1},\dots,v_{m}$
\end_inset
LN množica v
\begin_inset Formula $V$
\end_inset
.
Vemo,
da je
\begin_inset Formula $m\leq n$
\end_inset
.
Če
\begin_inset Formula $m=n$
\end_inset
,
je
\begin_inset Formula $v_{1},\dots,v_{m}$
\end_inset
baza po zgornji trditvi.
Sicer pa je
\begin_inset Formula $m<n$
\end_inset
:
Tedaj
\begin_inset Formula $v_{1},\dots,v_{m}$
\end_inset
ni ogrodje,
sicer bi imeli neko LN množico z več elementi kot neko ogrodje,
saj ima po popraj dokazanem vsako ogrodje vsaj toliko elementov kot vsaka LN množica.
Ker
\begin_inset Formula $v_{1},\dots,v_{m}$
\end_inset
ni ogrodje,
\begin_inset Formula $\exists v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset
.
Po lemi
\begin_inset CommandInset ref
LatexCommand ref
reference "lem:večja-ln"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
je torej
\begin_inset Formula $v_{1},\dots,v_{m+1}$
\end_inset
LN množica.
Če je
\begin_inset Formula $m+1=n$
\end_inset
,
je to že baza,
sicer ponavljamo dodajanje elementov,
dokler ne dodamo
\begin_inset Formula $k$
\end_inset
elementov in dosežemo
\begin_inset Formula $m+k=n$
\end_inset
.
Tedaj je to baza.
Naredili smo
\begin_inset Formula $k=m-n$
\end_inset
korakov.
\end_layout
\end_deeper
\begin_layout Proof
Uporabna vrednost tega izreka sta dva nova izreka o dimenzijah podprostorov:
\end_layout
\begin_layout Claim*
Če je
\begin_inset Formula $V$
\end_inset
je KRVP in
\begin_inset Formula $W$
\end_inset
njegov podprostor,
je
\begin_inset Formula $\dim W\leq\dim V$
\end_inset
.
\end_layout
\begin_layout Proof
PDDRAA
\begin_inset Formula $\dim W>\dim V$
\end_inset
.
Čim ima baza
\begin_inset Formula $W$
\end_inset
večjo moč kot baza
\begin_inset Formula $V$
\end_inset
,
obstaja v
\begin_inset Formula $W$
\end_inset
LN množica z večjo močjo kot baza
\begin_inset Formula $V$
\end_inset
.
Toda ker je ta LN množica LN tudi v
\begin_inset Formula $V$
\end_inset
,
obstaja v
\begin_inset Formula $V$
\end_inset
LN množica z več elementi kot baza
\begin_inset Formula $V$
\end_inset
,
kar je v
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
s trditvijo
\begin_inset CommandInset ref
LatexCommand vref
reference "enoličnost-moči-baze."
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
v razdelku
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:Obstoj-baze"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
.
\end_layout
\begin_layout Claim*
dimenzijska formula za podprostore.
Naj bo
\begin_inset Formula $V$
\end_inset
KRVP in
\begin_inset Formula $W_{1},W_{2}$
\end_inset
podprostora v
\begin_inset Formula $V$
\end_inset
.
Velja
\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim W_{1}+\dim W_{2}-\dim\left(W_{1}\cap W_{2}\right)$
\end_inset
.
Vsota vektorskih podprostorov je definirana v razdelku
\begin_inset CommandInset ref
LatexCommand vref
reference "subsec:Vsota-podprostorov"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
.
\end_layout
\begin_layout Proof
Izberimo bazo
\begin_inset Formula $w_{1},\dots,w_{m}$
\end_inset
za
\begin_inset Formula $W_{1}\cap W_{2}$
\end_inset
.
Naj bo
\begin_inset Formula $u_{1},\dots,u_{k}$
\end_inset
njena dopolnitev do baze
\begin_inset Formula $W_{1}$
\end_inset
in
\begin_inset Formula $v_{1},\dots,v_{l}$
\end_inset
njena dopolnitev do baze
\begin_inset Formula $W_{2}$
\end_inset
.
Trdimo,
da je
\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}$
\end_inset
baza za
\begin_inset Formula $W_{1}+W_{2}$
\end_inset
.
Tedaj bi namreč veljalo
\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=m+k+l$
\end_inset
,
\begin_inset Formula $\dim\left(W_{1}\cap W_{2}\right)=m$
\end_inset
,
\begin_inset Formula $\dim\left(W_{1}\right)=m+k$
\end_inset
in
\begin_inset Formula $\dim\left(W_{2}\right)=m+l$
\end_inset
.
Treba je dokazati še,
da je
\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}$
\end_inset
baza za
\begin_inset Formula $W_{1}+W_{2}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Itemize
Je ogrodje?
Vzemimo poljuben
\begin_inset Formula $v\in W_{1}+W_{2}$
\end_inset
.
Po definiciji
\begin_inset Formula $W_{1}+W_{2}\exists z_{1}\in W_{1},z_{2}\in W_{2}\ni:v=z_{1}+z_{2}$
\end_inset
.
Razvijmo
\begin_inset Formula $z_{1}$
\end_inset
po bazi
\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k}$
\end_inset
za
\begin_inset Formula $W_{1}$
\end_inset
in
\begin_inset Formula $z_{2}$
\end_inset
po bazi
\begin_inset Formula $w_{1},\dots,w_{m},v_{1},\dots,v_{k}$
\end_inset
za
\begin_inset Formula $W_{2}$
\end_inset
.
Takole:
\begin_inset Formula $z_{1}=\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}$
\end_inset
in
\begin_inset Formula $z_{2}=\gamma_{1}w_{1}+\cdots\gamma_{m}w_{m}+\delta_{1}v_{1}+\cdots\delta_{l}v_{l}$
\end_inset
.
Torej
\begin_inset Formula $v=z_{1}+z_{2}=\left(\alpha_{1}+\gamma_{1}\right)w_{1}+\cdots+\left(\alpha_{m}+\gamma_{m}\right)w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}+\delta_{1}v_{1}+\cdots+\delta_{l}v_{l}\in\Lin\left\{ w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}\right\} $
\end_inset
.
Je ogrodje.
\end_layout
\begin_layout Itemize
Je LN?
Naj bo
\begin_inset Formula
\[
\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}+\gamma_{1}v_{1}+\cdots+\gamma_{l}v_{l}=0
\]
\end_inset
\begin_inset Formula
\[
\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l}
\]
\end_inset
Leva stran enačbe je
\begin_inset Formula $\in W_{1}$
\end_inset
,
desna pa
\begin_inset Formula $\in W_{2}$
\end_inset
,
zatorej je element,
ki ga izraza na obeh straneh enačbe opisujeta,
\begin_inset Formula $\in W_{1}\cap W_{2}$
\end_inset
.
Torej je
\begin_inset Formula $v_{1},\dots,v_{l}$
\end_inset
baza za
\begin_inset Formula $W_{1}\cap W_{1}$
\end_inset
.
Toda baza od
\begin_inset Formula $W_{1}\cap W_{2}$
\end_inset
je tudi
\begin_inset Formula $w_{1},\dots,w_{m}$
\end_inset
,
zatorej lahko ta element razpišemo po njej:
\begin_inset Formula
\[
\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l}=\delta_{1}w_{1}+\cdots+\delta_{m}v_{m}
\]
\end_inset
\begin_inset Formula
\[
\delta_{1}w_{1}+\cdots+\delta_{m}w_{m}+\gamma_{1}v_{1}+\cdots+\gamma_{l}v_{l}=0
\]
\end_inset
Toda
\begin_inset Formula $w_{1},\dots,w_{m},v_{1},\dots,v_{l}$
\end_inset
je baza za
\begin_inset Formula $W_{2}$
\end_inset
po naši prejšnji definiciji,
torej je LN množica,
zato
\begin_inset Formula $\delta_{1}=\cdots=\delta_{m}=\gamma_{1}=\cdots=\gamma_{l}=0$
\end_inset
.
Ker
\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=0$
\end_inset
,
se lahko vrnemo k drugi enačbi te točke in to ugotovitev upoštevamo:
\begin_inset Formula
\[
\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l}
\]
\end_inset
\begin_inset Formula
\[
\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=0
\]
\end_inset
Toda
\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k}$
\end_inset
je baza za
\begin_inset Formula $W_{1}$
\end_inset
po naši prejšnji definiciji,
torej je LN množica,
zato
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{m}=\beta_{1}=\cdots=\beta_{k}=0$
\end_inset
.
Torej velja
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{m}=\beta_{1}=\cdots=\beta_{k}=\gamma_{1}=\cdots=\gamma_{l}=0$
\end_inset
,
torej je ta množica res LN.
\end_layout
\end_deeper
\begin_layout Corollary*
Velja torej
\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim\left(W_{1}\right)+\dim\left(W_{2}\right)$
\end_inset
.
Enačaj velja
\begin_inset Formula $\Leftrightarrow W_{1}\cap W_{2}=\left\{ 0\right\} $
\end_inset
,
kajti
\begin_inset Formula $\dim\left(\left\{ 0\right\} \right)=0$
\end_inset
.
\end_layout
\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:vsota-je-direktna"
\end_inset
Pravimo,
da je vsota
\begin_inset Formula $W_{1}+W_{2}$
\end_inset
direktna,
če velja
\begin_inset Formula $W_{1}\cap W_{2}=\left\{ 0\right\} $
\end_inset
oziroma ekvivalentno če je
\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim W_{1}+\dim W_{2}$
\end_inset
oziroma ekvivalentno
\begin_inset Formula $\forall w_{1}\in W_{1},w_{2}\in W_{2}:w_{1}+w_{2}=0\Rightarrow w_{1}=w_{2}=0$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Prehod na novo bazo
\end_layout
\begin_layout Standard
Naj bo
\begin_inset Formula $V$
\end_inset
vektorski prostor dimenzije
\begin_inset Formula $n$
\end_inset
.
Recimo,
da imamo dve bazi v
\begin_inset Formula $V$
\end_inset
.
\begin_inset Formula $B=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset
naj bo
\begin_inset Quotes gld
\end_inset
stara baza
\begin_inset Quotes grd
\end_inset
,
\begin_inset Formula $C=\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
pa naj bo
\begin_inset Quotes gld
\end_inset
nova baza
\begin_inset Quotes grd
\end_inset
.
\begin_inset Formula $\forall v\in V$
\end_inset
lahko razvijemo po
\begin_inset Formula $B$
\end_inset
in po
\begin_inset Formula $C$
\end_inset
.
Razvoj po
\begin_inset Formula $B$
\end_inset
:
\begin_inset Formula $v=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}$
\end_inset
,
razvoj po
\begin_inset Formula $C$
\end_inset
:
\begin_inset Formula $v=\gamma_{1}v_{1}+\cdots+\gamma_{n}v_{n}$
\end_inset
.
Kakšna je zveza med
\begin_inset Formula $\vec{\beta}$
\end_inset
in
\begin_inset Formula $\vec{\gamma}$
\end_inset
v obeh razvojih?
\end_layout
\begin_layout Standard
Uvedimo oznako
\begin_inset Formula $\left[v\right]_{B}$
\end_inset
,
to naj bodo koeficienti vektorja
\begin_inset Formula $v$
\end_inset
pri razvoju po
\begin_inset Formula $B$
\end_inset
.
\begin_inset Formula $\left[v\right]_{B}=\left[\begin{array}{c}
\beta_{1}\\
\vdots\\
\beta_{n}
\end{array}\right]$
\end_inset
.
\end_layout
\begin_layout Standard
Vsak vektor stare baze razvijmo po novi bazi,
kjer
\begin_inset Formula $\left[u_{i}\right]_{C}=\left[\begin{array}{c}
\alpha_{i1}\\
\vdots\\
\alpha_{in}
\end{array}\right]$
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\begin{array}{ccccccc}
u_{1} & = & \alpha_{11}v_{1} & + & \cdots & + & \alpha_{1n}v_{n}\\
\vdots & & \vdots & & & & \vdots\\
u_{n} & = & a_{n1}v_{1} & + & \cdots & + & a_{nn}v_{n}
\end{array}
\]
\end_inset
Koeficiente
\begin_inset Formula $\alpha$
\end_inset
zložimo v tako imenovanj prehodno matriko
\begin_inset Formula $P_{C\leftarrow B}$
\end_inset
:
\begin_inset Formula
\[
P_{C\leftarrow B}=\left[\begin{array}{ccc}
\left[u_{1}\right]_{C} & \cdots & \left[u_{n}\right]_{C}\end{array}\right]=\left[\begin{array}{ccc}
a_{11} & \cdots & a_{n1}\\
\vdots & & \vdots\\
a_{1n} & \cdots & a_{nn}
\end{array}\right]
\]
\end_inset
Sledi
\begin_inset Formula
\[
v=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}=\beta_{1}\left(\alpha_{11}v_{1}+\cdots+\alpha_{1n}v_{n}\right)+\cdots+\beta_{n}\left(\alpha_{n1}v_{1}+\cdots+\alpha_{nn}v_{n}\right)=
\]
\end_inset
\begin_inset Formula
\[
=v_{1}\left(\beta_{1}\alpha_{11}+\beta_{2}\alpha_{21}+\cdots+\beta_{n}\alpha_{n1}\right)+\cdots+v_{n}\left(\beta_{1}\alpha_{1n}+\beta_{2}\alpha_{2n}+\cdots+\beta_{n}\alpha_{nn}\right)=
\]
\end_inset
po drugi strani je
\begin_inset Formula $v$
\end_inset
tudi lahko razvit po novi bazi:
\begin_inset Formula
\[
=v=\gamma_{1}v_{1}+\cdots+\gamma_{n}v_{n}
\]
\end_inset
Iz česar,
ker je razvoj po bazi enoličen,
sledi
\begin_inset Formula
\[
\begin{array}{ccccccc}
\gamma_{1} & = & \beta_{1}\alpha_{11} & + & \cdots & + & \beta_{n}\alpha_{n1}\\
\vdots & & \vdots & & & & \vdots\\
\gamma_{n} & = & \beta_{1}a_{1n} & + & \cdots & + & \beta_{n}\alpha_{nn}
\end{array},
\]
\end_inset
kar v matrični obliki zapišemo
\begin_inset Formula
\[
\left[\begin{array}{ccc}
\alpha_{11} & \cdots & \alpha_{n1}\\
\vdots & & \vdots\\
\alpha_{1n} & \cdots & \alpha_{nn}
\end{array}\right]\left[\begin{array}{c}
\beta_{1}\\
\vdots\\
\beta_{n}
\end{array}\right]=\left[\begin{array}{c}
\gamma_{1}\\
\vdots\\
\gamma_{n}
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
P_{C\leftarrow B}\left[v\right]_{B}=\left[v\right]_{C}.
\]
\end_inset
\end_layout
\begin_layout Remark*
Velja:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $P_{B\leftarrow B}=I$
\end_inset
.
\end_layout
\begin_layout Itemize
Naj bodo prehodi med bazami takšnile:
\begin_inset Formula $B\overset{P_{C\leftarrow B}}{\longrightarrow}C\overset{P_{D\leftarrow C}}{\longrightarrow}D$
\end_inset
.
Potem je
\begin_inset Formula $P_{D\leftarrow B}=P_{C\leftarrow B}\cdot P_{C\leftarrow D}$
\end_inset
.
\end_layout
\begin_layout Itemize
\begin_inset Formula $P_{B\leftarrow C}\cdot P_{C\leftarrow B}=I$
\end_inset
,
\begin_inset Formula $\left(P_{B\leftarrow C}\right)^{-1}=P_{C\leftarrow B}$
\end_inset
.
\end_layout
\begin_layout Itemize
Naj bo
\begin_inset Formula $v\in F^{n}$
\end_inset
in
\begin_inset Formula $S$
\end_inset
standardna baza za
\begin_inset Formula $F^{n}$
\end_inset
.
Potem
\begin_inset Formula $\left[v\right]_{S}=\left[\begin{array}{c}
\alpha_{1}\\
\vdots\\
\alpha_{n}
\end{array}\right]=v$
\end_inset
.
Sledi
\begin_inset Formula $P_{S\leftarrow B}=\left[\begin{array}{ccc}
\left[u_{1}\right]_{S} & \cdots & \left[u_{n}\right]_{S}\end{array}\right]=\left[\begin{array}{ccc}
u_{1} & \cdots & u_{n}\end{array}\right]$
\end_inset
za
\begin_inset Formula $B=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset
.
Sledi tudi
\begin_inset Formula $P_{S\leftarrow C}=\left[\begin{array}{ccc}
v_{1} & \cdots & v_{n}\end{array}\right]$
\end_inset
,
kjer so
\begin_inset Formula $v,u,B,C$
\end_inset
kot prej (kot definirano na začetku tega razdelka).
\end_layout
\begin_layout Itemize
\begin_inset Formula $P_{C\leftarrow B}=P_{C\leftarrow S}\cdot P_{S\leftarrow B}$
\end_inset
(slednji dve točki veljata samo v
\begin_inset Formula $F^{n}$
\end_inset
,
kjer je standardna baza lepa in zapisljiva kot elementi v matriki)
\end_layout
\end_deeper
\begin_layout Section
Drugi semester
\end_layout
\begin_layout Subsection
Linearne preslikave
\end_layout
\begin_layout Standard
Radi bi definirali homomorfizem vektorskih prostorov.
Homomorfizem za abelove grupe smo že definirali,
vektorski prostor pa je le abelova grupa z dodatno strukturo (množenje s skalarjem).
\end_layout
\begin_layout Definition*
Preslikava
\begin_inset Formula $f:V_{1}\to V_{2}$
\end_inset
je homomorfizem vektorskih prostorov nad istim poljem oziroma linearna preslikava,
če je aditivna (homomorfizem) (
\begin_inset Formula $\forall u,v\in V_{1}:f\left(u+_{1}v\right)=fu+_{2}fv$
\end_inset
) in če je homogena:
\begin_inset Formula $\forall u\in V_{1},\alpha\in F:f\left(\alpha u\right)=\alpha f\left(u\right)$
\end_inset
.
\end_layout
\begin_layout Remark*
Ekvivalentno je preverjati oba pogoja hkrati.
Če za
\begin_inset Formula $L:U\to V$
\end_inset
velja
\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in F,u_{1},u_{2}\in U:L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=\alpha_{1}Lu_{1}+\alpha_{2}Lu_{2}$
\end_inset
,
je
\begin_inset Formula $L$
\end_inset
linearna preslikava.
\end_layout
\begin_layout Example*
Vrtež za kot
\begin_inset Formula $\tau$
\end_inset
v ravnini:
\begin_inset Formula
\[
\left[\begin{array}{c}
x\\
y
\end{array}\right]\to\left[\begin{array}{cc}
\cos\tau & -\sin\tau\\
\sin\tau & \cos\tau
\end{array}\right]\left[\begin{array}{c}
x\\
y
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Linearna funkcija iz analize ni linearna preslikava.
Premik za vektor
\begin_inset Formula $w$
\end_inset
ni linearna preslikava.
Odvajanje in integriranje sta linearni preslikavi.
\end_layout
\begin_layout Fact*
Vsaka linearna preslikava slika 0 v 0.
\end_layout
\begin_layout Definition*
Bijektivni linearni preslikavi pravimo linearni izomorfizem.
\end_layout
\begin_layout Claim
\begin_inset CommandInset label
LatexCommand label
name "claim:invLinIzJeLinIz"
\end_inset
Inverz linearnega izomorfizma je zopet linearni izomorfizem.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $L:U\to V$
\end_inset
bijektivna linearna preslikava med vektorskima prostoroma nad istim poljem
\begin_inset Formula $F$
\end_inset
.
Dokazati je treba,
da je
\begin_inset Formula $L^{-1}:V\to U$
\end_inset
spet linearna preslikava.
Ker je
\begin_inset Formula $L$
\end_inset
linearna,
velja
\begin_inset Formula
\[
\forall\alpha_{1},\alpha_{2}\in F,v_{1},v_{2}\in V:L\left(\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}\right)=\alpha_{1}LL^{-1}v_{1}+\alpha_{2}LL^{-1}v_{2}=LL^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)
\]
\end_inset
Ker je
\begin_inset Formula $L$
\end_inset
injektivna,
iz
\begin_inset Formula $L\left(\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}\right)=LL^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)$
\end_inset
sledi
\begin_inset Formula $\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}=L^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)$
\end_inset
.
\end_layout
\begin_layout Subsubsection
\begin_inset Formula $F^{n}$
\end_inset
je linearno izomorfen
\begin_inset Formula $n-$
\end_inset
razsežnem
\begin_inset Formula $V$
\end_inset
nad
\begin_inset Formula $F$
\end_inset
\end_layout
\begin_layout Claim*
Vsak
\begin_inset Formula $n-$
\end_inset
razsežen vektorski prostor nad
\begin_inset Formula $F$
\end_inset
je linearno izomorfen
\begin_inset Formula $F^{n}$
\end_inset
.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $V$
\end_inset
\begin_inset Formula $n-$
\end_inset
razsežen vektorski prostor nad
\begin_inset Formula $F$
\end_inset
in
\begin_inset Formula $B=\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
baza za
\begin_inset Formula $V$
\end_inset
.
Definirajmo preslikavo
\begin_inset Formula $\phi_{B}:F^{n}\to V$
\end_inset
s predpisom
\begin_inset Formula $\left(x_{1},\cdots,x_{1}\right)\mapsto x_{1}v_{1}+\cdots+x_{n}v_{n}$
\end_inset
.
Ker je
\begin_inset Formula $B$
\end_inset
ogrodje,
je
\begin_inset Formula $\phi_{B}$
\end_inset
surjektivna.
Ker je
\begin_inset Formula $B$
\end_inset
linearno neodvisna,
je
\begin_inset Formula $\phi_{B}$
\end_inset
injektivna.
Pokažimo še,
da je linearna presikava:
\begin_inset Formula
\[
\phi_{B}\left(\alpha\left(x_{1},\dots,x_{n}\right)+\beta\left(y_{1},\dots,y_{n}\right)\right)=\phi_{B}\left(\alpha x_{1}+\beta y_{1},\dots,\alpha x_{n}+\beta x_{n}\right)=v_{1}\left(\alpha x_{1}+\beta y_{1}\right)+\cdots+v_{n}\left(\alpha x_{n}+\beta x_{n}\right)=
\]
\end_inset
\begin_inset Formula
\[
=\alpha\left(v_{1}x_{1}+\cdots+v_{n}x_{n}\right)+\cdots+\beta\left(v_{1}y_{1}+\cdots+v_{n}y_{n}\right)=\alpha\phi_{B}\left(x_{1},\dots,x_{n}\right)+\beta\phi_{B}\left(y_{1},\dots y_{n}\right)
\]
\end_inset
\end_layout
\begin_layout Subsubsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Matrika-linearne-preslikave"
\end_inset
Matrika linearne preslikave —
linearni izomorfizem
\begin_inset Formula $M_{m,n}\left(F\right)\to\mathcal{L}\left(F^{n},F^{m}\right)$
\end_inset
\end_layout
\begin_layout Claim*
Naj bo
\begin_inset Formula $F$
\end_inset
polje in
\begin_inset Formula $m,n\in F$
\end_inset
.
\begin_inset Formula $\mathcal{L}\left(F^{n},F^{m}\right)$
\end_inset
je vektorski prostor linearnih preslikav iz
\begin_inset Formula $F^{n}\to F^{m}$
\end_inset
.
Seštevanje definiramo z
\begin_inset Formula $\left(L_{1}+L_{2}\right)u\coloneqq L_{1}u+L_{2}u$
\end_inset
,
množenje s skalarjem pa
\begin_inset Formula $\left(\alpha L\right)u=\alpha\left(Lu\right)$
\end_inset
.
Naj bo
\begin_inset Formula $M_{m,n}\left(F\right)$
\end_inset
vektorski prostor vseh
\begin_inset Formula $m\times n$
\end_inset
matrik nad
\begin_inset Formula $F$
\end_inset
z znanim seštevanjem in množenjem.
Obstaja linearni izomorfizem med tema dvema prostoroma.
\end_layout
\begin_layout Proof
Oznake kot v trditvi.
Za vsako
\begin_inset Formula $m\times n$
\end_inset
matriko
\begin_inset Formula $A=\left[a_{i,j}\right]$
\end_inset
definirajmo preslikavo
\begin_inset Formula $L_{A}$
\end_inset
iz
\begin_inset Formula $F^{n}$
\end_inset
v
\begin_inset Formula $F^{m}$
\end_inset
takole:
\begin_inset Formula $L_{A}\left(x_{1},\dots,x_{n}\right)=\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)$
\end_inset
.
Po definiciji matričnega množenja ta preslikava ustreza
\begin_inset Formula $L_{A}\vec{x}=A\vec{x}$
\end_inset
.
Dokažimo,
da je linearni izomorfizem.
\end_layout
\begin_deeper
\begin_layout Itemize
Linearnost:
\begin_inset Formula $L_{\alpha A+\beta B}\vec{x}=\left(\alpha A+\beta B\right)\vec{x}=\alpha A\vec{x}+\beta B\vec{x}=\alpha L_{A}\vec{x}+\beta L_{B}\vec{x}=\left(\alpha L_{A}+\beta L_{B}\right)\vec{x}$
\end_inset
\end_layout
\begin_layout Itemize
Bijektivnost:
Konstruirajmo inverzno preslikavo (iz trditve
\begin_inset CommandInset ref
LatexCommand ref
reference "claim:invLinIzJeLinIz"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
vemo,
da bo linearna).
Vsaki linearni presikavi
\begin_inset Formula $L:F^{n}\to F^{m}$
\end_inset
priredimo
\begin_inset Formula $m\times n$
\end_inset
matriko
\begin_inset Formula $\left[\begin{array}{ccc}
Le_{1} & \cdots & Le_{n}\end{array}\right]$
\end_inset
,
kjer je
\begin_inset Formula $e_{1},\dots,e_{n}$
\end_inset
standardna baza za
\begin_inset Formula $F^{n}$
\end_inset
.
Pokažimo,
da je ta preslikava res inverz,
torej preverimo,
da je kompozitum
\begin_inset Formula $A\mapsto L_{A}\mapsto\left[\begin{array}{ccc}
L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]$
\end_inset
identiteta in da je
\begin_inset Formula $\left[\begin{array}{ccc}
L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]\mapsto L_{A}\mapsto A$
\end_inset
tudi identiteta.
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
udensdash{$A
\backslash
mapsto L_{A}
\backslash
mapsto
\backslash
left[
\backslash
begin{array}{ccc}Le_{1} &
\backslash
cdots & Le_{n}
\backslash
end{array}
\backslash
right]
\backslash
overset{?}{=}id$}
\end_layout
\end_inset
:
\begin_inset Formula
\[
\left[\begin{array}{ccc}
L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]=\left[\begin{array}{ccc}
Ae_{1} & \cdots & Ae_{n}\end{array}\right]=A\left[\begin{array}{ccc}
e_{1} & \cdots & e_{n}\end{array}\right]=AI=A
\]
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
udensdash{$
\backslash
left[
\backslash
begin{array}{ccc}L_{A}e_{1} &
\backslash
cdots & L_{A}e_{n}
\backslash
end{array}
\backslash
right]
\backslash
mapsto L_{A}
\backslash
mapsto A
\backslash
overset{?}{=}id$}
\end_layout
\end_inset
:
\begin_inset Formula
\[
\forall x:L_{\left[\begin{array}{ccc}
Le_{1} & \cdots & Le_{n}\end{array}\right]}x=\left[\begin{array}{ccc}
Le_{1} & \cdots & Le_{n}\end{array}\right]\left[\begin{array}{c}
x_{1}\\
\vdots\\
x_{n}
\end{array}\right]=x_{1}Le_{1}+\cdots+x_{n}Le_{n}=L\left(x_{1}e_{1}+\cdots+x_{n}e_{n}\right)=
\]
\end_inset
\begin_inset Formula
\[
=L\left(x_{1}\left[\begin{array}{c}
1\\
0\\
\vdots\\
0
\end{array}\right]+\cdots+x_{n}\left[\begin{array}{c}
0\\
\vdots\\
0\\
1
\end{array}\right]\right)=Lx
\]
\end_inset
\end_layout
\end_deeper
\end_deeper
\begin_layout Proof
Vsaki linearni preslikavi med dvema vektorskima prostoroma sedaj lahko priredimo matriko.
Prirejanje je odvisno od izbire baz v obeh vektorskih prostorih.
Matrika namreč preslika koeficiente iz polja
\begin_inset Formula $F$
\end_inset
,
s katerimi je dan vektor,
ki ga z leve množimo z matriko,
razvit po
\begin_inset Quotes gld
\end_inset
vhodni
\begin_inset Quotes grd
\end_inset
bazi,
v koeficiente iz istega polja,
s katerimi je rezultantni vektor razvit po
\begin_inset Quotes gld
\end_inset
izhodni
\begin_inset Quotes grd
\end_inset
bazi.
\end_layout
\begin_layout Proof
Naj bosta
\begin_inset Formula $U$
\end_inset
in
\begin_inset Formula $V$
\end_inset
vektorska prostora nad istim poljem
\begin_inset Formula $F$
\end_inset
in naj bo
\begin_inset Formula $L:U\to V$
\end_inset
linearna preslikava.
Izberimo bazo
\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset
za
\begin_inset Formula $U$
\end_inset
in
\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset
za
\begin_inset Formula $V$
\end_inset
.
Razvijmo vektorje
\begin_inset Formula $Lu_{1},\dots,Lu_{n}$
\end_inset
po bazi
\begin_inset Formula $\mathcal{C}$
\end_inset
:
\begin_inset Formula
\[
\begin{array}{ccccccc}
Lu_{1} & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,m}v_{m}\\
\vdots & & \vdots & & & & \vdots\\
Lu_{n} & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,m}v_{m}
\end{array}
\]
\end_inset
Skalarje
\begin_inset Formula $\alpha_{i,j}$
\end_inset
sedaj zložimo v spodnjo matriko,
ki ji pravimo
\series bold
matrika linearne preslikave
\begin_inset Formula $L$
\end_inset
glede na bazi
\begin_inset Formula $\mathcal{B}$
\end_inset
in
\begin_inset Formula $\mathcal{C}$
\end_inset
\series default
.
\begin_inset Formula
\[
\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[\begin{array}{ccc}
a_{1,1} & \cdots & \alpha_{n,1}\\
\vdots & & \vdots\\
\alpha_{1,m} & \cdots & \alpha_{n,m}
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula $L:\mathbb{R}\left[x\right]_{\leq3}\to\mathbb{R}\left[x\right]_{\leq2}$
\end_inset
—
linearna preslikava iz realnih polinomov stopnje kvečjemu 3 v realne polinome stopnje kvečjemu 2,
ki predstavlja odvajanje polinomov.
Bazi sta
\begin_inset Formula $\mathcal{B}=\left\{ 1,x,x^{2},x^{2}\right\} $
\end_inset
in
\begin_inset Formula $\mathcal{C}=\left\{ 1,x,x^{2}\right\} $
\end_inset
.
\begin_inset Formula
\[
\begin{array}{ccccccc}
L\left(1\right) & = & 0 & + & 0x & + & 0x^{2}\\
L\left(x\right) & = & 1 & + & 0x & + & 0x^{2}\\
L\left(x^{2}\right) & = & 0 & + & 2x & + & 0x^{2}\\
L\left(x^{3}\right) & = & 0 & + & 0x & + & 3x^{2}
\end{array}
\]
\end_inset
Zapišimo matriko te linearne preslikave:
\begin_inset Formula
\[
\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{cccc}
0 & 1 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 3
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Prehodna matrika je poseben primer matrike linearne preslikave.
\begin_inset Formula $L:V\to V$
\end_inset
z bazama
\begin_inset Formula $\mathcal{B}=\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
in
\begin_inset Formula $\mathcal{C}=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset
,
kjer
\begin_inset Formula $L=id$
\end_inset
.
\begin_inset Formula
\[
\begin{array}{ccccccc}
id\left(u_{1}\right) & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,n}v_{n}\\
\vdots & & \vdots & & & & \vdots\\
id\left(u_{n}\right) & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,n}v_{n}
\end{array}
\]
\end_inset
Zapišimo matriko te linearne preslikave:
\begin_inset Formula $\left[id\right]_{\mathcal{C}\leftarrow\mathcal{B}}=P_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Lastnosti matrik linearnih preslikav
\end_layout
\begin_layout Standard
\begin_inset CommandInset counter
LatexCommand set
counter "theorem"
value "0"
lyxonly "false"
\end_inset
\end_layout
\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:osnovna-formula"
\end_inset
osnovna formula.
Posplošitev formule
\begin_inset Formula $\left[u\right]_{\mathcal{C}}=P_{\mathcal{C}\leftarrow\mathcal{B}}\cdot\left[u\right]_{\mathcal{B}}$
\end_inset
se glasi
\begin_inset Formula $\left[Lu\right]_{\mathcal{C}}=\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[u\right]_{\mathcal{B}}$
\end_inset
za linearno preslikavo
\begin_inset Formula $L:U\to V$
\end_inset
,
\begin_inset Formula $u\in U$
\end_inset
,
kjer je
\begin_inset Formula $\mathcal{\mathcal{B}}=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset
baza za
\begin_inset Formula $U$
\end_inset
in
\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset
baza za
\begin_inset Formula $V$
\end_inset
.
\end_layout
\begin_layout Proof
Po korakih:
\end_layout
\begin_deeper
\begin_layout Enumerate
Razvijmo
\begin_inset Formula $u$
\end_inset
po bazi
\begin_inset Formula $\mathcal{B}$
\end_inset
:
\begin_inset Formula $u=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:Uporabimo-L-na"
\end_inset
Uporabimo
\begin_inset Formula $L$
\end_inset
na obeh straneh:
\begin_inset Formula $Lu=L\left(\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}\right)=\beta_{1}Lu_{1}+\cdots+\beta_{n}Lu_{n}$
\end_inset
.
\end_layout
\begin_layout Enumerate
Razvijmo bazo
\begin_inset Formula $\mathcal{B}$
\end_inset
,
preslikano z
\begin_inset Formula $L$
\end_inset
,
po bazi
\begin_inset Formula $\mathcal{C}$
\end_inset
:
\begin_inset Formula
\[
\begin{array}{ccccccc}
Lu_{1} & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,m}v_{m}\\
\vdots & & \vdots & & & & \vdots\\
Lu_{n} & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,m}v_{m}
\end{array}
\]
\end_inset
\end_layout
\begin_layout Enumerate
Razvoj vstavimo v enačbo iz koraka
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:Uporabimo-L-na"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
in uredimo:
\begin_inset Formula
\[
Lu=\beta_{1}\left(\alpha_{1,1}v_{1}+\cdots+\alpha_{1,m}v_{m}\right)+\cdots+\beta_{n}\left(\alpha_{n,1}v_{1}+\cdots+\alpha_{n,m}v_{m}\right)=
\]
\end_inset
\begin_inset Formula
\[
=v_{1}\left(\beta_{1}\alpha_{1,1}+\cdots+\beta_{n}\alpha_{n,1}\right)+\cdots+v_{m}\left(\beta_{1}\alpha_{1,m}+\cdots+\beta_{n}\alpha_{n,m}v_{m}\right)
\]
\end_inset
\end_layout
\begin_layout Enumerate
Odtod sledi:
\begin_inset Formula
\[
\left[Lu\right]_{\mathcal{C}}=\left[\begin{array}{c}
\beta_{1}\alpha_{1,1}+\cdots+\beta_{n}\alpha_{n,1}\\
\vdots\\
\beta_{1}\alpha_{1,m}+\cdots+\beta_{n}\alpha_{n,m}v_{m}
\end{array}\right]=\left[\begin{array}{ccc}
\alpha_{1,1} & \cdots & \alpha_{n,1}\\
\vdots & & \vdots\\
\alpha_{1,m} & \cdots & \alpha_{n,m}
\end{array}\right]\left[\begin{array}{c}
\beta_{1}\\
\vdots\\
\beta_{n}
\end{array}\right]=\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[u\right]_{\mathcal{B}}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:matrika-kompozituma-linearnih"
\end_inset
matrika kompozituma linearnih preslikav.
Posplošitev formule
\begin_inset Formula $P_{\mathcal{\mathcal{D}\leftarrow\mathcal{B}}}=P_{\mathcal{D\leftarrow C}}\cdot P_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset
se glasi
\begin_inset Formula $\left[K\circ L\right]_{\mathcal{D\leftarrow B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\cdot\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset
.
Trdimo,
da je kompozitum linearnih preslikav spet linearna preslikava in da enačba velja.
\end_layout
\begin_layout Proof
Najprej dokažimo,
da je kompozitum linearnih preslikav spet linearna preslikava.
\begin_inset Formula
\[
\left(K\circ L\right)\left(\alpha u+\beta v\right)=K\left(L\left(\alpha u+\beta v\right)\right)=K\left(\alpha Lu+\beta Lv\right)=\alpha KLu+\beta KLv=\alpha\left(K\circ L\right)u+\beta\left(K\circ L\right)v
\]
\end_inset
Sedaj pa dokažimo še enačbo
\begin_inset Formula $\left[K\circ L\right]_{\mathcal{D\leftarrow B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\cdot\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset
.
Naj bosta
\begin_inset Formula $L:U\to V$
\end_inset
in
\begin_inset Formula $K:V\to W$
\end_inset
linearni preslikavi,
\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset
baza
\begin_inset Formula $U$
\end_inset
,
\begin_inset Formula $\mathcal{C}$
\end_inset
baza
\begin_inset Formula $V$
\end_inset
in
\begin_inset Formula $\mathcal{D}$
\end_inset
baza
\begin_inset Formula $W$
\end_inset
.
Od prej vemo,
da:
\begin_inset Formula
\[
\left[L\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
\left[Lu_{1}\right]_{\mathcal{D}} & \cdots & \left[Lu_{n}\right]_{\mathcal{D}}\end{array}\right],
\]
\end_inset
zato pišimo
\begin_inset Formula
\[
\left[K\circ L\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
\left[\left(K\circ L\right)u_{1}\right]_{\mathcal{D}} & \cdots & \left[\left(K\circ L\right)u_{n}\right]_{\mathcal{D}}\end{array}\right]=\left[\begin{array}{ccc}
\left[KLu_{1}\right]_{\mathcal{D}} & \cdots & \left[KLu_{n}\right]_{\mathcal{D}}\end{array}\right]\overset{\text{izrek \ref{thm:osnovna-formula}}}{=}
\]
\end_inset
\begin_inset Formula
\[
\overset{\text{izrek \ref{thm:osnovna-formula}}}{=}\left[\begin{array}{ccc}
\left[K\right]_{\mathcal{D}\leftarrow C}\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[K\right]_{\mathcal{D}\leftarrow C}\left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[\begin{array}{ccc}
\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}
\]
\end_inset
\end_layout
\begin_layout Subsubsection
Jedro in slika linearne preslikave
\end_layout
\begin_layout Definition*
Naj bosta
\begin_inset Formula $U$
\end_inset
in
\begin_inset Formula $V$
\end_inset
vektorska prostora nad istim poljem
\begin_inset Formula $F$
\end_inset
in
\begin_inset Formula $L:U\to V$
\end_inset
linearna preslikava.
