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| -rw-r--r-- | šola/la/teor.lyx | 5422 |
1 files changed, 5392 insertions, 30 deletions
diff --git a/šola/la/teor.lyx b/šola/la/teor.lyx index e25c8bf..1f99e2f 100644 --- a/šola/la/teor.lyx +++ b/šola/la/teor.lyx @@ -161,6 +161,15 @@ Povzeto po zapiskih s predavanj prof. Cimpriča. \end_layout +\begin_layout Standard +\begin_inset CommandInset toc +LatexCommand tableofcontents + +\end_inset + + +\end_layout + \begin_layout Part Teorija \end_layout @@ -6378,14 +6387,62 @@ Nekaj abelovih grup: . 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 - +\family roman +\series medium +\shape up +\size normal +\emph off +\nospellcheck off +\bar no +\strikeout off +\xout off +\uuline off +\uwave off +\noun off +\color none +vse obrnljive matrike fiksne dimenzije +\family default +\series default +\shape default +\size default +\emph default +\nospellcheck default +\bar default +\strikeout default +\xout default +\uuline default +\uwave default +\noun default +\color inherit +, + +\family roman +\series medium +\shape up +\size normal +\emph off +\nospellcheck off +\bar no +\strikeout off +\xout off +\uuline off +\uwave off +\noun off +\color none +vse permutacije neprazne končne množice +\family default +\series default +\shape default +\size default +\emph default +\nospellcheck default +\bar default +\strikeout default +\xout default +\uuline default +\uwave default +\noun default +\color inherit . \end_layout @@ -15217,7 +15274,7 @@ v_{1} & \cdots & v_{m_{i}} & v_{m_{i}+1} & \cdots & v_{n}\end{array}\right]$ ki je obrnljiva. \begin_inset Formula \[ -P^{-1}AP=\cdots=\left[\begin{array}{cc} +P^{-1}AP=\left[\begin{array}{cc} \lambda_{i}I_{m_{i}} & B\\ 0 & C \end{array}\right] @@ -15225,10 +15282,26 @@ P^{-1}AP=\cdots=\left[\begin{array}{cc} \end_inset +Ker je karakteristični polinom neodvisen od izbire baze, + velja +\begin_inset Formula +\[ +\det\left(A-xI_{n}\right)=\det\left(\lambda_{i}I_{m_{i}}-xI\right)\det\left(C-xI_{n-m_{i}}\right)=\left(\lambda_{i}-x\right)^{m_{i}}\det\left(C-xI_{n-m_{i}}\right) +\] -\series bold -Dokaza ne razumem. - Obupam. +\end_inset + +Ker +\begin_inset Formula $\left(\lambda-x\right)^{m_{i}}$ +\end_inset + + deli karakteristični polinom, + je algebraična večkratnost +\begin_inset Formula $\lambda_{i}$ +\end_inset + + vsaj tolikšna, + kot je geometrična. \end_layout \begin_layout Claim @@ -15875,36 +15948,42 @@ Glede na definicijo \end_inset , - kajti vsebuje determinante matrik, - katerih elementi so polinomi stopnje -\begin_inset Formula $\leq1$ -\end_inset - -, - torej takele oblike: -\begin_inset Formula -\[ -\tilde{\left(A-xI\right)}^{T}=B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1} -\] - + kajti vsebuje determinante kofaktorskih matrik, + torej je takele oblike ( +\begin_inset Formula $\forall i\in\left\{ 1..\left(n-1\right)\right\} :B_{i}\in M_{n}\left(\mathbb{C}\right)$ \end_inset - +): \begin_inset Foot status open \begin_layout Plain Layout -Ne razumem, - zakaj so tu matrike -\begin_inset Formula $B$ +To si predstavljamo tako, + da iz vsake celice matrike, + ki vsebuje polinom, + izpostavimo (homogenost) spremenljivko (torej +\begin_inset Formula $x$ +\end_inset + + na fiksno potenco) in nato koeficiente v celicah pred to spremenljivvko zložimo v eno matriko. + Slednje ponovimo za vsako potenco in dobimo te matrike +\begin_inset Formula $B_{i}$ \end_inset - in ne skalarji. +. \end_layout \end_inset +\begin_inset Formula +\[ +\tilde{\left(A-xI\right)}^{T}=B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1} +\] + +\end_inset + + \end_layout \begin_layout Proof @@ -16618,7 +16697,7 @@ LA1P FMF 2024-03-20.pdf je lahko dokazati, da je vsota cel prostor. V karakteristični polinom, - ki po Caylay-Hamiltonu anhilira + ki po Cayley-Hamiltonu anhilira \begin_inset Formula $A$ \end_inset @@ -19432,6 +19511,5289 @@ Dokazati je treba \end_layout +\begin_layout Theorem* +ortogonalni razcep. + Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVPSSP in +\begin_inset Formula $W$ +\end_inset + + vektorski podprostor +\begin_inset Formula $V$ +\end_inset + +. + Potem velja +\begin_inset Formula $V=W\oplus W^{\perp}$ +\end_inset + + (ortogonalni razcep glede na +\begin_inset Formula $W$ +\end_inset + +). +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $v\in V$ +\end_inset + + pojuben, + +\begin_inset Formula $V^{\perp}$ +\end_inset + + pa ortogonalna projekcija +\begin_inset Formula $V$ +\end_inset + + na +\begin_inset Formula $W$ +\end_inset + +. + Potem velja, + da je +\begin_inset Formula $v=v-v'+v'$ +\end_inset + +, + kjer je +\begin_inset Formula $v-v'$ +\end_inset + + pravokoten na +\begin_inset Formula $W$ +\end_inset + +, + +\begin_inset Formula $v'$ +\end_inset + + pa element +\begin_inset Formula $v'$ +\end_inset + +, + torej +\begin_inset Formula $v\in W\oplus W^{\perp}$ +\end_inset + +. + Vsota je direktna, + kajti +\begin_inset Formula $\forall v\in W\cap W^{\perp}:v\perp v\Leftrightarrow v\perp v\Leftrightarrow\left\langle v,v\right\rangle =0\Leftrightarrow v=0\Longrightarrow W\cap W^{\perp}=\left\{ 0\right\} $ +\end_inset + + (po karakterizaciji direktnih vsot). +\end_layout + +\begin_layout Claim* +Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVPSSP in +\begin_inset Formula $W$ +\end_inset + + vektorski podprostor v +\begin_inset Formula $V$ +\end_inset + +. + Velja +\begin_inset Formula $\left(W^{\perp}\right)^{\perp}=W$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Po definiciji ortogonalnega komplementa je +\begin_inset Formula $W\subseteq\left(W^{\perp}\right)^{\perp}$ +\end_inset + +, + ker +\begin_inset Formula $W\perp W^{\perp}$ +\end_inset + +. + Dokažimo +\begin_inset Formula $\dim W=\dim\left(W^{\perp}\right)^{\perp}$ +\end_inset + +. + Ortogonalni razcep glede na +\begin_inset Formula $W$ +\end_inset + + je +\begin_inset Formula $V=W\oplus W^{\perp}\Rightarrow\dim W+\dim W^{\perp}=\dim V$ +\end_inset + +, + ortogonalni razcep glede na +\begin_inset Formula $W^{\perp}$ +\end_inset + + pa je +\begin_inset Formula $V=W^{\perp}\oplus\left(W^{\perp}\right)^{\perp}\Rightarrow\dim W^{\perp}+\dim\left(W^{\perp}\right)^{\perp}=\dim V$ +\end_inset + +. +\begin_inset Formula +\[ +\dim V=\dim V +\] + +\end_inset + + +\begin_inset Formula +\[ +\dim W+\dim W^{\perp}=\dim W^{\perp}+\dim\left(W^{\perp}\right)^{\perp} +\] + +\end_inset + + +\begin_inset Formula +\[ +\dim W^{\perp}=\dim\left(W^{\perp}\right)^{\perp} +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +Alternativni dokaz: + Naj bodo +\begin_inset Formula $w_{1},\dots,w_{k}$ +\end_inset + + OB za +\begin_inset Formula $W$ +\end_inset + +. + Dopolnimo jo do OB za +\begin_inset Formula $V$ +\end_inset + + z GS z +\begin_inset Formula $w_{k+1},\dots,w_{n}$ +\end_inset + +. + Tedaj je +\begin_inset Formula $w_{k+1},\dots,w_{n}$ +\end_inset + + OB za +\begin_inset Formula $W^{\perp}$ +\end_inset + + in ker je +\begin_inset Formula $w_{1},\dots,w_{n}$ +\end_inset + + njena dopolnitev do OB +\begin_inset Formula $V$ +\end_inset + +, + je +\begin_inset Formula $w_{1},\dots,w_{k}$ +\end_inset + + OB za +\begin_inset Formula $\left(W^{\perp}\right)^{\perp}$ +\end_inset + +, + torej +\begin_inset Formula $W^{\perp}=\left(W^{\perp}\right)^{\perp}$ +\end_inset + +, + saj imata isti ortogonalni bazi. +\end_layout + +\begin_layout Subsection +Adjungirana linearna preslikava +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + vektorski prostor nad +\begin_inset Formula $F$ +\end_inset + +. + Vemo, + da je +\begin_inset Formula $F$ +\end_inset + + vekrorski prostor nad +\begin_inset Formula $F$ +\end_inset + +. + Linearnim preslikavam +\begin_inset Formula $V\to F$ +\end_inset + + pravimo linearni funkcionali na +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Naj bo +\begin_inset Formula $V$ +\end_inset + + VPSSP in +\begin_inset Formula $F\in\left\{ \mathbb{R},\mathbb{C}\right\} $ +\end_inset + +. + Naj bo +\begin_inset Formula $w\in V$ +\end_inset + +. + Naj bo +\begin_inset Formula $\varphi:V\to F$ +\end_inset + + (torej je +\begin_inset Formula $\varphi$ +\end_inset + + linearni funkcional), + ki slika +\begin_inset Formula $v\mapsto\left\langle v,w\right\rangle $ +\end_inset + +. + Preslikava je po aksiomu 3 za skalarni produkt linearna. +\end_layout + +\begin_layout Theorem* +Rieszov izrek o reprezentaciji linearnih funkcionalov. + Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVPSSP. + Za vsak linearen funkcional +\begin_inset Formula $\varphi$ +\end_inset + + na +\begin_inset Formula $V$ +\end_inset + + obstaja natanko en vektor +\begin_inset Formula $w\in V\ni:\forall v\in V:\varphi\left(v\right)=\left\langle v,w\right\rangle $ +\end_inset + +. + ZDB slednja konstrukcija nam da vse linearne funkcionale. +\end_layout + +\begin_layout Proof +Dokazujemo enolično eksistenco: +\end_layout + +\begin_deeper +\begin_layout Itemize +Eksistenca +\begin_inset Formula $w$ +\end_inset + +: + Vzemimo poljubno OB +\begin_inset Formula $w_{1},\dots,w_{n}$ +\end_inset + + za +\begin_inset Formula $V$ +\end_inset + +. + +\begin_inset Formula $\forall v\in V:v=\left\langle v,w_{1}\right\rangle w_{1}+\cdots+\left\langle v,w_{n}\right\rangle w_{n}$ +\end_inset + +. + (fourierov razvoj po OB). + Ker je +\begin_inset Formula $\varphi$ +\end_inset + + linearna, + velja +\begin_inset Formula +\[ +\varphi\left(v\right)=\varphi\left(\left\langle v,w_{1}\right\rangle w_{1}+\cdots+\left\langle v,w_{n}\right\rangle w_{n}\right)\overset{\text{linearna}}{=}\left\langle v,w_{1}\right\rangle \varphi w_{1}+\cdots+\left\langle v,w_{n}\right\rangle \varphi w_{n}\overset{\text{konj. hom. v 2. fakt.}}{=} +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left\langle v,\overline{\varphi w_{1}}w_{1}\right\rangle +\cdots+\left\langle v,\overline{\varphi w_{n}}w_{n}\right\rangle \overset{\text{konj. ad. v 2. fakt.