From 4fc141deb1fd41f46cf147765d4832b3b79e6ee9 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Anton=20Luka=20=C5=A0ijanec?= <anton@sijanec.eu>
Date: Sat, 29 Jun 2024 00:11:41 +0200
Subject: =?UTF-8?q?lateor,=201=20semester=20kon=C4=8Dan?=
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Content-Type: text/plain; charset=UTF-8
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---
 "\305\241ola/la/teor.lyx" | 9832 +++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 9832 insertions(+)

(limited to 'šola')

diff --git "a/\305\241ola/la/teor.lyx" "b/\305\241ola/la/teor.lyx"
index 440fc84..d84d320 100644
--- "a/\305\241ola/la/teor.lyx"
+++ "b/\305\241ola/la/teor.lyx"
@@ -27,6 +27,8 @@
 \DeclareMathOperator{\red}{red}
 \DeclareMathOperator{\karakteristika}{char}
 \DeclareMathOperator{\Ker}{Ker}
+\DeclareMathOperator{\sgn}{sgn}
+\DeclareMathOperator{\End}{End}
 \usepackage{algorithm,algpseudocode}
 \providecommand{\corollaryname}{Posledica}
 \end_preamble
@@ -1675,9 +1677,9839 @@ Označimo
 \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 Claim*
+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
+
+ Že dokazano zgoraj.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(1\Rightarrow3\right)$
+\end_inset
+
+ Že dokazano zgoraj.
+\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)
+\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.
+ 
+\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 
+\[
+\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det AB=\det B=\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A\det B
+\]
+
+\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
+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 
+\[
+X=A^{-1}=\left[\begin{array}{ccc}
+\frac{\det A_{11}\cdot\left(-1\right)^{1+1}}{\det A} & \cdots & \frac{\det A_{n1}\cdot\left(-1\right)^{n+1}}{\det A}\\
+\vdots &  & \vdots\\
+\frac{\det A_{1n}\cdot\left(-1\right)^{1+n}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A}
+\end{array}\right]=\frac{1}{\det A}\left[\begin{array}{ccc}
+\frac{\det A_{11}\cdot\left(-1\right)^{1+1}}{\det A} & \cdots & \frac{\det A_{1n}\cdot\left(-1\right)^{1+1}}{\det A}\\
+\vdots &  & \vdots\\
+\frac{\det A_{n1}\cdot\left(-1\right)^{n+1}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A}
+\end{array}\right]^{T}=\frac{1}{\det A}\tilde{A}^{T},
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+kjer 
+\begin_inset Formula $\tilde{A}$
+\end_inset
+
+ pravimo kofaktorska matrika.
+\end_layout
+
+\begin_layout Subsection
+Algenrske 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:
+ 
+\begin_inset Formula $\left(\text{vse obrnljive matrike fiksne dimenzije},\cdot\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(\text{vse permutacije neprazne končne množice},\circ\right)$
+\end_inset
+
+.
+\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
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Ta dokaz mi ni povsem jasen.
+ Zakaj je potrebno preverjati zgolj za 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}$
+\end_inset
+
+,
+ ne pa za vse elemente polja 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ torej tudi tiste,
+ ki niso v množici naših 
+\begin_inset Formula $\left\{ \alpha_{1},\dots,\alpha_{n}\right\} $
+\end_inset
+
+.
+ Če mi bralec zna razložiti,
+ naj mi piše na 
+\family typewriter
+anton@sijanec.eu
+\family default
+.
+\end_layout
+
+\end_inset
+
+
+\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.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Niti tega ne razumem.
+\end_layout
+
+\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*
+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
+
+.
+\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
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