Jedro
\begin_inset Formula $L$
\end_inset
naj bo
\begin_inset Formula $\Ker L\coloneqq\left\{ u\in U;Lu=0\right\} $
\end_inset
(angl.
kernel/null space) in Slika/zaloga vrednosti
\begin_inset Formula $L$
\end_inset
naj bo
\begin_inset Formula $\Slika L\coloneqq\left\{ Lu;\forall u\in U\right\} $
\end_inset
(angl.
image/range).
\end_layout
\begin_layout Claim*
Trdimo naslednje:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\Ker L$
\end_inset
je vektorski podprostor v
\begin_inset Formula $U$
\end_inset
(če vsebuje
\begin_inset Formula $\vec{a}$
\end_inset
in
\begin_inset Formula $\vec{b}$
\end_inset
,
vsebuje tudi vse LK
\begin_inset Formula $\vec{a}$
\end_inset
in
\begin_inset Formula $\vec{b}$
\end_inset
)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\Slika L$
\end_inset
je vektorski podprostor v
\begin_inset Formula $V$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Proof
Dokazujemo dve trditvi:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall u_{1},u_{2}\in\Ker L,\alpha_{1},\alpha_{2}\in F\overset{?}{\Longrightarrow}\alpha_{1}u_{1}+\alpha_{2}u_{2}\in\Ker L$
\end_inset
.
Po predpostavki velja
\begin_inset Formula $Lu_{1}=0$
\end_inset
in
\begin_inset Formula $Lu_{2}=0$
\end_inset
,
torej
\begin_inset Formula $\alpha_{1}Lu_{1}+\alpha_{2}Lu_{2}=0$
\end_inset
.
Iz linearnosti
\begin_inset Formula $L$
\end_inset
sledi
\begin_inset Formula $L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=0$
\end_inset
,
torej
\begin_inset Formula $\alpha_{1}u_{1}+\alpha_{2}u_{2}\in\Ker L$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall v_{1},v_{2}\in\Slika L,\beta_{1},\beta_{2}\in F\overset{?}{\Longrightarrow}\beta_{1}v_{1}+\beta_{2}v_{2}\in\Slika L$
\end_inset
.
Po predpostavki velja
\begin_inset Formula $\exists u_{1},u_{2}\in U\ni:v_{1}=Lu_{1}\wedge v_{2}=Lu_{2}$
\end_inset
.
Velja torej
\begin_inset Formula $\beta_{1}v_{1}+\beta_{2}v_{2}=\beta_{1}Lu_{1}+\beta_{2}Lu_{2}\overset{\text{linearnost}}{=}L\left(\beta_{1}u_{1}+\beta_{2}u_{2}\right)$
\end_inset
in
\begin_inset Formula $\beta_{1}u_{1}+\beta_{2}u_{2}\in U$
\end_inset
,
torej je
\begin_inset Formula $L\left(\beta_{1}u_{1}+\beta_{2}u_{2}\right)\in\Slika L$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Definition*
Ničnost
\begin_inset Formula $L$
\end_inset
je
\begin_inset Formula $\n\left(L\right)\coloneqq\dim\Ker L$
\end_inset
(angl.
nullity) in rang
\begin_inset Formula $L$
\end_inset
je
\begin_inset Formula $\rang\left(L\right)=\dim\Slika L$
\end_inset
(angl.
rank).
\end_layout
\begin_layout Remark*
Jedro in sliko smo definirali za linearne preslikave,
vendar ju lahko definiramo tudi za poljubno matriko
\begin_inset Formula $A$
\end_inset
nad poljem
\begin_inset Formula $F$
\end_inset
,
saj smo v
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:Matrika-linearne-preslikave"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
dokazali linearni izomorfizem med
\begin_inset Formula $m\times n$
\end_inset
matrikami nad
\begin_inset Formula $F$
\end_inset
in linearnimi preslikavami
\begin_inset Formula $F^{n}\to F^{m}$
\end_inset
.
\begin_inset Formula
\[
Au=\left[\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots & & \vdots\\
a_{m1} & \cdots & a_{mn}
\end{array}\right]\left[\begin{array}{c}
u_{1}\\
\vdots\\
u_{n}
\end{array}\right]=\left[\begin{array}{c}
a_{11}u_{1}+\cdots+a_{1n}u_{n}\\
\vdots\\
a_{m1}u_{1}+\cdots+a_{mn}u_{n}
\end{array}\right]=\left[\begin{array}{c}
a_{11}\\
\vdots\\
a_{m1}
\end{array}\right]u_{1}+\cdots+\left[\begin{array}{c}
a_{1n}\\
\vdots\\
a_{mn}
\end{array}\right]u_{n}
\]
\end_inset
Iz tega je razvidno,
da je
\begin_inset Formula $\Slika A$
\end_inset
torej linearna ogrinjača stolpcev matrike
\begin_inset Formula $A$
\end_inset
.
Pravimo tudi,
da je
\begin_inset Formula $\Slika A$
\end_inset
stolpični prostor
\begin_inset Formula $A$
\end_inset
oziroma
\begin_inset Formula $\Col A$
\end_inset
(angl.
column space).
\begin_inset Formula $\rang A=\dim\Slika A$
\end_inset
je torej največje število linearno neodvisnih stolpcev
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Claim*
Linearna preslikava
\begin_inset Formula $L$
\end_inset
je injektivna (
\begin_inset Formula $Lu_{1}=Lu_{2}\Rightarrow u_{1}=u_{2}$
\end_inset
)
\begin_inset Formula $\Leftrightarrow\Ker L=\left\{ 0\right\} $
\end_inset
.
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Predpostavimo,
da je
\begin_inset Formula $L$
\end_inset
injektivna,
torej
\begin_inset Formula $Lu_{1}=Lu_{2}\Rightarrow u_{1}=u_{2}$
\end_inset
.
Vzemimo poljuben
\begin_inset Formula $u\in\Ker L$
\end_inset
.
Zanj velja
\begin_inset Formula $Lu=0=L0\Rightarrow u=0$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
Predpostavimo,
\begin_inset Formula $\Ker L=\left\{ 0\right\} $
\end_inset
.
Računajmo:
\begin_inset Formula $Lu_{1}=Lu_{2}\Longrightarrow Lu_{1}-Lu_{2}=0\overset{\text{linearnost }}{\Longrightarrow}L\left(u_{1}-u_{2}\right)=0\Longrightarrow u_{1}-u_{2}=0\Longrightarrow u_{1}=u_{2}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Theorem*
osnovna formula.
Naj bo
\begin_inset Formula $L:U\to V$
\end_inset
linearna preslikava.
Tedaj je
\begin_inset Formula $\dim\Ker L+\dim\Slika L=\dim U$
\end_inset
,
torej
\begin_inset Formula $\n L+\rang L=\dim U$
\end_inset
.
Za matrike torej trdimo
\begin_inset Formula $\n A+\rang A=\dim F^{n}=n$
\end_inset
za
\begin_inset Formula $m\times n$
\end_inset
matriko
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Proof
Vemo,
da sta jedro in slika podprostora.
Naj bo
\begin_inset Formula $w_{1},\dots,w_{k}$
\end_inset
baza jedra in
\begin_inset Formula $u_{1},\dots,u_{l}$
\end_inset
njena dopolnitev do baze
\begin_inset Formula $U$
\end_inset
.
Torej
\begin_inset Formula $\dim U=k+l=\n L+l$
\end_inset
.
Treba je še dokazati,
da je
\begin_inset Formula $l=\rang A$
\end_inset
.
Konstruirajmo bazo za
\begin_inset Formula $\Slika L$
\end_inset
,
ki ima
\begin_inset Formula $l$
\end_inset
elementov in dokažimo,
da so
\begin_inset Formula $Lu_{1},\dots,Lu_{l}$
\end_inset
baza za
\begin_inset Formula $\Slika L$
\end_inset
:
\end_layout
\begin_deeper
\begin_layout Itemize
Je ogrodje?
Vzemimo poljuben
\begin_inset Formula $v\in\Slika L$
\end_inset
.
Zanj obstaja nek
\begin_inset Formula $u\in U\ni:Lu=v$
\end_inset
,
ki ga lahko razvijemo po bazi
\begin_inset Formula $U$
\end_inset
takole
\begin_inset Formula $u=\alpha_{1}w_{1}+\cdots+\alpha_{k}w_{k}+\beta_{1}u_{1}+\cdots+\beta_{l}u_{l}$
\end_inset
.
Sedaj na obeh straneh uporabimo
\begin_inset Formula $L$
\end_inset
in upoštevamo linearnost:
\begin_inset Formula
\[
v=Lu=L\left(\alpha_{1}w_{1}+\cdots+\alpha_{k}w_{k}+\beta_{1}u_{1}+\cdots+\beta_{l}u_{l}\right)=\alpha_{1}Lw_{1}+\cdots+\alpha_{k}Lw_{k}+\beta_{1}Lu_{1}+\cdots+\beta_{l}Lu_{l}
\]
\end_inset
\begin_inset Formula
\[
=\beta_{1}Lu_{1}+\cdots+\beta_{l}Lu_{l}
\]
\end_inset
Ker so
\begin_inset Formula $w_{i}$
\end_inset
baza
\begin_inset Formula $\Ker L$
\end_inset
,
so elementi
\begin_inset Formula $\Ker L$
\end_inset
,
torej je
\begin_inset Formula $Lw_{i}=0$
\end_inset
za vsak
\begin_inset Formula $i$
\end_inset
.
Tako poljuben
\begin_inset Formula $v\in\Slika L$
\end_inset
razpišemo z bazo velikosti
\begin_inset Formula $l$
\end_inset
.
\end_layout
\begin_layout Itemize
Je LN?
Računajmo:
\begin_inset Formula $\gamma_{1}Lu_{1}+\cdots+\gamma_{l}Lu_{l}=0\overset{\text{linearnost }}{\Longrightarrow}L\left(\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}\right)=0\Longrightarrow\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}\in\Ker L$
\end_inset
,
kar pomeni,
da ga je moč razviti po bazi
\begin_inset Formula $\Ker L$
\end_inset
:
\begin_inset Formula
\[
\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}=\delta_{1}w_{1}+\cdots+\delta_{k}w_{k}
\]
\end_inset
\begin_inset Formula
\[
\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}-\delta_{1}w_{1}-\cdots-\delta_{k}w_{k}=0
\]
\end_inset
Ker je
\begin_inset Formula $w_{1},\dots,w_{k},u_{1},\dots,u_{l}$
\end_inset
baza
\begin_inset Formula $U$
\end_inset
,
je LN,
zato velja
\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=w_{1}=\cdots=w_{k}=0$
\end_inset
,
kar pomeni,
da očitno velja
\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=0$
\end_inset
,
torej je res LN.
\end_layout
\end_deeper
\begin_layout Remark*
Bralcu prav pride skica s 3.
strani zapiskov predavanja
\begin_inset Quotes gld
\end_inset
LA1P FMF 2024-02-28
\begin_inset Quotes grd
\end_inset
.
\end_layout
\begin_layout Paragraph*
Do preproste matrike preslikave z ustreznimi bazami
\end_layout
\begin_layout Standard
Imenujmo sedaj
\begin_inset Formula $\mathcal{B}=\left\{ w_{1},\dots,w_{k},u_{1},\dots,u_{l}\right\} $
\end_inset
bazo za
\begin_inset Formula $U$
\end_inset
,
in
\begin_inset Formula $\mathcal{C}=\left\{ Lu_{1},\dots,Lu_{l},z_{1},\dots,z_{m}\right\} $
\end_inset
baza za
\begin_inset Formula $V$
\end_inset
,
kjer je
\begin_inset Formula $z_{1},\dots,z_{m}$
\end_inset
dopolnitev
\begin_inset Formula $Lu_{1},\dots,Lu_{l}$
\end_inset
do baze
\begin_inset Formula $V$
\end_inset
,
kajti
\begin_inset Formula $V$
\end_inset
je lahko večji kot samo
\begin_inset Formula $\Slika L$
\end_inset
,
in si oglejmo matriko naše preslikave
\begin_inset Formula $L:U\to V$
\end_inset
,
ki slika iz baze
\begin_inset Formula $\mathcal{B}$
\end_inset
v bazo
\begin_inset Formula $\mathcal{C}$
\end_inset
.
Najprej razpišimo preslikane elemente baze
\begin_inset Formula $\mathcal{B}$
\end_inset
po bazi
\begin_inset Formula $\mathcal{C}$
\end_inset
:
\begin_inset Formula
\[
\begin{array}{ccccccccccccc}
Lu_{1} & = & 1\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\
\vdots & & \vdots & & & & \vdots & & \vdots & & & & \vdots\\
Lu_{l} & = & 0\cdot Lu_{1} & + & \cdots & + & 1\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\
Lw_{1} & = & 0\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\
\vdots & & \vdots & & & & \vdots & & \vdots & & & & \vdots\\
Lw_{k} & = & 0\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}
\end{array}
\]
\end_inset
\begin_inset Formula
\[
\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{cc}
I_{l} & 0\\
0 & 0
\end{array}\right]
\]
\end_inset
S primerno izbiro baz
\begin_inset Formula $U$
\end_inset
in
\begin_inset Formula $V$
\end_inset
je torej matrika preslikave precej preprosta,
zgolj bločna matrika z identiteto,
veliko
\begin_inset Formula $\rang L$
\end_inset
in ničlami,
ki ustrezajo dimenzijam
\begin_inset Formula $U$
\end_inset
in
\begin_inset Formula $V$
\end_inset
.
\end_layout
\begin_layout Standard
Kaj pa,
če je
\begin_inset Formula $L$
\end_inset
matrika?
Recimo ji
\begin_inset Formula $A$
\end_inset
,
da je
\begin_inset Formula $L_{A}=L$
\end_inset
od prej.
Tedaj
\begin_inset Formula $A\in M_{p,n}\left(F\right)$
\end_inset
.
Naj bo
\begin_inset Formula $P=\left[\begin{array}{cccccc}
u_{1} & \cdots & u_{l} & w_{1} & \cdots & w_{k}\end{array}\right]$
\end_inset
matrika,
katere stolpci so baza
\begin_inset Formula $U$
\end_inset
in
\begin_inset Formula $Q=\left[\begin{array}{cccccc}
Au_{1} & \cdots & Au_{l} & z_{1} & \cdots & z_{m}\end{array}\right]$
\end_inset
matrika,
katere stolpci so baza
\begin_inset Formula $V$
\end_inset
.
Po karakterizaciji obrnljivih matrik sta obrnljivi.
Tedaj
\begin_inset Formula
\[
AP=\left[\begin{array}{cccccc}
Au_{1} & \cdots & Au_{l} & Aw_{1} & \cdots & Aw_{k}\end{array}\right]\overset{\text{jedro}}{=}\left[\begin{array}{cccccc}
Au_{1} & \cdots & Au_{l} & 0 & \cdots & 0\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
Q\left[\begin{array}{cc}
I_{l} & 0\\
0 & 0
\end{array}\right]=\left[\begin{array}{cccccc}
Au_{1} & \cdots & Au_{l} & z_{1} & \cdots & z_{m}\end{array}\right]\left[\begin{array}{cc}
I_{l} & 0\\
0 & 0
\end{array}\right]=\left[\begin{array}{cccccc}
Au_{1} & \cdots & Au_{l} & 0 & \cdots & 0\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
AP=Q\left[\begin{array}{cc}
I_{l} & 0\\
0 & 0
\end{array}\right]\Longrightarrow Q^{-1}AP=\left[\begin{array}{cc}
I_{l} & 0\\
0 & 0
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Subsubsection
Ekvivalentnost matrik
\end_layout
\begin_layout Definition*
Matriki
\begin_inset Formula $A$
\end_inset
in
\begin_inset Formula $B$
\end_inset
sta ekvivalentni (oznaka
\begin_inset Formula $A\sim B$
\end_inset
\begin_inset Foot
status open
\begin_layout Plain Layout
Isto oznako uporabljamo tudi za podobne matrike,
vendar podobnost ni enako kot ekvivalentnost.
\end_layout
\end_inset
)
\begin_inset Formula $\Leftrightarrow\exists$
\end_inset
obrnljivi
\begin_inset Formula $P,Q\ni:B=PAQ$
\end_inset
.
\end_layout
\begin_layout Example*
Dokazali smo,
da je vsaka matrika
\begin_inset Formula $A$
\end_inset
ekvivalentna matriki
\begin_inset Formula $\left[\begin{array}{cc}
I_{r} & 0\\
0 & 0
\end{array}\right]$
\end_inset
,
kjer je
\begin_inset Formula $r=\rang A$
\end_inset
.
\end_layout
\begin_layout Proof
Dokažimo,
da je relacija
\begin_inset Formula $\sim$
\end_inset
ekvivalenčna:
\end_layout
\begin_deeper
\begin_layout Itemize
refleksivnost:
\begin_inset Formula $A\sim A$
\end_inset
velja.
Naj bo
\begin_inset Formula $A\in M_{m,n}\left(F\right)$
\end_inset
.
Tedaj
\begin_inset Formula $A=I_{m}AI_{n}$
\end_inset
.
\end_layout
\begin_layout Itemize
simetričnost:
\begin_inset Formula $A\sim B\Rightarrow B\sim A$
\end_inset
,
kajti če velja
\begin_inset Formula $B=PAQ$
\end_inset
in sta
\begin_inset Formula $P$
\end_inset
in
\begin_inset Formula $Q$
\end_inset
obrnljivi,
velja
\begin_inset Formula $P^{-1}BQ^{-1}=A$
\end_inset
.
\end_layout
\begin_layout Itemize
tranzitivnost:
\begin_inset Formula $A\sim B\wedge B\sim C\Rightarrow A\sim C$
\end_inset
,
kajti,
če velja
\begin_inset Formula $B=PAQ$
\end_inset
in
\begin_inset Formula $C=SBT$
\end_inset
in so
\begin_inset Formula $P,Q,S,T$
\end_inset
obrnljive,
velja
\begin_inset Formula $C=\left(SP\right)A\left(QT\right)$
\end_inset
in produkt obrnljivih matrik je obrnljiva matrika.
\end_layout
\end_deeper
\begin_layout Theorem*
Dve matriki sta ekvivalentni natanko tedaj,
ko imata enako velikost in enak rang.
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
Po predpostavki imata
\begin_inset Formula $A$
\end_inset
in
\begin_inset Formula $B$
\end_inset
enako velikost in enak rang
\begin_inset Formula $r$
\end_inset
.
Od prej vemo,
da sta obe ekvivalentni
\begin_inset Formula $\left[\begin{array}{cc}
I_{r} & 0\\
0 & 0
\end{array}\right]$
\end_inset
,
ker pa je relacija ekvivalentnosti ekvivalenčna,
sta
\begin_inset Formula $A\sim B$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Po predpostavki
\begin_inset Formula $A\sim B$
\end_inset
,
torej
\begin_inset Formula $\exists P,Q\ni:B=PAQ$
\end_inset
.
Če je
\begin_inset Formula $A$
\end_inset
\begin_inset Formula $m\times n$
\end_inset
,
je
\begin_inset Formula $P$
\end_inset
\begin_inset Formula $m\times m$
\end_inset
in
\begin_inset Formula $Q$
\end_inset
\begin_inset Formula $n\times n$
\end_inset
,
zatorej je po definiciji matričnega množenja
\begin_inset Formula $B$
\end_inset
\begin_inset Formula $m\times n$
\end_inset
.
Dokazati je treba še
\begin_inset Formula $\rang A=\rang B$
\end_inset
.
\begin_inset Formula
\[
\rang B=\rang PAQ\overset{?}{=}\rang PA=\overset{?}{=}\rang A
\]
\end_inset
Dokažimo najprej
\begin_inset Formula $\rang PAQ=\rang PA$
\end_inset
oziroma
\begin_inset Formula $\rang CQ=\rang C$
\end_inset
za obrnljivo
\begin_inset Formula $Q$
\end_inset
in poljubno C.
Dokažemo lahko celo
\begin_inset Formula $\Slika CQ=\Slika C$
\end_inset
:
\begin_inset Formula
\[
\forall u:u\in\Slika CQ\Leftrightarrow\exists v\ni:u=\left(CQ\right)v\Leftrightarrow\exists v'\ni:u=Cv'\Leftrightarrow u\in\Slika C.
\]
\end_inset
Sedaj dokažimo še
\begin_inset Formula $\rang\left(PA\right)=\rang\left(A\right)$
\end_inset
.
Zadošča dokazati,
da je
\begin_inset Formula $\Ker\left(PA\right)=\Ker A$
\end_inset
,
kajti tedaj bi iz enakosti izrazov
\begin_inset Formula
\[
\dim\Slika A+\dim\Ker A=\dim F^{n}=n
\]
\end_inset
\begin_inset Formula
\[
\dim\Slika PA+\dim\Ker PA=\dim F^{n}=n
\]
\end_inset
dobili
\begin_inset Formula $\dim\Slika PA=\dim\Slika A$
\end_inset
.
Dokažimo torej
\begin_inset Formula $\Ker PA=\Ker A$
\end_inset
:
\begin_inset Formula
\[
\forall u:u\in\Ker PA\Leftrightarrow PAu=0\overset{P\text{ obrnljiva}}{\Longleftrightarrow}Au=0\Leftrightarrow u\in\Ker A.
\]
\end_inset
Torej je res
\begin_inset Formula $\Ker PA=\Ker A$
\end_inset
,
torej je res
\begin_inset Formula $\rang PA=\rang A$
\end_inset
,
torej je res
\begin_inset Formula $\rang A=\rang B$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsubsection
Podobnost matrik
\end_layout
\begin_layout Definition*
Kvadratni matriki
\begin_inset Formula $A$
\end_inset
in
\begin_inset Formula $B$
\end_inset
sta podobni,
če
\begin_inset Formula $\exists$
\end_inset
taka obrnljiva matrika
\begin_inset Formula $P\ni:B=PAP^{-1}$
\end_inset
.
\end_layout
\begin_layout Claim*
Podobnost je ekvivalenčna relacija.
\end_layout
\begin_layout Proof
Dokazujemo,
da je relacija ekvivalenčna,
torej:
\end_layout
\begin_deeper
\begin_layout Itemize
refleksivna:
\begin_inset Formula $A=IAI^{-1}=IAI=A$
\end_inset
\end_layout
\begin_layout Itemize
simetrična:
\begin_inset Formula $B=PAP^{-1}\Rightarrow P^{-1}BP=A$
\end_inset
\end_layout
\begin_layout Itemize
tranzitivna:
\begin_inset Formula $B=PAP^{-1}\wedge C=QBQ^{-1}\Rightarrow C=QPAP^{-1}Q^{-1}=\left(QP\right)A\left(QP\right)^{-1}$
\end_inset
\end_layout
\end_deeper
\begin_layout Remark
\begin_inset CommandInset label
LatexCommand label
name "rem:nista-podobni"
\end_inset
Očitno velja podobnost
\begin_inset Formula $\Rightarrow$
\end_inset
ekvivalentnost,
toda obrat ne velja vedno.
Na primer
\begin_inset Formula $\left[\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right]$
\end_inset
in
\begin_inset Formula $\left[\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right]$
\end_inset
sta ekvivalentni (sta enake velikosti in ranga),
toda nista podobni (dokaz kasneje).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Standard
Od prej vemo,
da je vsaka matrika ekvivalentna matriki
\begin_inset Formula $\left[\begin{array}{cc}
I_{r} & 0\\
0 & 0
\end{array}\right]$
\end_inset
,
kjer je
\begin_inset Formula $r$
\end_inset
njen rang.
A je vsaka kvadratna matrika podobna kakšni lepi matriki?
Ja.
Vsaka matrika je podobna zgornjetrikotni matriki in jordanski kanonični formi (več o tem kasneje).
Toda a je vsaka kvadratna matrika podobna diagonalni matriki?
Ne.
\end_layout
\begin_layout Definition*
Matrika
\begin_inset Formula $D$
\end_inset
je diagonalna
\begin_inset Formula $\sim d_{ij}\not=0\Rightarrow i=j$
\end_inset
.
\end_layout
\begin_layout Standard
Kdaj je matrika
\begin_inset Formula $A$
\end_inset
podobna neki diagonalni matriki?
Kdaj
\begin_inset Formula $\exists$
\end_inset
diagonalna
\begin_inset Formula $D$
\end_inset
in obrnljiva
\begin_inset Formula $P\ni:A=PDP^{-1}$
\end_inset
?
Izpeljimo iz nastavka.
\begin_inset Formula $D=\left[\begin{array}{ccc}
\lambda_{1} & & 0\\
& \ddots\\
0 & & \lambda n
\end{array}\right]$
\end_inset
in
\begin_inset Formula $P=\left[\begin{array}{ccc}
\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]$
\end_inset
,
kjer sta
\begin_inset Formula $D$
\end_inset
in
\begin_inset Formula $P$
\end_inset
neznani.
Ker mora biti
\begin_inset Formula $P$
\end_inset
obrnljiva,
so njeni stolpični vektorji LN.
\begin_inset Formula
\[
A=PDP^{-1}\Leftrightarrow AP=PD\Leftrightarrow A\left[\begin{array}{ccc}
\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc}
\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]\left[\begin{array}{ccc}
\lambda_{1} & & 0\\
& \ddots\\
0 & & \lambda_{n}
\end{array}\right]\text{ in }P\text{ obrnljiva}
\]
\end_inset
\begin_inset Formula
\[
\left[\begin{array}{ccc}
A\vec{v_{1}} & \cdots & A\vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc}
\lambda_{1}\vec{v_{1}} & \cdots & \lambda_{n}\vec{v_{n}}\end{array}\right]\text{ in }v_{i}\text{ so LN}
\]
\end_inset
\begin_inset Formula
\[
A\vec{v_{1}}=\lambda_{1}\vec{v_{1}},\dots,A\vec{v_{n}}=\lambda_{n}v_{n}\text{ in }\forall i:v_{i}\not=0
\]
\end_inset
\end_layout
\begin_layout Standard
Porodi se naloga,
imenovana
\begin_inset Quotes gld
\end_inset
Lastni problem
\begin_inset Quotes grd
\end_inset
.
Iščemo pare
\begin_inset Formula $\left(\lambda,\vec{v}\right)$
\end_inset
,
ki zadoščajo enačbi
\begin_inset Formula $A\vec{v}=\lambda\vec{v}$
\end_inset
.
\end_layout
\begin_layout Definition*
Pravimo,
da je
\begin_inset Formula $\lambda$
\end_inset
je lastna vrednost matrike
\begin_inset Formula $A$
\end_inset
,
če obstaja tak
\begin_inset Formula $\vec{v}\not=0$
\end_inset
,
da je
\begin_inset Formula $A\vec{v}=\lambda\vec{v}$
\end_inset
.
V tem primeru pravimo,
da je
\begin_inset Formula $\vec{v}$
\end_inset
lastni vektor,
ki pripada lastni vrednosti
\begin_inset Formula $\lambda$
\end_inset
.
Paru
\begin_inset Formula $\left(\lambda,\vec{v}\right)$
\end_inset
,
ki zadošča enačbi,
pravimo lastni par matrike
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Standard
Nalogo
\begin_inset Quotes gld
\end_inset
Lastni problem
\begin_inset Quotes grd
\end_inset
rešujemo v dveh korakih.
Najprej najdemo vse
\begin_inset Formula $\lambda$
\end_inset
,
nato za vsako poiščemo pripadajoče
\begin_inset Formula $\vec{v}$
\end_inset
,
ki za lastno vrednost obstajajo po definiciji.
\end_layout
\begin_layout Standard
Za nek
\begin_inset Formula $v\not=0$
\end_inset
pišimo
\begin_inset Formula $Av=\lambda v=\lambda Iv\Leftrightarrow Av-\lambda Iv=0\Leftrightarrow\left(A-\lambda I\right)v=0$
\end_inset
za nek
\begin_inset Formula $v\not=0\Leftrightarrow\Ker\left(A-\lambda I\right)\not=\left\{ 0\right\} \overset{\text{K.O.M.}}{\Longleftrightarrow}A-\lambda I$
\end_inset
ni obrnljiva
\begin_inset Formula $\Leftrightarrow\det\left(A-\lambda I\right)=0$
\end_inset
.
\end_layout
\begin_layout Definition*
Polinom
\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-xI\right)$
\end_inset
je karakteristični polinom matrike
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Definition*
Premislek zgoraj nam pove,
da so lastne vrednosti
\begin_inset Formula $A$
\end_inset
ničle
\begin_inset Formula $p_{A}\left(x\right)$
\end_inset
.
\end_layout
\begin_layout Remark*
Karakteristični polinom lahko nima nobene ničle:
\begin_inset Formula $A=\left[\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right]$
\end_inset
,
\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\det\left[\begin{array}{cc}
-\lambda & 1\\
-1 & -\lambda
\end{array}\right]=x^{2}+1$
\end_inset
,
katerega ničli sta
\begin_inset Formula $\lambda_{1}=i$
\end_inset
in
\begin_inset Formula $\lambda_{2}=-i$
\end_inset
,
ki nista realni števili.
V nadaljevanju se zato omejimo na kompleksne matrike in kompleksne lastne vrednosti,
saj ima po Osnovnem izreku Algebre polinom s kompleksnimi koeficienti vedno vsaj kompleksne ničle.
\end_layout
\begin_layout Standard
Kako pa iščemo lastne vektorje za lastno vrednost
\begin_inset Formula $\lambda$
\end_inset
?
Spomnimo se na
\begin_inset Formula $Av=\lambda v\Leftrightarrow v\in\Ker\left(A-\lambda I\right)$
\end_inset
.
Rešiti moramo homogen sistem linearnih enačb.
Po definiciji so lastni vektorji neničelni,
zato nas trivialna rešitev ne zanima.
\end_layout
\begin_layout Definition*
Množici
\begin_inset Formula $\Ker\left(A-\lambda I\right)$
\end_inset
pravimo lastni podprostor matrike
\begin_inset Formula $A$
\end_inset
,
ki pripada
\begin_inset Formula $\lambda$
\end_inset
.
Slednji vsebuje
\begin_inset Formula $\vec{0}$
\end_inset
in množico vektorjev,
ki so vsi lastni vektorji
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Exercise*
Izračunaj lastne vrednosti od
\begin_inset Formula $A=\left[\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right]$
\end_inset
.
Od prej vemo,
da
\begin_inset Formula $\lambda_{1}=i$
\end_inset
,
\begin_inset Formula $\lambda_{2}=-i$
\end_inset
.
Izračunajmo
\begin_inset Formula $\Ker\left(A-iI\right)$
\end_inset
in
\begin_inset Formula $\Ker\left(A+iI\right)$
\end_inset
:
\begin_inset Formula
\[
\Ker\left(A-iI\right):\quad\left[\begin{array}{cc}
-i & 1\\
-1 & -i
\end{array}\right]\left[\begin{array}{c}
x\\
y
\end{array}\right]=0\quad\Longrightarrow\quad-ix+y=0,-x-iy=0\quad\Longrightarrow\quad y=ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c}
1\\
i
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
\Ker\left(A+iI\right):\quad\left[\begin{array}{cc}
i & 1\\
-1 & i
\end{array}\right]\left[\begin{array}{c}
x\\
y
\end{array}\right]=0\quad\Longrightarrow\quad ix+y=0,-x+y=0\quad\Longrightarrow\quad y=-ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c}
1\\
-i
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
\Ker\left(A-iI\right)=\Lin\left\{ \left[\begin{array}{c}
1\\
i
\end{array}\right]\right\} ,\quad\Ker\left(A+iI\right)=\Lin\left\{ \left[\begin{array}{c}
1\\
-i
\end{array}\right]\right\}
\]
\end_inset
\end_layout
\begin_layout Exercise*
Vstavimo lastna vektorja v
\begin_inset Formula $P$
\end_inset
in lastne vrednosti v
\begin_inset Formula $D$
\end_inset
na pripadajoči mesti.
Dobimo obrnljivo
\begin_inset Formula $P$
\end_inset
in velja
\begin_inset Formula $A=PDP^{-1}$
\end_inset
\begin_inset Formula
\[
P=\left[\begin{array}{cc}
1 & 1\\
i & -i
\end{array}\right],\quad D=\left[\begin{array}{cc}
i & 0\\
0 & -i
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Exercise*
Temu početju pravimo
\begin_inset Quotes gld
\end_inset
diagonalizacija matrike
\begin_inset Formula $A$
\end_inset
\begin_inset Quotes grd
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Primer matrike,
ki ni diagonalizabilna:
\begin_inset Formula $A=\left[\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right]$
\end_inset
.
\begin_inset Formula $\det\left(A-\lambda I\right)=\left[\begin{array}{cc}
-\lambda & 1\\
0 & -\lambda
\end{array}\right]=\lambda^{2}$
\end_inset
.
Ničli/lastni vrednosti sta
\begin_inset Formula $\lambda_{1}=0$
\end_inset
in
\begin_inset Formula $\lambda_{2}=0$
\end_inset
.
Toda
\begin_inset Formula $\Ker\left(A-0I\right)=\Ker A=\Lin\left\{ \left[\begin{array}{c}
1\\
0
\end{array}\right]\right\} $
\end_inset
in
\begin_inset Formula $P=\left[\begin{array}{cc}
1 & 1\\
0 & 0
\end{array}\right]$
\end_inset
ni obrnljiva.
S tem dokažemo trditev v primeru
\begin_inset CommandInset ref
LatexCommand ref
reference "rem:nista-podobni"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
.
\begin_inset Formula $\left[\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right]$
\end_inset
in
\begin_inset Formula $\left[\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right]$
\end_inset
nista podobni,
ker je prva diagonalna,
druga pa ni podobna diagonalni matriki (ne da se je diagonalizirati).
\end_layout
\begin_layout Standard
Lastne vrednosti lahko definiramo tudi za linearne preslikave,
saj so linearne preslikave linearno izomorfne matrikam.
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $V$
\end_inset
vektorski prostor nad
\begin_inset Formula $F=\mathbb{C}$
\end_inset
in
\begin_inset Formula $L:V\to V$
\end_inset
linearna preslikava.
Število
\begin_inset Formula $\lambda\in F$
\end_inset
je lastna vrednost
\begin_inset Formula $L$
\end_inset
,
le obstaja tak neničelni
\begin_inset Formula $v\in V$
\end_inset
,
da velja
\begin_inset Formula $Lv=\lambda v$
\end_inset
.
\end_layout
\begin_layout Standard
Kako pa rešujemo
\begin_inset Quotes gld
\end_inset
Lastni problem
\begin_inset Quotes grd
\end_inset
za linearne preslikave?
\begin_inset Formula $Lv=\lambda v\Leftrightarrow Lv-\lambda\left(id\right)v=0\Leftrightarrow\left(L-\lambda\left(id\right)\right)v=0\Leftrightarrow v\in\Ker\left(L-\lambda\left(id\right)\right)\overset{v\not=0}{\Longleftrightarrow}\det\left(L-\lambda\left(id\right)\right)=0$
\end_inset
.
Toda determinante linearne preslikave nismo definirali.
Lahko pa determinanto izračunamo na matriki,
ki pripada tej linearni preslikavi.
Toda dvem različnim bazam pripadata različni matriki linearne preslikave.
Dokazati je treba,
da sta determinanti dveh matrik,
pripadajočih eni linearni preslikavi,
enaki,
četudi sta matriki v različnih bazah.
\end_layout
\begin_layout Lemma
\begin_inset CommandInset label
LatexCommand label
name "lem:Podobni-matriki-imata"
\end_inset
Podobni matriki imata isto determinanto.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $B=PAP^{-1}$
\end_inset
za neko obrnljivo
\begin_inset Formula $P$
\end_inset
.
Tedaj
\begin_inset Formula $\det B=\det PAP^{-1}=\det P\det A\det P^{-1}=\det P\det P^{-1}\det A=\det PP^{-1}\det A=\det I\det A=1\cdot\det A=\det A$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Proof
\begin_inset Formula $L:V\to V$
\end_inset
naj bo linearna preslikava,
\begin_inset Formula $V$
\end_inset
prostor nad
\begin_inset Formula $F=\mathbb{C}$
\end_inset
,
\begin_inset Formula $\mathcal{B}$
\end_inset
in
\begin_inset Formula $\mathcal{C}$
\end_inset
pa bazi
\begin_inset Formula $V$
\end_inset
.
Priredimo matriki
\begin_inset Formula $L_{\mathcal{B}\leftarrow\mathcal{B}}$
\end_inset
in
\begin_inset Formula $L_{\mathcal{C}\leftarrow\mathcal{C}}$
\end_inset
.
Spomnimo se izreka
\begin_inset CommandInset ref
LatexCommand vref
reference "thm:matrika-kompozituma-linearnih"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
:
\begin_inset Formula $\left[KL\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset
.
\begin_inset Formula $L=\left[id\circ L\circ id\right]$
\end_inset
,
zato
\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\circ L\circ id\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}\left[id\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{C}}}}=P\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}P^{-1}$
\end_inset
za neko obrnljivo
\begin_inset Formula $P$
\end_inset
.
Torej sta matriki
\begin_inset Formula $\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}$
\end_inset
in
\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}$
\end_inset
podobni,
torej imata po lemi
\begin_inset CommandInset ref
LatexCommand vref
reference "lem:Podobni-matriki-imata"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
isto determinanto.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Proof
Alternativen dokaz,
da imata podobni matriki iste lastne vrednosti:
\begin_inset Formula $A$
\end_inset
podobna
\begin_inset Formula
\[
B\Rightarrow B=PAP^{-1}\Rightarrow B-xI=P\left(A-xI\right)P^{-1}\Rightarrow\det\left(B-xI\right)=\det\left(A-xI\right)\Rightarrow p_{A}=p_{B},
\]
\end_inset
torej so lastne vrednosti enake.
Kaj pa lastni vektorji?
Naj bo
\begin_inset Formula $v$
\end_inset
lastni vektor
\begin_inset Formula $A$
\end_inset
,
torej
\begin_inset Formula
\[
Av=\lambda v\Rightarrow PAv=\lambda Pv\Rightarrow PAP^{-1}Pv=\lambda Pv\Rightarrow BPv=\lambda Pv,
\]
\end_inset
torej za
\begin_inset Formula $v$
\end_inset
lastni vektor
\begin_inset Formula $A$
\end_inset
sledi,
da je
\begin_inset Formula $Pv$
\end_inset
lastni vektor
\begin_inset Formula $B$
\end_inset
.
\end_layout
\begin_layout Standard
Linearni transformaciji torej priredimo tako matriko,
ki ima v začetnem in končnem prostoru isto bazo.
Tedaj lahko izračunamo lastne pare na tej matriki.
\end_layout
\begin_layout Theorem*
Schurov izrek.
Vsaka kompleksna kvadratna matrika je podobna zgornjetrikotni matriki.
\end_layout
\begin_layout Proof
Indukcija po velikosti matrike.
\end_layout
\begin_deeper
\begin_layout Itemize
Baza:
\begin_inset Formula $A_{1\times1}$
\end_inset
je zgornjetrikotna.