}}{=}\left\langle v,\left(\varphi w_{1}\right)w_{1}+\cdots+\left(\varphi w_{n}\right)w_{n}\right\rangle +\] + +\end_inset + +Za dan +\begin_inset Formula $\varphi$ +\end_inset + + smo konstruirali eksplicitno formulo za iskani +\begin_inset Formula $w$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Enoličnost +\begin_inset Formula $w$ +\end_inset + +: + PDDRAA +\begin_inset Formula $\forall v\in V:\varphi\left(v\right)=\left\langle v,w_{1}\right\rangle =\left\langle v,w_{2}\right\rangle $ +\end_inset + +. + Tedaj +\begin_inset Formula $\forall v\in V:\left\langle v,w_{1}-w_{2}\right\rangle =0$ +\end_inset + +. + Vzemimo konkreten +\begin_inset Formula $v=w_{1}-w_{2}$ +\end_inset + + in ga vstavimo v formulo +\begin_inset Formula $\left\langle v,w_{1}-w_{2}\right\rangle =0=\left\langle w_{1}-w_{2},w_{1}-w_{2}\right\rangle =0\Rightarrow w_{1}-w_{2}=0\Rightarrow w_{1}=w_{2}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Definition* +Naj bosta +\begin_inset Formula $U,V$ +\end_inset + + KRVPSSP in +\begin_inset Formula $L:U\to V$ +\end_inset + + linearna. + Adjungirana linearna preslikava, + pripadajoča +\begin_inset Formula $L$ +\end_inset + +, + je taka +\begin_inset Formula $L^{*}:V\to U$ +\end_inset + +, + da velja +\begin_inset Formula $\forall u\in U,v\in V:\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle $ +\end_inset + +. + Levi skalarni produkt je tisti iz +\begin_inset Formula $V$ +\end_inset + +, + desni pa tisti iz +\begin_inset Formula $U$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Da lahko pišemo +\begin_inset Formula $L^{*}$ +\end_inset + +, + trdimo, + da je +\begin_inset Formula $L^{*}$ +\end_inset + + vedno obstaja in to vselej enolično. +\end_layout + +\begin_layout Proof +Dokazujemo enolično eksistenco: +\end_layout + +\begin_deeper +\begin_layout Itemize +Enoličnost: + Naj bosta +\begin_inset Formula $L^{*}$ +\end_inset + + in +\begin_inset Formula $L^{\circ}$ +\end_inset + + dve adjungirani linearni preslikavi za +\begin_inset Formula $L$ +\end_inset + +, + torej +\begin_inset Formula $\forall u\in U,v\in V:\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle =\left\langle u,L^{\circ}v\right\rangle $ +\end_inset + +. + Torej +\begin_inset Formula +\[ +\left\langle u,L^{*}v\right\rangle =\left\langle u,L^{\circ}v\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +0=\left\langle u,L^{*}v-L^{\circ}v\right\rangle +\] + +\end_inset + +Za vsaka +\begin_inset Formula $u$ +\end_inset + + in +\begin_inset Formula $v$ +\end_inset + +. + Sedaj vstavimo +\begin_inset Formula $u=L^{*}v-L^{\circ}v$ +\end_inset + +: +\begin_inset Formula +\[ +0=\left\langle L^{*}v-L^{\circ}v,L^{*}v-L^{\circ}v\right\rangle \Longrightarrow L^{*}v-L^{\circ}v=0\Longrightarrow\forall v\in V:L^{*}v=L^{\circ}v +\] + +\end_inset + + +\end_layout + +\begin_layout Itemize +Eksistenca: + Naj bosta +\begin_inset Formula $U,V$ +\end_inset + + KRVPSSP in +\begin_inset Formula $L:U\to V$ +\end_inset + +. + Naj bo +\begin_inset Formula $v\in V$ +\end_inset + + poljuben. + Vpeljimo linearni funkcional +\begin_inset Formula $\varphi:U\to F$ +\end_inset + + s predpisom +\begin_inset Formula $u\mapsto\left\langle Lu,v\right\rangle $ +\end_inset + +. + Prepričajmo se, + da je ta funkcional linearna preslikava: + +\begin_inset Formula +\[ +\varphi\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=\left\langle L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right),v\right\rangle =\left\langle \alpha_{1}Lu_{1}+\alpha_{2}Lu_{2},v\right\rangle =\alpha_{1}\left\langle Lu_{1},v\right\rangle +\alpha_{2}\left\langle Lu_{2},v\right\rangle +\] + +\end_inset + +Uporabimo Rieszov izrek za funkcional +\begin_inset Formula $\varphi$ +\end_inset + +: + +\begin_inset Formula $\exists!w\in U\ni:\forall u\in U:\varphi u=\left\langle u,w\right\rangle $ +\end_inset + +. + Vpeljimo +\begin_inset Formula $L^{*}v=w$ +\end_inset + +, + s čimer za poljuben +\begin_inset Formula $v$ +\end_inset + + definiramo +\begin_inset Formula $L^{*}v$ +\end_inset + +. + Dokažimo, + da je dobljena preslikava linearna: + +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +udensdash{$L^{*} +\backslash +left( +\backslash +beta_{1}v_{1}+ +\backslash +beta_{2}v_{2} +\backslash +right)= +\backslash +beta_{1}L^{*}v_{1}+ +\backslash +beta_{2}L^{*}v_{2}$} +\end_layout + +\end_inset + +. + Vzemimo pojuben +\begin_inset Formula $u\in U$ +\end_inset + + in računajmo (fino bi bilo dobiti nič): +\begin_inset Formula +\[ +\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)\right\rangle =\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\beta_{1}L^{*}v_{1}-\beta_{2}L^{*}v_{2}\right\rangle \overset{\text{kl2f}}{=} +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle \overset{\text{lin }L^{*}}{=}\left\langle u,\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle = +\] + +\end_inset + + +\begin_inset Formula +\[ +=\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle +\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle =0 +\] + +\end_inset + +Ker to velja za vsak +\begin_inset Formula $u$ +\end_inset + +, + velja tudi za +\begin_inset Formula $u=L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)$ +\end_inset + +, + torej dobimo +\begin_inset Formula +\[ +\left\langle L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right),L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)\right\rangle =0 +\] + +\end_inset + +torej po prvem aksiomu za skalarni produtk velja linearnost: +\begin_inset Formula +\[ +L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)=\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2} +\] + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Example* +Naj bo +\begin_inset Formula $A\in M_{m\times n}\left(F\right)$ +\end_inset + + s pripadajočo linearno preslikavo +\begin_inset Formula $L_{A}=F^{n}\to F^{m}$ +\end_inset + +, + ki slika +\begin_inset Formula $v\mapsto Av$ +\end_inset + +. + Kako izgleda matrika +\begin_inset Formula $L_{A^{*}}$ +\end_inset + +? + Odgovor je odvisen od izbire skalarnega produkta. + Izberimo standardni skalarni produkt v +\begin_inset Formula $F^{n}$ +\end_inset + + in +\begin_inset Formula $F^{m}$ +\end_inset + + in +\begin_inset Formula $L_{A^{*}}:F^{m}\to F^{n}$ +\end_inset + + definiramo z +\begin_inset Formula $v\mapsto A^{*}v$ +\end_inset + +, + kjer je +\begin_inset Formula $A^{*}=\overline{A^{T}}$ +\end_inset + +, + torej transponiranka +\begin_inset Formula $A$ +\end_inset + + z vsemi elementi konjugiranimi. + Izkaže se, + da je potemtakem +\begin_inset Formula $L_{A^{*}}$ +\end_inset + + adjungirana linearna preslikava od +\begin_inset Formula $L_{A}$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Matrika adjungirane linearne preslikave +\end_layout + +\begin_layout Standard +Naj bosta +\begin_inset Formula $U,V$ +\end_inset + + KRVPSSP in naj bo +\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $ +\end_inset + + ONB za +\begin_inset Formula $U$ +\end_inset + + in +\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $ +\end_inset + + ONB za +\begin_inset Formula $V$ +\end_inset + +. + Vzemimo linearno preslikavo +\begin_inset Formula $L:U\to V$ +\end_inset + +. + Izpeljimo zvezo med +\begin_inset Formula $L$ +\end_inset + + in +\begin_inset Formula $L^{*}$ +\end_inset + + glede na bazi +\begin_inset Formula $\mathcal{B}$ +\end_inset + + in +\begin_inset Formula $\mathcal{C}$ +\end_inset + +. + Torej +\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$ +\end_inset + + za +\begin_inset Formula $L:U\to V$ +\end_inset + + in +\begin_inset Formula $\left[L^{*}\right]_{\mathcal{B}\leftarrow\mathcal{C}}$ +\end_inset + + za +\begin_inset Formula $L^{*}:V\to U$ +\end_inset + +. + Izračunajmo +\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$ +\end_inset + + tako, + da uporabimo fourierov razvoj: +\begin_inset Formula +\[ +\begin{array}{ccccccc} +Lu_{1} & = & \left\langle Lu_{1},v_{1}\right\rangle v_{1} & + & \cdots & + & \left\langle Lu_{1},v_{m}\right\rangle v_{m}\\ +\vdots & & \vdots & & & & \vdots\\ +Lu_{n} & = & \left\langle Lu_{n},v_{1}\right\rangle v_{1} & + & \cdots & + & \left\langle Lu_{n},v_{m}\right\rangle v_{m} +\end{array} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc} +\left\langle Lu_{1},v_{1}\right\rangle & \cdots & \left\langle Lu_{n},v_{1}\right\rangle \\ +\vdots & & \vdots\\ +\left\langle Lu_{1},v_{m}\right\rangle & \cdots & \left\langle Lu_{n},v_{m}\right\rangle +\end{array}\right]=\left[\begin{array}{ccc} +\left\langle u_{1},L^{*}v_{1}\right\rangle & \cdots & \left\langle u_{n},L^{*}v_{1}\right\rangle \\ +\vdots & & \vdots\\ +\left\langle u_{1},L^{*}v_{m}\right\rangle & \cdots & \left\langle u_{n},L^{*}v_{m}\right\rangle +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +Sedaj izračunajmo še +\begin_inset Formula $\left[L^{*}\right]_{\mathcal{B}\leftarrow\mathcal{C}}$ +\end_inset + + spet s fourierovim razvojem in primerjajmo istoležne koeficiente: +\begin_inset Formula +\[ +\begin{array}{ccccccc} +L^{*}v_{1} & = & \left\langle L^{*}v_{1},u_{1}\right\rangle u_{1} & + & \cdots & + & \left\langle L^{*}v_{1},u_{n}\right\rangle u_{n}\\ +\vdots & & \vdots & & & & \vdots\\ +L^{*}v_{m} & = & \left\langle Lv_{m},u_{1}\right\rangle u_{1} & + & \cdots & + & \left\langle Lv_{m},u_{n}\right\rangle u_{n} +\end{array} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left[L\right]_{\mathcal{B}\leftarrow\mathcal{C}}=\left[\begin{array}{ccc} +\left\langle L^{*}v_{1},u_{1}\right\rangle & \cdots & \left\langle L^{*}v_{m},u_{1}\right\rangle \\ +\vdots & & \vdots\\ +\left\langle L^{*}v_{1},u_{n}\right\rangle & \cdots & \left\langle L^{*}v_{m},u_{n}\right\rangle +\end{array}\right]=\left[\begin{array}{ccc} +\overline{\left\langle u_{1},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{1},L^{*}v_{m}\right\rangle }\\ +\vdots & & \vdots\\ +\overline{\left\langle u_{n},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{m}\right\rangle } +\end{array}\right]=\left[\begin{array}{ccc} +\overline{\left\langle u_{1},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{1}\right\rangle }\\ +\vdots & & \vdots\\ +\overline{\left\langle u_{1},L^{*}v_{m}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{m}\right\rangle } +\end{array}\right]^{T}= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\overline{\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}}^{T} +\] + +\end_inset + + +\end_layout + +\begin_layout Remark* +Kako izgleda lastnost +\begin_inset Formula $\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle $ +\end_inset + +? + Naj bo +\begin_inset Formula $u\in F^{n}$ +\end_inset + + in +\begin_inset Formula $v\in F^{m}$ +\end_inset + + in +\begin_inset Formula $A=m\times n$ +\end_inset + + matrika. + Ali za standardna skalarna produkta v +\begin_inset Formula $F^{n}$ +\end_inset + + in +\begin_inset Formula $F^{m}$ +\end_inset + + +\begin_inset Formula $\left\langle Au,v\right\rangle =\left\langle u,A^{*}v\right\rangle $ +\end_inset + + velja tudi za matrike, + če vzamemo +\begin_inset Formula $A^{*}=\overline{A}^{T}=\overline{A^{T}}$ +\end_inset + +? + Pa preverimo (ja, + velja): +\begin_inset Formula +\[ +\left\langle u,v\right\rangle =u_{1}\overline{v_{1}}+\cdots+u_{n}\overline{v_{n}}=\left[\begin{array}{ccc} +\overline{v_{1}} & \cdots & \overline{v_{n}}\end{array}\right]\left[\begin{array}{c} +u_{1}\\ +\vdots\\ +u_{n} +\end{array}\right]=v^{*}u +\] + +\end_inset + + +\begin_inset Formula +\[ +\left\langle Au,v\right\rangle =v^{*}Au +\] + +\end_inset + + +\begin_inset Formula +\[ +\left\langle u,A^{*}v\right\rangle =\left(A^{*}v\right)^{*}u=v^{*}\left(A^{*}\right)^{*}u=v^{*}Au +\] + +\end_inset + + +\end_layout + +\begin_layout Fact* +Lastnosti adjungiranja: + +\begin_inset Formula $\left(\alpha A+\beta B\right)^{*}=\overline{\alpha}A^{*}+\overline{\beta}B^{*}$ +\end_inset + +, + +\begin_inset Formula $\left(AB\right)^{*}=B^{*}A^{*}$ +\end_inset + +, + +\begin_inset Formula $\left(A^{*}\right)^{*}=A$ +\end_inset + + +\begin_inset Note Note +status open + +\begin_layout Plain Layout +TODO XXX FIXME DOKAŽI +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Jedro in slika adjungirane linearne preslikave +\end_layout + +\begin_layout Claim* +Naj bo +\begin_inset Formula $L:U\to V$ +\end_inset + + linearna. + Velja +\begin_inset Formula $\Ker\left(L^{*}\right)=\left(\Slika L\right)^{\perp}$ +\end_inset + + in +\begin_inset Formula $\Slika\left(L^{*}\right)=\left(\Ker L\right)^{\perp}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $v\in\Ker L^{*}$ +\end_inset + + za +\begin_inset Formula $L^{*}:V\to U$ +\end_inset + +. + Velja +\end_layout + +\begin_layout Proof +\begin_inset Formula +\[ +v\in\Ker\left(L^{*}\right)\Leftrightarrow L^{*}v=0\Leftrightarrow\forall u\in U:\left\langle u,L^{*}v\right\rangle =0\Leftrightarrow\forall u\in U:\left\langle Lu,v\right\rangle =0\Leftrightarrow\forall w\in\Slika L:\left\langle w,v\right\rangle =0\Leftrightarrow v\in\left(\Slika L\right)^{\perp} +\] + +\end_inset + +Velja torej +\begin_inset Formula $\Ker L^{*}=\left(\Slika L\right)^{\perp}\Rightarrow\Ker L=\left(\Slika L^{*}\right)^{\perp}\Rightarrow\left(\Ker L\right)^{\perp}=\Slika L^{*}\Rightarrow\left(\Ker L^{*}\right)^{\perp}=\Slika L$ +\end_inset + + +\end_layout + +\begin_layout Claim* +Za +\begin_inset Formula $L:U\to V$ +\end_inset + + velja +\begin_inset Formula $\Ker\left(L^{*}L\right)=\Ker L$ +\end_inset + + +\end_layout + +\begin_layout Proof +Vzemimo poljuben +\begin_inset Formula $u\in U$ +\end_inset + + in dokazujemo enakost množic (obe vsebovanosti): +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\supseteq\right)$ +\end_inset + + Če +\begin_inset Formula $u\in\Ker L\Rightarrow Lu=0\overset{\text{množimo z }L^{*}}{\Longrightarrow}L^{*}Lu=L^{*}u=0\Rightarrow u\in\Ker L^{*}L\Rightarrow\Ker L\subseteq\Ker L^{*}L$ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\subseteq\right)$ +\end_inset + + Če +\begin_inset Formula $u\in\Ker L^{*}L\Rightarrow L^{*}Lu=0\Rightarrow\left\langle u,L^{*}Lu\right\rangle =0\Rightarrow\left\langle Lu,Lu\right\rangle =0\Rightarrow Lu=0\Rightarrow u\in\Ker L\Rightarrow\Ker L^{*}L\subseteq\Ker L$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Corollary* +\begin_inset Formula $\Slika\left(L^{*}L\right)=\Slika\left(L\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +\begin_inset Formula $\Slika\left(L^{*}L\right)=\Slika\left(L^{*}\left(L^{*}\right)^{*}\right)=\Slika\left(\left(L^{*}L\right)^{*}\right)=\left(\Ker L^{*}L\right)^{\perp}=\left(\Ker L\right)^{\perp}=\Slika L^{*}$ +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Lastne vrednosti adjungirane linearne preslikave. +\end_layout + +\begin_layout Claim* +Če je +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost +\begin_inset Formula $A$ +\end_inset + +, + je +\begin_inset Formula $\overline{\lambda}$ +\end_inset + + lastna vrednost za +\begin_inset Formula $A^{*}$ +\end_inset + +. + ZDB +\begin_inset Formula $\det\left(A-\lambda I\right)=0\Rightarrow\det\left(A-\lambda\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $B=A-\lambda I$ +\end_inset + +. + Tedaj +\begin_inset Formula $B^{*}=A^{*}-\overline{\lambda}I^{*}=A^{*}-\overline{\lambda}I$ +\end_inset + +. + Radi bi dokazali +\begin_inset Formula $\det B=0\Rightarrow\det B^{*}=0$ +\end_inset + +. + Ker je +\begin_inset Formula $B^{*}=\overline{B}^{T}\Rightarrow\det B^{*}=\det\overline{B}^{T}=\det\overline{B}=\overline{\det B}\Rightarrow\det B=0\Rightarrow\det B^{*}=0$ +\end_inset + +. +\end_layout + +\begin_layout Corollary* +Iz te formule izvemo tudi karakteristični polimom +\begin_inset Formula $A^{*}$ +\end_inset + +. + +\begin_inset Formula $p_{A^{*}}\left(x\right)=\det\left(A^{*}-xI\right)\Rightarrow p_{A}\left(\overline{x}\right)=\det\left(A-\overline{x}I\right)=\det\left(A^{*}-xI\right)^{*}=\overline{\det\left(A^{*}-xI\right)}=\overline{p_{A^{*}}\left(x\right)}$ +\end_inset + +, + torej +\begin_inset Formula $p_{A^{*}}\left(x\right)=\overline{p_{A}\left(\overline{x}\right)}$ +\end_inset + +. + Torej, + če je +\begin_inset Formula $p_{A}\left(x\right)=c_{0}x^{0}+\cdots+x_{n}x^{n}$ +\end_inset + +, + je +\begin_inset Formula $p_{A^{*}}\left(x\right)=\overline{c_{0}\overline{x^{0}}+\cdots+x_{n}\overline{x^{n}}}=\overline{c_{0}}x^{0}+\cdots+\overline{c_{n}}x^{n}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Alternativen dokaz: + Najprej dokažimo +\begin_inset Formula $\dim\Ker B^{*}=\dim\Ker B$ +\end_inset + +. + Velja +\begin_inset Formula $\dim\Ker B^{*}=\dim\left(\Slika B\right)^{\perp}=n-\dim\Slika B=\dim\Ker B$ +\end_inset + +. + Torej +\begin_inset Formula $\Ker\left(B\right)\not=0\Leftrightarrow\Ker\left(B^{*}\right)\not=0$ +\end_inset + +, + torej so lastne vrednosti +\begin_inset Formula $A^{*}$ +\end_inset + + konjugirane lastne vrednosti +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Med lastnimi vektorji +\begin_inset Formula $A$ +\end_inset + + in lastnimi vektorji +\begin_inset Formula $A^{*}$ +\end_inset + + (žal) ni posebne zveze. + Primer: + +\begin_inset Formula $A=\left[\begin{array}{cc} +1 & 2\\ +i & 1 +\end{array}\right]$ +\end_inset + + ima lastne vektorje +\begin_inset Formula $\vec{v_{1}}=\left[\begin{array}{c} +1-i\\ +-1 +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $\vec{v_{2}}=\left[\begin{array}{c} +1-i\\ +1 +\end{array}\right]$ +\end_inset + +, + +\begin_inset Formula $A^{*}=\left[\begin{array}{cc} +1 & 1\\ +2 & -i +\end{array}\right]$ +\end_inset + + pa lastne vektorje +\begin_inset Formula $\vec{v_{1}'}=\left[\begin{array}{c} +1-i\\ +-2 +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $\vec{v_{2}'}=\left[\begin{array}{c} +1-i\\ +2 +\end{array}\right]$ +\end_inset + +. + Med temi vektorji ni nobenih kolinearnosti. + Obstajajo pa zveze v nekaterih zanimivih primerih: +\end_layout + +\begin_layout Claim* +Če matrika +\begin_inset Formula $A$ +\end_inset + + zadošča +\begin_inset Formula $A^{*}A=AA^{A}$ +\end_inset + + (pravimo +\begin_inset Formula $A$ +\end_inset + + je normalna), + iz +\begin_inset Formula $Av=\lambda v$ +\end_inset + + sledi +\begin_inset Formula $A^{*}v=\overline{\lambda}v$ +\end_inset + +, + torej imata +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $A^{*}$ +\end_inset + + iste lastne vrednosti. +\end_layout + +\begin_layout Proof +Če velja +\begin_inset Formula $Av=\lambda v$ +\end_inset + +, + velja +\begin_inset Formula $Av-\lambda v=\left(A-\lambda I\right)v=Bv=0\Rightarrow v\in\Ker B$ +\end_inset + +. + Če velja +\begin_inset Formula $A^{*}v=\overline{\lambda}v$ +\end_inset + +, + velja +\begin_inset Formula $A^{*}v-\overline{\lambda}v=\left(A^{*}-\overline{\lambda}I\right)v=B^{*}v=0\Rightarrow v\in\Ker B^{*}$ +\end_inset + +. + Dokazati je treba še +\begin_inset Formula $\Ker B=\Ker B^{*}$ +\end_inset + +: +\end_layout + +\begin_deeper +\begin_layout Enumerate +Ali velja +\begin_inset Formula $A^{*}A=AA^{*}\Rightarrow B^{*}B=BB^{*}$ +\end_inset + +? + Ja. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $B^{*}B=\left(A^{*}-\overline{\lambda}I\right)\left(A-\lambda I\right)=A^{*}A-\overline{\lambda}A-\lambda A^{*}+\overline{\lambda}\lambda I$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $BB^{*}=\left(A-\lambda I\right)\left(A^{*}-\overline{\lambda}I\right)=AA^{*}-\overline{\lambda}A-\lambda A^{*}+\lambda\overline{\lambda}I$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Enumerate +Ali velja +\begin_inset Formula $B^{*}B=BB^{*}\Rightarrow\Ker B=\Ker B^{*}$ +\end_inset + +? + Iz +\begin_inset Formula $B^{*}B=BB^{*}$ +\end_inset + + sledi +\begin_inset Formula $\Ker\left(B^{*}B\right)=\Ker\left(BB^{*}\right)\Rightarrow\Ker\left(B\right)=\Ker\left(B^{*}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\Ker B=\Ker B^{*}\Rightarrow\forall v\in V:Av=\lambda v\Leftrightarrow A^{*}v=\overline{\lambda}v$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Subsubsection +Normalne matrike +\end_layout + +\begin_layout Definition* +\begin_inset Formula $A$ +\end_inset + + je normalna +\begin_inset Formula $\Leftrightarrow A^{*}A=AA^{*}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Dokazali smo že, + da za normalne matrike velja, + da imata +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $A^{*}$ +\end_inset + + iste lastne vektorje, + kar v splošnem ne velja. +\end_layout + +\begin_layout Claim* +Lastni vektorji, + ki pripadajo različnim lastnim vrednostim normalne matrike, + so paroma ortogonalni. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A^{*}A=AA^{*}$ +\end_inset + + za neko +\begin_inset Formula $A$ +\end_inset + + in naj bo +\begin_inset Formula $Au=\lambda u$ +\end_inset + + in +\begin_inset Formula $Av=\mu v$ +\end_inset + + in +\begin_inset Formula $\mu\not=\lambda$ +\end_inset + +. + +\begin_inset Formula $u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$ +\end_inset + +. + Računajmo: +\begin_inset Formula +\[ +\mu\left\langle u,v\right\rangle =\left\langle u,\overline{\mu}v\right\rangle =\left\langle u,A^{*}v\right\rangle =\left\langle Au,v\right\rangle =\left\langle \lambda u,v\right\rangle =\lambda\left\langle u,v\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +\left(\mu-\lambda\right)\left\langle u,v\right\rangle =0\wedge u\not=\lambda\Rightarrow\left\langle u,v\right\rangle =0\Leftrightarrow u\perp v +\] + +\end_inset + + +\end_layout + +\begin_layout Claim* +Vsako normalno matriko se da diagonalizirati. +\end_layout + +\begin_layout Proof +Dokažimo, + da je jordanska forma normalne matrike diagonalna +\begin_inset Formula $\Leftrightarrow$ +\end_inset + + vsi korenski podprostori so lastni. + +\begin_inset Formula $\forall m,\lambda:\Ker\left(A-I\lambda\right)^{m}=\Ker\left(A-I\lambda\right)$ +\end_inset + +. + Zadošča dokazati za +\begin_inset Formula $m=2$ +\end_inset + +. + Naj bo +\begin_inset Formula $m=2$ +\end_inset + + in +\begin_inset Formula $B=A-I\lambda$ +\end_inset + +. + Dokažimo +\begin_inset Formula $\Ker B^{2}=\Ker B$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Enumerate +Če v +\begin_inset Formula $\Ker\left(A\right)=\Ker\left(A^{*}A\right)$ +\end_inset + + vstavimo +\begin_inset Formula $A=B^{2}$ +\end_inset + +, + dobimo +\begin_inset Formula $\Ker B^{2}=\Ker\left(\left(B^{2}\right)^{*}B^{2}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Ker je +\begin_inset Formula $A$ +\end_inset + + normalna, + je +\begin_inset Formula $B$ +\end_inset + + normalna, + torej +\begin_inset Formula $\left(B^{2}\right)^{*}B^{2}=B^{*}B^{*}BB=B^{*}BB^{*}B=\left(B^{*}B\right)^{2}$ +\end_inset + +. + Torej +\begin_inset Formula $\Ker\left(\left(B^{2}\right)^{*}B^{2}\right)=\Ker\left(B^{*}B\right)^{2}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Če v +\begin_inset Formula $\Ker A^{*}A=\Ker A$ +\end_inset + + vstavimo +\begin_inset Formula $A=B^{*}B$ +\end_inset + +, + dobimo +\begin_inset Formula $\Ker B^{*}BB^{*}B=\Ker\left(B^{*}B\right)^{2}=\Ker B^{*}B$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Zopet upoštevamo +\begin_inset Formula $\Ker A^{*}A=\Ker A$ +\end_inset + +, + torej +\begin_inset Formula $\Ker\left(B^{*}B\right)=\Ker B$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Proof +Ko dokažemo +\begin_inset Formula $B$ +\end_inset + + normalna +\begin_inset Formula $\Rightarrow B^{*}$ +\end_inset + + normalna, + bo iz +\begin_inset Formula $\Ker B^{2}=\Ker B$ +\end_inset + + sledilo +\begin_inset Formula $\Ker B^{4}=\Ker B$ +\end_inset + +. + Preverimo, + a je +\begin_inset Formula $B^{2}$ +\end_inset + + normalna, + če je +\begin_inset Formula $B$ +\end_inset + + normalna: + +\begin_inset Formula $\left(B^{2}\right)^{*}B^{2}=B^{*}B^{*}BB=B^{*}BB^{*}B=BB^{*}BB^{*}=BBB^{*}B^{*}=B^{2}\left(B^{2}\right)^{*}$ +\end_inset + +. + Sedaj vemo +\begin_inset Formula $\Ker B=\Ker B^{2}=\Ker B^{4}=\Ker B^{8}=\cdots$ +\end_inset + +. + Vemo pa tudi, + da +\begin_inset Formula +\[ +\Ker B\subseteq\Ker B^{2}\subseteq\Ker B^{3}\subseteq\Ker B^{4}\subseteq\Ker B^{5}\subseteq\Ker B^{6}\subseteq\cdots +\] + +\end_inset + + +\begin_inset Formula +\[ +\Ker B\subseteq\Ker B\subseteq\Ker B^{3}\subseteq\Ker B\subseteq\Ker B^{5}\subseteq\Ker B\subseteq\cdots +\] + +\end_inset + + +\begin_inset Formula +\[ +\Ker B=\Ker B^{2}=\Ker B^{3}=\Ker B^{4}=\Ker B^{5}=\cdots +\] + +\end_inset + + +\begin_inset Formula +\[ +\forall v:\Ker B^{m}=\Ker B +\] + +\end_inset + + +\end_layout + +\begin_layout Remark* +Torej za vsako normalno matriko +\begin_inset Formula $A\exists$ +\end_inset + + diagonalna +\begin_inset Formula $D$ +\end_inset + + in obrnljiva +\begin_inset Formula $P$ +\end_inset + + z ortonormiranimi stolpci, + da velja +\begin_inset Formula $AP=PD$ +\end_inset + +, + +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + +. + Diagonala +\begin_inset Formula $D$ +\end_inset + + so lastne vrednosti +\begin_inset Formula $A$ +\end_inset + +, + stolpci +\begin_inset Formula $P$ +\end_inset + + pa so njeni