\end_layout
\begin_layout Itemize
Korak:
Po I.
P.
trdimo,
da je vsaka
\begin_inset Formula $A_{\left(n-1\right)\times\left(n-1\right)}$
\end_inset
podobna kaki zgornjetrikotni matriki.
Dokažimo še za poljubno
\begin_inset Formula $A_{n\times n}$
\end_inset
.
Naj bo
\begin_inset Formula $\lambda$
\end_inset
lastna vrednost
\begin_inset Formula $A$
\end_inset
in
\begin_inset Formula $v_{1}$
\end_inset
pripadajoči lastni vektor ter
\begin_inset Formula $v_{2},\dots,v_{n}$
\end_inset
dopolnitev
\begin_inset Formula $v_{1}$
\end_inset
do baze
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
.
Potem je matrika
\begin_inset Formula $P=\left[\begin{array}{ccc}
v_{1} & \cdots & v_{n}\end{array}\right]$
\end_inset
obrnljiva.
\begin_inset Formula
\[
AP=\left[\begin{array}{ccc}
Av_{1} & \cdots & Av_{n}\end{array}\right]=\left[\begin{array}{ccc}
v_{1} & \cdots & v_{n}\end{array}\right]\left[\begin{array}{cccc}
\lambda & a_{1,2} & \cdots & a_{1,n}\\
0 & \vdots & & \vdots\\
\vdots & \vdots & & \vdots\\
0 & a_{m,n} & \cdots & a_{m.n}
\end{array}\right]=P\left[\begin{array}{cc}
\lambda & B\\
0 & C
\end{array}\right]
\]
\end_inset
Po I.
P.
obstaja taka zgornjetrikotna
\begin_inset Formula $T$
\end_inset
in obrnljiva
\begin_inset Formula $Q$
\end_inset
,
da
\begin_inset Formula $C=QTQ^{-1}$
\end_inset
.
\begin_inset Formula
\[
\left[\begin{array}{cc}
1 & 0\\
0 & Q
\end{array}\right]^{-1}P^{-1}AP\left[\begin{array}{cc}
1 & 0\\
0 & Q
\end{array}\right]=\left[\begin{array}{cc}
1 & 0\\
0 & Q
\end{array}\right]^{-1}\left[\begin{array}{cc}
\lambda & B\\
0 & C
\end{array}\right]\left[\begin{array}{cc}
1 & 0\\
0 & Q
\end{array}\right]=
\]
\end_inset
\begin_inset Formula
\[
=\left[\begin{array}{cc}
\lambda & C\\
0 & Q^{-1}B
\end{array}\right]\left[\begin{array}{cc}
1 & 0\\
0 & Q
\end{array}\right]=\left[\begin{array}{cc}
\lambda & CQ\\
0 & B
\end{array}\right]
\]
\end_inset
\begin_inset Formula $A$
\end_inset
je torej podobna
\begin_inset Formula $\left[\begin{array}{cc}
\lambda & CQ\\
0 & B
\end{array}\right]$
\end_inset
,
ki je zgornjetrikotna.
\end_layout
\end_deeper
\begin_layout Proof
\begin_inset Note Note
status open
\begin_layout Plain Layout
TODO karakterizacija linearnih preslikav
\begin_inset Quotes gld
\end_inset
LA1V FMF 2024-03-12
\begin_inset Quotes grd
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Subsubsection
Zadosten pogoj za diagonalizabilnost
\end_layout
\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:lave-razl-lavr-so-LN"
\end_inset
Lastni vektorji,
ki pripadajo različnim lastnim vrednostim,
so linearno neodvisni.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $A_{n\times n}$
\end_inset
matrika,
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset
njene lastne vrednosti in
\begin_inset Formula $v_{1},\dots,v_{k}$
\end_inset
njim pripadajoči lastni vektorji.
Dokazujemo
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset
paroma različni
\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$
\end_inset
LN.
Dokaz z indukcijo po
\begin_inset Formula $k$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Itemize
Baza
\begin_inset Formula $k=1$
\end_inset
:
Elementi
\begin_inset Formula $\left\{ \lambda_{1}\right\} $
\end_inset
so trivialno paroma različni in
\begin_inset Formula $v_{1}$
\end_inset
je kot neničen vektor LN.
\end_layout
\begin_layout Itemize
Korak:
Dokazujemo
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k+1}$
\end_inset
so paroma različne
\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$
\end_inset
so LN,
vedoč I.
P.
Denimo,
da
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$
\end_inset
.
Množimo z
\begin_inset Formula $A$
\end_inset
:
\begin_inset Formula
\[
A\left(\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}\right)=\alpha_{1}Av_{1}+\cdots+\alpha_{k+1}Av_{k+1}=\alpha_{1}\lambda_{1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0
\]
\end_inset
Množimo začetno enačbo z
\begin_inset Formula $\lambda_{k+1}$
\end_inset
(namesto z
\begin_inset Formula $A$
\end_inset
,
kot smo to storili zgoraj):
\begin_inset Formula
\[
\alpha_{1}\lambda_{k+1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0
\]
\end_inset
Odštejmo eno enačbo od druge,
dobiti moramo 0,
saj odštevamo 0 od 0:
\begin_inset Formula
\[
\alpha_{1}\left(\lambda_{1}-\lambda_{k+1}\right)v_{1}+\cdots+\alpha_{k}\left(\lambda_{k}-\lambda_{k+1}\right)v_{k}+\cancel{\alpha_{k+1}\left(\lambda_{k+1}-\lambda_{k+1}\right)v_{k+1}}=0
\]
\end_inset
Ker so lastne vrednosti paroma različne (
\begin_inset Formula $\lambda_{i}=\lambda_{j}\Rightarrow i=j$
\end_inset
),
so njihove razlike neničelne.
Ker so
\begin_inset Formula $v_{1},\dots,v_{k}$
\end_inset
po predpostavki LN,
sledi
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=0$
\end_inset
.
Vstavimo te konstante v
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$
\end_inset
in dobimo
\begin_inset Formula $\alpha_{k+1}v_{k+1}=0$
\end_inset
.
Ker je
\begin_inset Formula $v_{k+1}$
\end_inset
neničeln (je namreč lastni vektor),
sledi
\begin_inset Formula $\alpha_{k+1}=0$
\end_inset
,
torej
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=\alpha_{k+1}=0$
\end_inset
,
zatorej so
\begin_inset Formula $v_{1},\dots,v_{k+1}$
\end_inset
res LN.
\end_layout
\end_deeper
\begin_layout Corollary
\begin_inset CommandInset label
LatexCommand label
name "cor:vsota-lastnih-podpr-direktna"
\end_inset
Vsota vseh lastnih podprostorov matrike je direktna (definicija
\begin_inset CommandInset ref
LatexCommand vref
reference "def:vsota-je-direktna"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
).
\end_layout
\begin_layout Proof
Naj bodo
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset
vse paroma različne lastne vrednosti matrike
\begin_inset Formula $A\in M_{n}\left(\mathbb{C}\right)$
\end_inset
.
Pripadajoči lastni podprostori so torej
\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :V_{i}=\Ker\left(A-\lambda_{i}I\right)$
\end_inset
.
Trdimo,
da je vsota teh podprostorov direktna,
torej
\begin_inset Formula $\forall v_{1}\in V_{1},\dots,v_{k}\in V_{k}:v_{1}+\cdots+v_{k}=0\Rightarrow v_{1}=\cdots=v_{k}=0$
\end_inset
.
To sledi iz izreka
\begin_inset CommandInset ref
LatexCommand vref
reference "thm:lave-razl-lavr-so-LN"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
.
\end_layout
\begin_layout Corollary*
Če ima
\begin_inset Formula $n\times n$
\end_inset
matrika
\begin_inset Formula $n$
\end_inset
paroma različnih lastnih vrednosti,
je podobna diagonalni matriki.
\end_layout
\begin_layout Proof
Po posledici
\begin_inset CommandInset ref
LatexCommand vref
reference "cor:vsota-lastnih-podpr-direktna"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
je vsota lastnih podprostorov matrike
\begin_inset Formula $A_{n\times n}$
\end_inset
direktna.
Če je torej lastnih podprostorov
\begin_inset Formula $n$
\end_inset
,
je njihova vsota cel prostor
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
.
Matriko se da diagonalizirati,
kadar je vsota vseh lastnih podprostorov enaka podprostoru
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
(tedaj so namreč stolpci matrike
\begin_inset Formula $P$
\end_inset
linearno neodvisni,
zato je
\begin_inset Formula $P$
\end_inset
obrnljiva).
\end_layout
\begin_layout Subsubsection
Algebraične in geometrijske vekčratnosti
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $A_{n\times n}$
\end_inset
matrika.
\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\left(-1\right)^{n}\left(\lambda-\lambda_{1}\right)^{n_{1}}\cdots\left(\lambda-\lambda_{k}\right)^{n_{k}}$
\end_inset
,
kjer so
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset
vse paroma različne lastne vrednosti
\begin_inset Formula $A$
\end_inset
.
Stopnji ničle —
\begin_inset Formula $n_{i}$
\end_inset
—
rečemo algebraična večkratnost lastne vrednosti
\begin_inset Formula $\lambda_{i}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Geometrijska večkratnost lastne vrednosti
\begin_inset Formula $\lambda_{i}$
\end_inset
je
\begin_inset Formula $\dim\Ker\left(A-\lambda_{i}I\right)=\n$
\end_inset
\begin_inset Formula $\left(A-\lambda_{i}I\right)=m_{i}$
\end_inset
.
\end_layout
\begin_layout Standard
Algebraično večkratnost
\begin_inset Formula $\lambda_{i}$
\end_inset
označimo z
\begin_inset Formula $n_{i}$
\end_inset
in je večkratnost ničle
\begin_inset Formula $\lambda_{i}$
\end_inset
v
\begin_inset Formula $p_{A}\left(\lambda\right)$
\end_inset
(karakterističnem polinomu).
Geometrijsko večkratnost
\begin_inset Formula $\lambda_{i}$
\end_inset
pa označimo z
\begin_inset Formula $m_{i}$
\end_inset
in je dimenzija lastnega podprostora za
\begin_inset Formula $\lambda_{i}$
\end_inset
.
\end_layout
\begin_layout Claim
\begin_inset CommandInset label
LatexCommand label
name "claim:geom<=alg"
\end_inset
\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$
\end_inset
—
geometrijska večkratnost lastne vrednosti je kvečjemu tolikšna,
kot je algebraična večkratnost te lastne vrednosti.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $v_{1},\dots,v_{m_{i}}$
\end_inset
baza za lastni podprostor
\begin_inset Formula $V_{i}=\Ker\left(A-\lambda_{i}I\right)$
\end_inset
in naj bo
\begin_inset Formula $v_{m_{i}+1},\dots,v_{n}$
\end_inset
njena dopolnitev do baze
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
.
Tedaj velja:
\begin_inset Formula $Av_{1}=\lambda_{1}v_{1}$
\end_inset
,
...,
\begin_inset Formula $Av_{m_{i}}=\lambda_{m_{i}}v_{m_{i}}$
\end_inset
,
\begin_inset Formula $Av_{m_{i}+1}=$
\end_inset
linearna kombinacija
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
,
...,
\begin_inset Formula $Av_{n}=$
\end_inset
linearna kombinacija
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
.
Naj bo
\begin_inset Formula $P=\left[\begin{array}{cccccc}
v_{1} & \cdots & v_{m_{i}} & v_{m_{i}+1} & \cdots & v_{n}\end{array}\right]$
\end_inset
,
ki je obrnljiva.
\begin_inset Formula
\[
P^{-1}AP=\left[\begin{array}{cc}
\lambda_{i}I_{m_{i}} & B\\
0 & C
\end{array}\right]
\]
\end_inset
Ker je karakteristični polinom neodvisen od izbire baze,
velja
\begin_inset Formula
\[
\det\left(A-xI_{n}\right)=\det\left(\lambda_{i}I_{m_{i}}-xI\right)\det\left(C-xI_{n-m_{i}}\right)=\left(\lambda_{i}-x\right)^{m_{i}}\det\left(C-xI_{n-m_{i}}\right)
\]
\end_inset
Ker
\begin_inset Formula $\left(\lambda-x\right)^{m_{i}}$
\end_inset
deli karakteristični polinom,
je algebraična večkratnost
\begin_inset Formula $\lambda_{i}$
\end_inset
vsaj tolikšna,
kot je geometrična.
\end_layout
\begin_layout Claim
\begin_inset CommandInset label
LatexCommand label
name "claim:mi=ni=>diag"
\end_inset
Matriko s paroma različnimi lastnimi vrednostmi
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset
je moč diagonalizirati
\begin_inset Formula $\Leftrightarrow\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$
\end_inset
.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $V_{i}$
\end_inset
lastno podprostor lastne vrednosti
\begin_inset Formula $\lambda_{i}$
\end_inset
.
Vemo,
da se da
\begin_inset Formula $A_{n\times n}$
\end_inset
diagonalizirati
\begin_inset Formula $\Leftrightarrow A$
\end_inset
ima
\begin_inset Formula $n$
\end_inset
LN stolpičnih vektorjev
\begin_inset Formula $\Leftrightarrow\Ker\left(A-\lambda_{1}I\right)+\cdots+\Ker\left(A-\lambda_{k}I\right)=\mathbb{C}^{n}\Leftrightarrow\dim\left(V_{i}+\cdots+V_{k}\right)=\dim V_{i}+\cdots+\dim V_{k}\Leftrightarrow$
\end_inset
vsota lastnih podprostorov je direktna
\begin_inset Formula $\Leftrightarrow\dim\left(V_{1}+\cdots+V_{n}\right)=n\Leftrightarrow\dim V_{1}+\cdots+\dim V_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n_{1}+\cdots+n_{m}$
\end_inset
.
Toda ker po prejšnjem izreku
\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$
\end_inset
,
mora veljati
\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Minimalni polinom matrike
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+\cdots+c_{n}x^{n}\in\mathbb{C}\left[x\right]$
\end_inset
polinom in
\begin_inset Formula $A$
\end_inset
matrika.
\begin_inset Formula $p\left(A\right)\coloneqq c_{0}A^{0}+\cdots+c_{n}A^{n}=c_{0}I+\cdots+c_{n}A^{n}$
\end_inset
.
Če je
\begin_inset Formula $p\left(A\right)=0$
\end_inset
(ničelna matrika),
pravimo,
da polinom
\begin_inset Formula $p$
\end_inset
anhilira/uniči matriko
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Fact*
\begin_inset Formula $p\left(A\right)=0\Rightarrow p\left(P^{-1}AP\right)=0$
\end_inset
.
\end_layout
\begin_layout Standard
Izkaže se,
da karakteristični polimom anhilira matriko —
\begin_inset Formula $p_{A}\left(A\right)=0$
\end_inset
.
Dokaz kasneje.
\end_layout
\begin_layout Definition*
Polinom
\begin_inset Formula $m\left(x\right)$
\end_inset
je minimalen polinom
\begin_inset Formula $A$
\end_inset
,
če velja:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $m\left(A\right)=0$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $m$
\end_inset
ima vodilni koeficient 1
\end_layout
\begin_layout Enumerate
med vsemi polinomi,
ki zadoščajo prvi in drugi zahtevi,
ima
\begin_inset Formula $m$
\end_inset
najnižjo stopnjo
\end_layout
\end_deeper
\begin_layout Claim*
eksistenca minimalnega polinoma —
Minimalni polinom obstaja.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $A_{n\times n}$
\end_inset
matrika.
Očitno je
\begin_inset Formula $M_{n\times n}\left(\mathbb{C}\right)$
\end_inset
vektorski prostor dimenzije
\begin_inset Formula $n^{2}$
\end_inset
.
Matrike
\begin_inset Formula $\left\{ I,A,A^{2},\dots,A^{n^{2}}\right\} $
\end_inset
so linearno odvisne,
ker je moč te množice za 1 večja od moči vektorskega prostora.
Torej
\begin_inset Formula $\exists c_{0},\cdots,c_{n^{2}}\in\mathbb{C}$
\end_inset
,
ki niso vse 0
\begin_inset Formula $\ni:c_{0}I+c_{1}A+c_{2}A^{2}+\cdots+c_{n^{2}}A^{n^{2}}=0$
\end_inset
.
Torej polinom
\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+c_{1}x^{1}+c_{2}x^{2}+\cdots+c_{n^{2}}x^{n^{2}}$
\end_inset
anhilira
\begin_inset Formula $A$
\end_inset
.
Če ta polinom delimo z njegovim vodilnim koeficientom,
dobimo polinom,
ki ustreza prvima dvema zahevama za minimalni polinom.
Če med vsemi takimi izberemo takega z najnižjo stopnjo,
le-ta ustreza še tretji zahtevi.
\end_layout
\begin_layout Theorem*
Če je
\begin_inset Formula $m\left(x\right)$
\end_inset
minimalni polinom za
\begin_inset Formula $A$
\end_inset
in če
\begin_inset Formula $p\left(x\right)$
\end_inset
anhilira
\begin_inset Formula $A$
\end_inset
,
potem
\begin_inset Formula $m\left(x\right)\vert p\left(x\right)$
\end_inset
(
\begin_inset Formula $m\left(x\right)$
\end_inset
deli
\begin_inset Formula $p\left(x\right)$
\end_inset
).
\end_layout
\begin_layout Proof
Delimo
\begin_inset Formula $p$
\end_inset
z
\begin_inset Formula $m$
\end_inset
:
\begin_inset Formula $\exists k\left(x\right),r\left(x\right)\ni:p\left(x\right)=k\left(x\right)m\left(x\right)+r\left(x\right)\wedge\deg r\left(x\right)<\deg m\left(x\right)$
\end_inset
.
Vstavimo
\begin_inset Formula $A$
\end_inset
na obe strani:
\begin_inset Formula
\[
0=p\left(A\right)=k\left(A\right)m\left(A\right)+r\left(A\right)=k\left(A\right)\cdot0+r\left(A\right)=0+r\left(A\right)=r\left(A\right)=0
\]
\end_inset
Sledi
\begin_inset Formula $r\left(x\right)=0$
\end_inset
,
kajti če
\begin_inset Formula $r$
\end_inset
ne bi bil ničeln polinom,
bi ga lahko delili z vodilnim koeficientom in po predpostavki
\begin_inset Formula $\deg r\left(x\right)<\deg m\left(x\right)$
\end_inset
bi imel manjšo stopnjo kot
\begin_inset Formula $m\left(x\right)$
\end_inset
,
torej bi ustrezal zahtevam 1 in 2 za minimalni polinom in bi imel manjšo stopnjo od
\begin_inset Formula $m$
\end_inset
,
torej
\begin_inset Formula $m$
\end_inset
ne bi bil minimalni polinom,
kar bi vodilo v protislovje.
\end_layout
\begin_layout Corollary*
enoličnost minimalnega polinoma.
Naj bosta
\begin_inset Formula $m_{1}$
\end_inset
in
\begin_inset Formula $m_{2}$
\end_inset
minimalna polinoma matrike
\begin_inset Formula $A$
\end_inset
.
Ker
\begin_inset Formula $m$
\end_inset
po definiciji anhilira
\begin_inset Formula $A$
\end_inset
,
iz prejšnje trditve sledi,
če vstavimo
\begin_inset Formula $m=m_{1}$
\end_inset
in
\begin_inset Formula $p=m_{2}$
\end_inset
,
\begin_inset Formula $m_{1}\vert m_{2}$
\end_inset
.
Toda če vstavimo
\begin_inset Formula $m=m_{2}$
\end_inset
in
\begin_inset Formula $p=m_{1}$
\end_inset
,
\begin_inset Formula $m_{2}\vert m_{1}$
\end_inset
.
Iz
\begin_inset Formula $m_{1}\vert m_{2}\wedge m_{2}\vert m_{1}$
\end_inset
sledi,
da se
\begin_inset Formula $m_{1}$
\end_inset
in
\begin_inset Formula $m_{2}$
\end_inset
razlikujeta le za konstanten faktor,
ki pa je po definiciji minimalnega polinoma 1,
torej
\begin_inset Formula $m_{1}=m_{2}$
\end_inset
.
Zaradi enoličnosti lahko označimo minimalni polinom
\begin_inset Formula $A$
\end_inset
z
\begin_inset Formula $m_{A}\left(x\right)$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Ničle minimalnega polinoma
\end_layout
\begin_layout Claim*
\begin_inset Formula $m_{A}\left(x\right)$
\end_inset
in
\begin_inset Formula $p_{A}\left(x\right)$
\end_inset
imata iste ničle
\begin_inset Formula $\sim$
\end_inset
ničle
\begin_inset Formula $m_{A}\left(x\right)$
\end_inset
so lastne vrednosti
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Proof
Ker je
\begin_inset Formula $p_{A}\left(x\right)$
\end_inset
(dokaz kasneje),
velja po trditvi v dokazu enoličnosti,
da
\begin_inset Formula $m_{A}\vert p_{A}$
\end_inset
,
torej je vsaka ničla
\begin_inset Formula $m_{A}$
\end_inset
tudi ničla
\begin_inset Formula $p_{A}$
\end_inset
.
Treba je dokazati še,
da je vsaka ničla
\begin_inset Formula $p_{A}$
\end_inset
tudi ničla
\begin_inset Formula $m_{A}$
\end_inset
,
natančneje:
Treba je dokazati,
da če je
\begin_inset Formula $\lambda$
\end_inset
lastna vrednost matrike
\begin_inset Formula $A$
\end_inset
,
je
\begin_inset Formula $m_{A}\left(\lambda\right)=0$
\end_inset
.
Naj bo
\begin_inset Formula $v\not=0$
\end_inset
lastni vektor za
\begin_inset Formula $\lambda$
\end_inset
.
Tedaj
\begin_inset Formula $Av=\lambda v$
\end_inset
.
Potem velja
\begin_inset Formula $A^{2}v=AAv=A\lambda v=\lambda Av=\lambda\lambda v=\lambda^{2}v$
\end_inset
in splošneje
\begin_inset Formula $A^{n}v=\lambda^{n}v$
\end_inset
.
Sedaj recimo,
da je
\begin_inset Formula $m_{A}\left(x\right)=d_{0}x^{0}+\cdots+d_{r}x^{r}$
\end_inset
.
Potem je,
ker minimalni polinom anhilira
\begin_inset Formula $A$
\end_inset
,
\begin_inset Formula
\[
m_{A}\left(\lambda\right)v=\left(d_{0}+d_{1}\lambda+d_{2}\lambda^{2}+\cdots+d_{r}\lambda^{r}\right)v=d_{0}v+d_{1}\lambda v+d_{2}\lambda^{2}v+\cdots+d_{r}\lambda^{r}v=
\]
\end_inset
\begin_inset Formula
\[
=d_{0}v+d_{1}Av+d_{2}A^{2}v+\cdots+d_{r}A^{r}v=\left(d_{0}+d_{1}A+d_{2}A^{2}+\cdots+d_{r}A^{r}\right)v=m_{A}\left(A\right)v=0v=0
\]
\end_inset
Ker
\begin_inset Formula $m_{A}\left(\lambda\right)v=0$
\end_inset
in
\begin_inset Formula $v\not=0$
\end_inset
(je namreč lastni vektor),
velja
\begin_inset Formula $m_{A}\left(\lambda\right)=0$
\end_inset
.
\end_layout
\begin_layout Paragraph*
Lastnosti
\end_layout
\begin_layout Standard
Ker je
\begin_inset Formula $p_{A}\left(x\right)=\left(-1\right)^{n}\left(x-\lambda_{1}\right)^{n_{1}}\cdots\left(x-\lambda_{k}\right)^{n_{k}}$
\end_inset
in
\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{r_{1}}\cdots\left(x-\lambda_{k}\right)^{r_{k}}$
\end_inset
,
sledi iz
\begin_inset Formula $m_{A}\vert p_{A}\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\leq n_{i}$
\end_inset
.
Poleg tega,
ker
\begin_inset Formula $m_{A}\left(\lambda_{1}\right)=0\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\geq1$
\end_inset
.
Toda pozor:
\series bold
Ni
\series default
res,
da
\begin_inset Formula $r_{i}=m_{i}$
\end_inset
(stropnja lastnega podprostora).
\end_layout
\begin_layout Theorem*
Cayley-Hamilton.
\begin_inset Formula $p_{A}\left(A\right)=0$
\end_inset
—
karakteristični polinom matrike
\begin_inset Formula $A$
\end_inset
anhilira matriko
\begin_inset Formula $A$
\end_inset
\end_layout
\begin_layout Proof
Spomnimo se eksplicitne formule za celico inverza matrike (razdelek
\begin_inset CommandInset ref
LatexCommand vref
reference "subsec:Formula-za-inverz-matrike"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
),
ki pravi
\begin_inset Formula $B^{-1}=\frac{1}{\det B}\tilde{B}^{T}$
\end_inset
.
Računajmo in naposled vstavimo
\begin_inset Formula $B=A-xI$
\end_inset
:
\begin_inset Formula
\[
B^{-1}=\frac{1}{\det B}\tilde{B}^{T}\quad\quad\quad\quad/\cdot\left(\det B\right)B
\]
\end_inset
\begin_inset Formula
\[
\det B\cdot I=B\tilde{B}^{T}
\]
\end_inset
\begin_inset Formula
\[
\det\left(A-xI\right)\cdot I=p_{A}\left(x\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}
\]
\end_inset
Glede na definicijo
\begin_inset Formula $\tilde{A}$
\end_inset
je
\begin_inset Formula $\tilde{\left(A-xI\right)}^{T}$
\end_inset
matrika velikosti
\begin_inset Formula $n\times n$
\end_inset
,
ki vsebuje polinome stopnje
\begin_inset Formula $<n$
\end_inset
,
kajti vsebuje determinante kofaktorskih matrik,
torej je takele oblike (
\begin_inset Formula $\forall i\in\left\{ 1..\left(n-1\right)\right\} :B_{i}\in M_{n}\left(\mathbb{C}\right)$
\end_inset
):
\begin_inset Foot
status open
\begin_layout Plain Layout
To si predstavljamo tako,
da iz vsake celice matrike,
ki vsebuje polinom,
izpostavimo (homogenost) spremenljivko (torej
\begin_inset Formula $x$
\end_inset
na fiksno potenco) in nato koeficiente v celicah pred to spremenljivvko zložimo v eno matriko.
Slednje ponovimo za vsako potenco in dobimo te matrike
\begin_inset Formula $B_{i}$
\end_inset
.
\end_layout
\end_inset
\begin_inset Formula
\[
\tilde{\left(A-xI\right)}^{T}=B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}
\]
\end_inset
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-\lambda I\right)=c_{0}+c_{1}x+\cdots+c_{n}x^{n}$
\end_inset
.
Kot v enačbi množimo to z
\begin_inset Formula $I$
\end_inset
:
\begin_inset Formula $\det\left(A-\lambda I\right)\cdot I=c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}$
\end_inset
.
Oglejmo si še desno stran enačbe:
\begin_inset Formula
\[
\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}=\left(A-xI\right)\left(B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}\right)=AB_{0}+AB_{1}x+\cdots+AB_{n-1}x^{n-1}-B_{0}x-B_{1}x^{2}-\cdots-B_{n-1}x^{n}=
\]
\end_inset
\begin_inset Formula
\[
=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}x^{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n}
\]
\end_inset
In primerjajmo koeficiente v polinomih pred istoležnimi spremenljivkami na obeh straneh tele enačbe:
\begin_inset Formula
\[
\det\left(A-xI\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}
\]
\end_inset
\begin_inset Formula
\[
c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n}
\]
\end_inset
\begin_inset Formula
\[
\begin{array}{cccc}
1: & c_{0}I & = & AB_{0}\\
x: & c_{1}I & = & AB_{1}-B_{0}\\
x^{2}: & c_{2}I & = & AB_{2}x^{2}-B_{1}\\
\vdots\\
x^{n-1}: & c_{n-1}I & = & AB_{n-1}-B_{n-2}\\
x^{n}: & c_{n}I & = & -B_{n-1}
\end{array}
\]
\end_inset
Vstavimo sedaj
\begin_inset Formula $A$
\end_inset
v enačbo namesto
\begin_inset Formula $x$
\end_inset
:
\begin_inset Formula
\[
p_{A}\left(A\right)=c_{0}I+c_{1}IA+\cdots+c_{n}IA^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)A+\left(AB_{2}-B_{1}\right)A^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)A^{n-1}-B_{n-1}A^{n}=
\]
\end_inset
\begin_inset Formula
\[
=AB_{0}+A^{2}B_{1}-AB_{0}+A^{2}B^{2}-B_{1}A^{2}+\cdots+A^{n}B_{n-1}-A^{n-1}B_{n-2}-B_{n-1}A^{n}=0
\]
\end_inset
\begin_inset Formula
\[
p_{A}\left(A\right)=0
\]
\end_inset
\end_layout
\begin_layout Theorem*
Matriko
\begin_inset Formula $A$
\end_inset
se da diagonalizirati
\begin_inset Formula $\Leftrightarrow m_{A}\left(x\right)$
\end_inset
ima samo enostavne ničle (nima večkratnih —
potence so vse 1).
Torej
\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{1}\cdots\left(x-\lambda_{k}\right)^{1}$
\end_inset
za
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset
vse paroma različne lastne vrednosti
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Po predpostavki je
\begin_inset Formula $A$
\end_inset
podobna diagonalni matriki —
\begin_inset Formula $A=PDP^{-1}$
\end_inset
za diagonalno
\begin_inset Formula $D$
\end_inset
in obrnljivo
\begin_inset Formula $P$
\end_inset
.
BSŠ naj bo
\begin_inset Formula $D=\left[\begin{array}{ccc}
\lambda_{1} & 0 & 0\\
0 & \cdots & 0\\
0 & 0 & \lambda_{k}
\end{array}\right]$
\end_inset
in
\begin_inset Formula $\lambda_{1}\leq\cdots\leq\lambda_{k}$
\end_inset
.
Oglejmo si izraz
\begin_inset Formula
\[
\left(D-\lambda_{1}I\right)\cdots\left(D-\lambda_{k}I\right)=\left[\begin{array}{cccc}
0 & & & 0\\
& \lambda_{2}-\lambda_{1}\\
& & \ddots\\
0 & & & \lambda_{k}-\lambda_{1}
\end{array}\right]\cdots\left[\begin{array}{cccc}
\lambda_{1}-\lambda_{k} & & & 0\\
& \ddots\\
& & \lambda_{k-1}-\lambda_{k}\\
0 & & & 0
\end{array}\right]=0
\]
\end_inset
Sedaj pa še izraz
\begin_inset Formula
\[
\left(A-\lambda_{1}I\right)\cdots\left(A-\lambda_{k}I\right)=\left(PDP^{-1}-\lambda_{1}I\right)\cdots\left(PDP^{-1}-\lambda_{k}I\right)=\left(PDP^{-1}-\lambda_{1}PP^{-1}\right)\cdots\left(PDP^{-1}-\lambda_{k}PP^{-1}\right)=
\]
\end_inset
\begin_inset Formula
\[
=P\left(D-\lambda_{1}I\right)\cancel{P^{-1}}\cdots\cancel{P}\left(D-\lambda_{k}I\right)P^{-1}=P\left(D-\lambda_{1}I\right)\cdots\left(D-\lambda_{k}I\right)P^{-1}=P0P^{-1}=0,
\]
\end_inset
torej ta polinom anhilira
\begin_inset Formula $A$
\end_inset
.
Ker deli
\begin_inset Formula $m_{A}\left(x\right)$
\end_inset
—
vsebuje vse ničle
\begin_inset Formula $m_{A}\left(x\right)$
\end_inset
,
je prav to minimalen polinom
\begin_inset Formula $A$
\end_inset
—
ima najmanjšo stopnjo možno.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
Potrebujemo nekaj lem:
\begin_inset CommandInset counter
LatexCommand set
counter "theorem"
value "0"
lyxonly "false"
\end_inset
\end_layout
\begin_deeper
\begin_layout Lemma
Za vse matrike
\begin_inset Formula $A,B$
\end_inset
velja
\begin_inset Formula $\n\left(AB\right)\leq\n\left(A\right)+\n\left(B\right)\sim\dim\Ker\left(AB\right)\leq\dim\Ker\left(A\right)+\dim\Ker\left(B\right)$
\end_inset
.
\end_layout
\begin_layout Proof
Oglejmo si preslikavo
\begin_inset Formula $L:\Ker AB\to\Ker A$
\end_inset
,
ki slika
\begin_inset Formula $x\mapsto Bx$
\end_inset
.
Je dobro definirana,
kajti
\begin_inset Formula $x\in\Ker AB\Rightarrow ABx=0\Rightarrow Bx\in\Ker A$
\end_inset
.
Po osnovnem dimenzijskem izreku za preslikavo
\begin_inset Formula $L$
\end_inset
velja
\begin_inset Formula $\dim\Ker L+\dim\Slika L=\dim\Ker AB$
\end_inset
.
Ker velja
\begin_inset Formula $Lx=0\Rightarrow Bx=0$
\end_inset
,
velja
\begin_inset Formula $\Ker L\subseteq\Ker B$
\end_inset
in zato
\begin_inset Formula $\dim\Ker L\leq\dim\Ker B$
\end_inset
.
Poleg tega iz definicije velja
\begin_inset Formula $\Slika L\subseteq\Ker A$
\end_inset
in zato
\begin_inset Formula $\dim\Slika L\leq\dim\Ker A$
\end_inset
.
Vstavimo te neenakosti v enačbo iz dimenzijskega izreka:
\begin_inset Formula
\[
\dim\Ker L+\dim\Slika L=\dim\Ker AB
\]
\end_inset
\begin_inset Formula
\[
\dim\Ker B+\dim\Ker A\geq\dim\Ker AB
\]
\end_inset
Lemo lahko posplošimo na več faktorkev,
torej
\begin_inset Formula $\n\left(A_{1}\cdots A_{k}\right)\leq\n A_{1}+\cdots+\n A_{k}$
\end_inset
.
\end_layout
\begin_layout Standard
Nadaljujmo z dokazom
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
.
Denimo,
da
\begin_inset Formula $\left(x-\lambda_{1}\right)\cdots\left(x-\lambda_{k}\right)$
\end_inset
anhilira
\begin_inset Formula $A$
\end_inset
.
Upoštevamo
\begin_inset Formula
\[
\n\left(\left(A-\lambda_{1}\right)\cdots\left(A-\lambda_{k}\right)\right)\leq\n\left(A-\lambda_{1}\right)+\cdots+\n\left(A-\lambda_{k}\right)
\]
\end_inset
Členi na desni strani so geometrijske večkratnosti,
ker pa
\begin_inset Formula $A$
\end_inset
anhilira polinom po predpostavki,
je ta produkt ničelna preslikava in je dimenzija jedra dimenzija celega prostora.
\begin_inset Formula
\[
\n\left(0\right)=n=n_{1}+\cdots+n_{k}\leq m_{1}+\cdots+m_{k}
\]
\end_inset
Ker
\begin_inset Formula $\forall i:m_{i}\leq n_{i}$
\end_inset
(
\begin_inset CommandInset ref
LatexCommand vref
reference "claim:geom<=alg"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
),
velja v zgornji neenačbi enakost,
torej je matrika po
\begin_inset CommandInset ref
LatexCommand vref
reference "claim:mi=ni=>diag"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
diagonalizabilna.
\end_layout
\end_deeper
\end_deeper
\begin_layout Subsubsection
Korenski podprostori
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $A\in M_{n}$
\end_inset
in
\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{r_{1}}\cdots\left(x-\lambda_{k}\right)^{r_{k}}$
\end_inset
njen minimalni polinom.
\begin_inset Formula $\forall i\in\left\{ 1..k\right\} $
\end_inset
označimo z
\begin_inset Formula $W_{i}\coloneqq\Ker\left(A-\lambda_{i}I\right)^{r_{i}}$
\end_inset
korenski podprostor matrike
\begin_inset Formula $A$
\end_inset
za lastno vrednost
\begin_inset Formula $\lambda_{i}$
\end_inset
.
Vpeljimo še oznako
\begin_inset Formula $V_{i}\coloneqq\Ker\left(A-\lambda_{i}I\right)^{1}$
\end_inset
(tu potenca ni
\begin_inset Formula $r_{i}$
\end_inset
,
temveč je
\begin_inset Formula $1$
\end_inset
).
\end_layout
\begin_layout Definition*
\begin_inset CommandInset counter
LatexCommand set
counter "theorem"
value "0"
lyxonly "false"
\end_inset
\end_layout
\begin_layout Fact
Očitno je
\begin_inset Formula $\Ker\left(A-\lambda_{i}I\right)\subseteq\Ker\left(A-\lambda_{i}\right)^{2}\subseteq\Ker\left(A-\lambda_{i}I\right)^{3}\subseteq\cdots$
\end_inset
,
kajti če
\begin_inset Formula $x\in\Ker\left(A-\lambda_{i}I\right)^{m}\Rightarrow\left(A-\lambda_{i}I\right)^{m}x=0\Rightarrow\left(A-\lambda_{i}I\right)\left(A-\lambda_{i}I\right)^{m}x=0\Rightarrow x\in\Ker\left(A-\lambda_{i}I\right)^{m+1}$
\end_inset
.
Izkaže se,
da so vse inkluzije do
\begin_inset Formula $r_{i}-te$
\end_inset
potence stroge,
od
\begin_inset Formula $r_{i}-$
\end_inset
te potence dalje pa so vse inkluzije enačaji,
torej za
\begin_inset Formula $W_{i}=\Ker\left(A-\lambda_{i}I\right)^{r_{i}}$
\end_inset
velja
\begin_inset Formula
\[
\Ker\left(A-\lambda_{i}I\right)\subset\Ker\left(A-\lambda_{i}I\right)^{2}\subset\cdots\subset\Ker\left(A-\lambda_{i}I\right)^{r_{i}}=\Ker\left(A-\lambda_{i}I\right)^{r_{i}+1}=\cdots
\]
\end_inset
Poleg tega se izkaže,
da je
\begin_inset Formula $\dim W_{i}=n_{i}$
\end_inset
(algebraična večkratnost
\begin_inset Formula $\lambda_{i}$
\end_inset
).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Fact
\begin_inset Formula $\dim V_{i}=\dim\Ker\left(A-\lambda_{i}I\right)=m_{1}$
\end_inset
(geometrijska večkratnost
\begin_inset Formula $\lambda_{i}$
\end_inset
).
\end_layout
\begin_layout Claim
\begin_inset CommandInset label
LatexCommand label
name "claim:vsota-kor-podpr-je-vse"
\end_inset
\begin_inset Formula $\mathbb{C}^{n}=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{k}$
\end_inset
—
vsota vseh korenskih podprostorov je vse in ta vsota je direktna.
Tej vsoti pravimo
\begin_inset Quotes gld
\end_inset
korenski razcep matrike
\begin_inset Formula $A$
\end_inset
\begin_inset Quotes grd
\end_inset
.
\end_layout
\begin_layout Remark*
Dokazujmo:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Note Note
status open
\begin_layout Plain Layout
\begin_inset Formula $V_{1}+\cdots+V_{k}$
\end_inset
je tudi direktna,
ampak ni nujno enaka
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
.