lastni vektorji. + Lastni podprostori +\begin_inset Formula $\left(A-\lambda_{1}I\right),\dots,\left(A-\lambda_{n}I\right)$ +\end_inset + + so medsebojno pravokotni. + Izberimo ONB za vsak lasten podprostor. + Unija teh ONB je ONB za +\begin_inset Formula $F^{n}$ +\end_inset + +. + +\begin_inset Formula $F^{n}=\Ker\left(A-\lambda_{1}I\right)\oplus\cdots\oplus\Ker\left(A-\lambda_{n}I\right)$ +\end_inset + +. + Ta ONB so stolpci matrike +\begin_inset Formula $P$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Ortogonalne/unitarne matrike +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $A$ +\end_inset + + kvadratna z ON stolpci glede na standardni skalarni produkt. + Pravimo, + da je +\begin_inset Formula $A$ +\end_inset + + unitarna (v kompleksnem primer) oziroma ortogonalna (v realnem primeru). +\end_layout + +\begin_layout Claim* +Za unitarno +\begin_inset Formula $A$ +\end_inset + + velja +\begin_inset Formula $A^{*}A=AA^{*}=I$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokazujmo za unitarno. + Za ortogonalno je dokaz podoben. + Naj bo +\begin_inset Formula $A=\left[\begin{array}{ccc} +a_{11} & \cdots & a_{1n}\\ +\vdots & & \vdots\\ +a_{n1} & \cdots & a_{nn} +\end{array}\right]$ +\end_inset + + unitarna. + To pomeni, + da za vsaka stolpca +\begin_inset Formula $a_{i}=\left(a_{1i},\dots,a_{ni}\right)$ +\end_inset + + in +\begin_inset Formula $a_{j}=\left(a_{1j},\dots,a_{nj}\right)$ +\end_inset + + velja za vsak +\begin_inset Formula $i,j\in\left\{ 1..n\right\} $ +\end_inset + + velja +\begin_inset Formula $\left\langle \text{\left[\begin{array}{c} +a_{1i}\\ +\text{\ensuremath{\vdots}}\\ +a_{ni} +\end{array}\right],\left[\begin{array}{c} +a_{1j}\\ +\vdots\\ +a_{nj} +\end{array}\right]}\right\rangle =a_{1i}\overline{a_{1j}}+\cdots+a_{ni}\overline{a_{nj}}=\begin{cases} +0 & ;i\not=j\\ +1 & ;i=j +\end{cases}$ +\end_inset + +. + Oglejmo si +\begin_inset Formula +\[ +A^{*}A=\left[\begin{array}{ccc} +\overline{a_{11}} & \cdots & \overline{a_{n1}}\\ +\vdots & & \vdots\\ +\overline{a_{1n}} & \cdots & \overline{a_{nn}} +\end{array}\right]\left[\begin{array}{ccc} +a_{11} & \cdots & a_{1n}\\ +\vdots & & \vdots\\ +a_{n1} & \cdots & a_{nn} +\end{array}\right]=\left[\begin{array}{ccc} +1 & & 0\\ + & \ddots\\ +0 & & 1 +\end{array}\right] +\] + +\end_inset + +Očitno je res, + ker je vsak element +\begin_inset Formula $A^{*}A$ +\end_inset + + konstruiran s skalarnim množenjem vrstice leve matrike (konjugirani stolpci +\begin_inset Formula $A$ +\end_inset + +, + ker smo poprej matriko transponiralo) in stolpca desne, + za kar predpis smo poprej že razbrali. +\end_layout + +\begin_layout Remark* +Za nekvadratne unitarne velja le +\begin_inset Formula $A^{*}A=I$ +\end_inset + +, + +\begin_inset Formula $AA^{*}=I$ +\end_inset + + pa zaradi nezmožnosti množenja zaradi nepravilnih dimenzij seveda ne velja. +\end_layout + +\begin_layout Claim* +Naslednje trditve so za +\begin_inset Formula $P$ +\end_inset + + z ortonormiranimi stolpci ekvivalentne: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $P^{*}P=I$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall u,v:\left\langle Pu,Pv\right\rangle =\left\langle u,v\right\rangle $ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall u:\left|\left|Pu\right|\right|=\left|\left|u\right|\right|$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall$ +\end_inset + + ONB +\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} :\left\{ Pu_{1},\dots,Pu_{n}\right\} $ +\end_inset + + je ON množica +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\exists$ +\end_inset + + ONB +\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} :\left\{ Pu_{1},\dots,Pu_{n}\right\} $ +\end_inset + + je ON množica +\end_layout + +\end_deeper +\begin_layout Proof +Dokazujemo ekvivalenco: +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(1\Rightarrow2\right)$ +\end_inset + + +\begin_inset Formula $\left\langle Pu,Pv\right\rangle =\left\langle u,P^{*}Pv\right\rangle =\left\langle u,v\right\rangle $ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(2\Rightarrow3\right)$ +\end_inset + + +\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left\langle Pu,Pu\right\rangle =\left\langle u,u\right\rangle =\left|\left|u\right|\right|^{2}$ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(2\Rightarrow1\right)$ +\end_inset + + +\begin_inset Formula +\[ +\forall u,v:\left\langle Pu,Pv\right\rangle =\left\langle u,v\right\rangle \Rightarrow\left\langle u,P^{*}Pv\right\rangle -\left\langle u,v\right\rangle =0\Rightarrow\left\langle u,\left(P^{*}P-I\right)v\right\rangle =0 +\] + +\end_inset + + Sedaj izberimo +\begin_inset Formula $u=\left(P^{*}P-I\right)v$ +\end_inset + +: + +\begin_inset Formula $\left\langle \left(P^{*}P-I\right)v,\left(P^{*}P-I\right)v\right\rangle =0\Rightarrow P^{*}P-I=0\Rightarrow P^{*}P=0$ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(3\Rightarrow2\right)$ +\end_inset + + Po predpostavki +\begin_inset Formula $\forall u:\left|\left|Pu\right|\right|=\left|\left|u\right|\right|$ +\end_inset + + Izrazimo skalarni produkt z normo: + +\begin_inset Formula $\left\langle u,v\right\rangle =\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}$ +\end_inset + +, + torej +\begin_inset Formula +\[ +\left\langle Pu,Pv\right\rangle =\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|Pu+i^{k}Pv\right|\right|^{2}=\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|P\left(u+i^{k}v\right)\right|\right|^{2}\overset{\text{predpostavka}}{=}\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}=\left\langle u,v\right\rangle +\] + +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(5\Rightarrow4\right)$ +\end_inset + + Vzemimo poljuben +\begin_inset Formula $u$ +\end_inset + + in ga razvijmo po ONB +\begin_inset Formula $u_{1},\dots,u_{n}$ +\end_inset + +. + Tedaj +\begin_inset Formula $u=\alpha_{1}u_{1}+\cdots+\alpha_{n}u_{n}$ +\end_inset + +. + Ker so +\begin_inset Formula $u_{i}$ +\end_inset + + ONB, + velja +\begin_inset Formula $\left|\left|u\right|\right|^{2}=\left|\alpha_{1}\right|^{2}+\cdots\left|\alpha_{n}\right|^{2}$ +\end_inset + +. + Ker so +\begin_inset Formula $Pu_{1}$ +\end_inset + + ONB po predpostavki, + +\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left|\alpha_{1}\right|^{2}+\cdots+\left|\alpha_{n}\right|^{2}$ +\end_inset + +, + torej velja +\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left|\left|u\right|\right|^{2}$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(2\Rightarrow4\right)$ +\end_inset + + Ker so +\begin_inset Formula $u_{1},\dots,u_{n}$ +\end_inset + + ONM, + velja +\begin_inset Formula $\left\langle u_{i},u_{j}\right\rangle =\begin{cases} +1 & ;i=j\\ +0 & ;i\not=j +\end{cases}$ +\end_inset + +. + Tudi +\begin_inset Formula $Pu_{1},\dots,Pu_{n}$ +\end_inset + + ortonormirana, + kajti po predpostavki +\begin_inset Formula $2$ +\end_inset + + velja +\begin_inset Formula $\left\langle Pu_{i},Pu_{2}\right\rangle =\begin{cases} +1 & ;i=j\\ +0 & ;i\not=j +\end{cases}$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(4\Rightarrow5\right)$ +\end_inset + + Očitno. +\end_layout + +\end_deeper +\begin_layout Claim* +Lastne vrednosti unitarne matrike +\begin_inset Formula $A$ +\end_inset + + se nahajajo na enotski krožnici v +\begin_inset Formula $\Im$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A$ +\end_inset + + unitarna in naj bo +\begin_inset Formula $v$ +\end_inset + + tak, + da +\begin_inset Formula $Av=\lambda v$ +\end_inset + +. + Tedaj +\begin_inset Formula $\left\langle v,v\right\rangle =\left\langle Av,Av\right\rangle =\left\langle \lambda v,\lambda v\right\rangle =\lambda\overline{\lambda}\left\langle v,v\right\rangle \Rightarrow\lambda\overline{\lambda}=1\Rightarrow\left|\lambda\right|=1\Rightarrow\lambda=e^{i\varphi}$ +\end_inset + + za nek +\begin_inset Formula $\varphi$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Iz unitarnosti sledi normalnost, + zato so lastni vektorji unitarne matrike, + ki pripadajo paroma različnim lastnim vrednostim, + pravokotni (isto, + kot pri normalnih matrikah). +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +Prav tako kot pri normalnih matrikah lahko unitarne diagonalitziramo v tokrat ortogonalni bazi. + Pri unitarnih so stolpci +\begin_inset Formula $P$ +\end_inset + + še celo normirani. + +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + +, + kjer je +\begin_inset Formula $P$ +\end_inset + + unitarna, + torej +\begin_inset Formula $P^{*}=P^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +Očitno je, + da če je +\begin_inset Formula $A$ +\end_inset + + unitarna, + velja +\begin_inset Formula $A^{*}=A^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +\begin_inset CommandInset label +LatexCommand label +name "subsec:Simetrične/hermitske-matrike" + +\end_inset + +Simetrične/hermitske matrike +\end_layout + +\begin_layout Definition* +Matrika nad +\begin_inset Formula $\mathbb{R}$ +\end_inset + + je simetrična, + če zanjo velja +\begin_inset Formula $A^{*}=A$ +\end_inset + +. + Matrika nad +\begin_inset Formula $\mathbb{C}$ +\end_inset + + je hermitska, + če zanjo velja +\begin_inset Formula $A^{*}=A$ +\end_inset + +. + Linearni preslikavi, + pripadajoči hermitski/simetrični matriki, + pravimo sebiadjungirana. +\end_layout + +\begin_layout Fact* +Vsaka hermitska/simetrična matrika je normalna, + kajti +\begin_inset Formula $A^{*}A=AA=AA^{*}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Lastne vrednosti hermitskih/simetričnih matrik so realne. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A=A^{*}$ +\end_inset + + in naj bo +\begin_inset Formula $Av=\lambda v$ +\end_inset + + za nek neničeln +\begin_inset Formula $v$ +\end_inset + +. + Tedaj +\begin_inset Formula $\lambda\left\langle v,v\right\rangle =\left\langle \lambda v,v\right\rangle =\left\langle Av,v\right\rangle =\left\langle v,A^{*}v\right\rangle =\left\langle v,Av\right\rangle =\left\langle v,\lambda v\right\rangle =\overline{\lambda}\left\langle v,v\right\rangle $ +\end_inset + +. + Potemtakem +\begin_inset Formula $\lambda=\overline{\lambda}\Rightarrow\lambda\in\mathbb{R}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Diagonalizacija je zopet enaka kot pri normalnih matrikah z dodatkom — + vsaka hermitska matrika je podobna realni diagonalni, + kar za normalne ni res — + normalne so lahko podobne kompleksnim diagonalnim matrikam. +\end_layout + +\begin_layout Subsubsection +Pozitivno (semi)definitne matrike +\end_layout + +\begin_layout Definition* +\begin_inset Formula $A$ +\end_inset + + je pozitivno semidefinitna +\begin_inset Formula $\sim A\geq0\Leftrightarrow A=A^{*}\wedge\forall v:\left\langle Av,v\right\rangle \geq0$ +\end_inset + +. + +\begin_inset Formula $A$ +\end_inset + + je pozitivno definitna +\begin_inset Formula $\sim A>0\Leftrightarrow A=A^{*}\wedge\forall v\not=0:\left\langle Av,v\right\rangle >0$ +\end_inset + +. + S tem ko skalarni produkt primerjamo ( +\begin_inset Formula $>,\geq$ +\end_inset + +), + implicitno zahtevamo njegovo realnost. + Primerjalni operatorji namreč na kompleksnih številih niso definirani. +\end_layout + +\begin_layout Example* +Vzemimo poljubno nenujno kvadratno +\begin_inset Formula $B$ +\end_inset + + in definirajmo +\begin_inset Formula $A=B^{*}B$ +\end_inset + +. + Potem je +\begin_inset Formula $A$ +\end_inset + + pozitivno semidefinitna, + kajti +\begin_inset Formula $A^{*}=\left(B^{*}B\right)^{*}=B^{*}B=A$ +\end_inset + + in +\begin_inset Formula $\left\langle Av,v\right\rangle =\left\langle B^{*}Bv,v\right\rangle =\left\langle Bv,Bv\right\rangle \geq0$ +\end_inset + +. + Če pa bi bili stolpci +\begin_inset Formula $B$ +\end_inset + + linearno neodvisni, + pa bi veljalo +\begin_inset Formula $\forall v:v\not=0\Rightarrow\left\langle Av,v\right\rangle =\left\langle B^{*}Bv\right\rangle =\left\langle Bv,Bv\right\rangle >0$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $A\geq0\Rightarrow$ +\end_inset + + lastne vrednosti +\begin_inset Formula $A$ +\end_inset + + so +\begin_inset Formula $\geq0$ +\end_inset + +. + +\begin_inset Formula $A>0\Rightarrow$ +\end_inset + + lastne vrednosti +\begin_inset Formula $A$ +\end_inset + + so +\begin_inset Formula $>0$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $A\geq0$ +\end_inset + +. + Tedaj +\begin_inset Formula $Av=\lambda v$ +\end_inset + + za nek +\begin_inset Formula $v\not=0$ +\end_inset + +. + Torej +\begin_inset Formula $\left\langle Av,v\right\rangle =\left\langle \lambda v,v\right\rangle =\lambda\left\langle v,v\right\rangle $ +\end_inset + +. + Toda ker +\begin_inset Formula $\left\langle Av,v\right\rangle \geq0$ +\end_inset + +, + sledi +\begin_inset Formula $\lambda\left\langle v,v\right\rangle \geq0$ +\end_inset + +. + Ker je +\begin_inset Formula $\left\langle v,v\right\rangle >0$ +\end_inset + +, + sledi +\begin_inset Formula $\lambda\geq0$ +\end_inset + +. + Analogno za +\begin_inset Formula $A>0$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Diagonalizacija je ista kot za normalna, + s tem da za diagonalno +\begin_inset Formula $D$ +\end_inset + + velja še, + da je pozitivno (semi)definitna, + ko je +\begin_inset Formula $A$ +\end_inset + + pozitivno semidefinitna. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $\forall A\geq0\exists B=B^{*},B\geq0\ni:B^{2}=A$ +\end_inset + +. + ZDB Za vsako pozitivno semidefinitno matriko +\begin_inset Formula $A$ +\end_inset + + obstajaja taka unitarna pozitivno semidefinitna +\begin_inset Formula $B$ +\end_inset + +, + da velja +\begin_inset Formula $B^{2}=A$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + + in +\begin_inset Formula $P^{*}=P^{-1}$ +\end_inset + + in +\begin_inset Formula $D=\left[\begin{array}{ccc} +\lambda_{1} & & 0\\ + & \ddots\\ +0 & & \lambda_{n} +\end{array}\right]$ +\end_inset + +. + Definirajmo +\begin_inset Formula $E=\left[\begin{array}{ccc} +\sqrt{\lambda_{1}} & & 0\\ + & \ddots\\ +0 & & \sqrt{\lambda_{n}} +\end{array}\right]\geq0$ +\end_inset + +. + Naj bo +\begin_inset Formula $B=PEP^{-1}=PEP^{*}$ +\end_inset + +. + Opazimo +\begin_inset Formula $B^{*}=B$ +\end_inset + +, + kajti +\begin_inset Formula $\left(PEP^{-1}\right)^{*}=\left(PEP^{*}\right)^{*}=PE^{*}P^{*}=PEP^{-1}=PEP^{*}$ +\end_inset + +, + ker je +\begin_inset Formula $E^{*}=E$ +\end_inset + +, + ker je +\begin_inset Formula $\forall a\in\mathbb{R}:\sqrt{a}\in\mathbb{R}$ +\end_inset + +. + Oglejmo si +\begin_inset Formula $B^{2}=PEP^{-1}PEP^{-1}=PE^{2}P^{-1}=PDP^{-1}$ +\end_inset + +. + Tako definiramo +\begin_inset Formula $\sqrt{A}=B$ +\end_inset + + (tu +\begin_inset Formula $\sqrt{}$ +\end_inset + + ni funkcija, + kot pri JKF, + temveč nov operator). +\end_layout + +\begin_layout Claim* +Naslednje trditve so ekvivalentne (zamenjamo lahko +\begin_inset Formula $\geq$ +\end_inset + + in +\begin_inset Formula $>$ +\end_inset + +): +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $A\geq0$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A=A^{*}$ +\end_inset + + in vse lastne vrednosti so +\begin_inset Formula $\geq0$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + + za nek unitaren +\begin_inset Formula $P$ +\end_inset + + in diagonalen +\begin_inset Formula $D\geq0$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A=A^{*}$ +\end_inset + + in obstaja +\begin_inset Formula $\sqrt{A}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A=B^{*}B$ +\end_inset + + za neko nenujno kvadratno matriko +\begin_inset Formula $B$ +\end_inset + + (za pozitivno definitno zahtevamo, + da ima +\begin_inset Formula $B$ +\end_inset + + LN stolpce). +\end_layout + +\end_deeper +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Claim* +klasifikacija skalarnih produktov na +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + in +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Naj bo +\begin_inset Formula $\left\langle u,v\right\rangle $ +\end_inset + + standardni skalarni produkt na +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + +\begin_inset Formula $u=\left(\alpha_{1},\dots,\alpha_{n}\right)$ +\end_inset + + in +\begin_inset Formula $v=\left(\beta_{1},\dots,\beta_{n}\right)$ +\end_inset + + in velja +\begin_inset Formula $\left\langle u,v\right\rangle =\alpha_{1}\overline{\beta_{1}}+\cdots+\alpha_{n}\overline{\beta_{n}}=\left[\begin{array}{ccc} +\overline{\beta_{1}} & \cdots & \overline{\beta_{n}}\end{array}\right]\left[\begin{array}{c} +\alpha_{1}\\ +\vdots\\ +\alpha_{n} +\end{array}\right]=v^{*}\cdot u$ +\end_inset + +. + Za +\begin_inset Formula $A>0$ +\end_inset + + definirajmo +\begin_inset Formula $\left[u,v\right]=\left\langle Au,v\right\rangle =v^{*}Au$ +\end_inset + +. + Trdimo, + da je +\begin_inset Formula $\left[\cdot,\cdot\right]$ +\end_inset + + spet skalarni produkt na +\begin_inset Formula $\mathbb{R}^{n}/\mathbb{C}^{n}$ +\end_inset + + in da je vsak skalarni produkt v +\begin_inset Formula $\mathbb{R}^{n}/\mathbb{C}^{n}$ +\end_inset + + take oblike. +\end_layout + +\begin_layout Proof +Dokazujemo oba dela trditve: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $\left[\cdot,\cdot\right]$ +\end_inset + + je skalarni produkt +\end_layout + +\begin_deeper +\begin_layout Enumerate +pozitivna semidefinitnost: + +\begin_inset Formula $\forall u\not=0:\left[u,u\right]=\left\langle Au,u\right\rangle \geq0$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +konjutirana simetričnost: + +\begin_inset Formula $\forall u,v:\left[u,v\right]=\left\langle Au,v\right\rangle =\left\langle u,A^{*}v\right\rangle =\left\langle u,Av\right\rangle =\overline{\left\langle Av,u\right\rangle }=\overline{\left[v,u\right]}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Linearnost in homogenost: + +\begin_inset Formula $\forall\alpha_{1},\alpha_{2},u_{1}u_{2},v:\left[\alpha_{1}u_{1}+\alpha_{2}u_{2},v\right]=\left\langle A\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right),v\right\rangle =\left\langle \alpha_{1}Au_{1}+\alpha_{2}Au_{2},v\right\rangle =\alpha_{1}\left\langle Au_{1},v\right\rangle +\alpha_{2}\left\langle Au_{2},v\right\rangle =\alpha_{1}\left[u,v\right]+\alpha_{2}\left[u,v\right]$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Enumerate +Za vsak skalarni produkt +\begin_inset Formula $\left[\cdot,\cdot\right]$ +\end_inset + + na +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + + obstaja taka pozitivno definitna matrika +\begin_inset Formula $A$ +\end_inset + +, + da velja +\begin_inset Formula $\forall u,v\in\mathbb{C}^{n}:\left[u,v\right]=\left\langle Au,v\right\rangle =v^{*}Au$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Standard +Naj bo +\begin_inset Formula $e_{1},\dots,e_{n}$ +\end_inset + + standardna baza za +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Definirajmo +\begin_inset Formula $A=\left[\begin{array}{ccc} +\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\ +\vdots & & \vdots\\ +\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right] +\end{array}\right]$ +\end_inset + +. + Velja +\begin_inset Formula $A=A^{*}$ +\end_inset + +: +\begin_inset Formula +\[ +A^{*}=\left[\begin{array}{ccc} +\overline{\left[e_{1},e_{1}\right]} & \cdots & \overline{\left[e_{1},e_{n}\right]}\\ +\vdots & & \vdots\\ +\overline{\left[e_{n},e_{1}\right]} & \cdots & \overline{\left[e_{n},e_{n}\right]} +\end{array}\right]=\left[\begin{array}{ccc} +\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\ +\vdots & & \vdots\\ +\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right] +\end{array}\right]=A +\] + +\end_inset + + Preveriti je treba še +\begin_inset Formula $\forall u,v\in\mathbb{C}^{n}:\left[u,v\right]=v^{*}Au$ +\end_inset + +. + +\begin_inset Formula $u=\alpha_{1}e_{1}+\cdots+\alpha_{n}e_{n}$ +\end_inset + + in +\begin_inset Formula $v=\beta_{1}e_{1}+\cdots+\beta_{n}e_{n}$ +\end_inset + +. + Tedaj je +\begin_inset Formula +\[ +\left[u,v\right]=\left[\alpha_{1}e_{1}+\cdots+\alpha_{n}e_{n},\beta_{1}e_{1}+\cdots+\beta_{n}e_{n}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left(\alpha_{1}\overline{\beta_{1}}\left[e_{1},e_{1}\right]+\cdots+\alpha_{1}\overline{\beta_{n}}\left[e_{1},e_{n}\right]\right)+\cdots+\left(\alpha_{n}\overline{\beta_{1}}\left[e_{n},e_{1}\right]+\cdots+\alpha_{n}\overline{\beta_{n}}\left[e_{n},e_{n}\right]\right)= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left[\begin{array}{ccc} +\overline{\beta_{1}} & \cdots & \overline{\beta_{n}}\end{array}\right]\left[\begin{array}{ccc} +\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\ +\vdots & & \vdots\\ +\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right] +\end{array}\right]\left[\begin{array}{c} +\alpha_{1}\\ +\vdots\\ +\alpha_{n} +\end{array}\right]=v^{*}Au=\left\langle Au,v\right\rangle +\] + +\end_inset + +Da je +\begin_inset Formula $A$ +\end_inset + + pozitivno definitna sledi, + saj mora za vsak neničeln +\begin_inset Formula $u$ +\end_inset + + po aksiomu za pozitivno definitnost skalarnega produkta veljati +\begin_inset Formula $\left\langle Au,u\right\rangle >0$ +\end_inset + +. +\end_layout + +\end_deeper +\end_deeper +\begin_layout Subsubsection +Singularni razcep (angl. + singular value decomposition — + SVD) +\end_layout + +\begin_layout Standard +Naj bo +\begin_inset Formula $A_{n\times n}$ +\end_inset + + neka kompleksna ali realna matrika. + Tedaj je +\begin_inset Formula $A^{*}A$ +\end_inset + + hermitska ( +\begin_inset CommandInset ref +LatexCommand ref +reference "subsec:Simetrične/hermitske-matrike" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +) matrika dimenzij +\begin_inset Formula $n\times n$ +\end_inset + +. + Ker je +\begin_inset Formula $\forall u:\left\langle A^{*}Au,u\right\rangle =\left\langle Au,Au\right\rangle \geq0$ +\end_inset + +, + je +\begin_inset Formula $A^{*}A$ +\end_inset + + pozitivno semidefinitna, + torej so vse njene lastne vrednosti +\begin_inset Formula $\geq0$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Singularne vrednosti +\begin_inset Formula $A$ +\end_inset + + so kvadratni koreni lastnih vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Če je +\begin_inset Formula $A$ +\end_inset + + normalna in +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost +\begin_inset Formula $A$ +\end_inset + +, + obstaja tak +\begin_inset Formula $v\not=0\ni:Av=\lambda v\Rightarrow A^{*}v=\overline{\lambda}v$ +\end_inset + +. + Odtod sledi, + da je +\begin_inset Formula $A^{*}Av=A^{*}\lambda v=\lambda A^{*}v=\lambda\overline{\lambda}v$ +\end_inset + +, + torej je +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost matrike +\begin_inset Formula $A^{*}A$ +\end_inset + +. + Po definiciji singularne vrednosti je +\begin_inset Formula $\sqrt{\lambda\overline{\lambda}}=\sqrt{\left|\lambda\right|^{2}}=\left|\lambda\right|$ +\end_inset + + singularna vrednost matrike +\begin_inset Formula $A$ +\end_inset + +. + Potemtakem so singularne vrednostni normalnih matrik enake absolutnim vrednosti lastnih vrednosti. +\end_layout + +\begin_layout Standard +Nekatere lastne vrednosti so ničelne, + nekatere pa od nič strogo večje. + Koliko je katerih? + Število ničelnih singularnih vrednosti matrike +\begin_inset Formula $A$ +\end_inset + + je