Velja
\begin_inset Formula $\mathbb{C}^{n}=V_{1}+\cdots+V_{k}\Leftrightarrow A$
\end_inset
se da diagonalizirati (povedano prej).
Dokažimo trditev
\begin_inset CommandInset ref
LatexCommand vref
reference "claim:vsota-kor-podpr-je-vse"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
.
Dokažimo najprej,
da če
\begin_inset Formula $w_{1}\in W_{1},\dots,w_{k}\in W_{k}$
\end_inset
zadoščajo
\begin_inset Formula $w_{1}+\cdots+w_{k}=0$
\end_inset
,
velja
\begin_inset Formula $w_{1}=\cdots=w_{k}=0$
\end_inset
(direktna).
Delajmo indukcijo po številu členov:
\end_layout
\begin_layout Itemize
Baza:
\begin_inset Formula $w_{1}=0\Rightarrow w_{1}=0$
\end_inset
je očitno.
\end_layout
\begin_deeper
\begin_layout Itemize
Indukcijska predpostavka:
\begin_inset Formula $w_{1}+\cdots+w_{i}=0\Rightarrow w_{1}=\cdots=w_{i}=0$
\end_inset
.
\end_layout
\begin_layout Itemize
Korak:
Naj bodo
\begin_inset Formula $w_{1},\dots,w_{i+1}$
\end_inset
taki,
da
\begin_inset Formula
\[
w_{1}+\cdots+w_{i+1}=0\quad\quad\quad\quad/\cdot\left(A-\lambda_{i+1}I\right)^{r_{i+1}}
\]
\end_inset
\begin_inset Formula
\[
w_{1}'+\cdots+w_{i}'+0=0,
\]
\end_inset
kajti
\begin_inset Formula $w_{i+1}\in\Ker\left(A-\lambda_{i+1}I\right)^{r_{i+1}}$
\end_inset
.
Ker je vsak korenski prostor
\begin_inset Formula $W_{j}$
\end_inset
invarianten za
\begin_inset Formula $\left(A-\lambda_{i+1}I\right)^{r_{i+1}}$
\end_inset
,
...
Tega dokaza ne najdem,
zato
\series bold
tega dokaza ne razumem
\series default
.
Po definiciji je
\begin_inset Formula $V$
\end_inset
invarianten za
\begin_inset Formula $A$
\end_inset
,
če za vsak
\begin_inset Formula $v\in V$
\end_inset
velja
\begin_inset Formula $Av\in V$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Plain Layout
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Plain Layout
Nadaljuj
\begin_inset Quotes gld
\end_inset
LA1P FMF 2024-03-20.pdf
\begin_inset Quotes grd
\end_inset
na strani 3.
\end_layout
\end_inset
Če predpostavimo,
da je vsota direktna,
je lahko dokazati,
da je vsota cel prostor.
V karakteristični polinom,
ki po Cayley-Hamiltonu anhilira
\begin_inset Formula $A$
\end_inset
,
vstavimo
\begin_inset Formula $A$
\end_inset
in dobimo
\begin_inset Formula $0=\left(-1\right)^{n}\left(A-\lambda_{1}I\right)^{r_{1}}\cdots\left(A-\lambda_{k}I\right)^{r_{k}}=A_{1}\cdots A_{k}$
\end_inset
.
Ker je vsota direktna,
velja
\begin_inset Formula $\n\left(A_{1}\cdots A_{n}\right)=\n\left(0\right)=n=\n A_{1}\cdots\n A_{k}=\dim\left(W_{1}+\cdots+W_{k}\right)$
\end_inset
,
torej
\begin_inset Formula $W_{1}+\cdots+W_{k}=\mathbb{C}^{n}$
\end_inset
.
\end_layout
\begin_layout Itemize
Če predpostavimo,
da je
\begin_inset Formula $W_{i}\cap W_{j}=\left\{ 0\right\} $
\end_inset
za
\begin_inset Formula $i\not=j$
\end_inset
,
lahko od tod izpeljemo direktnost vsote korenskih podprostorov.
Dokaz z indukcijo:
\end_layout
\begin_deeper
\begin_layout Itemize
Baza:
\begin_inset Formula $W_{1}$
\end_inset
je direktna vsota.
Očitno (
\begin_inset Formula $\forall w_{1}\in W_{1}:w_{1}=0\Rightarrow w_{1}=0$
\end_inset
).
\end_layout
\begin_layout Itemize
Indukcijska predpostavka:
\begin_inset Formula $w_{1}+\cdots+w_{i}=0\Rightarrow w_{1}=\cdots=w_{i}=0$
\end_inset
.
\end_layout
\begin_layout Itemize
Korak:
Naj bodo
\begin_inset Formula $w_{1},\dots,w_{i+1}$
\end_inset
taki,
da
\begin_inset Formula
\[
w_{1}+\cdots+w_{i+1}=0\quad\quad\quad\quad/\cdot\left(A-\lambda_{i+1}I\right)^{r_{i+1}}
\]
\end_inset
\begin_inset Formula
\[
\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{1}+\cdots+\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{i}+0=0
\]
\end_inset
Ker
\begin_inset Formula $\left(A-\lambda_{h}I\right)^{r_{h}}$
\end_inset
in
\begin_inset Formula $\left(A-\lambda_{k}I\right)^{r_{k}}$
\end_inset
za vsaka
\begin_inset Formula $h,k$
\end_inset
komutirata (gre namreč za polinom,
v katerega je vstavljen
\begin_inset Formula $A$
\end_inset
),
velja za vsak
\begin_inset Formula $j$
\end_inset
iz
\begin_inset Formula $\left(A-\lambda_{j}I\right)^{r_{j}}w_{j}=0=\left(A-\lambda_{i+1}I\right)^{r_{i+1}}\left(A-\lambda_{j}I\right)^{r_{j}}w_{j}$
\end_inset
tudi
\begin_inset Formula
\[
\left(A-\lambda_{j}I\right)^{r_{j}}\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{j}=0
\]
\end_inset
Ker je po I.
P.
\begin_inset Formula $W_{1}+\cdots+W_{i}$
\end_inset
direktna,
velja za vsak
\begin_inset Formula $j$
\end_inset
\begin_inset Formula $w_{j}\in W_{j}$
\end_inset
,
toda zaradi našega množenja tudi
\begin_inset Formula $w_{j}\in W_{i+1}$
\end_inset
.
Zaradi predpostavke
\begin_inset Formula $m=n\Rightarrow W_{m}\cup W_{n}=\left\{ 0\right\} $
\end_inset
velja za vsak
\begin_inset Formula $j\in\left\{ 1..i\right\} :$
\end_inset
\begin_inset Formula $w_{j}=0$
\end_inset
.
V prvi enačbi ostane le še
\begin_inset Formula $w_{i+1}=0$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Itemize
Dokazati je treba še
\begin_inset Formula $i\not=j\Rightarrow W_{i}\cup W_{j}=\left\{ 0\right\} $
\end_inset
.
Dokažimo,
da je
\begin_inset Formula $W_{i}$
\end_inset
invarianten za
\begin_inset Formula $A$
\end_inset
,
t.
j.
\begin_inset Formula $v\in W_{i}\Rightarrow Av\in W_{i}$
\end_inset
.
Če je
\begin_inset Formula $v\in W_{i}$
\end_inset
,
tedaj
\begin_inset Formula
\[
\left(A-\lambda_{i}I\right)^{r_{i}}v=0\quad\quad\quad\quad/\cdot A
\]
\end_inset
\begin_inset Formula
\[
A\left(A-\lambda_{i}I\right)^{r_{i}}v=0
\]
\end_inset
\begin_inset Formula
\[
Av\in\Ker\left(A-\lambda_{i}I\right)^{r_{i}}
\]
\end_inset
\begin_inset Formula
\[
Av\in W_{i}
\]
\end_inset
Ker so vsi
\begin_inset Formula $W_{i}$
\end_inset
invariantni za
\begin_inset Formula $A$
\end_inset
,
so tudi njihovi preseki invariantni za
\begin_inset Formula $A$
\end_inset
.
Definirajmo torej linearno preslikavo
\begin_inset Formula $L:W_{i}\cap W_{j}\to W_{i}\cap W_{j}$
\end_inset
s predpisom
\begin_inset Formula $v\mapsto Av$
\end_inset
.
Vemo,
da ima
\begin_inset Formula $L$
\end_inset
vsaj eno lastno vrednost
\begin_inset Formula $\lambda$
\end_inset
in pripadajoči lastni vektor
\begin_inset Formula $w$
\end_inset
.
Torej
\begin_inset Formula $w\in W_{i}\cap W_{j}$
\end_inset
in
\begin_inset Formula $Lw=\lambda w$
\end_inset
,
toda
\begin_inset Formula $Lw=Aw=\lambda w$
\end_inset
.
Ker velja
\begin_inset Formula $Av=\lambda v\Rightarrow A^{q}v=\lambda^{q}v\Rightarrow p\left(A\right)v=p\left(\lambda\right)v$
\end_inset
za vsak polinom
\begin_inset Formula $p$
\end_inset
,
velja
\begin_inset Formula $p\left(A\right)w=p\left(\lambda\right)w$
\end_inset
za vsak polinom
\begin_inset Formula $p$
\end_inset
.
Uporabimo polinom
\begin_inset Formula $p\left(x\right)=\left(x-\lambda_{i}\right)^{r_{i}}$
\end_inset
in dobimo
\begin_inset Formula $\left(A-\lambda_{i}\right)^{r_{i}}w=\left(\lambda-\lambda_{i}\right)^{r_{i}}w$
\end_inset
.
Leva stran enačbe je 0,
PDDRAA
\begin_inset Formula $w$
\end_inset
ni 0,
torej
\begin_inset Formula $\left(\lambda-\lambda_{i}\right)^{r_{i}}=0$
\end_inset
,
torej
\begin_inset Formula $\lambda=\lambda_{i}$
\end_inset
.
Vendar lahko namesto tistega polimoma uporabimo polinom
\begin_inset Formula $p\left(x\right)=\left(x-\lambda_{j}\right)^{r_{j}}$
\end_inset
,
kar pokaže
\begin_inset Formula $\lambda=\lambda_{j}$
\end_inset
,
torej
\begin_inset Formula $\lambda_{j}=\lambda_{i}$
\end_inset
,
kar je v
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
s tem,
da so lastne vrednosti
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset
paroma različne.
Torej
\begin_inset Formula $w=0$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsubsection
Jordanska kanonična forma
\end_layout
\begin_layout Standard
Vsaka kvadratna matrika je podobna posebni zgornjetrikotni matriki,
ki ji pravimo JKF.
To je bločno diagonalna matrika,
ki ima za diagonalne bloke t.
i.
\begin_inset Quotes gld
\end_inset
jordanske kletke
\begin_inset Quotes grd
\end_inset
,
to so matrike oblike:
\begin_inset Formula
\[
\left[\begin{array}{ccccc}
\lambda & 1\\
& \lambda & 1\\
& & \ddots & \ddots\\
& & & \lambda & 1\\
& & & & \lambda
\end{array}\right].
\]
\end_inset
Jordanska matrika je sestavljena iz jordanskih kletk po diagonali (
\begin_inset Formula $J_{i}$
\end_inset
so jordanske kletke):
\begin_inset Formula
\[
\left[\begin{array}{ccc}
J_{1} & & 0\\
& \ddots\\
0 & & J_{m}
\end{array}\right].
\]
\end_inset
Običajno zahtevamo še,
da so JK,
ki imajo isto lastno vrednost,
skupaj,
ter da so JK padajoče urejene po lastni vrednosti od največje do najmanjše.
\end_layout
\begin_layout Theorem*
Za vsako kvadratno kompleksno matriko
\begin_inset Formula $A\in M_{n\times n}\left(\mathbb{C}\right)$
\end_inset
obstaja taka jordanska matrika
\begin_inset Formula $J$
\end_inset
in taka obrnljiva matrika
\begin_inset Formula $P$
\end_inset
,
da velja
\begin_inset Formula $A=PJP^{-1}$
\end_inset
.
ZDB vsaka
\begin_inset Formula $A\in M_{n\times n}\left(\mathbb{C}\right)$
\end_inset
je podobna neki Jordanski matriki.
\end_layout
\begin_layout Standard
Procesu iskanja jordanske matrike pravimo
\begin_inset Quotes gld
\end_inset
jordanifikacija
\begin_inset Quotes grd
\end_inset
.
Kako konstruiramo
\begin_inset Formula $J$
\end_inset
in
\begin_inset Formula $P$
\end_inset
?
Izračunamo lastne vrednosti in pripadajoče korenske podprostore.
\end_layout
\begin_layout Itemize
Naj bo
\begin_inset Formula $\lambda$
\end_inset
lastna vrednost
\begin_inset Formula $A$
\end_inset
.
Za preglednost pišimo
\begin_inset Formula $N\coloneqq A-\lambda I$
\end_inset
.
\end_layout
\begin_layout Itemize
Izračunamo lastne vektorje in lastni podprostor
\begin_inset Formula $\Ker N^{r}$
\end_inset
ter ga izrazimo z njegovo bazo,
recimo ji
\begin_inset Formula $B_{r}$
\end_inset
.
\end_layout
\begin_layout Itemize
Nato izberemo
\begin_inset Quotes gld
\end_inset
pomožne baze
\begin_inset Quotes grd
\end_inset
\begin_inset Formula $\mathcal{B}_{1},\dots,\mathcal{B}_{r}$
\end_inset
,
ki pripadajo prostorom
\begin_inset Formula $\Ker N^{1},\dots,\Ker N^{r}$
\end_inset
.
\end_layout
\begin_layout Itemize
Pomožno bazo
\begin_inset Formula $\mathcal{B}_{r-1}$
\end_inset
dopolnimo do baze
\begin_inset Formula $\mathcal{B}_{r}$
\end_inset
z elementi
\begin_inset Formula $\mathcal{B}_{r}$
\end_inset
\begin_inset Formula $u_{1},\dots,u_{k_{1}}$
\end_inset
.
Potem je
\begin_inset Formula $\mathcal{B}_{r-1}\cup$
\end_inset
\begin_inset Formula $\left\{ u_{1},\dots,u_{k_{1}}\right\} $
\end_inset
\begin_inset Quotes gld
\end_inset
popravek pomožne baze
\begin_inset Formula $\mathcal{B}_{r}$
\end_inset
\begin_inset Quotes grd
\end_inset
.
\end_layout
\begin_layout Itemize
Vektorje
\begin_inset Formula $\left\{ u_{1},\dots,u_{k_{1}}\right\} \in\Ker N^{r}$
\end_inset
pomnožimo z matriko
\begin_inset Formula $N$
\end_inset
,
dobljeni
\begin_inset Formula $Nu_{1},\dots,Nu_{k_{1}}$
\end_inset
ležijo v
\begin_inset Formula $\Ker N^{r-1}$
\end_inset
.
Množica
\begin_inset Formula $\mathcal{B}_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{k_{1}}\right\} $
\end_inset
je linearno neodvisna.
Izberemo take
\begin_inset Formula $v_{1},\dots,v_{k_{2}}\in B_{r-1}$
\end_inset
,
ki dopolnijo LN
\begin_inset Formula $B_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{2}\right\} $
\end_inset
do baze
\begin_inset Formula $\Ker N^{r-1}$
\end_inset
.
Potem je
\begin_inset Formula $\mathcal{B}_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{k_{1}}\right\} \cup\left\{ v_{1},\dots,v_{k_{2}}\right\} $
\end_inset
popravek pomožne baze
\begin_inset Formula $\mathcal{B}_{r-1}$
\end_inset
.
\end_layout
\begin_layout Itemize
Izberemo take
\begin_inset Formula $w_{1},\dots,w_{k_{3}}\in\mathcal{B}_{r-2}$
\end_inset
,
ki
\begin_inset Formula $\mathcal{B}_{r-3}\cup\left\{ N^{2}u_{1},\dots,N^{2}u_{k_{1}}Nv_{1},\dots,Nv_{k_{2}}\right\} $
\end_inset
dopolnijo do baze
\begin_inset Formula $\Ker N^{r-2}$
\end_inset
.
Tedaj je
\begin_inset Formula $\mathcal{B}_{r-3}\cup\left\{ N^{2}u_{1},\cdots,N^{2}u_{k_{1}},Nv_{1},\dots,Nv_{k_{2}},w_{1},\dots,w_{k_{3}}\right\} $
\end_inset
popravek pomožne baze
\begin_inset Formula $B_{r-2}$
\end_inset
.
\end_layout
\begin_layout Itemize
Postopek ponavljamo,
dokler ne popravimo vseh možnih baz.
\end_layout
\begin_layout Standard
Dobimo t.
i.
\begin_inset Quotes gld
\end_inset
jordanske verige
\begin_inset Quotes grd
\end_inset
.
Ena jordanska veriga je
\begin_inset Formula $\left(u,Nu,N^{2}u,\dots,N^{x}u\right)$
\end_inset
,
torej preslikanje elementa
\begin_inset Formula $u$
\end_inset
,
ki začne kot dopolnitev baze korenskega podprostora
\begin_inset Formula $\Ker N^{x+1}$
\end_inset
in je na koncu
\begin_inset Formula $x-$
\end_inset
krat preslikan z
\begin_inset Formula $N$
\end_inset
,
torej konča v korenskem podprostoru
\begin_inset Formula $\Ker N$
\end_inset
.
Nekatere verige se začno v največjem korenskem podprostoru
\begin_inset Formula $\Ker N^{r}$
\end_inset
,
nekatere šele kasneje,
v
\begin_inset Formula $\Ker N^{1}$
\end_inset
ali pa
\begin_inset Formula $\Ker N^{2}$
\end_inset
ali pa
\begin_inset Formula $\Ker N^{3}$
\end_inset
.
\end_layout
\begin_layout Standard
Imamo torej
\begin_inset Formula $k_{1}$
\end_inset
jordanskih verig dolžine
\begin_inset Formula $r$
\end_inset
,
\begin_inset Formula $k_{2}$
\end_inset
jordanskih verig dolžine
\begin_inset Formula $r-1$
\end_inset
,
\begin_inset Formula $k_{3}$
\end_inset
jordanskih verig dolžine
\begin_inset Formula $r-2$
\end_inset
,
...,
\begin_inset Formula $k_{r}$
\end_inset
jordanskih verig dolžine 1.
Skupaj je jordanskih verig
\begin_inset Formula $k_{1}+\cdots+k_{r}=\dim\Ker N$
\end_inset
.
Jordanskih verig za lastno vrednost
\begin_inset Formula $\lambda$
\end_inset
je torej toliko,
kot je njena geometrijska večkratnost.
\end_layout
\begin_layout Standard
Vsaki jordanski verigi dolžine
\begin_inset Formula $k$
\end_inset
pripada ena jordanska kletka velikosti
\begin_inset Formula $k\times k$
\end_inset
.
\begin_inset Formula $k-$
\end_inset
vektorjev iz verige zložimo v
\begin_inset Formula $P$
\end_inset
tako,
da je vektor z začetka verige (torej tisti iz popravljene baze večjega prostora) na levi strani v matriki.
\end_layout
\begin_layout Example*
Poišči jordansko kanonično formo matrike
\begin_inset Formula
\[
A=\left[\begin{array}{cccc}
0 & 1 & -1 & 2\\
0 & 2 & 2 & 2\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 2
\end{array}\right].
\]
\end_inset
\end_layout
\begin_layout Example*
Najprej izračunamo karakteristični polinom:
\begin_inset Formula $\det\left(A-\lambda I\right)=x\left(x-2\right)^{3}$
\end_inset
.
\begin_inset Formula $\lambda_{1}=0$
\end_inset
,
\begin_inset Formula $n_{1}=1$
\end_inset
,
\begin_inset Formula $\lambda_{2}=2$
\end_inset
,
\begin_inset Formula $n_{2}=3$
\end_inset
.
Lastni vektorji:
\begin_inset Formula $\Ker\left(A-0I\right)=\Lin\left\{ \left(1,0,0,0\right)\right\} $
\end_inset
,
\begin_inset Formula $\Ker\left(A-2I\right)=\Lin\left\{ \left(3,0,-2,2\right),\left(1,2,0,0\right)\right\} $
\end_inset
.
Če bi dobili 4 lastne vektorje,
bi lahko matriko diagonalizirali.
Tako je ne moremo.
Ker
\begin_inset Formula $n_{1}=1$
\end_inset
,
je
\begin_inset Formula $r_{1}$
\end_inset
največ
\begin_inset Formula $1$
\end_inset
,
torej
\begin_inset Formula $\Ker\left(A-0I\right)=\Ker\left(A-0I\right)^{2}=\cdots$
\end_inset
.
Izračunamo korenske podprostore
\end_layout
\begin_deeper
\begin_layout Itemize
za lastno vrednost 0:
\begin_inset Formula $\Ker\left(A-0I\right)=\Ker\left(A\right)=\Ker\left(A^{2}\right)$
\end_inset
.
Dobimo eno verigo
\begin_inset Formula $\left(\left(1,0,0,0\right)\right)$
\end_inset
dolžine
\begin_inset Formula $1$
\end_inset
za lastno vrednost
\begin_inset Formula $0$
\end_inset
.
\end_layout
\begin_layout Itemize
za lastno vrednost 2:
\begin_inset Formula $\Ker\left(A-2I\right)^{2}=\Lin\left\{ \left(1,0,0,2\right),\left(-1,0,1,0\right),\left(1,2,0,0\right)\right\} $
\end_inset
,
\begin_inset Formula $\Ker\left(A-2I\right)^{3}=\Ker\left(A-2I\right)^{2}$
\end_inset
.
Opazimo,
da je
\begin_inset Formula $\left(1,0,0,2\right)$
\end_inset
dopolnitev baze
\begin_inset Formula $\Ker\left(A-2I\right)$
\end_inset
do baze
\begin_inset Formula $\Ker\left(A-2I\right)^{2}$
\end_inset
.
Torej je
\begin_inset Formula $\left\{ \left(1,0,0,2\right)\right\} $
\end_inset
\begin_inset Quotes gld
\end_inset
popravljena baza
\begin_inset Quotes grd
\end_inset
\begin_inset Formula $N^{2}$
\end_inset
.
Preslikamo
\begin_inset Formula $\left(A-2I\right)\left(1,0,0,2\right)=\left(2,4,0,0\right)$
\end_inset
,
kar tvori verigo dolžine 2
\begin_inset Formula $\left(\left(1,0,0,2\right),\left(2,4,0,0\right)\right)$
\end_inset
.
Edini linearno neodvisen od
\begin_inset Formula $\left(2,4,0,0\right)$
\end_inset
v
\begin_inset Formula $\mathcal{B}_{1}$
\end_inset
je
\begin_inset Formula $\left(3,0,-2,2\right)$
\end_inset
,
zato je slednji začetek zadnje tretje verige dolžine 1
\begin_inset Formula $\left(\left(3,0,-2,2\right)\right)$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Example*
Tri verige,
ki jih dobimo,
so
\begin_inset Formula $\left(\left(1,0,0,0\right)\right)$
\end_inset
za lastno vrednost
\begin_inset Formula $0$
\end_inset
in
\begin_inset Formula $\left(\left(1,0,0,2\right),\left(2,4,0,0\right)\right)$
\end_inset
ter
\begin_inset Formula $\left(\left(3,0,-2,2\right)\right)$
\end_inset
obe za lastno vrednost 2.
Zložimo jih v matriko
\begin_inset Formula $P$
\end_inset
:
\begin_inset Formula
\[
P=\left[\begin{array}{cccc}
1 & 1 & 2 & 3\\
0 & 0 & 4 & 0\\
0 & 0 & 0 & -2\\
0 & 2 & 0 & 2
\end{array}\right]
\]
\end_inset
V matriko
\begin_inset Formula $J$
\end_inset
pa zložimo kletke pripadajočih velikosti:
\begin_inset Formula
\[
J=\left[\begin{array}{cccc}
0 & & & 0\\
& 2 & 1\\
& & 2\\
0 & & & 2
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Example*
In velja
\begin_inset Formula $A=PJP^{-1}$
\end_inset
(
\begin_inset Formula $P^{-1}$
\end_inset
izračunamo z Gaussom).
\end_layout
\begin_layout Subsubsection
Funkcije matrik
\end_layout
\begin_layout Standard
Če poznamo razcep
\begin_inset Formula $A=PJP^{-1}$
\end_inset
,
prevedemo računanje potenc
\begin_inset Formula $A$
\end_inset
na računanje potenc matrike
\begin_inset Formula $J$
\end_inset
,
kajti
\begin_inset Formula
\[
A^{n}=\left(PJP^{-1}\right)\left(PJP^{-1}\right)\cdots\left(PJP^{-1}\right)=PJP^{-1}PJP^{-1}\cdots PJP^{-1}=PJ^{n}P^{-1}.
\]
\end_inset
Ker je
\begin_inset Formula $J$
\end_inset
bločno diagonalna matrika,
sestavljena iz jordanskih kletk,
se potenciranje
\begin_inset Formula $J$
\end_inset
prevede na potenciranje kletk,
kajti
\begin_inset Formula
\[
J^{n}=\left[\begin{array}{ccc}
J_{1}^{n} & & 0\\
& \ddots\\
0 & & J_{m}^{n}
\end{array}\right].
\]
\end_inset
Potenciranje jordanske kletke:
\begin_inset Formula
\[
\left[\begin{array}{ccccc}
\lambda & 1 & & & 0\\
& \lambda & 1\\
& & \ddots & \ddots\\
& & & \lambda & 1\\
0 & & & & \lambda
\end{array}\right]^{n}=\left(\lambda I+\left[\begin{array}{ccc}
1 & & 0\\
& \ddots\\
0 & & 1
\end{array}\right]\right)^{n}=\left(\lambda I+N\right)^{n}=
\]
\end_inset
\begin_inset Formula
\[
=\binom{n}{0}\left(\lambda I\right)^{n}N^{0}+\binom{n}{1}\left(\lambda N\right)^{n-1}N^{1}+\cdots+\binom{n}{n}\left(\lambda N\right)^{0}N^{n}=\binom{n}{0}\lambda^{n}+\binom{n}{1}\lambda^{n-1}N^{1}+\cdots+\binom{n}{n}N^{n}
\]
\end_inset
Poraja se vprašanje,
kako potencirati
\begin_inset Formula $N=\left[\begin{array}{ccccc}
0 & 1 & & & 0\\
& \ddots & \ddots\\
& & & \ddots\\
& & & \ddots & 1\\
0 & & & & 0
\end{array}\right]$
\end_inset
.
Velja
\begin_inset Formula $N^{2}=\left[\begin{array}{ccccc}
0 & 0 & 1 & & 0\\
& \ddots & \ddots & \ddots\\
& & & \ddots & 1\\
& & & \ddots & 0\\
0 & & & & 0
\end{array}\right]$
\end_inset
in tako dalje (
\begin_inset Quotes gld
\end_inset
diagonalo
\begin_inset Quotes grd
\end_inset
enic pomikamo gor in desno).
Za
\begin_inset Formula $r\times r$
\end_inset
jordansko kletko,
kadar
\begin_inset Formula $n\geq r$
\end_inset
(sicer dobimo le prvih nekaj naddiagonal),
sledi
\begin_inset Formula
\[
\left[\begin{array}{ccccc}
\lambda & 1 & & & 0\\
& \lambda & 1\\
& & \ddots & \ddots\\
& & & \lambda & 1\\
0 & & & & \lambda
\end{array}\right]^{n}=\left[\begin{array}{ccccc}
\lambda^{n} & n\lambda^{n-1} & \cdots & \binom{n}{r-2}\lambda^{n-r+2} & \binom{n}{r-1}\lambda^{n-r+1}\\
& \lambda^{n} & n\lambda^{n-1} & \ddots & \binom{n}{r-2}\lambda^{n-r+2}\\
& & \ddots & \ddots & \vdots\\
& & & \lambda^{n} & n\lambda^{n-1}\\
0 & & & & \lambda^{n}
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
Za računanje poljubne funkcije jordanske kletke pa velja predpis
\begin_inset Formula
\[
f\left(\left[\begin{array}{ccccc}
\lambda & 1 & & & 0\\
& \lambda & 1\\
& & \ddots & \ddots\\
& & & \lambda & 1\\
0 & & & & \lambda
\end{array}\right]\right)=\left[\begin{array}{ccccc}
f\left(\lambda\right) & f'\left(\lambda\right) & \frac{f''\left(\lambda\right)}{2} & \cdots & \frac{f^{\left(k-1\right)\left(\lambda\right)}}{\left(k-1\right)!}\\
& f\left(\lambda\right) & f'\left(\lambda\right) & \ddots & \cdots\\
& & \ddots & \ddots & \frac{f''\left(\lambda\right)}{2}\\
& & & f\left(\lambda\right) & f'\left(\lambda\right)\\
0 & & & & f\left(\lambda\right)
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
In torej za računanje poljubne funkcije poljubne matrike
\begin_inset Formula $f\left(A\right)=f\left(PJP^{-1}\right)=Pf\left(J\right)P^{-1}$
\end_inset
.
\end_layout
\begin_layout Subsection
Vektorski prostori s skalarnim produktom
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $V$
\end_inset
vektorski prostor nad poljem
\begin_inset Formula $\mathbb{R}$
\end_inset
nenujno končno razsežen.
Preslikavi
\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle :V\times V\to\mathbb{R}$
\end_inset
pravimo skalarni produtkt,
če zadošča naslednjim lastnostim:
\end_layout
\begin_deeper
\begin_layout Enumerate
pozitivna definitnost:
\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle >0$
\end_inset
\end_layout
\begin_layout Enumerate
simetričnost:
\begin_inset Formula $\forall u,v\in V:\left\langle v,u\right\rangle =\left\langle u,v\right\rangle $
\end_inset
\end_layout
\begin_layout Enumerate
linearnost v prvem faktorju:
\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in\mathbb{R},v_{1},v_{2}\in V:\left\langle \alpha_{1}v_{1}+\alpha_{2}v_{2},v\right\rangle =\alpha_{1}\left\langle v_{1},v\right\rangle +\alpha_{2}\left\langle v_{2},v\right\rangle $
\end_inset
\end_layout
\end_deeper
\begin_layout Corollary*
linearnost v drugem faktorju.
\begin_inset Formula $\left\langle u,\beta_{1}v_{1}+\beta_{2}v_{2}\right\rangle =\left\langle \beta_{1}v_{1}+\beta_{2}v_{2},v\right\rangle =\beta_{1}\left\langle v_{1},v\right\rangle +\beta_{2}\left\langle v_{2},v\right\rangle =\beta_{1}\left\langle v,v_{1}\right\rangle +\beta_{2}\left\langle v,v_{2}\right\rangle $
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Corollary*
Skalarni produkt z 0:
\begin_inset Formula $\left\langle 0,v\right\rangle =\left\langle 0\cdot v+0\cdot v,v\right\rangle =0\left\langle v,v\right\rangle +0\left\langle v,v\right\rangle =0\Rightarrow\left\langle v,0\right\rangle =0$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Corollary*
Alternativna formulacija 1:
\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$
\end_inset
.
\begin_inset Note Note
status open
\begin_layout Plain Layout
Dokazujemo ekvivalenco:
alternativna formulacija 1
\begin_inset Formula $\Leftrightarrow$
\end_inset
originalna definicija 1.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
Predpostavimo
\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle \geq0$
\end_inset
in izjavo negirajmo:
\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \leq0\Rightarrow v=0$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Predpostavimo
\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$
\end_inset
\end_layout
\end_deeper
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Primeri vektorskih prostorov s skalarnim produktom:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
s standardnim skalarnim produktom:
\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\alpha_{1}\beta_{1}+\cdots+\alpha_{n}\beta_{n}$
\end_inset
.
\end_layout
\begin_layout Itemize
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
z nestandardnim skalarnim produktom:
Za pojubne
\begin_inset Formula $\gamma_{1}>0,\dots,\gamma_{n}>0$
\end_inset
definirajmo
\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\gamma_{1}\alpha_{1}\beta_{1}+\cdots+\gamma_{n}\alpha_{n}\beta_{n}$
\end_inset
.
\end_layout
\begin_layout Itemize
neskončno razsežen primer s standardnim skalarnim produktom:
\begin_inset Formula $V=C\left[a,b\right]\sim$
\end_inset
zvezne
\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
\end_inset
.
Definirajmo
\begin_inset Formula $\forall f,g\in V:\left\langle f,g\right\rangle =\int_{a}^{b}f\left(x\right)g\left(x\right)dx$
\end_inset
.
Zveznost je potrebna za dokaz aksioma 1,
sicer za neznano neničelno funkcijo
\begin_inset Formula $f\left(x\right)=\begin{cases}
1 & ;x=0\\
0 & ;\text{drugače}
\end{cases}$
\end_inset
velja
\begin_inset Formula $\int_{a}^{b}f\left(x\right)g\left(x\right)dx=0$
\end_inset
.
Temu pravimo standardni skalarni produkt v
\begin_inset Formula $C\left[a,b\right]$
\end_inset
.
\end_layout
\begin_layout Itemize
neskončno razsežen primer z nestandardnim skalarnim produktom:
Naj bo
\begin_inset Formula $w:\left[a,b\right]\to\mathbb{R}$
\end_inset
zvezna,
ki zadošča
\begin_inset Formula $\forall x\in\left[a,b\right]:w\left(x\right)>0$
\end_inset
.
Ostalo kot prej.
\begin_inset Formula $\forall f,g\in V:\left\langle f,g\right\rangle _{w}=\int_{a}^{b}f\left(x\right)g\left(x\right)w\left(x\right)dx$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Remark*
Vektorski prostor s skalarnnim produktom je tak par
\begin_inset Formula $\left(V,\left\langle \cdot,\cdot\right\rangle \right)$
\end_inset
,
kjer je
\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle $
\end_inset
skalarni produkt na
\begin_inset Formula $V$
\end_inset
.
To je torej vektorski prostor,
za katerega izberemo in fiksiramo skalarni produkt.
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $V$
\end_inset
vektorski prostor nad poljem
\begin_inset Formula $\mathbb{C}$
\end_inset
nenujno končno razsežen.
Preslikavi
\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle :V\times V\to\mathbb{C}$
\end_inset
pravimo skalarni produtkt,
če zadošča naslednjim lastnostim:
\end_layout
\begin_deeper
\begin_layout Enumerate
pozitivna definitnost:
\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle \in\mathbb{R}\wedge\left\langle v,v\right\rangle >0$
\end_inset
\end_layout
\begin_layout Enumerate
konjugirana simetričnost:
\begin_inset Formula $\forall u,v\in V:\left\langle v,u\right\rangle =\overline{\left\langle u,v\right\rangle }$
\end_inset
\end_layout
\begin_layout Enumerate
linearnost v prvem faktorju:
\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in\mathbb{R},v_{1},v_{2}\in V:\left\langle \alpha_{1}v_{1}+\alpha_{2}v_{2},v\right\rangle =\alpha_{1}\left\langle v_{1},v\right\rangle +\alpha_{2}\left\langle v_{2},v\right\rangle $
\end_inset
\end_layout
\end_deeper
\begin_layout Corollary*
konjugirana linearnost v drugem faktorju.
\begin_inset Formula $\left\langle u,\beta_{1}v_{1}+\beta_{2}v_{2}\right\rangle =\overline{\left\langle \beta_{1}v_{1}+\beta_{2}v_{2},v\right\rangle }=\overline{\beta_{1}\left\langle v_{1},v\right\rangle +\beta_{2}\left\langle v_{2},v\right\rangle }=\overline{\beta_{1}}\overline{\left\langle v_{1},v\right\rangle }+\overline{\beta_{2}}\overline{\left\langle v_{2},v\right\rangle }=\overline{\beta_{1}}\left\langle v,v_{1}\right\rangle +\overline{\beta_{2}}\left\langle v,v_{2}\right\rangle $
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Corollary*
Skalarni produkt z 0:
\begin_inset Formula $\left\langle 0,v\right\rangle =\left\langle 0\cdot v+0\cdot v,v\right\rangle =0\left\langle v,v\right\rangle +0\left\langle v,v\right\rangle =0\Rightarrow\left\langle v,0\right\rangle =0$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Corollary*
Alternativna formulacija 1:
\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \in\mathbb{R}\wedge\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$
\end_inset
.
\end_layout
\begin_layout Example*
Primeri vektorskih prostorov s skalarnim produktom:
\end_layout
\begin_deeper
\begin_layout Itemize
standardni skalarni produkt na
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
:
\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\alpha_{1}\overline{\beta_{1}}+\cdots+\alpha_{n}\overline{\beta_{n}}$
\end_inset
.
\end_layout
\begin_layout Itemize
nestandardni skalarni produkt na
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
:
Za neke
\begin_inset Formula $\gamma_{1}\in\mathbb{\mathbb{R}}^{+},\dots,\gamma_{n}\in\mathbb{R}^{+}$
\end_inset
definiramo
\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\gamma_{1}\alpha_{1}\overline{\beta_{1}}+\cdots+\gamma_{n}\alpha_{n}\overline{\beta_{n}}$
\end_inset
.
\end_layout
\begin_layout Itemize
neskončno razsežen vektorski prostor na
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
s standardnim skalarnim produktom:
Naj bo
\begin_inset Formula $V=C\left(\left[a,b\right],\mathbb{C}\right)$
\end_inset
—
\begin_inset Formula $f=g+ih$
\end_inset
za
\begin_inset Formula $g,h\in C\left[a,b\right]$
\end_inset
(zvezni funkciji iz
\begin_inset Formula $\left[a,b\right]$
\end_inset
v
\begin_inset Formula $\mathbb{R}$
\end_inset
).
Definiramo
\begin_inset Formula $\left\langle f_{1},f_{2}\right\rangle =\int_{a}^{b}f_{1}\left(x\right)\overline{f_{2}\left(x\right)}dx=\int_{a}^{b}\left(g_{1}+ih_{1}\right)\left(x\right)\left(g_{2}-ih_{2}\right)\left(x\right)dx=\int_{a}^{b}\left(g_{1}g_{2}+g_{1}g_{2}\right)\left(x\right)dx+i\int_{a}^{b}\left(h_{1}g_{2}-g_{1}h_{2}\right)xdx$
\end_inset
.
\end_layout
\begin_layout Itemize
neskončno razsežen vektorski prostor na
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
z nestandardnim skalarnim produktom:
Isto kot zgoraj,
le da spet množimo z nekimi funkcijami,
kot pri realnem skalarnem produktu.
\end_layout
\end_deeper
\begin_layout Subsubsection
Norma
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $V$
\end_inset
vektorski prostor s skalarnim produktom.
\begin_inset Formula $\forall v\in V:\left|\left|v\right|\right|=\sqrt{\left\langle v,v\right\rangle }$
\end_inset
je norma
\begin_inset Formula $v$
\end_inset
.
\end_layout
\begin_layout Paragraph
Osnovne lastnosti norme
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\left(\left|\left|v\right|\right|>0\Leftrightarrow v\not=0\right)\wedge\left|\left|0\right|\right|=0$
\end_inset
sledi iz prvega aksioma skalarnega produkta
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall\alpha\in F,v\in V:\left|\left|\alpha v\right|\right|=\left|\alpha\right|\left|\left|v\right|\right|$
\end_inset
\end_layout
\begin_layout Enumerate
trikotniška neenakost:
\begin_inset Formula $\forall u,v\in V:\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$
\end_inset
sledi iz Cauchy-Schwarzove neenakosti na običajen način.
\end_layout
\begin_layout Claim*
Cauchy-Schwarz.
Za
\begin_inset Formula $V$
\end_inset
vektorski prostor s skalarnim produktom velja
\begin_inset Formula $\forall v\in V:\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$
\end_inset
.
\end_layout
\begin_layout Proof
Za
\begin_inset Formula $v=0$
\end_inset
očitno velja
\begin_inset Formula $0=0$
\end_inset
.
Za
\begin_inset Formula $v\not=0$
\end_inset
definirajmo
\begin_inset Formula
\[
w=u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v
\]
\end_inset
po prvi lastnosti velja
\begin_inset Formula
\[
0\leq\left\langle w,w\right\rangle =\left\langle w,u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v\right\rangle =\left\langle w,u\right\rangle -\frac{\overline{\left\langle u,v\right\rangle }}{\left\langle v,v\right\rangle }\left\langle w,v\right\rangle
\]
\end_inset
Oglejmo si
\begin_inset Formula
\[
\left\langle w,v\right\rangle =\left\langle u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,v\right\rangle =\left\langle u,v\right\rangle -\left\langle \frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,v\right\rangle =\cancel{\left\langle u,v\right\rangle }-\frac{\cancel{\left\langle u,v\right\rangle }}{\cancel{\left\langle v,v\right\rangle }}\cancel{\left\langle v,v\right\rangle }=0
\]
\end_inset
In se vrnimo k prejšnji enačbi:
\begin_inset Formula
\[
0\leq\left\langle w,w\right\rangle =\left\langle w,u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v\right\rangle =\left\langle w,u\right\rangle -\frac{\overline{\left\langle u,v\right\rangle }}{\left\langle v,v\right\rangle }\left\langle w,v\right\rangle =\left\langle w,u\right\rangle -0=\left\langle w,u\right\rangle =
\]
\end_inset
\begin_inset Formula
\[
=\left\langle u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,u\right\rangle =\left\langle u,u\right\rangle -\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }\left\langle v,u\right\rangle =\left|\left|u\right|\right|^{2}-\frac{\left\langle u,v\right\rangle \overline{\left\langle u,v\right\rangle }}{\left|\left|v\right|\right|^{2}}=\left|\left|u\right|\right|^{2}-\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}}
\]
\end_inset
\begin_inset Formula
\[
0\leq\left|\left|u\right|\right|^{2}-\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}}
\]
\end_inset
\begin_inset Formula
\[
\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}}\leq\left|\left|u\right|\right|^{2}
\]
\end_inset
\begin_inset Formula
\[
\left|\left\langle u,v\right\rangle \right|^{2}\leq\left|\left|u\right|\right|^{2}\left|\left|v\right|\right|^{2}
\]
\end_inset
\begin_inset Formula
\[
\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|
\]
\end_inset
\end_layout
\begin_layout Claim*
Z normo lahko izrazimo skalarni produkt:
\end_layout
\begin_deeper
\begin_layout Itemize
V
\begin_inset Formula $\mathbb{R}$
\end_inset
:
\begin_inset Formula $\left\langle u,v\right\rangle =\frac{1}{4}\left(\left|\left|u+v\right|\right|^{2}-\left|\left|u-v\right|\right|^{2}\right)$
\end_inset
\end_layout
\begin_layout Itemize
V
\begin_inset Formula $\mathbb{C}$
\end_inset
:
\begin_inset Formula $\left\langle u,v\right\rangle =\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}$
\end_inset
\end_layout
\end_deeper
\begin_layout Proof
Dokaz v
\begin_inset Formula $\mathbb{C}$
\end_inset
.
Oglejmo si
\begin_inset Formula
\[
\left|\left|u+i^{k}v\right|\right|^{2}=\left\langle u+i^{k}v,u+i^{k}v\right\rangle =\left\langle u,u+i^{k}v\right\rangle +i^{k}\left\langle v,u+i^{k}v\right\rangle =\overline{\left\langle u+i^{k}v,u\right\rangle }+i^{k}\overline{\left\langle u+i^{k}v,v\right\rangle }=
\]
\end_inset
\begin_inset Formula
\[
=\overline{\left\langle u,u\right\rangle }+\overline{\left\langle i^{k}v,u\right\rangle }+i^{k}\overline{\left\langle u,v\right\rangle }+i^{k}\overline{\left\langle i^{k}v,v\right\rangle }=\left\langle u,u\right\rangle +\left\langle u,i^{k}v\right\rangle +i^{k}\left\langle v,u\right\rangle +i^{k}\left\langle v,i^{k}v\right\rangle =
\]
\end_inset
\begin_inset Formula
\[
=\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +i^{k}\left(-\left(i^{k}\right)\right)\left\langle v,v\right\rangle =
\]
\end_inset
\begin_inset Formula
\[
=\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +1\left\langle v,v\right\rangle
\]
\end_inset
Dodajmo vsoto:
\begin_inset Formula
\[
\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}=\sum_{k=0}^{3}i^{k}\left(\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +1\left\langle v,v\right\rangle \right)=
\]
\end_inset
\begin_inset Formula
\[
=\sum_{k=0}^{3}i^{k}\left\langle u,u\right\rangle +\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle +\sum_{k=0}^{3}i^{k}i^{k}\left\langle v,u\right\rangle +\sum_{k=0}^{3}i^{k}\left\langle v,v\right\rangle =0+\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle +0+0,
\]
\end_inset
kajti
\begin_inset Formula $\sum_{k=0}^{3}i^{k}=1+i+\left(-1\right)+\left(-i\right)=0$
\end_inset
in
\begin_inset Formula $\sum_{k=0}^{3}i^{2k}=1+\left(-1\right)+1+\left(-1\right)=0$
\end_inset
.
Nadaljujmo:
\end_layout
\begin_layout Proof
\begin_inset Formula
\[
=\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle =\sum_{k=0}^{3}1\left\langle u,v\right\rangle =4\left\langle u,v\right\rangle
\]
\end_inset
\end_layout
\begin_layout Subsubsection
Ortogonalne množice in ortogonalne baze
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $V$
\end_inset
VPSSP
\begin_inset Formula $\forall u,v\in V:u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$
\end_inset
.
\end_layout
\begin_layout Remark*
trivialne opombe.
\begin_inset Formula $\forall v\in V:v\perp\vec{0}$
\end_inset
,
\begin_inset Formula $\forall v\in V:v\not=0\Leftrightarrow v\not\perp v$
\end_inset
(prvi aksiom skalarnega produkta),
\begin_inset Formula $\forall v\in V:u\perp v\Leftrightarrow v\perp u$
\end_inset
.
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $V$
\end_inset
VPSSP in
\begin_inset Formula $v_{1},\dots,v_{k}\in V$
\end_inset
.
Množica
\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
\end_inset
je:
\end_layout
\begin_deeper
\begin_layout Itemize
ortogonalna,
če
\begin_inset Formula $v_{1}\not=0\wedge\cdots\wedge v_{k}\not=0$
\end_inset
in
\begin_inset Formula $\forall i,j\in\left\{ 1..k\right\} :i\not=j\Rightarrow v_{i}\perp v_{j}$
\end_inset
.
\end_layout
\begin_layout Itemize
normirana,
če
\begin_inset Formula $\forall v\in\left\{ v_{1},\dots,v_{k}\right\} :\left|\left|v\right|\right|=1$
\end_inset
.
\end_layout
\begin_layout Itemize
ortonormirana,
če je ortogonalna in ortonormirana hkrati.
\end_layout
\end_deeper
\begin_layout Remark*
Iz (ortogonalne) množice
\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
\end_inset
dobimo (orto)normirano tako,
da vsak element delimo z njegovo normo.
\begin_inset Formula $\left\{ \frac{v_{1}}{\left|\left|v_{1}\right|\right|},\dots,\frac{v_{k}}{\left|\left|v_{k}\right|\right|}\right\} $
\end_inset
je vedno normirana.
\end_layout
\begin_layout Claim*
Vsaka ortogonalna množica je linearno neodvisna.
\end_layout
\begin_layout Proof
Denimo,
da je
\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
\end_inset
ortogonalna.
Vzemimo take
\begin_inset Formula $\alpha_{1},\dots,\alpha_{k}\ni:\alpha_{1}v_{1}+\cdots+\alpha_{k}v_{k}=0$
\end_inset
.
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
udensdash{$
\backslash
alpha_1=
\backslash
cdots=
\backslash
alpha_k=0$}
\end_layout
\end_inset
.
\begin_inset Formula
\[
\forall i\in\left\{ 1..k\right\} :0=\left\langle 0,v_{i}\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{k}v_{k},v\right\rangle =\alpha_{1}\left\langle v_{1},v_{i}\right\rangle +\cdots+\alpha_{i}\left\langle v_{i},v_{i}\right\rangle +\cdots+\alpha_{k}\left\langle v_{k},v_{i}\right\rangle =\cdots
\]
\end_inset
Ker je množica ortogonalna,
je
\begin_inset Formula $\left\langle v_{l},v_{k}\right\rangle =0\Leftrightarrow l\not=k$
\end_inset
.
Nadaljujmo ...
\begin_inset Formula
\[
\cdots=\alpha_{i}\left\langle v_{i},v_{i}\right\rangle
\]
\end_inset
Ker
\begin_inset Formula $\left\langle v_{i},v_{i}\right\rangle $
\end_inset
ni 0,
ker je
\begin_inset Formula $v_{i}$
\end_inset
neničeln (da,
tudi to je del definicije ortogonalnosti),
je
\begin_inset Formula $\alpha_{i}=0$
\end_inset
.
In to za vsak
\begin_inset Formula $i$
\end_inset
.
\end_layout
\begin_layout Standard
Ni pa vsaka ortogonalna množica ogrodje.
Ortogonalni množici,
ki je ogrodje,
rečemo ortogonalna baza (LN sledi iz ortogonalnost).
\end_layout
\begin_layout Subsubsection
Fourierov razvoj
\end_layout
\begin_layout Standard
Naj bo
\begin_inset Formula $V$
\end_inset
KRVPSSP,
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} =\mathcal{B}$
\end_inset
ortogonalna baza za
\begin_inset Formula $V$
\end_inset
in
\begin_inset Formula $v\in V$
\end_inset
poljuben element.
Kako razvijemo
\begin_inset Formula $v$
\end_inset
po
\begin_inset Formula $\mathcal{B}$
\end_inset
,
vedoč,
da je ta baza ortogonalna?
Postopek imenujemo Fourierov razvoj.
\end_layout
\begin_layout Standard
Ker je
\begin_inset Formula $\mathcal{B}$
\end_inset
ogrodje,
\begin_inset Formula $\exists\alpha_{1},\dots,\alpha_{n}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
\end_inset
.
Množimo skalarno z
\begin_inset Formula $v_{i}$
\end_inset
:
\begin_inset Formula
\[
v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}\quad\quad\quad\quad/\cdot v_{i}
\]
\end_inset
\begin_inset Formula
\[
\left\langle v,v_{i}\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n},v_{i}\right\rangle
\]
\end_inset
\begin_inset Formula
\[
\left\langle v,v_{i}\right\rangle =\cancel{\alpha_{1}\left\langle v_{1},v_{i}\right\rangle }+\cancel{\cdots}+\alpha_{i}\left\langle v_{i},v_{i}\right\rangle +\cancel{\cdots}+\cancel{\alpha_{n}\left\langle v_{n},v_{i}\right\rangle }=\alpha_{i}\left\langle v_{i},v_{i}\right\rangle
\]
\end_inset
\begin_inset Formula
\[
\frac{\left\langle v,v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }=\alpha_{i}
\]
\end_inset
Torej
\begin_inset Formula $\forall v\in V$
\end_inset
velja
\begin_inset Formula $v=\sum_{i=1}^{n}\frac{\left\langle v,v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }v_{i}$
\end_inset
.
Koeficientu
\begin_inset Formula $\frac{\left\langle v,v_{i}\right\rangle }{\left|\left|v_{i}\right|\right|^{2}}$
\end_inset
pravimo Fourierov koeficient.
Če je baza ortonormirana,
je Fourierov koeficient
\begin_inset Formula $\frac{\left\langle v,v_{i}\right\rangle }{\cancel{\left|\left|v_{i}\right|\right|^{2}}}=\left\langle v,v_{i}\right\rangle $
\end_inset
.
\end_layout
\begin_layout Subsubsection
Parsevalova identiteta
\end_layout
\begin_layout Theorem*
Parsevalova identiteta.
Naj bo
\begin_inset Formula $V$
\end_inset
VPSSP in
\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
\end_inset
njegova ortogonalna baza.
Tedaj
\begin_inset Formula $\forall v\in V:$
\end_inset
\begin_inset Formula
\[
\left|\left|v\right|\right|^{2}=\sum_{i=1}^{n}\frac{\left|\left\langle v,v_{i}\right\rangle \right|^{2}}{\left\langle v_{i},v_{i}\right\rangle }.
\]
\end_inset
Če je baza ortonormirana,
se enačba očitno poenostavi v
\begin_inset Formula
\[
\left|\left|v\right|\right|^{2}=\sum_{i=1}^{n}\left|\left\langle v,v_{i}\right\rangle \right|^{2}.
\]
\end_inset
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
\end_inset
.
Tedaj
\begin_inset Formula $\left|\left|v\right|\right|^{2}=\left\langle v,v\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n},\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}\right\rangle =$
\end_inset
(uporabimo linearnost v 1.
in konjugirano linearnost v 2.
faktorju)
\begin_inset Formula
\[
\begin{array}{ccccccc}
= & \alpha_{1}\overline{\alpha_{1}}\left\langle v_{1},v_{1}\right\rangle & + & \cancel{\cdots} & + & \cancel{\alpha_{1}\overline{\alpha_{n}}\left\langle v_{1},v_{n}\right\rangle } & +\\
& \vdots & & & & \vdots\\
+ & \cancel{\alpha_{n}\overline{\alpha_{1}}\left\langle v_{n},v_{1}\right\rangle } & + & \cancel{\cdots} & + & \alpha_{n}\overline{\alpha_{n}}\left\langle v_{n},v_{n}\right\rangle & =
\end{array}
\]
\end_inset
\begin_inset Formula
\[
=\alpha_{1}\overline{\alpha_{1}}\left\langle v_{1},v_{1}\right\rangle +\cdots+\alpha_{n}\overline{\alpha_{n}}\left\langle v_{n},v_{n}\right\rangle =\left|\alpha_{1}\right|^{2}\left|\left|v_{1}\right|\right|^{2}+\cdots+\left|\alpha_{n}\right|^{2}\left|\left|v_{n}\right|\right|^{2}=
\]
\end_inset
Vstavimo formule za koeficiente po Fourierjevem razvoju:
\begin_inset Formula
\[
=\left|\frac{\left\langle v,v_{1}\right\rangle }{\left|\left|v_{1}\right|\right|^{\cancel{2}}}\right|^{2}\cancel{\left|\left|v_{1}\right|\right|^{2}}+\cdots+\left|\frac{\left\langle v,v_{n}\right\rangle }{\left|\left|v_{n}\right|\right|^{\cancel{2}}}\right|^{2}\cancel{\left|\left|v_{n}\right|\right|^{2}}=\frac{\left|\left\langle v,v_{1}\right\rangle \right|^{2}}{\left|\left|v_{1}\right|\right|}+\cdots+\frac{\left|\left\langle v,v_{n}\right\rangle \right|^{2}}{\left|\left|v_{n}\right|\right|}=\sum_{i=1}^{n}\frac{\left|\left\langle v_{i},v\right\rangle \right|^{2}}{\left\langle v_{i},v_{i}\right\rangle }
\]
\end_inset
\end_layout
\begin_layout Subsubsection
Projekcija na podprostor
\end_layout
\begin_layout Standard
Naj bo
\begin_inset Formula $V$
\end_inset
KRVPSSP in
\begin_inset Formula $W$
\end_inset
podprostor
\begin_inset Formula $V$
\end_inset
.
Za vsak
\begin_inset Formula $v\in V$
\end_inset
želimo izračunati njegovo ortogonalno projekcijo na
\begin_inset Formula $W$
\end_inset
.
\end_layout
\begin_layout Definition*
Vektor
\begin_inset Formula $v'\in W$
\end_inset
je ortogonalna projekcija vektorja
\begin_inset Formula $v\in V$
\end_inset
,
če
\begin_inset Formula $\forall w\in W:\left|\left|v-v'\right|\right|\leq\left|\left|v-w\right|\right|$
\end_inset
.
ZDB
\begin_inset Formula $v'$
\end_inset
je najbližje
\begin_inset Formula $v$
\end_inset
izmed vseh elementov
\begin_inset Formula $W$
\end_inset
.
\end_layout
\begin_layout Remark*
Zadošča preveriti,
da je
\begin_inset Formula $v-v'$
\end_inset
ortogonalen na vse elemente
\begin_inset Formula $W$
\end_inset
(pitagorov izrek),
kajti v tem primeru (če predpostavimo
\begin_inset Formula $\left(v'-w\right)\perp\left(v-v'\right)$
\end_inset
) velja
\begin_inset Formula
\[
\left|\left|v-w\right|\right|^{2}=\left|\left|v-v'+v'-w\right|\right|=\left|\left|v-v'\right|\right|^{2}+\left|\left|v'-w\right|\right|^{2}\geq\left|\left|v-v'\right|\right|^{2}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
Dokaz pitagovorega izreka:
\begin_inset Formula $\left|\left|a+b\right|\right|^{2}=\left\langle a+b,a+b\right\rangle =\left\langle a,a\right\rangle +\cancel{\left\langle a,b\right\rangle +\left\langle b,a\right\rangle }+\left\langle b,b\right\rangle =\left|\left|a\right|\right|^{2}+\left|\left|b\right|\right|^{2}$
\end_inset
.
\end_layout
\begin_layout Claim*
Naj bo
\begin_inset Formula $\left\{ w_{1},\dots,w_{k}\right\} $
\end_inset
ortogonalna baza za
\begin_inset Formula $W$
\end_inset
.
Formula za ortogonalno projekcijo se glasi:
\begin_inset Formula
\[
v'=\frac{\left\langle v,w_{1}\right\rangle }{\left\langle w_{1},w_{1}\right\rangle }w_{1}+\cdots+\frac{\left\langle v,w_{k}\right\rangle }{\left\langle w_{k},w_{k}\right\rangle }=\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i}
\]
\end_inset
\end_layout
\begin_layout Proof
Dokažimo,
da je
\begin_inset Formula $v-\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i}$
\end_inset
pravokoten na vse elemente
\begin_inset Formula $W$
\end_inset
.
Zaradi linearnosti skalarnega produkta zadošča preveriti,
da je pravokoten na bazo
\begin_inset Formula $W$
\end_inset
.
\begin_inset Formula $\forall j\in\left\{ 1..k\right\} $
\end_inset
velja (spomnimo se,
da je
\begin_inset Formula $\left\langle w_{i},w_{j}\right\rangle =0\Leftrightarrow i\not=j$
\end_inset
,
zato po drugem enačaju ostane le še en člen vsote):
\begin_inset Formula
\[
\left\langle v-\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i},w_{j}\right\rangle =\left\langle v,w_{j}\right\rangle -\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }\left\langle w_{i},w_{j}\right\rangle =\left\langle v,w_{j}\right\rangle -\frac{\left\langle v,w_{j}\right\rangle }{\cancel{\left\langle w_{j},w_{j}\right\rangle }}\cancel{\left\langle w_{j},w_{j}\right\rangle }=
\]
\end_inset
\begin_inset Formula
\[
=\left\langle v,w_{j}\right\rangle -\left\langle v,w_{j}\right\rangle =0
\]
\end_inset
\end_layout
\begin_layout Subsubsection
Obstoj ortogonalne baze —
Gram-Schmidtova ortogonalizacija
\end_layout
\begin_layout Standard
Radi bi dokazali,
da ima vsak KRVPSSP ortogonalno bazo in da je moč vsako ortogonalno množico dopolniti do ortogonalne baze.
Konstruktiven dokaz
\begin_inset Formula $\ddot{\smile}!$
\end_inset
—
postopek,
imenovan Gram-Schmidtova ortogonalizacija,
iz poljubne baze naredi ortogonalno.
\end_layout
\begin_layout Standard
Naj bo
\begin_inset Formula $V$
\end_inset
KRVPSSP in
\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset
njegova poljubna baza.
Naj bo
\begin_inset Formula $v_{1}\coloneqq u_{1}$
\end_inset
,
\begin_inset Formula
\[
v_{2}\coloneqq u_{2}-\frac{\left\langle u_{2},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}=u_{2}-u_{2}'
\]
\end_inset
\begin_inset Formula
\[
v_{3}\coloneqq u_{3}-\frac{\left\langle u_{3},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}-\frac{\left\langle u_{3},v_{2}\right\rangle }{\left\langle v_{2},v_{2}\right\rangle }v_{2}=u_{3}-u_{3}'
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\cdots
\]
\end_inset
\begin_inset Formula
\[
v_{n}\coloneqq u_{n}-\sum_{i=1}^{n-1}\frac{\left\langle u_{n},v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }v_{i}=u_{n}-u_{n}'
\]
\end_inset
\end_layout
\begin_layout Standard
Trdimo,
da je
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
ortogonalna baza za
\begin_inset Formula $V$
\end_inset
.
\end_layout
\begin_layout Standard
Opazimo,
da je
\begin_inset Formula $u_{2}'$
\end_inset
ortogonalna projekcija
\begin_inset Formula $u_{2}$
\end_inset
na
\begin_inset Formula $\Lin\left\{ v_{1}\right\} $
\end_inset
,
\begin_inset Formula $u_{3}'$
\end_inset
ortogonalna projekcija
\begin_inset Formula $u_{3}$
\end_inset
na
\begin_inset Formula $\Lin\left\{ v_{1},v_{2}\right\} $
\end_inset
,
...,
\begin_inset Formula $u_{n}'$
\end_inset
pa ortogonalna projekcija na
\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n-1}\right\} $
\end_inset
,
torej
\begin_inset Formula
\[
v_{2}=u_{2}-u_{2}'\perp\Lin\left\{ v_{1}\right\} \text{, torej }v_{2}\perp v_{1}
\]
\end_inset
\begin_inset Formula
\[
v_{3}=u_{3}-u_{3}'\perp\Lin\left\{ v_{1},v_{2}\right\} \text{, torej }v_{3}\perp v_{1},v_{3}\perp v_{2}
\]
\end_inset
\begin_inset Formula
\[
\cdots
\]
\end_inset
\begin_inset Formula
\[
v_{n}=u_{n}-u_{n}'\perp\Lin\left\{ v_{1},\dots,v_{n-1}\right\} \text{, torej }v_{n}\perp v_{1},\dots,v_{n}\perp v_{n-1},
\]
\end_inset
kar pomeni,
da so
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
paroma ortogonalni.
Toda vprašanje je,
ali so neničelni,
kajti to je,
ne boste verjeli,
prav tako pogoj za ortogonalno množico.
\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :$
\end_inset
dokažimo neničelnost
\begin_inset Formula $v_{i}$
\end_inset
:-)
\end_layout
\begin_layout Standard
\begin_inset Formula $v_{1}$
\end_inset
je neničeln,
ker je enak
\begin_inset Formula $u_{1}$
\end_inset
,
ki je element baze
\begin_inset Formula $V$
\end_inset
.
\begin_inset Formula $v_{2}$
\end_inset
je neničeln,
ker je
\begin_inset Formula $v_{2}=u_{2}-\alpha v_{1}$
\end_inset
in
\begin_inset Formula $u_{2}\not=\alpha v_{1}$
\end_inset
,
ker sta linearno neodvisna,
ker tvorita ortogonalno množico.
\begin_inset Formula $v_{3}$
\end_inset
je neničeln,
ker
\begin_inset Formula $v_{3}=u_{3}-\left(\beta v_{1}+\gamma v_{2}\right)$
\end_inset
in ker so
\begin_inset Formula $v_{1},v_{2},u_{3}$
\end_inset
LN,
\begin_inset Formula $u_{3}\not=\left(\beta v_{1}+\gamma v_{2}\right)$
\end_inset
.
In tako dalje.
\end_layout
\begin_layout Paragraph*
Dopolnitev ortogonalne množice do baze
\end_layout
\begin_layout Standard
Naj bo
\begin_inset Formula $\left\{ u_{1},\dots,u_{k}\right\} $
\end_inset
ortogonalna množica,
torej je linearno neodvisna,
torej jo lahko dopolnimo do baze.
\begin_inset Formula $\left\{ u_{k+1},\dots,u_{n}\right\} $
\end_inset
je dopolnitev do baze.
Toda slednja še ni ortogonalna.
A nič ne de,
uporabimo lahko Gram-Schmidtovo ortogonalizacijo na
\begin_inset Formula $\left\{ u_{1},\dots,u_{k},u_{k+1},\dots,u_{n}\right\} $
\end_inset
in dobimo ortogonalno bazo
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset
.
Opazimo,
da ker so po predpostavki
\begin_inset Formula $u_{1},\dots,u_{k}$
\end_inset
ortogonalni,
velja
\begin_inset Formula $v_{1}=u_{1},\dots,v_{k}=u_{k}$
\end_inset
(po GS).
\end_layout
\begin_layout Example*
primer GS ortogonalizacije iz analize.
Naj bo
\begin_inset Formula $V=\mathbb{R}\left[x\right]_{\leq3}$
\end_inset
.
Baza:
\begin_inset Formula $u_{1}=1,u_{2}=x,u_{3}=x^{2},u_{4}=x^{3}$
\end_inset
,
skalarni produkt naj bo
\begin_inset Formula $\left\langle p,q\right\rangle =\int_{-1}^{1}p\left(x\right)q\left(x\right)dx$
\end_inset
.
Konstruirajmo pripadajočo ortogonalno bazo
\begin_inset Formula $v_{1},\dots,v_{4}$
\end_inset
:
\begin_inset Formula
\[
v_{1}\coloneqq u_{1}=1
\]
\end_inset
\begin_inset Formula
\[
v_{2}\coloneqq u_{2}-\frac{\left\langle u_{2},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}=x-\frac{\int_{-1}^{1}xdx}{\int_{-1}^{1}dx}=x-0=x=u_{2}
\]
\end_inset
\begin_inset Formula
\[
v_{3}\coloneqq u_{3}-\frac{\left\langle u_{3},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}-\frac{\left\langle u_{3},v_{2}\right\rangle }{\left\langle v_{2},v_{2}\right\rangle }=x^{2}-\frac{\int_{-1}^{1}x^{2}dx}{\int_{-1}^{1}dx}-\frac{\int_{-1}^{1}x^{3}dx}{\int_{-1}^{1}x^{2}dx}x=\cdots=x²-\frac{1}{3}
\]
\end_inset
\begin_inset Formula
\[
v_{4}\coloneqq\cdots=x^{2}-\frac{3}{5}x
\]
\end_inset
Sklep:
\begin_inset Formula $\left\{ 1,x,x^{2}-\frac{1}{3},x^{2}-\frac{3}{5}x\right\} $
\end_inset
je ortogonalna baza za ta vektorski prostor s tem skalarnim produktom.
Normirajmo jo!
Norme teh baznih vektorjev po vrsti so
\begin_inset Formula $\sqrt{2},\sqrt{\frac{2}{3}},\sqrt{\frac{8}{45}},\sqrt{\frac{8}{175}}$
\end_inset
.
Pripadajoča ortonormirana baza je torej
\begin_inset Formula $\left\{ \frac{1}{\sqrt{2}},\frac{x}{\sqrt{\frac{2}{3}}},\frac{x^{2}-\frac{1}{3}}{\sqrt{\frac{8}{45}}},\frac{x^{2}-\frac{3}{5}x}{\sqrt{\frac{8}{175}}}\right\} .$
\end_inset
Normiranje bi sicer prineslo lepše formule,
vendar bi v račune prineslo te objektivno grde konstante.
\end_layout
\begin_layout Subsubsection
Ortogonalni komplement
\end_layout
\begin_layout Definition
Naj bo
\begin_inset Formula $V$
\end_inset
KRVPSSP nad
\begin_inset Formula $F$
\end_inset
in
\begin_inset Formula $S\subseteq V$
\end_inset
.
Ortogonalni komplement
\begin_inset Formula $S$
\end_inset
je množica
\begin_inset Formula $S^{\perp}$
\end_inset
.
Vsebuje vse tiste vektorje iz
\begin_inset Formula $V$
\end_inset
,
ki so ortogonalni na
\begin_inset Formula $S$
\end_inset
.
ZDB
\begin_inset Formula $S^{\perp}\coloneqq\left\{ v\in V;\forall s\in S:v\perp s\right\} =\left\{ v\in V;v\perp S\right\} $
\end_inset
.
\end_layout
\begin_layout Claim*
\begin_inset Formula $\forall S\subseteq V:S^{\perp}$
\end_inset
je podprostor
\begin_inset Formula $V$
\end_inset
.
\end_layout
\begin_layout Proof
Dokazati je treba
\begin_inset Formula $\forall u_{1},u_{2}\in S^{\perp},\alpha_{1},\alpha_{2}\in F:\alpha_{1}v_{1}+\alpha_{2}v_{2}\in S^{\perp}$
\end_inset
.
Po definiciji ortogonalnega komplementa velja
\begin_inset Formula
\[
\forall s\in S:\left\langle u_{1},s\right\rangle =0\wedge\left\langle u_{2},s\right\rangle =0\Longrightarrow0=\alpha_{1}\left\langle u_{1},s\right\rangle +\alpha_{2}\left\langle u_{2},s\right\rangle =\left\langle \alpha_{1}u_{1}+\alpha_{2}u_{2},s\right\rangle \Longrightarrow\alpha_{1}u_{1}+\alpha_{2}u_{2}\in S^{\perp}
\]
\end_inset
\end_layout
\begin_layout Theorem*
ortogonalni razcep.
Naj bo
\begin_inset Formula $V$
\end_inset
KRVPSSP in
\begin_inset Formula $W$
\end_inset
vektorski podprostor
\begin_inset Formula $V$
\end_inset
.
Potem velja
\begin_inset Formula $V=W\oplus W^{\perp}$
\end_inset
(ortogonalni razcep glede na
\begin_inset Formula $W$
\end_inset
).
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $v\in V$
\end_inset
pojuben,
\begin_inset Formula $V^{\perp}$
\end_inset
pa ortogonalna projekcija
\begin_inset Formula $V$
\end_inset
na
\begin_inset Formula $W$
\end_inset
.
Potem velja,
da je
\begin_inset Formula $v=v-v'+v'$
\end_inset
,
kjer je
\begin_inset Formula $v-v'$
\end_inset
pravokoten na
\begin_inset Formula $W$
\end_inset
,
\begin_inset Formula $v'$
\end_inset
pa element
\begin_inset Formula $v'$
\end_inset
,
torej
\begin_inset Formula $v\in W\oplus W^{\perp}$
\end_inset
.
Vsota je direktna,
kajti
\begin_inset Formula $\forall v\in W\cap W^{\perp}:v\perp v\Leftrightarrow v\perp v\Leftrightarrow\left\langle v,v\right\rangle =0\Leftrightarrow v=0\Longrightarrow W\cap W^{\perp}=\left\{ 0\right\} $
\end_inset
(po karakterizaciji direktnih vsot).
\end_layout
\begin_layout Claim*
Naj bo
\begin_inset Formula $V$
\end_inset
KRVPSSP in
\begin_inset Formula $W$
\end_inset
vektorski podprostor v
\begin_inset Formula $V$
\end_inset
.
Velja
\begin_inset Formula $\left(W^{\perp}\right)^{\perp}=W$
\end_inset
.
\end_layout
\begin_layout Proof
Po definiciji ortogonalnega komplementa je
\begin_inset Formula $W\subseteq\left(W^{\perp}\right)^{\perp}$
\end_inset
,
ker
\begin_inset Formula $W\perp W^{\perp}$
\end_inset
.
Dokažimo
\begin_inset Formula $\dim W=\dim\left(W^{\perp}\right)^{\perp}$
\end_inset
.
Ortogonalni razcep glede na
\begin_inset Formula $W$
\end_inset
je
\begin_inset Formula $V=W\oplus W^{\perp}\Rightarrow\dim W+\dim W^{\perp}=\dim V$
\end_inset
,
ortogonalni razcep glede na
\begin_inset Formula $W^{\perp}$
\end_inset
pa je
\begin_inset Formula $V=W^{\perp}\oplus\left(W^{\perp}\right)^{\perp}\Rightarrow\dim W^{\perp}+\dim\left(W^{\perp}\right)^{\perp}=\dim V$
\end_inset
.
\begin_inset Formula
\[
\dim V=\dim V
\]
\end_inset
\begin_inset Formula
\[
\dim W+\dim W^{\perp}=\dim W^{\perp}+\dim\left(W^{\perp}\right)^{\perp}
\]
\end_inset
\begin_inset Formula
\[
\dim W^{\perp}=\dim\left(W^{\perp}\right)^{\perp}
\]
\end_inset
\end_layout
\begin_layout Proof
Alternativni dokaz:
Naj bodo
\begin_inset Formula $w_{1},\dots,w_{k}$
\end_inset
OB za
\begin_inset Formula $W$
\end_inset
.
Dopolnimo jo do OB za
\begin_inset Formula $V$
\end_inset
z GS z
\begin_inset Formula $w_{k+1},\dots,w_{n}$
\end_inset
.
Tedaj je
\begin_inset Formula $w_{k+1},\dots,w_{n}$
\end_inset
OB za
\begin_inset Formula $W^{\perp}$
\end_inset
in ker je
\begin_inset Formula $w_{1},\dots,w_{n}$
\end_inset
njena dopolnitev do OB
\begin_inset Formula $V$
\end_inset
,
je
\begin_inset Formula $w_{1},\dots,w_{k}$
\end_inset
OB za
\begin_inset Formula $\left(W^{\perp}\right)^{\perp}$
\end_inset
,
torej
\begin_inset Formula $W^{\perp}=\left(W^{\perp}\right)^{\perp}$
\end_inset
,
saj imata isti ortogonalni bazi.
\end_layout
\begin_layout Subsection
Adjungirana linearna preslikava
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $V$
\end_inset
vektorski prostor nad
\begin_inset Formula $F$
\end_inset
.
Vemo,
da je
\begin_inset Formula $F$
\end_inset
vekrorski prostor nad
\begin_inset Formula $F$
\end_inset
.
Linearnim preslikavam
\begin_inset Formula $V\to F$
\end_inset
pravimo linearni funkcionali na
\begin_inset Formula $V$
\end_inset
.
\end_layout
\begin_layout Example*
Naj bo
\begin_inset Formula $V$
\end_inset
VPSSP in
\begin_inset Formula $F\in\left\{ \mathbb{R},\mathbb{C}\right\} $
\end_inset
.
Naj bo
\begin_inset Formula $w\in V$
\end_inset
.
Naj bo
\begin_inset Formula $\varphi:V\to F$
\end_inset
(torej je
\begin_inset Formula $\varphi$
\end_inset
linearni funkcional),
ki slika
\begin_inset Formula $v\mapsto\left\langle v,w\right\rangle $
\end_inset
.
Preslikava je po aksiomu 3 za skalarni produkt linearna.
\end_layout
\begin_layout Theorem*
Rieszov izrek o reprezentaciji linearnih funkcionalov.
Naj bo
\begin_inset Formula $V$
\end_inset
KRVPSSP.
Za vsak linearen funkcional
\begin_inset Formula $\varphi$
\end_inset
na
\begin_inset Formula $V$
\end_inset
obstaja natanko en vektor
\begin_inset Formula $w\in V\ni:\forall v\in V:\varphi\left(v\right)=\left\langle v,w\right\rangle $
\end_inset
.
ZDB slednja konstrukcija nam da vse linearne funkcionale.
\end_layout
\begin_layout Proof
Dokazujemo enolično eksistenco:
\end_layout
\begin_deeper
\begin_layout Itemize
Eksistenca
\begin_inset Formula $w$
\end_inset
:
Vzemimo poljubno OB
\begin_inset Formula $w_{1},\dots,w_{n}$
\end_inset
za
\begin_inset Formula $V$
\end_inset
.
\begin_inset Formula $\forall v\in V:v=\left\langle v,w_{1}\right\rangle w_{1}+\cdots+\left\langle v,w_{n}\right\rangle w_{n}$
\end_inset
.
(fourierov razvoj po OB).
Ker je
\begin_inset Formula $\varphi$
\end_inset
linearna,
velja
\begin_inset Formula
\[
\varphi\left(v\right)=\varphi\left(\left\langle v,w_{1}\right\rangle w_{1}+\cdots+\left\langle v,w_{n}\right\rangle w_{n}\right)\overset{\text{linearna}}{=}\left\langle v,w_{1}\right\rangle \varphi w_{1}+\cdots+\left\langle v,w_{n}\right\rangle \varphi w_{n}\overset{\text{konj. hom. v 2. fakt.}}{=}
\]
\end_inset
\begin_inset Formula
\[
=\left\langle v,\overline{\varphi w_{1}}w_{1}\right\rangle +\cdots+\left\langle v,\overline{\varphi w_{n}}w_{n}\right\rangle \overset{\text{konj. ad. v 2. fakt.}}{=}\left\langle v,\left(\varphi w_{1}\right)w_{1}+\cdots+\left(\varphi w_{n}\right)w_{n}\right\rangle
\]
\end_inset
Za dan
\begin_inset Formula $\varphi$
\end_inset
smo konstruirali eksplicitno formulo za iskani
\begin_inset Formula $w$
\end_inset
.
\end_layout
\begin_layout Itemize
Enoličnost
\begin_inset Formula $w$
\end_inset
:
PDDRAA
\begin_inset Formula $\forall v\in V:\varphi\left(v\right)=\left\langle v,w_{1}\right\rangle =\left\langle v,w_{2}\right\rangle $
\end_inset
.
Tedaj
\begin_inset Formula $\forall v\in V:\left\langle v,w_{1}-w_{2}\right\rangle =0$
\end_inset
.
Vzemimo konkreten
\begin_inset Formula $v=w_{1}-w_{2}$
\end_inset
in ga vstavimo v formulo
\begin_inset Formula $\left\langle v,w_{1}-w_{2}\right\rangle =0=\left\langle w_{1}-w_{2},w_{1}-w_{2}\right\rangle =0\Rightarrow w_{1}-w_{2}=0\Rightarrow w_{1}=w_{2}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Definition*
Naj bosta
\begin_inset Formula $U,V$
\end_inset
KRVPSSP in
\begin_inset Formula $L:U\to V$
\end_inset
linearna.
Adjungirana linearna preslikava,
pripadajoča
\begin_inset Formula $L$
\end_inset
,
je taka
\begin_inset Formula $L^{*}:V\to U$
\end_inset
,
da velja
\begin_inset Formula $\forall u\in U,v\in V:\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle $
\end_inset
.
Levi skalarni produkt je tisti iz
\begin_inset Formula $V$
\end_inset
,
desni pa tisti iz
\begin_inset Formula $U$
\end_inset
.
\end_layout
\begin_layout Claim*
Da lahko pišemo
\begin_inset Formula $L^{*}$
\end_inset
,
trdimo,
da je
\begin_inset Formula $L^{*}$
\end_inset
vedno obstaja in to vselej enolično.
\end_layout
\begin_layout Proof
Dokazujemo enolično eksistenco:
\end_layout
\begin_deeper
\begin_layout Itemize
Enoličnost:
Naj bosta
\begin_inset Formula $L^{*}$
\end_inset
in
\begin_inset Formula $L^{\circ}$
\end_inset
dve adjungirani linearni preslikavi za
\begin_inset Formula $L$
\end_inset
,
torej
\begin_inset Formula $\forall u\in U,v\in V:\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle =\left\langle u,L^{\circ}v\right\rangle $
\end_inset
.
Torej
\begin_inset Formula
\[
\left\langle u,L^{*}v\right\rangle =\left\langle u,L^{\circ}v\right\rangle
\]
\end_inset
\begin_inset Formula
\[
0=\left\langle u,L^{*}v-L^{\circ}v\right\rangle
\]
\end_inset
Za vsaka
\begin_inset Formula $u$
\end_inset
in
\begin_inset Formula $v$
\end_inset
.
Sedaj vstavimo
\begin_inset Formula $u=L^{*}v-L^{\circ}v$
\end_inset
:
\begin_inset Formula
\[
0=\left\langle L^{*}v-L^{\circ}v,L^{*}v-L^{\circ}v\right\rangle \Longrightarrow L^{*}v-L^{\circ}v=0\Longrightarrow\forall v\in V:L^{*}v=L^{\circ}v
\]
\end_inset
\end_layout
\begin_layout Itemize
Eksistenca:
Naj bosta
\begin_inset Formula $U,V$
\end_inset
KRVPSSP in
\begin_inset Formula $L:U\to V$
\end_inset
.
Naj bo
\begin_inset Formula $v\in V$
\end_inset
poljuben.
Vpeljimo linearni funkcional
\begin_inset Formula $\varphi:U\to F$
\end_inset
s predpisom
\begin_inset Formula $u\mapsto\left\langle Lu,v\right\rangle $
\end_inset
.
Prepričajmo se,
da je ta funkcional linearna preslikava:
\begin_inset Formula
\[
\varphi\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=\left\langle L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right),v\right\rangle =\left\langle \alpha_{1}Lu_{1}+\alpha_{2}Lu_{2},v\right\rangle =\alpha_{1}\left\langle Lu_{1},v\right\rangle +\alpha_{2}\left\langle Lu_{2},v\right\rangle
\]
\end_inset
Uporabimo Rieszov izrek za funkcional
\begin_inset Formula $\varphi$
\end_inset
:
\begin_inset Formula $\exists!w\in U\ni:\forall u\in U:\varphi u=\left\langle u,w\right\rangle $
\end_inset
.
Vpeljimo
\begin_inset Formula $L^{*}v=w$
\end_inset
,
s čimer za poljuben
\begin_inset Formula $v$
\end_inset
definiramo
\begin_inset Formula $L^{*}v$
\end_inset
.
Dokažimo,
da je dobljena preslikava linearna:
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
udensdash{$L^{*}
\backslash
left(
\backslash
beta_{1}v_{1}+
\backslash
beta_{2}v_{2}
\backslash
right)=
\backslash
beta_{1}L^{*}v_{1}+
\backslash
beta_{2}L^{*}v_{2}$}
\end_layout
\end_inset
.
Vzemimo pojuben
\begin_inset Formula $u\in U$
\end_inset
in računajmo (fino bi bilo dobiti nič):
\begin_inset Formula
\[
\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)\right\rangle =\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\beta_{1}L^{*}v_{1}-\beta_{2}L^{*}v_{2}\right\rangle \overset{\text{kl2f}}{=}
\]
\end_inset
\begin_inset Formula
\[
=\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle \overset{\text{lin }L^{*}}{=}\left\langle u,\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle =
\]
\end_inset
\begin_inset Formula
\[
=\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle +\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle =0
\]
\end_inset
Ker to velja za vsak
\begin_inset Formula $u$
\end_inset
,
velja tudi za
\begin_inset Formula $u=L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)$
\end_inset
,
torej dobimo
\begin_inset Formula
\[
\left\langle L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right),L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)\right\rangle =0
\]
\end_inset
torej po prvem aksiomu za skalarni produtk velja linearnost:
\begin_inset Formula
\[
L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)=0
\]
\end_inset
\begin_inset Formula
\[
L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)=\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Example*
Naj bo
\begin_inset Formula $A\in M_{m\times n}\left(F\right)$
\end_inset
s pripadajočo linearno preslikavo
\begin_inset Formula $L_{A}=F^{n}\to F^{m}$
\end_inset
,
ki slika
\begin_inset Formula $v\mapsto Av$
\end_inset
.
Kako izgleda matrika
\begin_inset Formula $L_{A^{*}}$
\end_inset
?
Odgovor je odvisen od izbire skalarnega produkta.
Izberimo standardni skalarni produkt v
\begin_inset Formula $F^{n}$
\end_inset
in
\begin_inset Formula $F^{m}$
\end_inset
in
\begin_inset Formula $L_{A^{*}}:F^{m}\to F^{n}$
\end_inset
definiramo z
\begin_inset Formula $v\mapsto A^{*}v$
\end_inset
,
kjer je
\begin_inset Formula $A^{*}=\overline{A^{T}}$
\end_inset
,
torej transponiranka
\begin_inset Formula $A$
\end_inset
z vsemi elementi konjugiranimi.
Izkaže se,
da je potemtakem
\begin_inset Formula $L_{A^{*}}$
\end_inset
adjungirana linearna preslikava od
\begin_inset Formula $L_{A}$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Matrika adjungirane linearne preslikave
\end_layout
\begin_layout Standard
Naj bosta
\begin_inset Formula $U,V$
\end_inset
KRVPSSP in naj bo
\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset
ONB za
\begin_inset Formula $U$
\end_inset
in
\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset
ONB za
\begin_inset Formula $V$
\end_inset
.
Vzemimo linearno preslikavo
\begin_inset Formula $L:U\to V$
\end_inset
.
Izpeljimo zvezo med
\begin_inset Formula $L$
\end_inset
in
\begin_inset Formula $L^{*}$
\end_inset
glede na bazi
\begin_inset Formula $\mathcal{B}$
\end_inset
in
\begin_inset Formula $\mathcal{C}$
\end_inset
.
Torej
\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset
za
\begin_inset Formula $L:U\to V$
\end_inset
in
\begin_inset Formula $\left[L^{*}\right]_{\mathcal{B}\leftarrow\mathcal{C}}$
\end_inset
za
\begin_inset Formula $L^{*}:V\to U$
\end_inset
.
Izračunajmo
\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset
tako,
da uporabimo fourierov razvoj:
\begin_inset Formula
\[
\begin{array}{ccccccc}
Lu_{1} & = & \left\langle Lu_{1},v_{1}\right\rangle v_{1} & + & \cdots & + & \left\langle Lu_{1},v_{m}\right\rangle v_{m}\\
\vdots & & \vdots & & & & \vdots\\
Lu_{n} & = & \left\langle Lu_{n},v_{1}\right\rangle v_{1} & + & \cdots & + & \left\langle Lu_{n},v_{m}\right\rangle v_{m}
\end{array}
\]
\end_inset
\begin_inset Formula
\[
\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
\left\langle Lu_{1},v_{1}\right\rangle & \cdots & \left\langle Lu_{n},v_{1}\right\rangle \\
\vdots & & \vdots\\
\left\langle Lu_{1},v_{m}\right\rangle & \cdots & \left\langle Lu_{n},v_{m}\right\rangle
\end{array}\right]=\left[\begin{array}{ccc}
\left\langle u_{1},L^{*}v_{1}\right\rangle & \cdots & \left\langle u_{n},L^{*}v_{1}\right\rangle \\
\vdots & & \vdots\\
\left\langle u_{1},L^{*}v_{m}\right\rangle & \cdots & \left\langle u_{n},L^{*}v_{m}\right\rangle
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
Sedaj izračunajmo še
\begin_inset Formula $\left[L^{*}\right]_{\mathcal{B}\leftarrow\mathcal{C}}$
\end_inset
spet s fourierovim razvojem in primerjajmo istoležne koeficiente:
\begin_inset Formula
\[
\begin{array}{ccccccc}
L^{*}v_{1} & = & \left\langle L^{*}v_{1},u_{1}\right\rangle u_{1} & + & \cdots & + & \left\langle L^{*}v_{1},u_{n}\right\rangle u_{n}\\
\vdots & & \vdots & & & & \vdots\\
L^{*}v_{m} & = & \left\langle Lv_{m},u_{1}\right\rangle u_{1} & + & \cdots & + & \left\langle Lv_{m},u_{n}\right\rangle u_{n}
\end{array}
\]
\end_inset
\begin_inset Formula
\[
\left[L\right]_{\mathcal{B}\leftarrow\mathcal{C}}=\left[\begin{array}{ccc}
\left\langle L^{*}v_{1},u_{1}\right\rangle & \cdots & \left\langle L^{*}v_{m},u_{1}\right\rangle \\
\vdots & & \vdots\\
\left\langle L^{*}v_{1},u_{n}\right\rangle & \cdots & \left\langle L^{*}v_{m},u_{n}\right\rangle
\end{array}\right]=\left[\begin{array}{ccc}
\overline{\left\langle u_{1},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{1},L^{*}v_{m}\right\rangle }\\
\vdots & & \vdots\\
\overline{\left\langle u_{n},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{m}\right\rangle }
\end{array}\right]=\left[\begin{array}{ccc}
\overline{\left\langle u_{1},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{1}\right\rangle }\\
\vdots & & \vdots\\
\overline{\left\langle u_{1},L^{*}v_{m}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{m}\right\rangle }
\end{array}\right]^{T}=
\]
\end_inset
\begin_inset Formula
\[
=\overline{\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}}^{T}
\]
\end_inset
\end_layout
\begin_layout Remark*
Kako izgleda lastnost
\begin_inset Formula $\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle $
\end_inset
?
Naj bo
\begin_inset Formula $u\in F^{n}$
\end_inset
in
\begin_inset Formula $v\in F^{m}$
\end_inset
in
\begin_inset Formula $A=m\times n$
\end_inset
matrika.
Ali za standardna skalarna produkta v
\begin_inset Formula $F^{n}$
\end_inset
in
\begin_inset Formula $F^{m}$
\end_inset
\begin_inset Formula $\left\langle Au,v\right\rangle =\left\langle u,A^{*}v\right\rangle $
\end_inset
velja tudi za matrike,
če vzamemo
\begin_inset Formula $A^{*}=\overline{A}^{T}=\overline{A^{T}}$
\end_inset
?
Pa preverimo (ja,
velja):
\begin_inset Formula
\[
\left\langle u,v\right\rangle =u_{1}\overline{v_{1}}+\cdots+u_{n}\overline{v_{n}}=\left[\begin{array}{ccc}
\overline{v_{1}} & \cdots & \overline{v_{n}}\end{array}\right]\left[\begin{array}{c}
u_{1}\\
\vdots\\
u_{n}
\end{array}\right]=v^{*}u
\]
\end_inset
\begin_inset Formula
\[
\left\langle Au,v\right\rangle =v^{*}Au
\]
\end_inset
\begin_inset Formula
\[
\left\langle u,A^{*}v\right\rangle =\left(A^{*}v\right)^{*}u=v^{*}\left(A^{*}\right)^{*}u=v^{*}Au
\]
\end_inset
\end_layout
\begin_layout Fact*
Lastnosti adjungiranja:
\begin_inset Formula $\left(\alpha A+\beta B\right)^{*}=\overline{\alpha}A^{*}+\overline{\beta}B^{*}$
\end_inset
,
\begin_inset Formula $\left(AB\right)^{*}=B^{*}A^{*}$
\end_inset
,
\begin_inset Formula $\left(A^{*}\right)^{*}=A$
\end_inset
\begin_inset Note Note
status open
\begin_layout Plain Layout
TODO XXX FIXME DOKAŽI
\end_layout
\end_inset
\end_layout
\begin_layout Subsubsection
Jedro in slika adjungirane linearne preslikave
\end_layout
\begin_layout Claim*
Naj bo
\begin_inset Formula $L:U\to V$
\end_inset
linearna.
Velja
\begin_inset Formula $\Ker\left(L^{*}\right)=\left(\Slika L\right)^{\perp}$
\end_inset
in
\begin_inset Formula $\Slika\left(L^{*}\right)=\left(\Ker L\right)^{\perp}$
\end_inset
.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $v\in\Ker L^{*}$
\end_inset
za
\begin_inset Formula $L^{*}:V\to U$
\end_inset
.
Velja
\end_layout
\begin_layout Proof
\begin_inset Formula
\[
v\in\Ker\left(L^{*}\right)\Leftrightarrow L^{*}v=0\Leftrightarrow\forall u\in U:\left\langle u,L^{*}v\right\rangle =0\Leftrightarrow\forall u\in U:\left\langle Lu,v\right\rangle =0\Leftrightarrow\forall w\in\Slika L:\left\langle w,v\right\rangle =0\Leftrightarrow v\in\left(\Slika L\right)^{\perp}
\]
\end_inset
Velja torej
\begin_inset Formula $\Ker L^{*}=\left(\Slika L\right)^{\perp}\Rightarrow\Ker L=\left(\Slika L^{*}\right)^{\perp}\Rightarrow\left(\Ker L\right)^{\perp}=\Slika L^{*}\Rightarrow\left(\Ker L^{*}\right)^{\perp}=\Slika L$
\end_inset
\end_layout
\begin_layout Claim*
Za
\begin_inset Formula $L:U\to V$
\end_inset
velja
\begin_inset Formula $\Ker\left(L^{*}L\right)=\Ker L$
\end_inset
\end_layout
\begin_layout Proof
Vzemimo poljuben
\begin_inset Formula $u\in U$
\end_inset
in dokazujemo enakost množic (obe vsebovanosti):
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\supseteq\right)$
\end_inset
Če
\begin_inset Formula $u\in\Ker L\Rightarrow Lu=0\overset{\text{množimo z }L^{*}}{\Longrightarrow}L^{*}Lu=L^{*}u=0\Rightarrow u\in\Ker L^{*}L\Rightarrow\Ker L\subseteq\Ker L^{*}L$
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\subseteq\right)$
\end_inset
Če
\begin_inset Formula $u\in\Ker L^{*}L\Rightarrow L^{*}Lu=0\Rightarrow\left\langle u,L^{*}Lu\right\rangle =0\Rightarrow\left\langle Lu,Lu\right\rangle =0\Rightarrow Lu=0\Rightarrow u\in\Ker L\Rightarrow\Ker L^{*}L\subseteq\Ker L$
\end_inset
\end_layout
\end_deeper
\begin_layout Corollary*
\begin_inset Formula $\Slika\left(L^{*}L\right)=\Slika\left(L\right)$
\end_inset
.
\end_layout
\begin_layout Proof
\begin_inset Formula $\Slika\left(L^{*}L\right)=\Slika\left(L^{*}\left(L^{*}\right)^{*}\right)=\Slika\left(\left(L^{*}L\right)^{*}\right)=\left(\Ker L^{*}L\right)^{\perp}=\left(\Ker L\right)^{\perp}=\Slika L^{*}$
\end_inset
\end_layout
\begin_layout Subsubsection
Lastne vrednosti adjungirane linearne preslikave.
\end_layout
\begin_layout Claim*
Če je
\begin_inset Formula $\lambda$
\end_inset
lastna vrednost
\begin_inset Formula $A$
\end_inset
,
je
\begin_inset Formula $\overline{\lambda}$
\end_inset
lastna vrednost za
\begin_inset Formula $A^{*}$
\end_inset
.
ZDB
\begin_inset Formula $\det\left(A-\lambda I\right)=0\Rightarrow\det\left(A-\lambda\right)$
\end_inset
.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $B=A-\lambda I$
\end_inset
.
Tedaj
\begin_inset Formula $B^{*}=A^{*}-\overline{\lambda}I^{*}=A^{*}-\overline{\lambda}I$
\end_inset
.
Radi bi dokazali
\begin_inset Formula $\det B=0\Rightarrow\det B^{*}=0$
\end_inset
.
Ker je
\begin_inset Formula $B^{*}=\overline{B}^{T}\Rightarrow\det B^{*}=\det\overline{B}^{T}=\det\overline{B}=\overline{\det B}\Rightarrow\det B=0\Rightarrow\det B^{*}=0$
\end_inset
.
\end_layout
\begin_layout Corollary*
Iz te formule izvemo tudi karakteristični polimom
\begin_inset Formula $A^{*}$
\end_inset
.
\begin_inset Formula $p_{A^{*}}\left(x\right)=\det\left(A^{*}-xI\right)\Rightarrow p_{A}\left(\overline{x}\right)=\det\left(A-\overline{x}I\right)=\det\left(A^{*}-xI\right)^{*}=\overline{\det\left(A^{*}-xI\right)}=\overline{p_{A^{*}}\left(x\right)}$
\end_inset
,
torej
\begin_inset Formula $p_{A^{*}}\left(x\right)=\overline{p_{A}\left(\overline{x}\right)}$
\end_inset
.
Torej,
če je
\begin_inset Formula $p_{A}\left(x\right)=c_{0}x^{0}+\cdots+x_{n}x^{n}$
\end_inset
,
je
\begin_inset Formula $p_{A^{*}}\left(x\right)=\overline{c_{0}\overline{x^{0}}+\cdots+x_{n}\overline{x^{n}}}=\overline{c_{0}}x^{0}+\cdots+\overline{c_{n}}x^{n}$
\end_inset
.
\end_layout
\begin_layout Proof
Alternativen dokaz:
Najprej dokažimo
\begin_inset Formula $\dim\Ker B^{*}=\dim\Ker B$
\end_inset
.
Velja
\begin_inset Formula $\dim\Ker B^{*}=\dim\left(\Slika B\right)^{\perp}=n-\dim\Slika B=\dim\Ker B$
\end_inset
.
Torej
\begin_inset Formula $\Ker\left(B\right)\not=0\Leftrightarrow\Ker\left(B^{*}\right)\not=0$
\end_inset
,
torej so lastne vrednosti
\begin_inset Formula $A^{*}$
\end_inset
konjugirane lastne vrednosti
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Remark*
Med lastnimi vektorji
\begin_inset Formula $A$
\end_inset
in lastnimi vektorji
\begin_inset Formula $A^{*}$
\end_inset
(žal) ni posebne zveze.
Primer:
\begin_inset Formula $A=\left[\begin{array}{cc}
1 & 2\\
i & 1
\end{array}\right]$
\end_inset
ima lastne vektorje
\begin_inset Formula $\vec{v_{1}}=\left[\begin{array}{c}
1-i\\
-1
\end{array}\right]$
\end_inset
in
\begin_inset Formula $\vec{v_{2}}=\left[\begin{array}{c}
1-i\\
1
\end{array}\right]$
\end_inset
,
\begin_inset Formula $A^{*}=\left[\begin{array}{cc}
1 & 1\\
2 & -i
\end{array}\right]$
\end_inset
pa lastne vektorje
\begin_inset Formula $\vec{v_{1}'}=\left[\begin{array}{c}
1-i\\
-2
\end{array}\right]$
\end_inset
in
\begin_inset Formula $\vec{v_{2}'}=\left[\begin{array}{c}
1-i\\
2
\end{array}\right]$
\end_inset
.
Med temi vektorji ni nobenih kolinearnosti.
Obstajajo pa zveze v nekaterih zanimivih primerih:
\end_layout
\begin_layout Claim*
Če matrika
\begin_inset Formula $A$
\end_inset
zadošča
\begin_inset Formula $A^{*}A=AA^{A}$
\end_inset
(pravimo
\begin_inset Formula $A$
\end_inset
je normalna),
iz
\begin_inset Formula $Av=\lambda v$
\end_inset
sledi
\begin_inset Formula $A^{*}v=\overline{\lambda}v$
\end_inset
,
torej imata
\begin_inset Formula $A$
\end_inset
in
\begin_inset Formula $A^{*}$
\end_inset
iste lastne vrednosti.
\end_layout
\begin_layout Proof
Če velja
\begin_inset Formula $Av=\lambda v$
\end_inset
,
velja
\begin_inset Formula $Av-\lambda v=\left(A-\lambda I\right)v=Bv=0\Rightarrow v\in\Ker B$
\end_inset
.
Če velja
\begin_inset Formula $A^{*}v=\overline{\lambda}v$
\end_inset
,
velja
\begin_inset Formula $A^{*}v-\overline{\lambda}v=\left(A^{*}-\overline{\lambda}I\right)v=B^{*}v=0\Rightarrow v\in\Ker B^{*}$
\end_inset
.
Dokazati je treba še
\begin_inset Formula $\Ker B=\Ker B^{*}$
\end_inset
:
\end_layout
\begin_deeper
\begin_layout Enumerate
Ali velja
\begin_inset Formula $A^{*}A=AA^{*}\Rightarrow B^{*}B=BB^{*}$
\end_inset
?
Ja.
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $B^{*}B=\left(A^{*}-\overline{\lambda}I\right)\left(A-\lambda I\right)=A^{*}A-\overline{\lambda}A-\lambda A^{*}+\overline{\lambda}\lambda I$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $BB^{*}=\left(A-\lambda I\right)\left(A^{*}-\overline{\lambda}I\right)=AA^{*}-\overline{\lambda}A-\lambda A^{*}+\lambda\overline{\lambda}I$
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
Ali velja
\begin_inset Formula $B^{*}B=BB^{*}\Rightarrow\Ker B=\Ker B^{*}$
\end_inset
?
Iz
\begin_inset Formula $B^{*}B=BB^{*}$
\end_inset
sledi
\begin_inset Formula $\Ker\left(B^{*}B\right)=\Ker\left(BB^{*}\right)\Rightarrow\Ker\left(B\right)=\Ker\left(B^{*}\right)$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\Ker B=\Ker B^{*}\Rightarrow\forall v\in V:Av=\lambda v\Leftrightarrow A^{*}v=\overline{\lambda}v$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsubsection
Normalne matrike
\end_layout
\begin_layout Definition*
\begin_inset Formula $A$
\end_inset
je normalna
\begin_inset Formula $\Leftrightarrow A^{*}A=AA^{*}$
\end_inset
.
\end_layout
\begin_layout Remark*
Dokazali smo že,
da za normalne matrike velja,
da imata
\begin_inset Formula $A$
\end_inset
in
\begin_inset Formula $A^{*}$
\end_inset
iste lastne vektorje,
kar v splošnem ne velja.
\end_layout
\begin_layout Claim*
Lastni vektorji,
ki pripadajo različnim lastnim vrednostim normalne matrike,
so paroma ortogonalni.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $A^{*}A=AA^{*}$
\end_inset
za neko
\begin_inset Formula $A$
\end_inset
in naj bo
\begin_inset Formula $Au=\lambda u$
\end_inset
in
\begin_inset Formula $Av=\mu v$
\end_inset
in
\begin_inset Formula $\mu\not=\lambda$
\end_inset
.
\begin_inset Formula $u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$
\end_inset
.
Računajmo:
\begin_inset Formula
\[
\mu\left\langle u,v\right\rangle =\left\langle u,\overline{\mu}v\right\rangle =\left\langle u,A^{*}v\right\rangle =\left\langle Au,v\right\rangle =\left\langle \lambda u,v\right\rangle =\lambda\left\langle u,v\right\rangle
\]
\end_inset
\begin_inset Formula
\[
\left(\mu-\lambda\right)\left\langle u,v\right\rangle =0\wedge u\not=\lambda\Rightarrow\left\langle u,v\right\rangle =0\Leftrightarrow u\perp v
\]
\end_inset
\end_layout
\begin_layout Claim*
Vsako normalno matriko se da diagonalizirati.
\end_layout
\begin_layout Proof
Dokažimo,
da je jordanska forma normalne matrike diagonalna
\begin_inset Formula $\Leftrightarrow$
\end_inset
vsi korenski podprostori so lastni.
\begin_inset Formula $\forall m,\lambda:\Ker\left(A-I\lambda\right)^{m}=\Ker\left(A-I\lambda\right)$
\end_inset
.
Zadošča dokazati za
\begin_inset Formula $m=2$
\end_inset
.
Naj bo
\begin_inset Formula $m=2$
\end_inset
in
\begin_inset Formula $B=A-I\lambda$
\end_inset
.
Dokažimo
\begin_inset Formula $\Ker B^{2}=\Ker B$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Enumerate
Če v
\begin_inset Formula $\Ker\left(A\right)=\Ker\left(A^{*}A\right)$
\end_inset
vstavimo
\begin_inset Formula $A=B^{2}$
\end_inset
,
dobimo
\begin_inset Formula $\Ker B^{2}=\Ker\left(\left(B^{2}\right)^{*}B^{2}\right)$
\end_inset
.
\end_layout
\begin_layout Enumerate
Ker je
\begin_inset Formula $A$
\end_inset
normalna,
je
\begin_inset Formula $B$
\end_inset
normalna,
torej
\begin_inset Formula $\left(B^{2}\right)^{*}B^{2}=B^{*}B^{*}BB=B^{*}BB^{*}B=\left(B^{*}B\right)^{2}$
\end_inset
.
Torej
\begin_inset Formula $\Ker\left(\left(B^{2}\right)^{*}B^{2}\right)=\Ker\left(B^{*}B\right)^{2}$
\end_inset
.
\end_layout
\begin_layout Enumerate
Če v
\begin_inset Formula $\Ker A^{*}A=\Ker A$
\end_inset
vstavimo
\begin_inset Formula $A=B^{*}B$
\end_inset
,
dobimo
\begin_inset Formula $\Ker B^{*}BB^{*}B=\Ker\left(B^{*}B\right)^{2}=\Ker B^{*}B$
\end_inset
.
\end_layout
\begin_layout Enumerate
Zopet upoštevamo
\begin_inset Formula $\Ker A^{*}A=\Ker A$
\end_inset
,
torej
\begin_inset Formula $\Ker\left(B^{*}B\right)=\Ker B$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Proof
Ko dokažemo
\begin_inset Formula $B$
\end_inset
normalna
\begin_inset Formula $\Rightarrow B^{*}$
\end_inset
normalna,
bo iz
\begin_inset Formula $\Ker B^{2}=\Ker B$
\end_inset
sledilo
\begin_inset Formula $\Ker B^{4}=\Ker B$
\end_inset
.
Preverimo,
a je
\begin_inset Formula $B^{2}$
\end_inset
normalna,
če je
\begin_inset Formula $B$
\end_inset
normalna:
\begin_inset Formula $\left(B^{2}\right)^{*}B^{2}=B^{*}B^{*}BB=B^{*}BB^{*}B=BB^{*}BB^{*}=BBB^{*}B^{*}=B^{2}\left(B^{2}\right)^{*}$
\end_inset
.
Sedaj vemo
\begin_inset Formula $\Ker B=\Ker B^{2}=\Ker B^{4}=\Ker B^{8}=\cdots$
\end_inset
.
Vemo pa tudi,
da
\begin_inset Formula
\[
\Ker B\subseteq\Ker B^{2}\subseteq\Ker B^{3}\subseteq\Ker B^{4}\subseteq\Ker B^{5}\subseteq\Ker B^{6}\subseteq\cdots
\]
\end_inset
\begin_inset Formula
\[
\Ker B\subseteq\Ker B\subseteq\Ker B^{3}\subseteq\Ker B\subseteq\Ker B^{5}\subseteq\Ker B\subseteq\cdots
\]
\end_inset
\begin_inset Formula
\[
\Ker B=\Ker B^{2}=\Ker B^{3}=\Ker B^{4}=\Ker B^{5}=\cdots
\]
\end_inset
\begin_inset Formula
\[
\forall v:\Ker B^{m}=\Ker B
\]
\end_inset
\end_layout
\begin_layout Remark*
Torej za vsako normalno matriko
\begin_inset Formula $A\exists$
\end_inset
diagonalna
\begin_inset Formula $D$
\end_inset
in obrnljiva
\begin_inset Formula $P$
\end_inset
z ortonormiranimi stolpci,
da velja
\begin_inset Formula $AP=PD$
\end_inset
,
\begin_inset Formula $A=PDP^{-1}$
\end_inset
.
Diagonala
\begin_inset Formula $D$
\end_inset
so lastne vrednosti
\begin_inset Formula $A$
\end_inset
,
stolpci
\begin_inset Formula $P$
\end_inset
pa so njeni lastni vektorji.
Lastni podprostori
\begin_inset Formula $\left(A-\lambda_{1}I\right),\dots,\left(A-\lambda_{n}I\right)$
\end_inset
so medsebojno pravokotni.
Izberimo ONB za vsak lasten podprostor.
Unija teh ONB je ONB za
\begin_inset Formula $F^{n}$
\end_inset
.
\begin_inset Formula $F^{n}=\Ker\left(A-\lambda_{1}I\right)\oplus\cdots\oplus\Ker\left(A-\lambda_{n}I\right)$
\end_inset
.
Ta ONB so stolpci matrike
\begin_inset Formula $P$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Ortogonalne/unitarne matrike
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $A$
\end_inset
kvadratna z ON stolpci glede na standardni skalarni produkt.
Pravimo,
da je
\begin_inset Formula $A$
\end_inset
unitarna (v kompleksnem primer) oziroma ortogonalna (v realnem primeru).
\end_layout
\begin_layout Claim*
Za unitarno
\begin_inset Formula $A$
\end_inset
velja
\begin_inset Formula $A^{*}A=AA^{*}=I$
\end_inset
.
\end_layout
\begin_layout Proof
Dokazujmo za unitarno.
Za ortogonalno je dokaz podoben.
Naj bo
\begin_inset Formula $A=\left[\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots & & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{array}\right]$
\end_inset
unitarna.
To pomeni,
da za vsaka stolpca
\begin_inset Formula $a_{i}=\left(a_{1i},\dots,a_{ni}\right)$
\end_inset
in
\begin_inset Formula $a_{j}=\left(a_{1j},\dots,a_{nj}\right)$
\end_inset
velja za vsak
\begin_inset Formula $i,j\in\left\{ 1..n\right\} $
\end_inset
velja
\begin_inset Formula $\left\langle \text{\left[\begin{array}{c}
a_{1i}\\
\text{\ensuremath{\vdots}}\\
a_{ni}
\end{array}\right],\left[\begin{array}{c}
a_{1j}\\
\vdots\\
a_{nj}
\end{array}\right]}\right\rangle =a_{1i}\overline{a_{1j}}+\cdots+a_{ni}\overline{a_{nj}}=\begin{cases}
0 & ;i\not=j\\
1 & ;i=j
\end{cases}$
\end_inset
.
Oglejmo si
\begin_inset Formula
\[
A^{*}A=\left[\begin{array}{ccc}
\overline{a_{11}} & \cdots & \overline{a_{n1}}\\
\vdots & & \vdots\\
\overline{a_{1n}} & \cdots & \overline{a_{nn}}
\end{array}\right]\left[\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots & & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{array}\right]=\left[\begin{array}{ccc}
1 & & 0\\
& \ddots\\
0 & & 1
\end{array}\right]
\]
\end_inset
Očitno je res,
ker je vsak element
\begin_inset Formula $A^{*}A$
\end_inset
konstruiran s skalarnim množenjem vrstice leve matrike (konjugirani stolpci
\begin_inset Formula $A$
\end_inset
,
ker smo poprej matriko transponiralo) in stolpca desne,
za kar predpis smo poprej že razbrali.
\end_layout
\begin_layout Remark*
Za nekvadratne unitarne velja le
\begin_inset Formula $A^{*}A=I$
\end_inset
,
\begin_inset Formula $AA^{*}=I$
\end_inset
pa zaradi nezmožnosti množenja zaradi nepravilnih dimenzij seveda ne velja.
\end_layout
\begin_layout Claim*
Naslednje trditve so za
\begin_inset Formula $P$
\end_inset
z ortonormiranimi stolpci ekvivalentne:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $P^{*}P=I$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall u,v:\left\langle Pu,Pv\right\rangle =\left\langle u,v\right\rangle $
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall u:\left|\left|Pu\right|\right|=\left|\left|u\right|\right|$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset
ONB
\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} :\left\{ Pu_{1},\dots,Pu_{n}\right\} $
\end_inset
je ON množica
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\exists$
\end_inset
ONB
\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} :\left\{ Pu_{1},\dots,Pu_{n}\right\} $
\end_inset
je ON množica
\end_layout
\end_deeper
\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(1\Rightarrow2\right)$
\end_inset
\begin_inset Formula $\left\langle Pu,Pv\right\rangle =\left\langle u,P^{*}Pv\right\rangle =\left\langle u,v\right\rangle $
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(2\Rightarrow3\right)$
\end_inset
\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left\langle Pu,Pu\right\rangle =\left\langle u,u\right\rangle =\left|\left|u\right|\right|^{2}$
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(2\Rightarrow1\right)$
\end_inset
\begin_inset Formula
\[
\forall u,v:\left\langle Pu,Pv\right\rangle =\left\langle u,v\right\rangle \Rightarrow\left\langle u,P^{*}Pv\right\rangle -\left\langle u,v\right\rangle =0\Rightarrow\left\langle u,\left(P^{*}P-I\right)v\right\rangle =0
\]
\end_inset
Sedaj izberimo
\begin_inset Formula $u=\left(P^{*}P-I\right)v$
\end_inset
:
\begin_inset Formula $\left\langle \left(P^{*}P-I\right)v,\left(P^{*}P-I\right)v\right\rangle =0\Rightarrow P^{*}P-I=0\Rightarrow P^{*}P=0$
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(3\Rightarrow2\right)$
\end_inset
Po predpostavki
\begin_inset Formula $\forall u:\left|\left|Pu\right|\right|=\left|\left|u\right|\right|$
\end_inset
Izrazimo skalarni produkt z normo:
\begin_inset Formula $\left\langle u,v\right\rangle =\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}$
\end_inset
,
torej
\begin_inset Formula
\[
\left\langle Pu,Pv\right\rangle =\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|Pu+i^{k}Pv\right|\right|^{2}=\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|P\left(u+i^{k}v\right)\right|\right|^{2}\overset{\text{predpostavka}}{=}\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}=\left\langle u,v\right\rangle
\]
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(5\Rightarrow4\right)$
\end_inset
Vzemimo poljuben
\begin_inset Formula $u$
\end_inset
in ga razvijmo po ONB
\begin_inset Formula $u_{1},\dots,u_{n}$
\end_inset
.
Tedaj
\begin_inset Formula $u=\alpha_{1}u_{1}+\cdots+\alpha_{n}u_{n}$
\end_inset
.
Ker so
\begin_inset Formula $u_{i}$
\end_inset
ONB,
velja
\begin_inset Formula $\left|\left|u\right|\right|^{2}=\left|\alpha_{1}\right|^{2}+\cdots\left|\alpha_{n}\right|^{2}$
\end_inset
.
Ker so
\begin_inset Formula $Pu_{1}$
\end_inset
ONB po predpostavki,
\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left|\alpha_{1}\right|^{2}+\cdots+\left|\alpha_{n}\right|^{2}$
\end_inset
,
torej velja
\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left|\left|u\right|\right|^{2}$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(2\Rightarrow4\right)$
\end_inset
Ker so
\begin_inset Formula $u_{1},\dots,u_{n}$
\end_inset
ONM,
velja
\begin_inset Formula $\left\langle u_{i},u_{j}\right\rangle =\begin{cases}
1 & ;i=j\\
0 & ;i\not=j
\end{cases}$
\end_inset
.
Tudi
\begin_inset Formula $Pu_{1},\dots,Pu_{n}$
\end_inset
ortonormirana,
kajti po predpostavki
\begin_inset Formula $2$
\end_inset
velja
\begin_inset Formula $\left\langle Pu_{i},Pu_{2}\right\rangle =\begin{cases}
1 & ;i=j\\
0 & ;i\not=j
\end{cases}$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(4\Rightarrow5\right)$
\end_inset
Očitno.
\end_layout
\end_deeper
\begin_layout Claim*
Lastne vrednosti unitarne matrike
\begin_inset Formula $A$
\end_inset
se nahajajo na enotski krožnici v
\begin_inset Formula $\Im$
\end_inset
.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $A$
\end_inset
unitarna in naj bo
\begin_inset Formula $v$
\end_inset
tak,
da
\begin_inset Formula $Av=\lambda v$
\end_inset
.
Tedaj
\begin_inset Formula $\left\langle v,v\right\rangle =\left\langle Av,Av\right\rangle =\left\langle \lambda v,\lambda v\right\rangle =\lambda\overline{\lambda}\left\langle v,v\right\rangle \Rightarrow\lambda\overline{\lambda}=1\Rightarrow\left|\lambda\right|=1\Rightarrow\lambda=e^{i\varphi}$
\end_inset
za nek
\begin_inset Formula $\varphi$
\end_inset
.
\end_layout
\begin_layout Remark*
Iz unitarnosti sledi normalnost,
zato so lastni vektorji unitarne matrike,
ki pripadajo paroma različnim lastnim vrednostim,
pravokotni (isto,
kot pri normalnih matrikah).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
Prav tako kot pri normalnih matrikah lahko unitarne diagonalitziramo v tokrat ortogonalni bazi.
Pri unitarnih so stolpci
\begin_inset Formula $P$
\end_inset
še celo normirani.
\begin_inset Formula $A=PDP^{-1}$
\end_inset
,
kjer je
\begin_inset Formula $P$
\end_inset
unitarna,
torej
\begin_inset Formula $P^{*}=P^{-1}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
Očitno je,
da če je
\begin_inset Formula $A$
\end_inset
unitarna,
velja
\begin_inset Formula $A^{*}=A^{-1}$
\end_inset
.
\end_layout
\begin_layout Subsubsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Simetrične/hermitske-matrike"
\end_inset
Simetrične/hermitske matrike
\end_layout
\begin_layout Definition*
Matrika nad
\begin_inset Formula $\mathbb{R}$
\end_inset
je simetrična,
če zanjo velja
\begin_inset Formula $A^{*}=A$
\end_inset
.
Matrika nad
\begin_inset Formula $\mathbb{C}$
\end_inset
je hermitska,
če zanjo velja
\begin_inset Formula $A^{*}=A$
\end_inset
.
Linearni preslikavi,
pripadajoči hermitski/simetrični matriki,
pravimo sebiadjungirana.
\end_layout
\begin_layout Fact*
Vsaka hermitska/simetrična matrika je normalna,
kajti
\begin_inset Formula $A^{*}A=AA=AA^{*}$
\end_inset
.
\end_layout
\begin_layout Claim*
Lastne vrednosti hermitskih/simetričnih matrik so realne.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $A=A^{*}$
\end_inset
in naj bo
\begin_inset Formula $Av=\lambda v$
\end_inset
za nek neničeln
\begin_inset Formula $v$
\end_inset
.
Tedaj
\begin_inset Formula $\lambda\left\langle v,v\right\rangle =\left\langle \lambda v,v\right\rangle =\left\langle Av,v\right\rangle =\left\langle v,A^{*}v\right\rangle =\left\langle v,Av\right\rangle =\left\langle v,\lambda v\right\rangle =\overline{\lambda}\left\langle v,v\right\rangle $
\end_inset
.
Potemtakem
\begin_inset Formula $\lambda=\overline{\lambda}\Rightarrow\lambda\in\mathbb{R}$
\end_inset
.
\end_layout
\begin_layout Remark*
Diagonalizacija je zopet enaka kot pri normalnih matrikah z dodatkom —
vsaka hermitska matrika je podobna realni diagonalni,
kar za normalne ni res —
normalne so lahko podobne kompleksnim diagonalnim matrikam.
\end_layout
\begin_layout Subsubsection
Pozitivno (semi)definitne matrike
\end_layout
\begin_layout Definition*
\begin_inset Formula $A$
\end_inset
je pozitivno semidefinitna
\begin_inset Formula $\sim A\geq0\Leftrightarrow A=A^{*}\wedge\forall v:\left\langle Av,v\right\rangle \geq0$
\end_inset
.
\begin_inset Formula $A$
\end_inset
je pozitivno definitna
\begin_inset Formula $\sim A>0\Leftrightarrow A=A^{*}\wedge\forall v\not=0:\left\langle Av,v\right\rangle >0$
\end_inset
.
S tem ko skalarni produkt primerjamo (
\begin_inset Formula $>,\geq$
\end_inset
),
implicitno zahtevamo njegovo realnost.
Primerjalni operatorji namreč na kompleksnih številih niso definirani.
\end_layout
\begin_layout Example*
Vzemimo poljubno nenujno kvadratno
\begin_inset Formula $B$
\end_inset
in definirajmo
\begin_inset Formula $A=B^{*}B$
\end_inset
.
Potem je
\begin_inset Formula $A$
\end_inset
pozitivno semidefinitna,
kajti
\begin_inset Formula $A^{*}=\left(B^{*}B\right)^{*}=B^{*}B=A$
\end_inset
in
\begin_inset Formula $\left\langle Av,v\right\rangle =\left\langle B^{*}Bv,v\right\rangle =\left\langle Bv,Bv\right\rangle \geq0$
\end_inset
.
Če pa bi bili stolpci
\begin_inset Formula $B$
\end_inset
linearno neodvisni,
pa bi veljalo
\begin_inset Formula $\forall v:v\not=0\Rightarrow\left\langle Av,v\right\rangle =\left\langle B^{*}Bv\right\rangle =\left\langle Bv,Bv\right\rangle >0$
\end_inset
.
\end_layout
\begin_layout Claim*
\begin_inset Formula $A\geq0\Rightarrow$
\end_inset
lastne vrednosti
\begin_inset Formula $A$
\end_inset
so
\begin_inset Formula $\geq0$
\end_inset
.
\begin_inset Formula $A>0\Rightarrow$
\end_inset
lastne vrednosti
\begin_inset Formula $A$
\end_inset
so
\begin_inset Formula $>0$
\end_inset
.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $\lambda$
\end_inset
lastna vrednost
\begin_inset Formula $A$
\end_inset
in
\begin_inset Formula $A\geq0$
\end_inset
.
Tedaj
\begin_inset Formula $Av=\lambda v$
\end_inset
za nek
\begin_inset Formula $v\not=0$
\end_inset
.
Torej
\begin_inset Formula $\left\langle Av,v\right\rangle =\left\langle \lambda v,v\right\rangle =\lambda\left\langle v,v\right\rangle $
\end_inset
.
Toda ker
\begin_inset Formula $\left\langle Av,v\right\rangle \geq0$
\end_inset
,
sledi
\begin_inset Formula $\lambda\left\langle v,v\right\rangle \geq0$
\end_inset
.
Ker je
\begin_inset Formula $\left\langle v,v\right\rangle >0$
\end_inset
,
sledi
\begin_inset Formula $\lambda\geq0$
\end_inset
.
Analogno za
\begin_inset Formula $A>0$
\end_inset
.
\end_layout
\begin_layout Remark*
Diagonalizacija je ista kot za normalna,
s tem da za diagonalno
\begin_inset Formula $D$
\end_inset
velja še,
da je pozitivno (semi)definitna,
ko je
\begin_inset Formula $A$
\end_inset
pozitivno semidefinitna.
\end_layout
\begin_layout Claim*
\begin_inset Formula $\forall A\geq0\exists B=B^{*},B\geq0\ni:B^{2}=A$
\end_inset
.
ZDB Za vsako pozitivno semidefinitno matriko
\begin_inset Formula $A$
\end_inset
obstajaja taka unitarna pozitivno semidefinitna
\begin_inset Formula $B$
\end_inset
,
da velja
\begin_inset Formula $B^{2}=A$
\end_inset
.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $A=PDP^{-1}$
\end_inset
in
\begin_inset Formula $P^{*}=P^{-1}$
\end_inset
in
\begin_inset Formula $D=\left[\begin{array}{ccc}
\lambda_{1} & & 0\\
& \ddots\\
0 & & \lambda_{n}
\end{array}\right]$
\end_inset
.
Definirajmo
\begin_inset Formula $E=\left[\begin{array}{ccc}
\sqrt{\lambda_{1}} & & 0\\
& \ddots\\
0 & & \sqrt{\lambda_{n}}
\end{array}\right]\geq0$
\end_inset
.
Naj bo
\begin_inset Formula $B=PEP^{-1}=PEP^{*}$
\end_inset
.
Opazimo
\begin_inset Formula $B^{*}=B$
\end_inset
,
kajti
\begin_inset Formula $\left(PEP^{-1}\right)^{*}=\left(PEP^{*}\right)^{*}=PE^{*}P^{*}=PEP^{-1}=PEP^{*}$
\end_inset
,
ker je
\begin_inset Formula $E^{*}=E$
\end_inset
,
ker je
\begin_inset Formula $\forall a\in\mathbb{R}:\sqrt{a}\in\mathbb{R}$
\end_inset
.
Oglejmo si
\begin_inset Formula $B^{2}=PEP^{-1}PEP^{-1}=PE^{2}P^{-1}=PDP^{-1}$
\end_inset
.
Tako definiramo
\begin_inset Formula $\sqrt{A}=B$
\end_inset
(tu
\begin_inset Formula $\sqrt{}$
\end_inset
ni funkcija,
kot pri JKF,
temveč nov operator).
\end_layout
\begin_layout Claim*
Naslednje trditve so ekvivalentne (zamenjamo lahko
\begin_inset Formula $\geq$
\end_inset
in
\begin_inset Formula $>$
\end_inset
):
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $A\geq0$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A=A^{*}$
\end_inset
in vse lastne vrednosti so
\begin_inset Formula $\geq0$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A=PDP^{-1}$
\end_inset
za nek unitaren
\begin_inset Formula $P$
\end_inset
in diagonalen
\begin_inset Formula $D\geq0$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A=A^{*}$
\end_inset
in obstaja
\begin_inset Formula $\sqrt{A}$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A=B^{*}B$
\end_inset
za neko nenujno kvadratno matriko
\begin_inset Formula $B$
\end_inset
(za pozitivno definitno zahtevamo,
da ima
\begin_inset Formula $B$
\end_inset
LN stolpce).
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Claim*
klasifikacija skalarnih produktov na
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
in
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
.
Naj bo
\begin_inset Formula $\left\langle u,v\right\rangle $
\end_inset
standardni skalarni produkt na
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
.
\begin_inset Formula $u=\left(\alpha_{1},\dots,\alpha_{n}\right)$
\end_inset
in
\begin_inset Formula $v=\left(\beta_{1},\dots,\beta_{n}\right)$
\end_inset
in velja
\begin_inset Formula $\left\langle u,v\right\rangle =\alpha_{1}\overline{\beta_{1}}+\cdots+\alpha_{n}\overline{\beta_{n}}=\left[\begin{array}{ccc}
\overline{\beta_{1}} & \cdots & \overline{\beta_{n}}\end{array}\right]\left[\begin{array}{c}
\alpha_{1}\\
\vdots\\
\alpha_{n}
\end{array}\right]=v^{*}\cdot u$
\end_inset
.
Za
\begin_inset Formula $A>0$
\end_inset
definirajmo
\begin_inset Formula $\left[u,v\right]=\left\langle Au,v\right\rangle =v^{*}Au$
\end_inset
.
Trdimo,
da je
\begin_inset Formula $\left[\cdot,\cdot\right]$
\end_inset
spet skalarni produkt na
\begin_inset Formula $\mathbb{R}^{n}/\mathbb{C}^{n}$
\end_inset
in da je vsak skalarni produkt v
\begin_inset Formula $\mathbb{R}^{n}/\mathbb{C}^{n}$
\end_inset
take oblike.
\end_layout
\begin_layout Proof
Dokazujemo oba dela trditve:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\left[\cdot,\cdot\right]$
\end_inset
je skalarni produkt
\end_layout
\begin_deeper
\begin_layout Enumerate
pozitivna semidefinitnost:
\begin_inset Formula $\forall u\not=0:\left[u,u\right]=\left\langle Au,u\right\rangle \geq0$
\end_inset
.
\end_layout
\begin_layout Enumerate
konjutirana simetričnost:
\begin_inset Formula $\forall u,v:\left[u,v\right]=\left\langle Au,v\right\rangle =\left\langle u,A^{*}v\right\rangle =\left\langle u,Av\right\rangle =\overline{\left\langle Av,u\right\rangle }=\overline{\left[v,u\right]}$
\end_inset
.
\end_layout
\begin_layout Enumerate
Linearnost in homogenost:
\begin_inset Formula $\forall\alpha_{1},\alpha_{2},u_{1}u_{2},v:\left[\alpha_{1}u_{1}+\alpha_{2}u_{2},v\right]=\left\langle A\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right),v\right\rangle =\left\langle \alpha_{1}Au_{1}+\alpha_{2}Au_{2},v\right\rangle =\alpha_{1}\left\langle Au_{1},v\right\rangle +\alpha_{2}\left\langle Au_{2},v\right\rangle =\alpha_{1}\left[u,v\right]+\alpha_{2}\left[u,v\right]$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Enumerate
Za vsak skalarni produkt
\begin_inset Formula $\left[\cdot,\cdot\right]$
\end_inset
na
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
obstaja taka pozitivno definitna matrika
\begin_inset Formula $A$
\end_inset
,
da velja
\begin_inset Formula $\forall u,v\in\mathbb{C}^{n}:\left[u,v\right]=\left\langle Au,v\right\rangle =v^{*}Au$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Standard
Naj bo
\begin_inset Formula $e_{1},\dots,e_{n}$
\end_inset
standardna baza za
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
.
Definirajmo
\begin_inset Formula $A=\left[\begin{array}{ccc}
\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\
\vdots & & \vdots\\
\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right]
\end{array}\right]$
\end_inset
.
Velja
\begin_inset Formula $A=A^{*}$
\end_inset
:
\begin_inset Formula
\[
A^{*}=\left[\begin{array}{ccc}
\overline{\left[e_{1},e_{1}\right]} & \cdots & \overline{\left[e_{1},e_{n}\right]}\\
\vdots & & \vdots\\
\overline{\left[e_{n},e_{1}\right]} & \cdots & \overline{\left[e_{n},e_{n}\right]}
\end{array}\right]=\left[\begin{array}{ccc}
\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\
\vdots & & \vdots\\
\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right]
\end{array}\right]=A
\]
\end_inset
Preveriti je treba še
\begin_inset Formula $\forall u,v\in\mathbb{C}^{n}:\left[u,v\right]=v^{*}Au$
\end_inset
.
\begin_inset Formula $u=\alpha_{1}e_{1}+\cdots+\alpha_{n}e_{n}$
\end_inset
in
\begin_inset Formula $v=\beta_{1}e_{1}+\cdots+\beta_{n}e_{n}$
\end_inset
.
Tedaj je
\begin_inset Formula
\[
\left[u,v\right]=\left[\alpha_{1}e_{1}+\cdots+\alpha_{n}e_{n},\beta_{1}e_{1}+\cdots+\beta_{n}e_{n}\right]=
\]
\end_inset
\begin_inset Formula
\[
=\left(\alpha_{1}\overline{\beta_{1}}\left[e_{1},e_{1}\right]+\cdots+\alpha_{1}\overline{\beta_{n}}\left[e_{1},e_{n}\right]\right)+\cdots+\left(\alpha_{n}\overline{\beta_{1}}\left[e_{n},e_{1}\right]+\cdots+\alpha_{n}\overline{\beta_{n}}\left[e_{n},e_{n}\right]\right)=
\]
\end_inset
\begin_inset Formula
\[
=\left[\begin{array}{ccc}
\overline{\beta_{1}} & \cdots & \overline{\beta_{n}}\end{array}\right]\left[\begin{array}{ccc}
\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\
\vdots & & \vdots\\
\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right]
\end{array}\right]\left[\begin{array}{c}
\alpha_{1}\\
\vdots\\
\alpha_{n}
\end{array}\right]=v^{*}Au=\left\langle Au,v\right\rangle
\]
\end_inset
Da je
\begin_inset Formula $A$
\end_inset
pozitivno definitna sledi,
saj mora za vsak neničeln
\begin_inset Formula $u$
\end_inset
po aksiomu za pozitivno definitnost skalarnega produkta veljati
\begin_inset Formula $\left\langle Au,u\right\rangle >0$
\end_inset
.
\end_layout
\end_deeper
\end_deeper
\begin_layout Subsubsection
Singularni razcep (angl.
singular value decomposition —
SVD)
\end_layout
\begin_layout Standard
Naj bo
\begin_inset Formula $A_{n\times n}$
\end_inset
neka kompleksna ali realna matrika.
Tedaj je
\begin_inset Formula $A^{*}A$
\end_inset
hermitska (
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:Simetrične/hermitske-matrike"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
) matrika dimenzij
\begin_inset Formula $n\times n$
\end_inset
.
Ker je
\begin_inset Formula $\forall u:\left\langle A^{*}Au,u\right\rangle =\left\langle Au,Au\right\rangle \geq0$
\end_inset
,
je
\begin_inset Formula $A^{*}A$
\end_inset
pozitivno semidefinitna,
torej so vse njene lastne vrednosti
\begin_inset Formula $\geq0$
\end_inset
.
\end_layout
\begin_layout Definition*
Singularne vrednosti
\begin_inset Formula $A$
\end_inset
so kvadratni koreni lastnih vrednosti
\begin_inset Formula $A^{*}A$
\end_inset
.
\end_layout
\begin_layout Example*
Če je
\begin_inset Formula $A$
\end_inset
normalna in
\begin_inset Formula $\lambda$
\end_inset
lastna vrednost
\begin_inset Formula $A$
\end_inset
,
obstaja tak
\begin_inset Formula $v\not=0\ni:Av=\lambda v\Rightarrow A^{*}v=\overline{\lambda}v$
\end_inset
.
Odtod sledi,
da je
\begin_inset Formula $A^{*}Av=A^{*}\lambda v=\lambda A^{*}v=\lambda\overline{\lambda}v$
\end_inset
,
torej je
\begin_inset Formula $\lambda$
\end_inset
lastna vrednost matrike
\begin_inset Formula $A^{*}A$
\end_inset
.
Po definiciji singularne vrednosti je
\begin_inset Formula $\sqrt{\lambda\overline{\lambda}}=\sqrt{\left|\lambda\right|^{2}}=\left|\lambda\right|$
\end_inset
singularna vrednost matrike
\begin_inset Formula $A$
\end_inset
.
Potemtakem so singularne vrednostni normalnih matrik enake absolutnim vrednosti lastnih vrednosti.
\end_layout
\begin_layout Standard
Nekatere lastne vrednosti so ničelne,
nekatere pa od nič strogo večje.
Koliko je katerih?
Število ničelnih singularnih vrednosti matrike
\begin_inset Formula $A$
\end_inset
je število ničelnih lastnih vrednosti matrike
\begin_inset Formula $A^{*}A$
\end_inset
.
Ker je
\begin_inset Formula $A^{*}A$
\end_inset
hermitska,
je diagonalizabilna,
zato je algebraična večkratnost lastne vrednosti 0 enaka geometrijski večkratnosti lastne vrednosti 0,
slednja pa je definirana kot
\begin_inset Formula $\dim\Ker A^{*}A$
\end_inset
.
Ko upoštevamo
\begin_inset Formula $\dim\Ker A^{*}A=\dim\Ker A$
\end_inset
,
izvemo,
da je število ničelnih singularnih vrednosti matrike
\begin_inset Formula $A$
\end_inset
njena ničnost (
\begin_inset Formula $\n A$
\end_inset
).
Ker je
\begin_inset Formula $A^{*}A$
\end_inset
velikosti
\begin_inset Formula $n\times n$
\end_inset
,
ima
\begin_inset Formula $A$
\end_inset
\begin_inset Formula $n$
\end_inset
singularnih vrednosti,
torej je število neničelnih singularnih vrednosti
\begin_inset Formula $A$
\end_inset
enako
\begin_inset Formula $n-\Ker A$
\end_inset
.
Upoštevajoč osnovni dimenzijski izrek jedra in slike,
velja
\begin_inset Formula $\n A+\rang A=n$
\end_inset
,
torej je neničelnih singularnih vrednosti
\begin_inset Formula $=\rang A=\dim\Slika A$
\end_inset
.
\end_layout
\begin_layout Remark*
Za
\begin_inset Formula $m\times n$
\end_inset
matriko velja
\begin_inset Formula $\rang A\le\min\left\{ m,n\right\} $
\end_inset
.
\end_layout
\begin_layout Definition*
posplošitev pojma diagonalne matrike na nekvadratne matrike.
Matrika
\begin_inset Formula $D_{m\times n}$
\end_inset
je diagonalna,
če velja
\begin_inset Formula $\forall i:\left\{ 1..m\right\} ,j\in\left\{ 1..n\right\} :i\not=j\Rightarrow D_{i.j}=0$
\end_inset
.
\end_layout
\begin_layout Example*
Primeri pravokotnih diagonalnih matrik:
\begin_inset Formula
\[
\left[\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 2 & 0 & 0\\
0 & 0 & 3 & 0
\end{array}\right],\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 3\\
0 & 0 & 0
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Theorem*
singularni razcep.
Naj bo
\begin_inset Formula $A$
\end_inset
kompleksna
\begin_inset Formula $m\times n$
\end_inset
matrika.
Potem obstajata taki unitarni
\begin_inset Formula $Q_{1},Q_{2}$
\end_inset
in taka diagonalna
\begin_inset Formula $D$
\end_inset
z diagonalci
\begin_inset Formula $\geq0\ni:A=Q_{1}DQ_{2}^{-1}$
\end_inset
.
\end_layout
\begin_layout Remark*
Diagonalci
\begin_inset Formula $D$
\end_inset
so ravno singularne vrednosti matrike
\begin_inset Formula $A$
\end_inset
.
Ker je
\begin_inset Formula $Q_{2}$
\end_inset
unitarna,
je
\begin_inset Formula $Q_{2}^{*}=Q_{2}^{-1}\Rightarrow A=Q_{1}DQ_{2}^{*}$
\end_inset
.
Če
\begin_inset Formula $A=Q_{1}DQ_{2}^{*}$
\end_inset
,
je
\begin_inset Formula $A^{*}=Q_{2}^{**}D^{*}Q_{1}^{*}=Q_{2}D^{*}Q_{1}^{*}$
\end_inset
in
\begin_inset Formula $A^{*}A=Q_{2}D^{*}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{2}D^{*}DQ_{2}^{*}$
\end_inset
,
torej je
\begin_inset Formula $A^{*}A$
\end_inset
podobna
\begin_inset Formula $D^{*}D$
\end_inset
,
diagonalci
\begin_inset Formula $D^{*}D$
\end_inset
so lastne vrednosti
\begin_inset Formula $A^{*}A$
\end_inset
in stolpci
\begin_inset Formula $Q_{2}$
\end_inset
so lastni vektorji
\begin_inset Formula $A^{*}A$
\end_inset
.
Diagonalci
\begin_inset Formula $D$
\end_inset
so bodisi 0 bodisi kvadratni koreni od diagonalcev
\begin_inset Formula $D^{*}D$
\end_inset
,
torej kvadratni koreni lastnih vrednosti
\begin_inset Formula $A^{*}A$
\end_inset
,
torej singularne vrednosti od
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Proof
obstoj singularnega razcepa.
Konstruirajmo
\begin_inset Formula $Q_{1},D,Q_{2}$
\end_inset
in dokažimo veljavnost.
\end_layout
\begin_deeper
\begin_layout Itemize
Konstrukcija
\begin_inset Formula $Q_{2}$
\end_inset
:
\begin_inset Formula $A$
\end_inset
je
\begin_inset Formula $m\times n$
\end_inset
kompleksna.
Tvorimo
\begin_inset Formula $n\times n$
\end_inset
matriko
\begin_inset Formula $A^{*}A$
\end_inset
.
Izračunajmo lastne vrednosti
\begin_inset Formula $A^{*}A$
\end_inset
in jih uredimo padajoče —
\begin_inset Formula $\lambda_{1}\geq\cdots\geq\lambda_{n}$
\end_inset
.
Naj bodo
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
pripadajoči lastni vektorji —
\begin_inset Formula $A^{*}Av_{i}=\lambda_{i}v_{i}$
\end_inset
.
Te
\begin_inset Formula $v_{i}$
\end_inset
izberimo tako,
da so ortonormirani.
Lastni podprostori
\begin_inset Formula $A^{*}A$
\end_inset
so namreč paroma pravokotni,
saj je
\begin_inset Formula $A^{*}A$
\end_inset
normalna,
saj je hermitska.
V vsakem podprostoru vzamemo ONB in
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
je unija teh ON baz.
Definiramo
\begin_inset Formula $Q_{2}=\left[\begin{array}{ccc}
v_{1} & \cdots & v_{n}\end{array}\right]$
\end_inset
.
Ker so
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset
ON,
je
\begin_inset Formula $Q_{2}$
\end_inset
unitarna.
\end_layout
\begin_layout Itemize
Konstrukcija
\begin_inset Formula $D$
\end_inset
:
Naj bo
\begin_inset Formula $r\coloneqq\rang A$
\end_inset
(število ničelnih singularnih vrednosti
\begin_inset Formula $A$
\end_inset
).
Oglejmo si zaporedje lastnih vrednosti
\begin_inset Formula $A^{*}A$
\end_inset
\begin_inset Formula $\lambda_{1}\geq\cdots>\lambda_{r+1}=\cdots=\lambda_{n}$
\end_inset
.
Lastne vrednosti po
\begin_inset Formula $r$
\end_inset
so ničelne,
ostale pa večje od 0.
Lastne vrednosti
\begin_inset Formula $A^{*}A$
\end_inset
v tem vrstnem redu so singularne vrednosti matrike
\begin_inset Formula $A$
\end_inset
:
\begin_inset Formula $\sigma_{1}^{2}=\lambda_{1}\geq\cdots\geq\sigma_{r}^{2}=\lambda_{r}>\sigma_{r+1}^{2}=\cdots=\sigma_{n}^{2}=0$
\end_inset
.
Definiramo
\begin_inset Formula $D$
\end_inset
kot
\begin_inset Formula $m\times n$
\end_inset
diagonalno matriko takole:
\begin_inset Formula
\[
D=\left[\begin{array}{cccccc}
\sigma_{1} & & & & & 0\\
& \ddots\\
& & \sigma_{r}\\
& & & \sigma_{r+1}=0\\
& & & & \ddots\\
0 & & & & & \sigma_{n}=0
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Itemize
Konstrukcija
\begin_inset Formula $Q_{1}$
\end_inset
:
\begin_inset Formula $\forall i\in\left\{ 1..r\right\} :u_{i}\coloneqq\frac{1}{\sigma_{1}}Av_{i}$
\end_inset
za
\begin_inset Formula $v_{i}$
\end_inset
lastne vektorje
\begin_inset Formula $A^{*}A$
\end_inset
,
torej
\begin_inset Formula $A^{*}Av_{i}=\lambda_{i}v_{i}=\sigma_{i}^{2}v_{i}$
\end_inset
.
Pokažimo,
da je
\begin_inset Formula $u_{1},\dots,u_{r}$
\end_inset
ON množica.
\begin_inset Formula $\forall i,j\in\left\{ 1..r\right\} :$
\end_inset
\begin_inset Formula
\[
\left\langle u_{i},u_{j}\right\rangle =\left\langle \frac{1}{\sigma_{j}}Av_{i},\frac{1}{\sigma_{j}}Av_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle Av_{i},Av_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle A^{*}Av_{i},v_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle \lambda_{i}v_{i},v_{j}\right\rangle =\frac{\lambda_{i}=\sigma_{i}^{\cancel{2}}}{\cancel{\sigma}_{i}\sigma_{j}}\left\langle v_{i},v_{j}\right\rangle =
\]
\end_inset
\begin_inset Formula
\[
=\frac{\sigma_{i}}{\sigma_{j}}\left\langle v_{i},v_{j}\right\rangle =\begin{cases}
0 & ;i\not=j\\
1 & ;i=j
\end{cases}
\]
\end_inset
Sedaj ONM
\begin_inset Formula $u_{1},\dots,u_{r}$
\end_inset
z
\begin_inset Formula $u_{r+1},\dots,u_{n}$
\end_inset
dopolnimo do ONB za
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
(GS).
Definiramo
\begin_inset Formula $Q_{1}=\left[\begin{array}{ccc}
u_{1} & \cdots & u_{n}\end{array}\right]$
\end_inset
.
Ker so stolpci ONB,
je matrika unitarna.
\end_layout
\begin_layout Itemize
Sedaj preverimo,
da velja
\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}\Leftrightarrow AQ_{2}=Q_{1}D$
\end_inset
.
\begin_inset Formula
\[
AQ_{2}=A\left[\begin{array}{ccc}
v_{1} & \cdots & v_{n}\end{array}\right]=\left[\begin{array}{ccc}
Av_{1} & \cdots & Av_{n}\end{array}\right]=\cdots
\]
\end_inset
Upoštevamo,
da
\begin_inset Formula $i>r\Rightarrow\lambda_{i}=0\Rightarrow A^{*}Av_{i}=\lambda_{i}v_{i}=0\Rightarrow v_{i}\in\Ker A^{*}A\Rightarrow v_{i}\in\Ker A\Leftrightarrow Av_{i}=0$
\end_inset
:
\begin_inset Formula
\[
\cdots=\left[\begin{array}{ccc}
Av_{1} & \cdots & Av_{n}\end{array}\right]=\left[\begin{array}{cccccc}
Av_{1} & \cdots & Av_{r} & 0 & \cdots & 0\end{array}\right]
\]
\end_inset
Sedaj izračunajmo še
\begin_inset Formula
\[
Q_{1}D=\left[\begin{array}{ccc}
u_{1} & \cdots & u_{n}\end{array}\right]\left[\begin{array}{cccccc}
\sigma_{1}\\
& \ddots\\
& & \sigma_{r}\\
& & & 0\\
& & & & \ddots\\
& & & & & 0
\end{array}\right]=\left[\begin{array}{cccccc}
\sigma_{1}u_{1} & \cdots & \sigma_{r}u_{r} & 0 & \cdots & 0\end{array}\right]=
\]
\end_inset
\begin_inset Formula
\[
=\left[\begin{array}{cccccc}
\cancel{\sigma_{1}\frac{1}{\sigma_{1}}}Av_{i} & \cdots & \cancel{\sigma_{r}\frac{1}{\sigma_{r}}}Av_{r} & 0 & \cdots & 0\end{array}\right]=\left[\begin{array}{cccccc}
Av_{i} & \cdots & Av_{r} & 0 & \cdots & 0\end{array}\right]=AQ_{2}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Example*
Poišči singularni razcep
\begin_inset Formula $A=\left[\begin{array}{cccc}
1 & 1 & -1 & -1\\
-1 & 0 & 1 & 0\\
0 & -1 & 0 & 1
\end{array}\right]$
\end_inset
.
Izračunajmo
\begin_inset Formula
\[
A^{*}A=\left[\begin{array}{cccc}
1 & 1 & -1 & -1\\
-1 & 0 & 1 & 0\\
0 & -1 & 0 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & -1 & 0\\
1 & 0 & -1\\
-1 & 1 & 0\\
-1 & 0 & 1
\end{array}\right]=\cdots=\left[\begin{array}{cccc}
2 & 1 & -2 & -1\\
1 & 2 & -1 & -2\\
-2 & -1 & 2 & 1\\
-1 & -2 & 1 & 2
\end{array}\right]
\]
\end_inset
Izračunajmo
\begin_inset Formula $p_{A^{*}A}\left(x\right)=\det\left(A^{*}A-xI\right)=\cdots=x^{2}\left(x-2\right)\left(x-6\right)$
\end_inset
,
torej
\begin_inset Formula $\lambda_{1}=6$
\end_inset
,
\begin_inset Formula $\lambda_{2}=2$
\end_inset
,
\begin_inset Formula $\lambda_{3}=0$
\end_inset
,
\begin_inset Formula $\lambda_{4}=0$
\end_inset
,
torej
\begin_inset Formula $\sigma_{1}=\sqrt{6}$
\end_inset
in
\begin_inset Formula $\sigma_{2}=\sqrt{2}$
\end_inset
ter
\begin_inset Formula $\sigma_{3}=\sigma_{4}=0$
\end_inset
.
(ujema se z dejstvom,
da je
\begin_inset Formula $\rang A=2$
\end_inset
).
Izračunajmo lastne vektorje
\begin_inset Formula $A^{*}A$
\end_inset
:
\begin_inset Formula
\[
\lambda_{1}=6:\quad v_{1}'=\left[\begin{array}{c}
1\\
1\\
-1\\
-1
\end{array}\right],\quad\left|\left|v_{1}'\right|\right|=6
\]
\end_inset
\begin_inset Formula
\[
\lambda_{2}=2:\quad v_{2}'=\left[\begin{array}{c}
1\\
-1\\
-1\\
1
\end{array}\right],\quad\left|\left|v_{2}'\right|\right|=2
\]
\end_inset
\begin_inset Formula
\[
\lambda_{3}=\lambda_{4}:\quad v_{3}'=\left[\begin{array}{c}
1\\
0\\
1\\
0
\end{array}\right],v_{4}'=\left[\begin{array}{c}
0\\
1\\
0\\
2
\end{array}\right],\quad\left|\left|v_{3}'\right|\right|=\sqrt{2},\left|\left|v_{4}'\right|\right|=\sqrt{2}
\]
\end_inset
\end_layout
\begin_layout Example*
Z Gram-Schmidtom naredimo ortogonalno množico (v tem primeru so že ortogonalni) in jih normirajmo:
\begin_inset Formula
\[
v_{1}=\frac{1}{2}\left[\begin{array}{c}
1\\
1\\
-1\\
-1
\end{array}\right],\quad v_{2}=\frac{1}{2}\left[\begin{array}{c}
1\\
-1\\
-1\\
1
\end{array}\right],\quad v_{3}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}
1\\
0\\
1\\
0
\end{array}\right],\quad v_{4}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}
0\\
1\\
0\\
1
\end{array}\right].
\]
\end_inset
Sestavimo
\begin_inset Formula
\[
Q_{2}=\left[\begin{array}{cccc}
\frac{1}{2} & \frac{1}{2} & \frac{1}{\sqrt{2}} & 0\\
\frac{1}{2} & -\frac{1}{2} & 0 & \frac{1}{\sqrt{2}}\\
-\frac{1}{2} & -\frac{1}{2} & \frac{1}{\sqrt{2}} & 0\\
-\frac{1}{2} & \frac{1}{2} & 0 & \frac{1}{\sqrt{2}}
\end{array}\right],\quad D=\left[\begin{array}{cccc}
\sqrt{6} & & & 0\\
& \sqrt{2}\\
& & 0\\
0 & & & 0
\end{array}\right]
\]
\end_inset
Izračunamo
\begin_inset Formula $u_{1},\dots,u_{r}$
\end_inset
za
\begin_inset Formula $Q_{1}$
\end_inset
:
\begin_inset Formula
\[
u_{1}=\frac{1}{\sigma_{1}}Av_{1}=\frac{1}{\sqrt{6}}\left[\begin{array}{c}
2\\
-1\\
-1
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
u_{2}=\frac{1}{\sigma_{2}}Av_{2}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}
0\\
-1\\
1
\end{array}\right]
\]
\end_inset
Dopolnimo ju do ONB za
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset
z Gram-Schmidtom (oz.
uganemo
\begin_inset Formula $\left[\begin{array}{c}
1\\
1\\
1
\end{array}\right]$
\end_inset
).
Dopolnitev normiramo:
\begin_inset Formula $u_{3}=\frac{1}{\sqrt{3}}\left[\begin{array}{c}
1\\
1\\
1
\end{array}\right]$
\end_inset
in vektorje vstavimo v
\begin_inset Formula $Q_{1}$
\end_inset
:
\begin_inset Formula
\[
Q_{1}=\left[\begin{array}{ccc}
\frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}}\\
-\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}}\\
-\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}}
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Example*
Iskani razcep je
\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}$
\end_inset
(Izračunati je potrebno še en inverz —
\begin_inset Formula $Q_{2}^{-1}$
\end_inset
namreč).
\end_layout
\begin_layout Subsubsection
Psevdoinverz —
Moore-Penroseov inverz
\end_layout
\begin_layout Standard
Psevdoinverz je posplošitev inverza na nenujno kvadratne nenujno obrnljive matrike.
Najprej diagonalne matrike:
Njihov navaden inverz je takšen:
\begin_inset Formula
\[
\left[\begin{array}{ccc}
d_{11} & & 0\\
& \ddots\\
0 & & d_{nn}
\end{array}\right]^{-1}=\left[\begin{array}{ccc}
d_{11}^{-1} & & 0\\
& \ddots\\
0 & & d_{nn}^{-1}
\end{array}\right]
\]
\end_inset
Kadar je diagonalec ničeln,
kot element polja nima multiplikativnega inverza.
Ideja za posplošeni inverz diagonelne matrike:
take diagonalce pustimo na 0,
torej na primer:
\begin_inset Formula
\[
\left[\begin{array}{ccc}
1 & & 0\\
& 2\\
0 & & 0
\end{array}\right]^{+}=\left[\begin{array}{ccc}
1 & & 0\\
& \frac{1}{2}\\
0 & & 0
\end{array}\right]
\]
\end_inset
Za nekvadratne diagonalne matrike pa takole:
\begin_inset Formula
\[
\left[\begin{array}{cccc}
1 & & & 0\\
& 2\\
0 & & 0
\end{array}\right]^{+}=\left[\begin{array}{ccc}
1 & & 0\\
& \frac{1}{2}\\
& & 0\\
0
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Definition*
posplošeni inverz diagonalne matrike.
Naj bo
\begin_inset Formula $D$
\end_inset
diagonalna
\begin_inset Formula $m\times n$
\end_inset
z neničelnimi diagonalci
\begin_inset Formula $d_{1},\dots d_{r}$
\end_inset
,
je
\begin_inset Formula $D^{+}$
\end_inset
diagonalna
\begin_inset Formula $n\times m$
\end_inset
z neničelnimi diagonalci
\begin_inset Formula $\frac{1}{d_{1}^{-1}},\dots,\frac{1}{d_{r}^{-1}}$
\end_inset
.
\end_layout
\begin_layout Remark*
Za diagonalno
\begin_inset Formula $D$
\end_inset
opazimo
\begin_inset Formula $D^{++}=D$
\end_inset
in za obrnljivo diagonalno
\begin_inset Formula $D$
\end_inset
opazimo
\begin_inset Formula $D^{+}=D^{-1}$
\end_inset
.
\end_layout
\begin_layout Standard
Sedaj bi radi pojem posplošili na nediagonalne matrike —
to storimo s pomočjo SVD.
\begin_inset Formula $A=Q_{1}DQ_{2}^{*}\ni:D$
\end_inset
diagonalna in
\begin_inset Formula $Q_{1},Q_{2}$
\end_inset
unitarni.
Tedaj velja
\begin_inset Formula $A$
\end_inset
obrnljiva
\begin_inset Formula $\Leftrightarrow D$
\end_inset
obrnljiva,
kajti
\begin_inset Formula $A^{-1}=Q_{2}D^{-1}D_{1}^{-1}$
\end_inset
.
\end_layout
\begin_layout Definition*
Za splošen nenujno obrnljiv
\begin_inset Formula $A$
\end_inset
definiramo
\begin_inset Formula $A^{+}\coloneqq Q_{2}D^{+}Q_{1}^{-1}$
\end_inset
.
\end_layout
\begin_layout Fact*
Opazimo:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $A^{++}=\left(Q_{2}DQ_{1}^{-1}\right)^{+}=Q_{1}D^{++}Q_{2}^{-1}=Q_{1}DQ_{2}^{-1}=A$
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $A$
\end_inset
obrnljiva:
\begin_inset Formula $A^{+}=A^{-1}$
\end_inset
\end_layout
\end_deeper
\begin_layout Claim*
osnovne lastnosti psevdoinverza.
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $AA^{+}A=A$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\left(A^{+}A\right)^{*}=A^{+}A$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A^{+}AA^{+}=A^{+}$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\left(AA^{+}\right)^{*}=AA^{+}$
\end_inset
\end_layout
\end_deeper
\begin_layout Proof
Dokažimo te 4 lastnosti najprej za
\begin_inset Formula $D$
\end_inset
in nato za SVD.
Pri
\begin_inset Formula $D$
\end_inset
predpostavimo,
da so ničle spodaj desno,
sicer obstaja permutacijska matrika,
ki je ortogonalna,
s katero lahko množimo
\begin_inset Formula $D$
\end_inset
,
da jo pretvorimo v željeno obliko (in potem dokaz take
\begin_inset Formula $D$
\end_inset
pade v primer SVD):
\end_layout
\begin_deeper
\begin_layout Itemize
Diagonalen primer.
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $DD^{+}D=$
\end_inset
\begin_inset Formula
\[
\left[\begin{array}{cccccc}
d_{1} & & & & & 0\\
& \ddots\\
& & d_{r}\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right]\left[\begin{array}{cccccc}
d_{1}^{-1} & & & & & 0\\
& \ddots\\
& & d_{r}^{-1}\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right]\left[\begin{array}{cccccc}
d_{1} & & & & & 0\\
& \ddots\\
& & d_{r}\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right]=
\]
\end_inset
\begin_inset Formula
\[
=\left[\begin{array}{cccccc}
1 & & & & & 0\\
& \ddots\\
& & 1\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right]\left[\begin{array}{cccccc}
d_{1} & & & & & 0\\
& \ddots\\
& & d_{r}\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right]=D
\]
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $D^{+}DD^{+}=\cdots=D^{+}$
\end_inset
na podoben način
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\left(DD^{+}\right)^{*}=\left[\begin{array}{cccccc}
1 & & & & & 0\\
& \ddots\\
& & 1\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right]^{*}=\left[\begin{array}{cccccc}
1 & & & & & 0\\
& \ddots\\
& & 1\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right]=DD^{+}$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\left(D^{+}D\right)^{*}=\cdots=D^{+}D$
\end_inset
podobno
\end_layout
\end_deeper
\begin_layout Itemize
Splošen primer
\begin_inset Formula $A$
\end_inset
—
vstavimo
\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $AA^{+}A=Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{1}DD^{+}DQ_{2}^{*}=Q_{1}DQ_{2}^{*}=A$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A^{+}AA^{+}=\cdots=A^{+}$
\end_inset
na podoben način
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\left(AA^{+}\right)^{*}=\left(Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}\right)^{*}=\left(Q_{1}DD^{+}Q_{1}^{*}\right)^{*}=Q_{1}\left(DD^{+}\right)^{*}Q_{1}^{*}=Q_{1}DD^{+}Q_{1}^{*}=Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}=AA^{+}$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\left(A^{+}A\right)^{*}=\cdots=A^{+}A$
\end_inset
podobno
\end_layout
\end_deeper
\end_deeper
\begin_layout Remark*
\begin_inset Formula $A$
\end_inset
obrnljiva
\begin_inset Formula $\Leftrightarrow D$
\end_inset
obrnljiva,
torej
\begin_inset Formula $A^{+}=Q_{2}D^{+}Q_{1}^{-1}=Q_{2}D^{-1}Q_{1}^{-1}=A^{-1}$
\end_inset
.
Potemtakem za obrnljivo
\begin_inset Formula $A$
\end_inset
velja
\begin_inset Formula $A^{+}=A^{-1}$
\end_inset
.
\end_layout
\begin_layout Proof
Da je definicija dobra,
je treba dokazati,
da je
\begin_inset Formula $A^{+}$
\end_inset
enoličen ne glede na SVD,
kajti SVD za
\begin_inset Formula $A$
\end_inset
ni enoličen.
Naj bo
\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}=Q_{3}EQ_{4}^{-1}$
\end_inset
,
njen prvi psevsoinverz
\begin_inset Formula $B=Q_{2}D^{+}Q_{1}^{-1}$
\end_inset
in njen drugi psevdoinverz
\begin_inset Formula $C=Q_{4}D^{+}Q_{3}^{-1}$
\end_inset
.
Ali velja
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
udensdash{$B
\backslash
overset{?}{=}C$}
\end_layout
\end_inset
?
Velja
\begin_inset Formula
\[
AB=\left(ACA\right)B=ACAB=\left(AC\right)^{*}\left(AB\right)^{*}=C^{*}A^{*}B^{*}A^{*}=C^{*}\left(ABA\right)^{*}=C^{*}A^{*}=\left(AC\right)^{*}=AC
\]
\end_inset
in
\begin_inset Formula
\[
BA=B\left(ACA\right)=BACA=\left(BA\right)^{*}\left(CA\right)^{*}=A^{*}B^{*}A^{*}C^{*}=\left(ABA\right)^{*}C^{*}=A^{*}C^{*}=\left(CA\right)^{*}=CA
\]
\end_inset
ter nazadnje še
\begin_inset Formula
\[
B=BAB=CAB=CAC=C.
\]
\end_inset
\end_layout
\begin_layout Paragraph
Kako izračunamo
\begin_inset Formula $A^{+}$
\end_inset
brez SVD?
\end_layout
\begin_layout Standard
Če je
\begin_inset Formula $A$
\end_inset
pozitivno semidefinitna,
jo lahko diagonaliziramo v ortonormirani bazi:
\begin_inset Formula $A=PDP^{-1}$
\end_inset
,
da ima
\begin_inset Formula $D$
\end_inset
pozitivne diagonalce in da je
\begin_inset Formula $P^{-1}=P^{*}$
\end_inset
.
Opazimo,
da je to SVD od
\begin_inset Formula $A$
\end_inset
,
kajti
\begin_inset Formula $Q_{1}=P,Q_{2}=P,D=D$
\end_inset
in tedaj
\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}=PDP^{-1}$
\end_inset
.
Potemtakem je
\begin_inset Formula $A^{+}=PD^{+}P^{-1}$
\end_inset
.
\end_layout
\begin_layout Claim*
Za splošno matriko
\begin_inset Formula $A$
\end_inset
(nenujno pozitivno semidefinitno) pa velja
\begin_inset Formula $A^{+}=\left(A^{*}A\right)^{+}A^{*}=A^{*}\left(AA^{*}\right)^{+}$
\end_inset
.
\begin_inset Formula $A^{*}A$
\end_inset
in
\begin_inset Formula $AA^{*}$
\end_inset
sta pozitivno semidefinitni.
\end_layout
\begin_layout Proof
Najprej bomo preverili za diagonalno,
nato za SVD:
\end_layout
\begin_deeper
\begin_layout Itemize
Diagonalna
\begin_inset Formula $D_{n\times m}$
\end_inset
:
\begin_inset Formula
\[
D=\left[\begin{array}{cccccc}
d_{1} & & & & & 0\\
& \ddots\\
& & d_{r}\\
& & & 0\\
& & & & \ddots\\
& & & & & 0
\end{array}\right],\quad D^{*}=\left[\begin{array}{cccccc}
\overline{d_{1}} & & & & & 0\\
& \ddots\\
& & \overline{d_{r}}\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right],\quad D^{*}D=\left[\begin{array}{cccccc}
\frac{1}{\overline{d_{1}}d_{1}} & & & & & 0\\
& \ddots\\
& & \frac{1}{\overline{d_{r}}d_{r}}\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right],
\]
\end_inset
\begin_inset Formula
\[
\left(D^{*}D\right)^{+}=\left[\begin{array}{cccccc}
\frac{1}{\overline{d_{1}}d_{1}} & & & & & 0\\
& \ddots\\
& & \frac{1}{\overline{d_{r}}d_{r}}\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right],\quad\left(D^{*}D\right)^{+}D^{*}=\left[\begin{array}{cccccc}
\frac{\cancel{\overline{d_{1}}}}{\cancel{\overline{d_{1}}}d_{1}} & & & & & 0\\
& \ddots\\
& & \frac{\cancel{\overline{d_{r}}}}{\cancel{\overline{d_{r}}}d_{r}}\\
& & & 0\\
& & & & \ddots\\
0 & & & & & 0
\end{array}\right]=D^{+}
\]
\end_inset
\end_layout
\begin_layout Itemize
Za splošen
\begin_inset Formula $A$
\end_inset
uporabimo SVD,
da to dokažemo:
\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}$
\end_inset
.
Velja
\begin_inset Formula $A^{*}A=Q_{2}D^{*}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{2}D^{*}DQ_{2}^{*}$
\end_inset
in
\begin_inset Formula $\left(A^{*}A\right)^{+}=Q_{2}\left(D^{*}D\right)^{+}Q_{2}^{*}$
\end_inset
.
Torej
\begin_inset Formula $\left(A^{*}A\right)^{+}A^{*}=Q_{2}\left(D^{*}D\right)^{+}Q_{2}^{*}Q_{2}D^{*}Q_{1}^{*}=Q_{2}\left(D^{*}D\right)^{+}D^{*}Q_{1}^{*}=Q_{2}DQ_{1}^{*}=A^{+}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Remark*
V posebnih primerih lahko poenostavljamo dalje.
Recimo,
da ima
\begin_inset Formula $A$
\end_inset
LN stolpce in je kvadratna
\begin_inset Formula $\Rightarrow\Ker A=\left\{ 0\right\} =\Ker A^{*}A$
\end_inset
,
torej
\begin_inset Formula $A^{*}A$
\end_inset
je obrnljiva in velja
\begin_inset Formula $\left(A^{*}A\right)^{-1}=\left(A^{*}A\right)^{+}$
\end_inset
.
Takrat torej velja
\begin_inset Formula $A^{+}=\left(A^{*}A\right)^{-1}A^{*}$
\end_inset
.
\end_layout
\begin_layout Remark*
To smo uporabili pri iskanju posplošene rešitve predoločenega sistema:
Za sistem
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
iščemo
\begin_inset Formula $\vec{x}$
\end_inset
,
da je
\begin_inset Formula $\left|\left|A\vec{x}-\vec{b}\right|\right|$
\end_inset
minimalen,
tedaj bo tak
\begin_inset Formula $\vec{x}$
\end_inset
posplošena rešitev sistema.
Vemo,
da je posplošena reštev
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
enaka rešitvi od
\begin_inset Formula $A^{*}A\vec{x}=A^{*}\vec{b}$
\end_inset
,
kajti,
če ima
\begin_inset Formula $A$
\end_inset
LN stolpce,
je
\begin_inset Formula $A^{*}A$
\end_inset
obrnljiva (s tem dokažemo,
da ima ta sistem vedno rešitev):
\begin_inset Formula
\[
A^{*}A\vec{x}=A^{*}\vec{b}\quad\quad\quad\quad/\cdot\left(A^{*}A\right)^{-1}
\]
\end_inset
\begin_inset Formula
\[
\vec{x}=\left(A^{*}A\right)^{-1}A^{*}\vec{b}
\]
\end_inset
\begin_inset Formula
\[
\vec{x}=A^{+}\vec{b}
\]
\end_inset
\end_layout
\begin_layout Paragraph
Uporaba psevdoinverza
\end_layout
\begin_layout Standard
Vemo,
kaj je posplošena rešitev sistema
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset
.
Problem je,
da ima sistem lahko več posplošenih rešitev (to se lahko zgodi,
če
\begin_inset Formula $A$
\end_inset
nima LN stolpcev).
Med vsemi rešitvami iščemo tisto,
ki je najkrajša po normi —
\begin_inset Formula $\left|\left|\vec{x}\right|\right|$
\end_inset
.
\end_layout
\begin_layout Claim*
Najkrajša posplošena rešitev sistema
\begin_inset Formula $Ax=b$
\end_inset
je ravno
\begin_inset Formula $x=A^{+}b$
\end_inset
.
\end_layout
\begin_layout Proof
Dokažimo najprej za diagonalno matriko koeficientov,
nato pa še za splošen primer:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $Dx=b$
\end_inset
\begin_inset Formula
\[
D_{m\times n}=\left[\begin{array}{cccccc}
d_{1} & & & & & 0\\
& \ddots\\
& & d_{r}\\
& & & 0\\
& & & & \ddots\\
& & & & & 0
\end{array}\right],\quad x=\left[\begin{array}{c}
x_{1}\\
\vdots\\
x_{n}
\end{array}\right],\quad b=\left[\begin{array}{c}
b_{1}\\
\vdots\\
b_{m}
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
\left|\left|Dx-b\right|\right|^{2}=\left|\left|\left[\begin{array}{cccccc}
d_{1} & & & & & 0\\
& \ddots\\
& & d_{r}\\
& & & 0\\
& & & & \ddots\\
& & & & & 0
\end{array}\right]\left[\begin{array}{c}
x_{1}\\
\vdots\\
x_{n}
\end{array}\right]-\left[\begin{array}{c}
b_{1}\\
\vdots\\
b_{m}
\end{array}\right]\right|\right|^{2}=\left(d_{1}x_{1}-b_{1}\right)^{2}+\cdots+\left(d_{r}x_{r}-b_{r}\right)^{2}+b_{r+1}^{2}+\cdots+b_{m}^{2}
\]
\end_inset
Ta izraz doseže minimum,
ko
\begin_inset Formula $\left(d_{1}x_{1}-b_{1}\right)^{2}+\cdots+\left(d_{r}x_{r}-b_{r}\right)^{2}=0$
\end_inset
,
torej
\begin_inset Formula $x_{1}=\frac{b_{1}}{d_{1}},\dots,x_{r}=\frac{b_{r}}{d_{r}},x_{r+1}=\times,\dots,x_{n}=\times$
\end_inset
,
kjer
\begin_inset Formula $\times$
\end_inset
predstavlja poljubno vrednost.
Najkrajša rešitev bo torej tista,
kjer
\begin_inset Formula $x_{r+1}=\cdots=x_{n}=0$
\end_inset
.
Trdimo,
da je
\begin_inset Formula $\left(\frac{b_{1}}{d_{1}},\cdots,\frac{b_{r}}{d_{r}},0,\cdots,0\right)=D^{+}b$
\end_inset
.
Preverimo:
\begin_inset Formula
\[
D_{n\times m}^{+}=\left[\begin{array}{cccccc}
d_{1}^{-1} & & & & & 0\\
& \ddots\\
& & d_{r}^{-1}\\
& & & 0\\
& & & & \ddots\\
& & & & & 0
\end{array}\right],\quad b=\left[\begin{array}{c}
b_{1}\\
\vdots\\
b_{m}
\end{array}\right],\quad D^{+}b=\left[\begin{array}{c}
\frac{b_{1}}{d_{1}}\\
\vdots\\
\frac{b_{m}}{d_{m}}\\
0\\
\vdots\\
0
\end{array}\right]
\]
\end_inset
Res je!
\end_layout
\begin_layout Itemize
Splošen primer s SVD:
\begin_inset Formula $A_{m\times n}=Q_{1}DQ_{2}^{*}$
\end_inset
,
kjer sta
\begin_inset Formula $Q_{1},Q_{2}$
\end_inset
ortogonalni in
\begin_inset Formula $D$
\end_inset
diagonalna.
Za tretji enačaj uporabimo dejstvo,
da množenje z ortogonalno matriko ohranja normo.
\begin_inset Foot
status open
\begin_layout Plain Layout
\begin_inset Formula $\left|\left|Q_{2}^{*}x\right|\right|^{2}=\left\langle Q_{2}^{*}x,Q_{2}^{*}x\right\rangle =\left\langle x,Q_{2}Q_{2}^{*}x\right\rangle =\left\langle x,x\right\rangle =\left|\left|x\right|\right|^{2}$
\end_inset
\end_layout
\end_inset
\begin_inset Formula $\left|\left|Ax-b\right|\right|=\left|\left|Q_{1}DQ_{2}^{*}x-b\right|\right|=\left|\left|Q_{1}\left(DQ_{2}^{*}x-Q_{1}^{-1}b\right)\right|\right|=\left|\left|DQ_{2}^{*}x-Q_{1}^{-1}b\right|\right|=\left|\left|Dx'-c\right|\right|$
\end_inset
za
\begin_inset Formula $x'=Q_{2}^{*}x$
\end_inset
in
\begin_inset Formula $c=Q_{1}^{-1}b$
\end_inset
.
Ker je
\begin_inset Formula $Q_{2}$
\end_inset
obrnljiva,
velja,
da če
\begin_inset Formula $x$
\end_inset
preteče vse vektorje v
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
,
tudi
\begin_inset Formula $x'$
\end_inset
preteče vse vektorje v
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Standard
Potemtakem je
\begin_inset Formula $\min\left|\left|Ax-b\right|\right|=\min\left|\left|Dx'-c\right|\right|$
\end_inset
.
Če
\begin_inset Formula $\left|\left|Ax-b\right|\right|$
\end_inset
zavzame minimum v
\begin_inset Formula $x_{0}$
\end_inset
,
potem
\begin_inset Formula $\left|\left|Dx'-c\right|\right|$
\end_inset
zavzame minimum v
\begin_inset Formula $x_{0}'=Q_{2}^{-1}x_{0}$
\end_inset
in obratno,
če
\begin_inset Formula $\left|\left|Dx'-c\right|\right|$
\end_inset
zavzame minimum v
\begin_inset Formula $x_{0}'$
\end_inset
,
potem
\begin_inset Formula $\left|\left|Ax-b\right|\right|$
\end_inset
zavzame minimum v
\begin_inset Formula $x_{0}=Q_{2}x_{0}'$
\end_inset
.
Torej je
\begin_inset Formula $x\mapsto Q_{2}^{-1}x$
\end_inset
bijektivna korespondenca med posplošenimi rešitvami
\begin_inset Formula $Ax-b$
\end_inset
in posplošenimi rešitvami
\begin_inset Formula $Dx'-c$
\end_inset
.
Opazimo,
da preslikava ohranja normo,
torej
\begin_inset Formula $\left|\left|x_{0}'\right|\right|=\left|\left|Q_{2}x_{0}'\right|\right|=\left|\left|x_{0}\right|\right|$
\end_inset
.
\end_layout
\begin_layout Standard
Od prej vemmo,
da je najkrajša posplošena rešitev
\begin_inset Formula $Dx_{0}-c$
\end_inset
prav
\begin_inset Formula $x_{0}=D^{+}c$
\end_inset
.
Po zgornjem odstavku sledi,
da je
\begin_inset Formula $x_{0}=Q_{2}x_{0}'$
\end_inset
najkrajša posplošena rešitev od
\begin_inset Formula $Ax=b$
\end_inset
.
Dobimo namreč
\begin_inset Formula $x_{0}=Q_{2}x_{0}'=Q_{2}D^{+}c=Q_{2}D^{+}Q_{1}^{-1}b=A^{+}b$
\end_inset
.
\end_layout
\end_deeper
\end_deeper
\begin_layout Subsection
Kvadratne forme
\end_layout
\begin_layout Definition*
Forma je homogen polinom,
torej tak,
v katerem imajo vsi monomi isto stopnjo.
Stopnja monoma je
\begin_inset Formula $\deg\left(\beta x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}\right)\coloneqq\alpha_{1}+\cdots+\alpha_{n}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Polinom je vsota monomov.
Stopnja polinoma je najvišja stopnja monoma v njem.
\end_layout
\begin_layout Example*
Linearna forma v treh spremenljivkah:
\begin_inset Formula $ax+by+cz=\left[\begin{array}{ccc}
a & b & c\end{array}\right]\left[\begin{array}{c}
x\\
y\\
z
\end{array}\right]$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Kvadratna forma je homogen polinom stopnje 2.
Primer kvadratne forme:
\begin_inset Formula
\[
ax^{2}+bxy+cy^{2}=\left[\begin{array}{cc}
x & y\end{array}\right]\left[\begin{array}{cc}
a & b/2\\
b/2 & a
\end{array}\right]\left[\begin{array}{c}
x\\
y
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Kubična forma v treh spremenljivkah:
\begin_inset Formula
\[
ax^{3}+by^{3}+cz^{3}+dx^{2}y+ex^{2}z+fy^{2}x+gy^{2}x+iz^{2}x+jz^{2}y+kxyz
\]
\end_inset
\end_layout
\begin_layout Definition*
Pravimo,
da sta matriki
\begin_inset Formula $A$
\end_inset
in
\begin_inset Formula $B$
\end_inset
kongruentni,
če obstaja obrnljiva
\begin_inset Formula $P\ni:B=PAP^{T}$
\end_inset
.
\end_layout
\begin_layout Standard
Radi bi naredili klasifikacijo kvadratnih form.
Če naredimo primerno linearno zamenjavo koordinat,
se kvadratna forma poenostavi v
\begin_inset Formula $ex^{2}+fy^{2}$
\end_inset
(mešani členi izginejo).
\begin_inset Formula
\[
x=\alpha x'+\beta y'
\]
\end_inset
\begin_inset Formula
\[
y=\gamma x'+\delta y'
\]
\end_inset
zapišemo kot
\begin_inset Formula
\[
\left[\begin{array}{c}
x\\
y
\end{array}\right]=\left[\begin{array}{cc}
\alpha & \beta\\
\gamma & \delta
\end{array}\right]\left[\begin{array}{c}
x'\\
y'
\end{array}\right],\quad\quad\overset{\text{transponiranje}}{\Longrightarrow}\quad\quad\left[\begin{array}{cc}
x & y\end{array}\right]=\left[\begin{array}{cc}
x' & y'\end{array}\right]\left[\begin{array}{cc}
\alpha & \gamma\\
\beta & \delta
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
ax^{2}+bxy+cy^{2}=\left[\begin{array}{cc}
x & y\end{array}\right]\left[\begin{array}{cc}
a & b/2\\
b/2 & c
\end{array}\right]\left[\begin{array}{c}
x\\
y
\end{array}\right]=\left[\begin{array}{cc}
x' & y'\end{array}\right]\left[\begin{array}{cc}
\alpha & \gamma\\
\beta & \delta
\end{array}\right]\left[\begin{array}{cc}
a & b/2\\
b/2 & a
\end{array}\right]\left[\begin{array}{cc}
\alpha & \beta\\
\gamma & \delta
\end{array}\right]\left[\begin{array}{c}
x'\\
y'
\end{array}\right]=
\]
\end_inset
\begin_inset Formula
\[
=\left[\begin{array}{cc}
x' & y'\end{array}\right]P^{T}AP\left[\begin{array}{c}
x'\\
y'
\end{array}\right]
\]
\end_inset
Ker je
\begin_inset Formula $A$
\end_inset
simetrična,
lahko izberemo tako ortogonalno
\begin_inset Formula $P$
\end_inset
,
da je
\begin_inset Formula $P^{T}AP$
\end_inset
diagonalna,
recimo
\begin_inset Formula $\left[\begin{array}{cc}
d_{1} & 0\\
0 & d_{2}
\end{array}\right]$
\end_inset
,
torej
\begin_inset Formula
\[
\left[\begin{array}{cc}
x' & y'\end{array}\right]\left[\begin{array}{cc}
d_{1} & 0\\
0 & d_{2}
\end{array}\right]\left[\begin{array}{c}
x'\\
y'
\end{array}\right]=d_{1}\left(x'\right)^{2}+d_{2}\left(y'\right)^{2}
\]
\end_inset
Kaj vemo o
\begin_inset Formula $2\times2$
\end_inset
ortogonalnih matrikah?
\begin_inset Formula $P=\left[\begin{array}{cc}
a & b\\
c & d
\end{array}\right]\Rightarrow P^{T}P=\left[\begin{array}{cc}
a & c\\
b & d
\end{array}\right]\left[\begin{array}{cc}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{cc}
a^{2}+c^{2} & ab+cd\\
ab+cd & b^{2}+d^{2}
\end{array}\right]$
\end_inset
.
Da je
\begin_inset Formula $P^{T}P=I$
\end_inset
,
mora veljati
\begin_inset Formula $ab+cd=0$
\end_inset
in
\begin_inset Formula $a^{2}+c^{2}=b^{2}+d^{2}=1$
\end_inset
,
torej
\begin_inset Formula $a=\cos\varphi$
\end_inset
,
\begin_inset Formula $c=\sin\varphi$
\end_inset
,
\begin_inset Formula $b=\cos\tau$
\end_inset
,
\begin_inset Formula $d=\sin\tau$
\end_inset
.
Iz
\begin_inset Formula $\cos\left(\varphi+\tau\right)=0$
\end_inset
sledi
\begin_inset Formula $\tau=\varphi\pm\frac{\pi}{2}$
\end_inset
,
torej je
\begin_inset Formula $P_{1}=\left[\begin{array}{cc}
\cos\varphi & -\sin\varphi\\
\sin\varphi & \cos\varphi
\end{array}\right]$
\end_inset
(vrtež za
\begin_inset Formula $\varphi$
\end_inset
) ali
\begin_inset Formula $P_{2}=\left[\begin{array}{cc}
\cos\varphi & \sin\varphi\\
\sin\varphi & -\cos\varphi
\end{array}\right]$
\end_inset
(zrcaljenje žez
\begin_inset Formula $\varphi/2$
\end_inset
).
\begin_inset Formula $\det P_{1}=1$
\end_inset
,
\begin_inset Formula $\det P_{2}=-1$
\end_inset
.
\end_layout
\begin_layout Standard
Če je
\begin_inset Formula $A=\left[\begin{array}{cc}
v_{1} & v_{2}\end{array}\right]\left[\begin{array}{cc}
d_{1} & 0\\
0 & d_{2}
\end{array}\right]\left[\begin{array}{cc}
v_{1} & v_{2}\end{array}\right]^{-1}$
\end_inset
,
je tudi
\begin_inset Formula $A=\left[\begin{array}{cc}
v_{1} & -v_{2}\end{array}\right]\left[\begin{array}{cc}
d_{1} & 0\\
0 & d_{2}
\end{array}\right]\left[\begin{array}{cc}
v_{1} & -v_{2}\end{array}\right]^{-1}$
\end_inset
.
Če je
\begin_inset Formula $\left[\begin{array}{cc}
v_{1} & v_{2}\end{array}\right]$
\end_inset
ortogonalna,
je tudi
\begin_inset Formula $\left[\begin{array}{cc}
v_{1} & -v_{2}\end{array}\right]$
\end_inset
ortogonalna.
Če je
\begin_inset Formula $A$
\end_inset
\begin_inset Formula $2\times2$
\end_inset
simetrična matrika,
lahko poiščemo tak vrtež
\begin_inset Formula $P=\left[\begin{array}{cc}
\cos\varphi & -\sin\varphi\\
\sin\varphi & \cos\varphi
\end{array}\right]$
\end_inset
,
da je
\begin_inset Formula $A=P\left[\begin{array}{cc}
d_{1} & 0\\
0 & d_{2}
\end{array}\right]P^{-1}$
\end_inset
.
\end_layout
\begin_layout Standard
Povzetek:
\begin_inset Formula $ax^{2}+bxy+cy^{2}\overset{\text{vrtež}}{\longrightarrow}d_{1}x^{2}+d_{2}y^{2}$
\end_inset
\end_layout
\begin_layout Example*
Nariši krivuljo
\begin_inset Formula $4x^{2}+4xy+7y^{2}=1$
\end_inset
.
Pripadajoča kvadratna forma:
\begin_inset Formula
\[
\left[\begin{array}{cc}
x & y\end{array}\right]\left[\begin{array}{cc}
4 & 2\\
2 & 7
\end{array}\right]\left[\begin{array}{c}
x\\
y
\end{array}\right]=1=\cdots
\]
\end_inset
Radi bi se znebili mešanega člena:
\begin_inset Formula
\[
\cdots=\left[\begin{array}{cc}
x' & y'\end{array}\right]P^{T}\left[\begin{array}{cc}
4 & 2\\
2 & 7
\end{array}\right]P\left[\begin{array}{c}
x'\\
y'
\end{array}\right]=1=\cdots
\]
\end_inset
Iščemo tak vrtež
\begin_inset Formula $P$
\end_inset
,
da bo
\begin_inset Formula $P^{T}AP$
\end_inset
diagonalna.
Izračunamo lastne vrednosti
\begin_inset Formula $A=\left[\begin{array}{cc}
4 & 2\\
2 & 7
\end{array}\right]$
\end_inset
.
Lastni vrednosti sta
\begin_inset Formula $\left\{ 3,8\right\} $
\end_inset
.
Izračunamo lastna vektorja:
\begin_inset Formula $\left\{ \left[\begin{array}{c}
-2\\
1
\end{array}\right],\left[\begin{array}{c}
1\\
2
\end{array}\right]\right\} $
\end_inset
.
Sta že ortogonalna,
treba ju je še normirati:
\begin_inset Formula $\left\{ \left[\begin{array}{c}
-\frac{2}{\sqrt{5}}\\
\frac{1}{\sqrt{5}}
\end{array}\right],\left[\begin{array}{c}
\frac{1}{\sqrt{5}}\\
\frac{2}{\sqrt{5}}
\end{array}\right]\right\} $
\end_inset
.
Izdelamo vrzež:
\begin_inset Formula $P=\left[\begin{array}{cc}
\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}}\\
\frac{2}{\sqrt{5}} & \frac{2}{\sqrt{5}}
\end{array}\right]$
\end_inset
.
Izračunamo kot vrteža:
\begin_inset Formula $\frac{\sin\varphi}{\cos\varphi}=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}=2$
\end_inset
,
\begin_inset Formula $\arctan2\approx63,4^{\circ}$
\end_inset
.
\end_layout
\begin_layout Standard
Ogledamo si torej kvadratno formo
\begin_inset Formula $\left[\begin{array}{cc}
x' & y'\end{array}\right]\left[\begin{array}{cc}
8 & 0\\
0 & 3
\end{array}\right]\left[\begin{array}{c}
x'\\
y'
\end{array}\right]=8x'^{2}+3y'^{2}=1$
\end_inset
,
kar je elipsa (
\begin_inset Formula $\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$
\end_inset
) s polosema
\begin_inset Formula $\frac{1}{\sqrt{8}}$
\end_inset
in
\begin_inset Formula $\frac{1}{\sqrt{3}}$
\end_inset
.
\end_layout
\begin_layout Standard
Elipso narišemo in jo v koordinatnem sistemu zavrtimo v negativno smer za
\begin_inset Formula $63,4^{\circ}$
\end_inset
.
Po zasuku je risba te krivulje risba naše prvotne kvadratne forme.
\end_layout
\begin_layout Part
Vaja za ustni izpit
\end_layout
\begin_layout Standard
Ustni izpit je sestavljen iz treh vprašanj.
Sekcije so zaporedna vprašanja na izpitu,
podsekcije so učiteljevi naslovi iz Primerov vprašanj,
podpodsekcije pa so dejanska vprašanja,
kot so se pojavila na dosedanjih izpitih.
\end_layout
\begin_layout Section
Prvo vprašanje
\end_layout
\begin_layout Standard
Prvo vprašanje je iz 1.
semestra.
\end_layout
\begin_layout Subsubsection
\begin_inset Formula $\det AB=\det A\det B$
\end_inset
\end_layout
\begin_layout Subsection
Baze vektorskega prostora
\end_layout
\begin_layout Subsubsection
Linearno neodvisne množice
\end_layout
\begin_layout Subsubsection
Ogrodje
\end_layout
\begin_layout Subsubsection
Definicija baze
\end_layout
\begin_layout Subsubsection
Dimenzija prostora
\end_layout
\begin_layout Subsection
Cramerovo pravilo
\end_layout
\begin_layout Subsubsection
Trditev in dokaz
\end_layout
\begin_layout Subsection
Obrnljive matrike
\end_layout
\begin_layout Subsubsection
Definicija obrnljivosti
\end_layout
\begin_layout Subsubsection
Produkt obrnljivih matrik je obrnljiva matrika
\end_layout
\begin_layout Subsubsection
Karakterizacija obrnljivih matrik z dokazom
\end_layout
\begin_layout Subsubsection
\begin_inset Formula $\Ker A=\left\{ 0\right\} \Leftrightarrow A$
\end_inset
obrnljiva
\end_layout
\begin_layout Subsubsection
\begin_inset Formula $A$
\end_inset
ima desni inverz
\begin_inset Formula $\Rightarrow A$
\end_inset
obrnljiva
\end_layout
\begin_layout Subsubsection
Formula za inverz matrike z dokazom
\end_layout
\begin_layout Subsection
Vektorski podprostori
\end_layout
\begin_layout Subsection
Elementarne matrike
\end_layout
\begin_layout Subsection
Pod-/predoločeni sistem
\end_layout
\begin_layout Subsubsection
Definicija,
iskanje posplošene rešitve z izpeljavo
\end_layout
\begin_layout Subsubsection
Moč ogrodja
\begin_inset Formula $\geq$
\end_inset
moč LN množice
\end_layout
\begin_layout Subsubsection
Vsak poddoločen sistem ima netrivialno rešitev
\end_layout
\begin_layout Standard
Posledica prejšnje trditve.
\end_layout
\begin_layout Subsection
Regresijska premica
\end_layout
\begin_layout Subsubsection
Definicija
\end_layout
\begin_layout Subsection
Vektorski/mešani produkt
\end_layout
\begin_layout Subsection
Grupe/polgrupe
\end_layout
\begin_layout Subsubsection
Definicija in lastnosti grupe
\end_layout
\begin_layout Subsubsection
Definicija homomorfizma
\end_layout
\begin_layout Subsubsection
Primeri homomorfizmov z dokazi
\end_layout
\begin_layout Subsubsection
Definicija permutacijske grupe in dokaz,
da je grupa
\end_layout
\begin_layout Subsubsection
Primeri grup
\end_layout
\begin_layout Subsubsection
Dokaz,
da so ortogonalne matrike podgrupa v grupi obrnljivih matrik
\end_layout
\begin_layout Subsubsection
Matrika permutacije
\end_layout
\begin_layout Subsubsection
Dokaz,
da je preslikava,
ki permutaciji priredi matriko,
homomorfizem
\end_layout
\begin_layout Subsection
Projekcija točke na premico/ravnino
\end_layout
\begin_layout Subsection
\begin_inset Formula $\det A=\det A^{T}$
\end_inset
\end_layout
\begin_layout Subsection
Formula za inverz
\end_layout
\begin_layout Subsection
Homogeni sistemi enačb
\end_layout
\begin_layout Section
Drugo vprašanje
\end_layout
\begin_layout Standard
Drugo vprašanje zajema snov linearnih preslikav/lastnih vrednosti.
\end_layout
\begin_layout Subsection
Diagonalizacija
\end_layout
\begin_layout Subsubsection
Definicija,
trditve
\end_layout
\begin_layout Subsection
Prehod na novo bazo
\end_layout
\begin_layout Subsubsection
Prehodna matrika in njene lastnosti
\end_layout
\begin_layout Subsubsection
Predstavitev vektorjev in linearnih preslikav z različnimi bazami
\end_layout
\begin_layout Subsubsection
Razvoj vektorja po eni in drugi bazi (prehod vektorja na drugo bazo)
\end_layout
\begin_layout Subsection
Matrika linearne preslikave
\end_layout
\begin_layout Subsection
Rang matrike
\end_layout
\begin_layout Subsubsection
Definicija
\end_layout
\begin_layout Subsubsection
Dokaz,
da je rang število LN stolpcev
\end_layout
\begin_layout Subsubsection
Dimenzijska formula za podprostore
\end_layout
\begin_layout Subsection
\begin_inset Formula $\rang A=\rang A^{T}$
\end_inset
\end_layout
\begin_layout Subsection
Ekvivalentnost matrik
\end_layout
\begin_layout Subsubsection
Definicija
\end_layout
\begin_layout Subsubsection
Dokaz,
da je relacija ekvivalenčna
\end_layout
\begin_layout Subsubsection
Dokaz,
da je vsaka matrika ekvivalentna matriki
\begin_inset Formula $I_{r}$
\end_inset
,
t.
j.
bločni matriki,
katere zgornji levi blok je
\begin_inset Formula $I$
\end_inset
dimenzije
\begin_inset Formula $r$
\end_inset
,
drugi trije bloki pa so ničelne matrike.
\end_layout
\begin_layout Subsection
Jedro/slika
\end_layout
\begin_layout Subsection
Minimalni poinom
\end_layout
\begin_layout Subsubsection
Definicija karakterističnega in minimalnega polinoma
\end_layout
\begin_layout Subsection
Cayley-Hamiltonov izrek
\end_layout
\begin_layout Subsubsection
Trditev in dokaz
\end_layout
\begin_layout Subsection
Korenski razcep
\end_layout
\begin_layout Subsubsection
Definicija korenskih podprostorov
\end_layout
\begin_layout Subsubsection
Presek različnih korenskih podprostorov je trivialen
\end_layout
\begin_layout Subsubsection
Vsota korenskih podprostorov je direktna (se sklicuje na zgornjo trditev)
\end_layout
\begin_layout Subsection
Osnovna formula rang
\begin_inset Formula $+$
\end_inset
ničnost
\end_layout
\begin_layout Subsubsection
Definicija
\end_layout
\begin_layout Subsection
Funkcije matrik
\end_layout
\begin_layout Section
Tretje vprašanje
\end_layout
\begin_layout Standard
Tretje vprašanje zajema naslednje snovi:
\end_layout
\begin_layout Itemize
vektorski prostori s skalarnim produktom,
\end_layout
\begin_layout Itemize
adjungirana preslikava,
\end_layout
\begin_layout Itemize
singularni razcep,
\end_layout
\begin_layout Itemize
kvadratne forme.
\end_layout
\begin_layout Subsubsection
Singularni razcep:
Konstrukcija
\begin_inset Formula $Q_{1},Q_{2},D$
\end_inset
in dokaz
\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}$
\end_inset
.
\end_layout
\begin_layout Subsection
Ortogonalne/unitarne matrike
\end_layout
\begin_layout Subsubsection
Definicija
\end_layout
\begin_layout Subsubsection
Dokaz
\begin_inset Formula $AA^{*}=I$
\end_inset
\end_layout
\begin_layout Subsubsection
Lastne vrednosti
\end_layout
\begin_layout Subsubsection
Prehodna matrika iz ONB v drugo ONB ima ortogonalne stolpce (dokaz)
\end_layout
\begin_layout Subsection
Kvadratne krivulje
\end_layout
\begin_layout Subsection
Psevdoinverz
\end_layout
\begin_layout Subsubsection
Definicija
\end_layout
\begin_layout Subsection
Najkrajša posplošena rešitev sistema
\end_layout
\begin_layout Subsubsection
Definicija,
trditev in dokaz
\end_layout
\begin_layout Subsection
Simetrične matrike
\end_layout
\begin_layout Subsubsection
Vse o simetričnih matrikah
\end_layout
\begin_layout Subsection
Adjungirana linearna preslikava
\end_layout
\begin_layout Subsubsection
Definicija in celotna formulacija
\end_layout
\begin_layout Subsubsection
Rieszov izrek
\end_layout
\begin_layout Subsubsection
Dokaz obstoja in enoličnosti kot posledica Rieszovega izreka
\end_layout
\begin_layout Subsubsection
Formula za matriko linearne preslikave in
\begin_inset Formula $\left\langle Au,v\right\rangle =v^{*}Au=\left\langle u,A^{*}v\right\rangle $
\end_inset
\end_layout
\begin_layout Subsubsection
Lastne vrednosti adjungirane matrike
\end_layout
\begin_layout Subsection
Klasifikacija skalarnih produktov
\end_layout
\begin_layout Subsection
Normalne matrike
\end_layout
\begin_layout Subsubsection
Definicija,
lastnosti,
izreki,
dokazi
\end_layout
\begin_layout Subsubsection
\begin_inset Formula $A$
\end_inset
normalna
\begin_inset Formula $\Rightarrow A$
\end_inset
in
\begin_inset Formula $A^{*}$
\end_inset
imata isto množico lastnih vrednosti
\end_layout
\begin_layout Subsubsection
\begin_inset Formula $\Ker\left(A-xI\right)=\Ker\left(A-\overline{x}I\right)$
\end_inset
za normalno
\begin_inset Formula $A$
\end_inset
\end_layout
\begin_layout Subsection
Ortogonalni komplement
\end_layout
\begin_layout Subsubsection
Formula za ortogonalno projekcijo
\end_layout
\begin_layout Subsection
Izrek o reprezentaciji linearnih funkcionalov
\end_layout
\begin_layout Subsection
Pozitivno semidefinitne matrike
\end_layout
\begin_layout Subsubsection
Definicija,
lastnosti.
\end_layout
\begin_layout Subsubsection
Dokaz,
da imajo nenegativne lastne vrednosti.
\end_layout
\begin_layout Subsubsection
Kvadratni koren pozitivno semidefinitne matrike.
\end_layout
\begin_layout Subsubsection
\begin_inset Formula $A\geq0\Rightarrow A$
\end_inset
sebiadjungirana
\end_layout
\begin_layout Subsection
Ortogonalne in ortonormirane baze/Gram-Schmidt
\end_layout
\end_body
\end_document