število ničelnih lastnih vrednosti matrike +\begin_inset Formula $A^{*}A$ +\end_inset + +. + Ker je +\begin_inset Formula $A^{*}A$ +\end_inset + + hermitska, + je diagonalizabilna, + zato je algebraična večkratnost lastne vrednosti 0 enaka geometrijski večkratnosti lastne vrednosti 0, + slednja pa je definirana kot +\begin_inset Formula $\dim\Ker A^{*}A$ +\end_inset + +. + Ko upoštevamo +\begin_inset Formula $\dim\Ker A^{*}A=\dim\Ker A$ +\end_inset + +, + izvemo, + da je število ničelnih singularnih vrednosti matrike +\begin_inset Formula $A$ +\end_inset + + njena ničnost ( +\begin_inset Formula $\n A$ +\end_inset + +). + Ker je +\begin_inset Formula $A^{*}A$ +\end_inset + + velikosti +\begin_inset Formula $n\times n$ +\end_inset + +, + ima +\begin_inset Formula $A$ +\end_inset + + +\begin_inset Formula $n$ +\end_inset + + singularnih vrednosti, + torej je število neničelnih singularnih vrednosti +\begin_inset Formula $A$ +\end_inset + + enako +\begin_inset Formula $n-\Ker A$ +\end_inset + +. + Upoštevajoč osnovni dimenzijski izrek jedra in slike, + velja +\begin_inset Formula $\n A+\rang A=n$ +\end_inset + +, + torej je neničelnih singularnih vrednosti +\begin_inset Formula $=\rang A=\dim\Slika A$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Za +\begin_inset Formula $m\times n$ +\end_inset + + matriko velja +\begin_inset Formula $\rang A\le\min\left\{ m,n\right\} $ +\end_inset + +. +\end_layout + +\begin_layout Definition* +posplošitev pojma diagonalne matrike na nekvadratne matrike. + Matrika +\begin_inset Formula $D_{m\times n}$ +\end_inset + + je diagonalna, + če velja +\begin_inset Formula $\forall i:\left\{ 1..m\right\} ,j\in\left\{ 1..n\right\} :i\not=j\Rightarrow D_{i.j}=0$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Primeri pravokotnih diagonalnih matrik: +\begin_inset Formula +\[ +\left[\begin{array}{cccc} +1 & 0 & 0 & 0\\ +0 & 2 & 0 & 0\\ +0 & 0 & 3 & 0 +\end{array}\right],\left[\begin{array}{ccc} +1 & 0 & 0\\ +0 & 2 & 0\\ +0 & 0 & 3\\ +0 & 0 & 0 +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Theorem* +singularni razcep. + Naj bo +\begin_inset Formula $A$ +\end_inset + + kompleksna +\begin_inset Formula $m\times n$ +\end_inset + + matrika. + Potem obstajata taki unitarni +\begin_inset Formula $Q_{1},Q_{2}$ +\end_inset + + in taka diagonalna +\begin_inset Formula $D$ +\end_inset + + z diagonalci +\begin_inset Formula $\geq0\ni:A=Q_{1}DQ_{2}^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Diagonalci +\begin_inset Formula $D$ +\end_inset + + so ravno singularne vrednosti matrike +\begin_inset Formula $A$ +\end_inset + +. + Ker je +\begin_inset Formula $Q_{2}$ +\end_inset + + unitarna, + je +\begin_inset Formula $Q_{2}^{*}=Q_{2}^{-1}\Rightarrow A=Q_{1}DQ_{2}^{*}$ +\end_inset + +. + Če +\begin_inset Formula $A=Q_{1}DQ_{2}^{*}$ +\end_inset + +, + je +\begin_inset Formula $A^{*}=Q_{2}^{**}D^{*}Q_{1}^{*}=Q_{2}D^{*}Q_{1}^{*}$ +\end_inset + + in +\begin_inset Formula $A^{*}A=Q_{2}D^{*}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{2}D^{*}DQ_{2}^{*}$ +\end_inset + +, + torej je +\begin_inset Formula $A^{*}A$ +\end_inset + + podobna +\begin_inset Formula $D^{*}D$ +\end_inset + +, + diagonalci +\begin_inset Formula $D^{*}D$ +\end_inset + + so lastne vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + + in stolpci +\begin_inset Formula $Q_{2}$ +\end_inset + + so lastni vektorji +\begin_inset Formula $A^{*}A$ +\end_inset + +. + Diagonalci +\begin_inset Formula $D$ +\end_inset + + so bodisi 0 bodisi kvadratni koreni od diagonalcev +\begin_inset Formula $D^{*}D$ +\end_inset + +, + torej kvadratni koreni lastnih vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + +, + torej singularne vrednosti od +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Proof +obstoj singularnega razcepa. + Konstruirajmo +\begin_inset Formula $Q_{1},D,Q_{2}$ +\end_inset + + in dokažimo veljavnost. +\end_layout + +\begin_deeper +\begin_layout Itemize +Konstrukcija +\begin_inset Formula $Q_{2}$ +\end_inset + +: + +\begin_inset Formula $A$ +\end_inset + + je +\begin_inset Formula $m\times n$ +\end_inset + + kompleksna. + Tvorimo +\begin_inset Formula $n\times n$ +\end_inset + + matriko +\begin_inset Formula $A^{*}A$ +\end_inset + +. + Izračunajmo lastne vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + + in jih uredimo padajoče — + +\begin_inset Formula $\lambda_{1}\geq\cdots\geq\lambda_{n}$ +\end_inset + +. + Naj bodo +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + pripadajoči lastni vektorji — + +\begin_inset Formula $A^{*}Av_{i}=\lambda_{i}v_{i}$ +\end_inset + +. + Te +\begin_inset Formula $v_{i}$ +\end_inset + + izberimo tako, + da so ortonormirani. + Lastni podprostori +\begin_inset Formula $A^{*}A$ +\end_inset + + so namreč paroma pravokotni, + saj je +\begin_inset Formula $A^{*}A$ +\end_inset + + normalna, + saj je hermitska. + V vsakem podprostoru vzamemo ONB in +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + je unija teh ON baz. + Definiramo +\begin_inset Formula $Q_{2}=\left[\begin{array}{ccc} +v_{1} & \cdots & v_{n}\end{array}\right]$ +\end_inset + +. + Ker so +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + ON, + je +\begin_inset Formula $Q_{2}$ +\end_inset + + unitarna. +\end_layout + +\begin_layout Itemize +Konstrukcija +\begin_inset Formula $D$ +\end_inset + +: + Naj bo +\begin_inset Formula $r\coloneqq\rang A$ +\end_inset + + (število ničelnih singularnih vrednosti +\begin_inset Formula $A$ +\end_inset + +). + Oglejmo si zaporedje lastnih vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + + +\begin_inset Formula $\lambda_{1}\geq\cdots>\lambda_{r+1}=\cdots=\lambda_{n}$ +\end_inset + +. + Lastne vrednosti po +\begin_inset Formula $r$ +\end_inset + + so ničelne, + ostale pa večje od 0. + Lastne vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + + v tem vrstnem redu so singularne vrednosti matrike +\begin_inset Formula $A$ +\end_inset + +: + +\begin_inset Formula $\sigma_{1}^{2}=\lambda_{1}\geq\cdots\geq\sigma_{r}^{2}=\lambda_{r}>\sigma_{r+1}^{2}=\cdots=\sigma_{n}^{2}=0$ +\end_inset + +. + Definiramo +\begin_inset Formula $D$ +\end_inset + + kot +\begin_inset Formula $m\times n$ +\end_inset + + diagonalno matriko takole: +\begin_inset Formula +\[ +D=\left[\begin{array}{cccccc} +\sigma_{1} & & & & & 0\\ + & \ddots\\ + & & \sigma_{r}\\ + & & & \sigma_{r+1}=0\\ + & & & & \ddots\\ +0 & & & & & \sigma_{n}=0 +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Itemize +Konstrukcija +\begin_inset Formula $Q_{1}$ +\end_inset + +: + +\begin_inset Formula $\forall i\in\left\{ 1..r\right\} :u_{i}\coloneqq\frac{1}{\sigma_{1}}Av_{i}$ +\end_inset + + za +\begin_inset Formula $v_{i}$ +\end_inset + + lastne vektorje +\begin_inset Formula $A^{*}A$ +\end_inset + +, + torej +\begin_inset Formula $A^{*}Av_{i}=\lambda_{i}v_{i}=\sigma_{i}^{2}v_{i}$ +\end_inset + +. + Pokažimo, + da je +\begin_inset Formula $u_{1},\dots,u_{r}$ +\end_inset + + ON množica. + +\begin_inset Formula $\forall i,j\in\left\{ 1..r\right\} :$ +\end_inset + + +\begin_inset Formula +\[ +\left\langle u_{i},u_{j}\right\rangle =\left\langle \frac{1}{\sigma_{j}}Av_{i},\frac{1}{\sigma_{j}}Av_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle Av_{i},Av_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle A^{*}Av_{i},v_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle \lambda_{i}v_{i},v_{j}\right\rangle =\frac{\lambda_{i}=\sigma_{i}^{\cancel{2}}}{\cancel{\sigma}_{i}\sigma_{j}}\left\langle v_{i},v_{j}\right\rangle = +\] + +\end_inset + + +\begin_inset Formula +\[ +=\frac{\sigma_{i}}{\sigma_{j}}\left\langle v_{i},v_{j}\right\rangle =\begin{cases} +0 & ;i\not=j\\ +1 & ;i=j +\end{cases} +\] + +\end_inset + +Sedaj ONM +\begin_inset Formula $u_{1},\dots,u_{r}$ +\end_inset + + z +\begin_inset Formula $u_{r+1},\dots,u_{n}$ +\end_inset + + dopolnimo do ONB za +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + + (GS). + Definiramo +\begin_inset Formula $Q_{1}=\left[\begin{array}{ccc} +u_{1} & \cdots & u_{n}\end{array}\right]$ +\end_inset + +. + Ker so stolpci ONB, + je matrika unitarna. +\end_layout + +\begin_layout Itemize +Sedaj preverimo, + da velja +\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}\Leftrightarrow AQ_{2}=Q_{1}D$ +\end_inset + +. +\begin_inset Formula +\[ +AQ_{2}=A\left[\begin{array}{ccc} +v_{1} & \cdots & v_{n}\end{array}\right]=\left[\begin{array}{ccc} +Av_{1} & \cdots & Av_{n}\end{array}\right]=\cdots +\] + +\end_inset + +Upoštevamo, + da +\begin_inset Formula $i>r\Rightarrow\lambda_{i}=0\Rightarrow A^{*}Av_{i}=\lambda_{i}v_{i}=0\Rightarrow v_{i}\in\Ker A^{*}A\Rightarrow v_{i}\in\Ker A\Leftrightarrow Av_{i}=0$ +\end_inset + +: +\begin_inset Formula +\[ +\cdots=\left[\begin{array}{ccc} +Av_{1} & \cdots & Av_{n}\end{array}\right]=\left[\begin{array}{cccccc} +Av_{1} & \cdots & Av_{r} & 0 & \cdots & 0\end{array}\right] +\] + +\end_inset + +Sedaj izračunajmo še +\begin_inset Formula +\[ +Q_{1}D=\left[\begin{array}{ccc} +u_{1} & \cdots & u_{n}\end{array}\right]\left[\begin{array}{cccccc} +\sigma_{1}\\ + & \ddots\\ + & & \sigma_{r}\\ + & & & 0\\ + & & & & \ddots\\ + & & & & & 0 +\end{array}\right]=\left[\begin{array}{cccccc} +\sigma_{1}u_{1} & \cdots & \sigma_{r}u_{r} & 0 & \cdots & 0\end{array}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left[\begin{array}{cccccc} +\cancel{\sigma_{1}\frac{1}{\sigma_{1}}}Av_{i} & \cdots & \cancel{\sigma_{r}\frac{1}{\sigma_{r}}}Av_{r} & 0 & \cdots & 0\end{array}\right]=\left[\begin{array}{cccccc} +Av_{i} & \cdots & Av_{r} & 0 & \cdots & 0\end{array}\right]=AQ_{2} +\] + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Example* +Poišči singularni razcep +\begin_inset Formula $A=\left[\begin{array}{cccc} +1 & 1 & -1 & -1\\ +-1 & 0 & 1 & 0\\ +0 & -1 & 0 & 1 +\end{array}\right]$ +\end_inset + +. + Izračunajmo +\begin_inset Formula +\[ +A^{*}A=\left[\begin{array}{cccc} +1 & 1 & -1 & -1\\ +-1 & 0 & 1 & 0\\ +0 & -1 & 0 & 1 +\end{array}\right]\left[\begin{array}{ccc} +1 & -1 & 0\\ +1 & 0 & -1\\ +-1 & 1 & 0\\ +-1 & 0 & 1 +\end{array}\right]=\cdots=\left[\begin{array}{cccc} +2 & 1 & -2 & -1\\ +1 & 2 & -1 & -2\\ +-2 & -1 & 2 & 1\\ +-1 & -2 & 1 & 2 +\end{array}\right] +\] + +\end_inset + +Izračunajmo +\begin_inset Formula $p_{A^{*}A}\left(x\right)=\det\left(A^{*}A-xI\right)=\cdots=x^{2}\left(x-2\right)\left(x-6\right)$ +\end_inset + +, + torej +\begin_inset Formula $\lambda_{1}=6$ +\end_inset + +, + +\begin_inset Formula $\lambda_{2}=2$ +\end_inset + +, + +\begin_inset Formula $\lambda_{3}=0$ +\end_inset + +, + +\begin_inset Formula $\lambda_{4}=0$ +\end_inset + +, + torej +\begin_inset Formula $\sigma_{1}=\sqrt{6}$ +\end_inset + + in +\begin_inset Formula $\sigma_{2}=\sqrt{2}$ +\end_inset + + ter +\begin_inset Formula $\sigma_{3}=\sigma_{4}=0$ +\end_inset + +. + (ujema se z dejstvom, + da je +\begin_inset Formula $\rang A=2$ +\end_inset + +). + Izračunajmo lastne vektorje +\begin_inset Formula $A^{*}A$ +\end_inset + +: +\begin_inset Formula +\[ +\lambda_{1}=6:\quad v_{1}'=\left[\begin{array}{c} +1\\ +1\\ +-1\\ +-1 +\end{array}\right],\quad\left|\left|v_{1}'\right|\right|=6 +\] + +\end_inset + + +\begin_inset Formula +\[ +\lambda_{2}=2:\quad v_{2}'=\left[\begin{array}{c} +1\\ +-1\\ +-1\\ +1 +\end{array}\right],\quad\left|\left|v_{2}'\right|\right|=2 +\] + +\end_inset + + +\begin_inset Formula +\[ +\lambda_{3}=\lambda_{4}:\quad v_{3}'=\left[\begin{array}{c} +1\\ +0\\ +1\\ +0 +\end{array}\right],v_{4}'=\left[\begin{array}{c} +0\\ +1\\ +0\\ +2 +\end{array}\right],\quad\left|\left|v_{3}'\right|\right|=\sqrt{2},\left|\left|v_{4}'\right|\right|=\sqrt{2} +\] + +\end_inset + + +\end_layout + +\begin_layout Example* +Z Gram-Schmidtom naredimo ortogonalno množico (v tem primeru so že ortogonalni) in jih normirajmo: +\begin_inset Formula +\[ +v_{1}=\frac{1}{2}\left[\begin{array}{c} +1\\ +1\\ +-1\\ +-1 +\end{array}\right],\quad v_{2}=\frac{1}{2}\left[\begin{array}{c} +1\\ +-1\\ +-1\\ +1 +\end{array}\right],\quad v_{3}=\frac{1}{\sqrt{2}}\left[\begin{array}{c} +1\\ +0\\ +1\\ +0 +\end{array}\right],\quad v_{4}=\frac{1}{\sqrt{2}}\left[\begin{array}{c} +0\\ +1\\ +0\\ +1 +\end{array}\right]. +\] + +\end_inset + +Sestavimo +\begin_inset Formula +\[ +Q_{2}=\left[\begin{array}{cccc} +\frac{1}{2} & \frac{1}{2} & \frac{1}{\sqrt{2}} & 0\\ +\frac{1}{2} & -\frac{1}{2} & 0 & \frac{1}{\sqrt{2}}\\ +-\frac{1}{2} & -\frac{1}{2} & \frac{1}{\sqrt{2}} & 0\\ +-\frac{1}{2} & \frac{1}{2} & 0 & \frac{1}{\sqrt{2}} +\end{array}\right],\quad D=\left[\begin{array}{cccc} +\sqrt{6} & & & 0\\ + & \sqrt{2}\\ + & & 0\\ +0 & & & 0 +\end{array}\right] +\] + +\end_inset + +Izračunamo +\begin_inset Formula $u_{1},\dots,u_{r}$ +\end_inset + + za +\begin_inset Formula $Q_{1}$ +\end_inset + +: +\begin_inset Formula +\[ +u_{1}=\frac{1}{\sigma_{1}}Av_{1}=\frac{1}{\sqrt{6}}\left[\begin{array}{c} +2\\ +-1\\ +-1 +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +u_{2}=\frac{1}{\sigma_{2}}Av_{2}=\frac{1}{\sqrt{2}}\left[\begin{array}{c} +0\\ +-1\\ +1 +\end{array}\right] +\] + +\end_inset + +Dopolnimo ju do ONB za +\begin_inset Formula $\mathbb{R}^{3}$ +\end_inset + + z Gram-Schmidtom (oz. + uganemo +\begin_inset Formula $\left[\begin{array}{c} +1\\ +1\\ +1 +\end{array}\right]$ +\end_inset + +). + Dopolnitev normiramo: + +\begin_inset Formula $u_{3}=\frac{1}{\sqrt{3}}\left[\begin{array}{c} +1\\ +1\\ +1 +\end{array}\right]$ +\end_inset + + in vektorje vstavimo v +\begin_inset Formula $Q_{1}$ +\end_inset + +: +\begin_inset Formula +\[ +Q_{1}=\left[\begin{array}{ccc} +\frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}}\\ +-\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}}\\ +-\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Example* +Iskani razcep je +\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}$ +\end_inset + + (Izračunati je potrebno še en inverz — + +\begin_inset Formula $Q_{2}^{-1}$ +\end_inset + + namreč). +\end_layout + +\begin_layout Subsubsection +Psevdoinverz — + Moore-Penroseov inverz +\end_layout + +\begin_layout Standard +Psevdoinverz je posplošitev inverza na nenujno kvadratne nenujno obrnljive matrike. + Najprej diagonalne matrike: + Njihov navaden inverz je takšen: +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +d_{11} & & 0\\ + & \ddots\\ +0 & & d_{nn} +\end{array}\right]^{-1}=\left[\begin{array}{ccc} +d_{11}^{-1} & & 0\\ + & \ddots\\ +0 & & d_{nn}^{-1} +\end{array}\right] +\] + +\end_inset + +Kadar je diagonalec ničeln, + kot element polja nima multiplikativnega inverza. + Ideja za posplošeni inverz diagonelne matrike: + take diagonalce pustimo na 0, + torej na primer: +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +1 & & 0\\ + & 2\\ +0 & & 0 +\end{array}\right]^{+}=\left[\begin{array}{ccc} +1 & & 0\\ + & \frac{1}{2}\\ +0 & & 0 +\end{array}\right] +\] + +\end_inset + +Za nekvadratne diagonalne matrike pa takole: +\begin_inset Formula +\[ +\left[\begin{array}{cccc} +1 & & & 0\\ + & 2\\ +0 & & 0 +\end{array}\right]^{+}=\left[\begin{array}{ccc} +1 & & 0\\ + & \frac{1}{2}\\ + & & 0\\ +0 +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Definition* +posplošeni inverz diagonalne matrike. + Naj bo +\begin_inset Formula $D$ +\end_inset + + diagonalna +\begin_inset Formula $m\times n$ +\end_inset + + z neničelnimi diagonalci +\begin_inset Formula $d_{1},\dots d_{r}$ +\end_inset + +, + je +\begin_inset Formula $D^{+}$ +\end_inset + + diagonalna +\begin_inset Formula $n\times m$ +\end_inset + + z neničelnimi diagonalci +\begin_inset Formula $\frac{1}{d_{1}^{-1}},\dots,\frac{1}{d_{r}^{-1}}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Za diagonalno +\begin_inset Formula $D$ +\end_inset + + opazimo +\begin_inset Formula $D^{++}=D$ +\end_inset + + in za obrnljivo diagonalno +\begin_inset Formula $D$ +\end_inset + + opazimo +\begin_inset Formula $D^{+}=D^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Sedaj bi radi pojem posplošili na nediagonalne matrike — + to storimo s pomočjo SVD. + +\begin_inset Formula $A=Q_{1}DQ_{2}^{*}\ni:D$ +\end_inset + + diagonalna in +\begin_inset Formula $Q_{1},Q_{2}$ +\end_inset + + unitarni. + Tedaj velja +\begin_inset Formula $A$ +\end_inset + + obrnljiva +\begin_inset Formula $\Leftrightarrow D$ +\end_inset + + obrnljiva, + kajti +\begin_inset Formula $A^{-1}=Q_{2}D^{-1}D_{1}^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Za splošen nenujno obrnljiv +\begin_inset Formula $A$ +\end_inset + + definiramo +\begin_inset Formula $A^{+}\coloneqq Q_{2}D^{+}Q_{1}^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Fact* +Opazimo: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $A^{++}=\left(Q_{2}DQ_{1}^{-1}\right)^{+}=Q_{1}D^{++}Q_{2}^{-1}=Q_{1}DQ_{2}^{-1}=A$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $A$ +\end_inset + + obrnljiva: + +\begin_inset Formula $A^{+}=A^{-1}$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Claim* +osnovne lastnosti psevdoinverza. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $AA^{+}A=A$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(A^{+}A\right)^{*}=A^{+}A$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A^{+}AA^{+}=A^{+}$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(AA^{+}\right)^{*}=AA^{+}$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Proof +Dokažimo te 4 lastnosti najprej za +\begin_inset Formula $D$ +\end_inset + + in nato za SVD. + Pri +\begin_inset Formula $D$ +\end_inset + + predpostavimo, + da so ničle spodaj desno, + sicer obstaja permutacijska matrika, + ki je ortogonalna, + s katero lahko množimo +\begin_inset Formula $D$ +\end_inset + +, + da jo pretvorimo v željeno obliko (in potem dokaz take +\begin_inset Formula $D$ +\end_inset + + pade v primer SVD): +\end_layout + +\begin_deeper +\begin_layout Itemize +Diagonalen primer. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $DD^{+}D=$ +\end_inset + + +\begin_inset Formula +\[ +\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]\left[\begin{array}{cccccc} +d_{1}^{-1} & & & & & 0\\ + & \ddots\\ + & & d_{r}^{-1}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left[\begin{array}{cccccc} +1 & & & & & 0\\ + & \ddots\\ + & & 1\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]=D +\] + +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $D^{+}DD^{+}=\cdots=D^{+}$ +\end_inset + + na podoben način +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(DD^{+}\right)^{*}=\left[\begin{array}{cccccc} +1 & & & & & 0\\ + & \ddots\\ + & & 1\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]^{*}=\left[\begin{array}{cccccc} +1 & & & & & 0\\ + & \ddots\\ + & & 1\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]=DD^{+}$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(D^{+}D\right)^{*}=\cdots=D^{+}D$ +\end_inset + + podobno +\end_layout + +\end_deeper +\begin_layout Itemize +Splošen primer +\begin_inset Formula $A$ +\end_inset + + — + vstavimo +\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $AA^{+}A=Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{1}DD^{+}DQ_{2}^{*}=Q_{1}DQ_{2}^{*}=A$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A^{+}AA^{+}=\cdots=A^{+}$ +\end_inset + + na podoben način +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(AA^{+}\right)^{*}=\left(Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}\right)^{*}=\left(Q_{1}DD^{+}Q_{1}^{*}\right)^{*}=Q_{1}\left(DD^{+}\right)^{*}Q_{1}^{*}=Q_{1}DD^{+}Q_{1}^{*}=Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}=AA^{+}$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(A^{+}A\right)^{*}=\cdots=A^{+}A$ +\end_inset + + podobno +\end_layout + +\end_deeper +\end_deeper +\begin_layout Remark* +\begin_inset Formula $A$ +\end_inset + + obrnljiva +\begin_inset Formula $\Leftrightarrow D$ +\end_inset + + obrnljiva, + torej +\begin_inset Formula $A^{+}=Q_{2}D^{+}Q_{1}^{-1}=Q_{2}D^{-1}Q_{1}^{-1}=A^{-1}$ +\end_inset + +. + Potemtakem za obrnljivo +\begin_inset Formula $A$ +\end_inset + + velja +\begin_inset Formula $A^{+}=A^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Da je definicija dobra, + je treba dokazati, + da je +\begin_inset Formula $A^{+}$ +\end_inset + + enoličen ne glede na SVD, + kajti SVD za +\begin_inset Formula $A$ +\end_inset + + ni enoličen. + Naj bo +\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}=Q_{3}EQ_{4}^{-1}$ +\end_inset + +, + njen prvi psevsoinverz +\begin_inset Formula $B=Q_{2}D^{+}Q_{1}^{-1}$ +\end_inset + + in njen drugi psevdoinverz +\begin_inset Formula $C=Q_{4}D^{+}Q_{3}^{-1}$ +\end_inset + +. + Ali velja +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +udensdash{$B +\backslash +overset{?}{=}C$} +\end_layout + +\end_inset + +? + Velja +\begin_inset Formula +\[ +AB=\left(ACA\right)B=ACAB=\left(AC\right)^{*}\left(AB\right)^{*}=C^{*}A^{*}B^{*}A^{*}=C^{*}\left(ABA\right)^{*}=C^{*}A^{*}=\left(AC\right)^{*}=AC +\] + +\end_inset + +in +\begin_inset Formula +\[ +BA=B\left(ACA\right)=BACA=\left(BA\right)^{*}\left(CA\right)^{*}=A^{*}B^{*}A^{*}C^{*}=\left(ABA\right)^{*}C^{*}=A^{*}C^{*}=\left(CA\right)^{*}=CA +\] + +\end_inset + +ter nazadnje še +\begin_inset Formula +\[ +B=BAB=CAB=CAC=C. +\] + +\end_inset + + +\end_layout + +\begin_layout Paragraph +Kako izračunamo +\begin_inset Formula $A^{+}$ +\end_inset + + brez SVD? +\end_layout + +\begin_layout Standard +Če je +\begin_inset Formula $A$ +\end_inset + + pozitivno semidefinitna, + jo lahko diagonaliziramo v ortonormirani bazi: + +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + +, + da ima +\begin_inset Formula $D$ +\end_inset + + pozitivne diagonalce in da je +\begin_inset Formula $P^{-1}=P^{*}$ +\end_inset + +. + Opazimo, + da je to SVD od +\begin_inset Formula $A$ +\end_inset + +, + kajti +\begin_inset Formula $Q_{1}=P,Q_{2}=P,D=D$ +\end_inset + + in tedaj +\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}=PDP^{-1}$ +\end_inset + +. + Potemtakem je +\begin_inset Formula $A^{+}=PD^{+}P^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Za splošno matriko +\begin_inset Formula $A$ +\end_inset + + (nenujno pozitivno semidefinitno) pa velja +\begin_inset Formula $A^{+}=\left(A^{*}A\right)^{+}A^{*}=A^{*}\left(AA^{*}\right)^{+}$ +\end_inset + +. + +\begin_inset Formula $A^{*}A$ +\end_inset + + in +\begin_inset Formula $AA^{*}$ +\end_inset + + sta pozitivno semidefinitni. +\end_layout + +\begin_layout Proof +Najprej bomo preverili za diagonalno, + nato za SVD: +\end_layout + +\begin_deeper +\begin_layout Itemize +Diagonalna +\begin_inset Formula $D_{n\times m}$ +\end_inset + +: + +\begin_inset Formula +\[ +D=\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ + & & & & & 0 +\end{array}\right],\quad D^{*}=\left[\begin{array}{cccccc} +\overline{d_{1}} & & & & & 0\\ + & \ddots\\ + & & \overline{d_{r}}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right],\quad D^{*}D=\left[\begin{array}{cccccc} +\frac{1}{\overline{d_{1}}d_{1}} & & & & & 0\\ + & \ddots\\ + & & \frac{1}{\overline{d_{r}}d_{r}}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right], +\] + +\end_inset + + +\begin_inset Formula +\[ +\left(D^{*}D\right)^{+}=\left[\begin{array}{cccccc} +\frac{1}{\overline{d_{1}}d_{1}} & & & & & 0\\ + & \ddots\\ + & & \frac{1}{\overline{d_{r}}d_{r}}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right],\quad\left(D^{*}D\right)^{+}D^{*}=\left[\begin{array}{cccccc} +\frac{\cancel{\overline{d_{1}}}}{\cancel{\overline{d_{1}}}d_{1}} & & & & & 0\\ + & \ddots\\ + & & \frac{\cancel{\overline{d_{r}}}}{\cancel{\overline{d_{r}}}d_{r}}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]=D^{+} +\] + +\end_inset + + +\end_layout + +\begin_layout Itemize +Za splošen +\begin_inset Formula $A$ +\end_inset + + uporabimo SVD, + da to dokažemo: + +\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}$ +\end_inset + +. + Velja +\begin_inset Formula $A^{*}A=Q_{2}D^{*}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{2}D^{*}DQ_{2}^{*}$ +\end_inset + + in +\begin_inset Formula $\left(A^{*}A\right)^{+}=Q_{2}\left(D^{*}D\right)^{+}Q_{2}^{*}$ +\end_inset + +. + Torej +\begin_inset Formula $\left(A^{*}A\right)^{+}A^{*}=Q_{2}\left(D^{*}D\right)^{+}Q_{2}^{*}Q_{2}D^{*}Q_{1}^{*}=Q_{2}\left(D^{*}D\right)^{+}D^{*}Q_{1}^{*}=Q_{2}DQ_{1}^{*}=A^{+}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Remark* +V posebnih primerih lahko poenostavljamo dalje. + Recimo, + da ima +\begin_inset Formula $A$ +\end_inset + + LN stolpce in je kvadratna +\begin_inset Formula $\Rightarrow\Ker A=\left\{ 0\right\} =\Ker A^{*}A$ +\end_inset + +, + torej +\begin_inset Formula $A^{*}A$ +\end_inset + + je obrnljiva in velja +\begin_inset Formula $\left(A^{*}A\right)^{-1}=\left(A^{*}A\right)^{+}$ +\end_inset + +. + Takrat torej velja +\begin_inset Formula $A^{+}=\left(A^{*}A\right)^{-1}A^{*}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +To smo uporabili pri iskanju posplošene rešitve predoločenega sistema: + Za sistem +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + + iščemo +\begin_inset Formula $\vec{x}$ +\end_inset + +, + da je +\begin_inset Formula $\left|\left|A\vec{x}-\vec{b}\right|\right|$ +\end_inset + + minimalen, + tedaj bo tak +\begin_inset Formula $\vec{x}$ +\end_inset + + posplošena rešitev sistema. + Vemo, + da je posplošena reštev +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + + enaka rešitvi od +\begin_inset Formula $A^{*}A\vec{x}=A^{*}\vec{b}$ +\end_inset + +, + kajti, + če ima +\begin_inset Formula $A$ +\end_inset + + LN stolpce, + je +\begin_inset Formula $A^{*}A$ +\end_inset + + obrnljiva (s tem dokažemo, + da ima ta sistem vedno rešitev): +\begin_inset Formula +\[ +A^{*}A\vec{x}=A^{*}\vec{b}\quad\quad\quad\quad/\cdot\left(A^{*}A\right)^{-1} +\] + +\end_inset + + +\begin_inset Formula +\[ +\vec{x}=\left(A^{*}A\right)^{-1}A^{*}\vec{b} +\] + +\end_inset + + +\begin_inset Formula +\[ +\vec{x}=A^{+}\vec{b} +\] + +\end_inset + + +\end_layout + +\begin_layout Paragraph +Uporaba psevdoinverza +\end_layout + +\begin_layout Standard +Vemo, + kaj je posplošena rešitev sistema +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + +. + Problem je, + da ima sistem lahko več posplošenih rešitev (to se lahko zgodi, + če +\begin_inset Formula $A$ +\end_inset + + nima LN stolpcev). + Med vsemi rešitvami iščemo tisto, + ki je najkrajša po normi — + +\begin_inset Formula $\left|\left|\vec{x}\right|\right|$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Najkrajša posplošena rešitev sistema +\begin_inset Formula $Ax=b$ +\end_inset + + je ravno +\begin_inset Formula $x=A^{+}b$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokažimo najprej za diagonalno matriko koeficientov, + nato pa še za splošen primer: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $Dx=b$ +\end_inset + + +\begin_inset Formula +\[ +D_{m\times n}=\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ + & & & & & 0 +\end{array}\right],\quad x=\left[\begin{array}{c} +x_{1}\\ +\vdots\\ +x_{n} +\end{array}\right],\quad b=\left[\begin{array}{c} +b_{1}\\ +\vdots\\ +b_{m} +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +\left|\left|Dx-b\right|\right|^{2}=\left|\left|\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ + & & & & & 0 +\end{array}\right]\left[\begin{array}{c} +x_{1}\\ +\vdots\\ +x_{n} +\end{array}\right]-\left[\begin{array}{c} +b_{1}\\ +\vdots\\ +b_{m} +\end{array}\right]\right|\right|^{2}=\left(d_{1}x_{1}-b_{1}\right)^{2}+\cdots+\left(d_{r}x_{r}-b_{r}\right)^{2}+b_{r+1}^{2}+\cdots+b_{m}^{2} +\] + +\end_inset + +Ta izraz doseže minimum, + ko +\begin_inset Formula $\left(d_{1}x_{1}-b_{1}\right)^{2}+\cdots+\left(d_{r}x_{r}-b_{r}\right)^{2}=0$ +\end_inset + +, + torej +\begin_inset Formula $x_{1}=\frac{b_{1}}{d_{1}},\dots,x_{r}=\frac{b_{r}}{d_{r}},x_{r+1}=\times,\dots,x_{n}=\times$ +\end_inset + +, + kjer +\begin_inset Formula $\times$ +\end_inset + + predstavlja poljubno vrednost. + Najkrajša rešitev bo torej tista, + kjer +\begin_inset Formula $x_{r+1}=\cdots=x_{n}=0$ +\end_inset + +. + Trdimo, + da je +\begin_inset Formula $\left(\frac{b_{1}}{d_{1}},\cdots,\frac{b_{r}}{d_{r}},0,\cdots,0\right)=D^{+}b$ +\end_inset + +. + Preverimo: +\begin_inset Formula +\[ +D_{n\times m}^{+}=\left[\begin{array}{cccccc} +d_{1}^{-1} & & & & & 0\\ + & \ddots\\ + & & d_{r}^{-1}\\ + & & & 0\\ + & & & & \ddots\\ + & & & & & 0 +\end{array}\right],\quad b=\left[\begin{array}{c} +b_{1}\\ +\vdots\\ +b_{m} +\end{array}\right],\quad D^{+}b=\left[\begin{array}{c} +\frac{b_{1}}{d_{1}}\\ +\vdots\\ +\frac{b_{m}}{d_{m}}\\ +0\\ +\vdots\\ +0 +\end{array}\right] +\] + +\end_inset + +Res je! +\end_layout + +\begin_layout Itemize +Splošen primer s SVD: + +\begin_inset Formula $A_{m\times n}=Q_{1}DQ_{2}^{*}$ +\end_inset + +, + kjer sta +\begin_inset Formula $Q_{1},Q_{2}$ +\end_inset + + ortogonalni in +\begin_inset Formula $D$ +\end_inset + + diagonalna. + Za tretji enačaj uporabimo dejstvo, + da množenje z ortogonalno matriko ohranja normo. +\begin_inset Foot +status open + +\begin_layout Plain Layout +\begin_inset Formula $\left|\left|Q_{2}^{*}x\right|\right|^{2}=\left\langle Q_{2}^{*}x,Q_{2}^{*}x\right\rangle =\left\langle x,Q_{2}Q_{2}^{*}x\right\rangle =\left\langle x,x\right\rangle =\left|\left|x\right|\right|^{2}$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Formula $\left|\left|Ax-b\right|\right|=\left|\left|Q_{1}DQ_{2}^{*}x-b\right|\right|=\left|\left|Q_{1}\left(DQ_{2}^{*}x-Q_{1}^{-1}b\right)\right|\right|=\left|\left|DQ_{2}^{*}x-Q_{1}^{-1}b\right|\right|=\left|\left|Dx'-c\right|\right|$ +\end_inset + + za +\begin_inset Formula $x'=Q_{2}^{*}x$ +\end_inset + + in +\begin_inset Formula $c=Q_{1}^{-1}b$ +\end_inset + +. + Ker je +\begin_inset Formula $Q_{2}$ +\end_inset + + obrnljiva, + velja, + da če +\begin_inset Formula $x$ +\end_inset + + preteče vse vektorje v +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +, + tudi +\begin_inset Formula $x'$ +\end_inset + + preteče vse vektorje v +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Standard +Potemtakem je +\begin_inset Formula $\min\left|\left|Ax-b\right|\right|=\min\left|\left|Dx'-c\right|\right|$ +\end_inset + +. + Če +\begin_inset Formula $\left|\left|Ax-b\right|\right|$ +\end_inset + + zavzame minimum v +\begin_inset Formula $x_{0}$ +\end_inset + +, + potem +\begin_inset Formula $\left|\left|Dx'-c\right|\right|$ +\end_inset + + zavzame minimum v +\begin_inset Formula $x_{0}'=Q_{2}^{-1}x_{0}$ +\end_inset + + in obratno, + če +\begin_inset Formula $\left|\left|Dx'-c\right|\right|$ +\end_inset + + zavzame minimum v +\begin_inset Formula $x_{0}'$ +\end_inset + +, + potem +\begin_inset Formula $\left|\left|Ax-b\right|\right|$ +\end_inset + + zavzame minimum v +\begin_inset Formula $x_{0}=Q_{2}x_{0}'$ +\end_inset + +. + Torej je +\begin_inset Formula $x\mapsto Q_{2}^{-1}x$ +\end_inset + + bijektivna korespondenca med posplošenimi rešitvami +\begin_inset Formula $Ax-b$ +\end_inset + + in posplošenimi rešitvami +\begin_inset Formula $Dx'-c$ +\end_inset + +. + Opazimo, + da preslikava ohranja normo, + torej +\begin_inset Formula $\left|\left|x_{0}'\right|\right|=\left|\left|Q_{2}x_{0}'\right|\right|=\left|\left|x_{0}\right|\right|$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Od prej vemmo, + da je najkrajša posplošena rešitev +\begin_inset Formula $Dx_{0}-c$ +\end_inset + + prav +\begin_inset Formula $x_{0}=D^{+}c$ +\end_inset + +. + Po zgornjem odstavku sledi, + da je +\begin_inset Formula $x_{0}=Q_{2}x_{0}'$ +\end_inset + + najkrajša posplošena rešitev od +\begin_inset Formula $Ax=b$ +\end_inset + +. + Dobimo namreč +\begin_inset Formula $x_{0}=Q_{2}x_{0}'=Q_{2}D^{+}c=Q_{2}D^{+}Q_{1}^{-1}b=A^{+}b$ +\end_inset + +. +\end_layout + +\end_deeper +\end_deeper +\begin_layout Subsection +Kvadratne forme +\end_layout + +\begin_layout Definition* +Forma je homogen polinom, + torej tak, + v katerem imajo vsi monomi isto stopnjo. + Stopnja monoma je +\begin_inset Formula $\deg\left(\beta x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}\right)\coloneqq\alpha_{1}+\cdots+\alpha_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Polinom je vsota monomov. + Stopnja polinoma je najvišja stopnja monoma v njem. +\end_layout + +\begin_layout Example* +Linearna forma v treh spremenljivkah: + +\begin_inset Formula $ax+by+cz=\left[\begin{array}{ccc} +a & b & c\end{array}\right]\left[\begin{array}{c} +x\\ +y\\ +z +\end{array}\right]$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Kvadratna forma je homogen polinom stopnje 2. + Primer kvadratne forme: +\begin_inset Formula +\[ +ax^{2}+bxy+cy^{2}=\left[\begin{array}{cc} +x & y\end{array}\right]\left[\begin{array}{cc} +a & b/2\\ +b/2 & a +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Kubična forma v treh spremenljivkah: +\begin_inset Formula +\[ +ax^{3}+by^{3}+cz^{3}+dx^{2}y+ex^{2}z+fy^{2}x+gy^{2}x+iz^{2}x+jz^{2}y+kxyz +\] + +\end_inset + + +\end_layout + +\begin_layout Definition* +Pravimo, + da sta matriki +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $B$ +\end_inset + + kongruentni, + če obstaja obrnljiva +\begin_inset Formula $P\ni:B=PAP^{T}$ +\end_inset + +. + +\end_layout + +\begin_layout Standard +Radi bi naredili klasifikacijo kvadratnih form. + Če naredimo primerno linearno zamenjavo koordinat, + se kvadratna forma poenostavi v +\begin_inset Formula $ex^{2}+fy^{2}$ +\end_inset + + (mešani členi izginejo). +\begin_inset Formula +\[ +x=\alpha x'+\beta y' +\] + +\end_inset + + +\begin_inset Formula +\[ +y=\gamma x'+\delta y' +\] + +\end_inset + +zapišemo kot +\begin_inset Formula +\[ +\left[\begin{array}{c} +x\\ +y +\end{array}\right]=\left[\begin{array}{cc} +\alpha & \beta\\ +\gamma & \delta +\end{array}\right]\left[\begin{array}{c} +x'\\ +y' +\end{array}\right],\quad\quad\overset{\text{transponiranje}}{\Longrightarrow}\quad\quad\left[\begin{array}{cc} +x & y\end{array}\right]=\left[\begin{array}{cc} +x' & y'\end{array}\right]\left[\begin{array}{cc} +\alpha & \gamma\\ +\beta & \delta +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +ax^{2}+bxy+cy^{2}=\left[\begin{array}{cc} +x & y\end{array}\right]\left[\begin{array}{cc} +a & b/2\\ +b/2 & c +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right]=\left[\begin{array}{cc} +x' & y'\end{array}\right]\left[\begin{array}{cc} +\alpha & \gamma\\ +\beta & \delta +\end{array}\right]\left[\begin{array}{cc} +a & b/2\\ +b/2 & a +\end{array}\right]\left[\begin{array}{cc} +\alpha & \beta\\ +\gamma & \delta +\end{array}\right]\left[\begin{array}{c} +x'\\ +y' +\end{array}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left[\begin{array}{cc} +x' & y'\end{array}\right]P^{T}AP\left[\begin{array}{c} +x'\\ +y' +\end{array}\right] +\] + +\end_inset + +Ker je +\begin_inset Formula $A$ +\end_inset + + simetrična, + lahko izberemo tako ortogonalno +\begin_inset Formula $P$ +\end_inset + +, + da je +\begin_inset Formula $P^{T}AP$ +\end_inset + + diagonalna, + recimo +\begin_inset Formula $\left[\begin{array}{cc} +d_{1} & 0\\ +0 & d_{2} +\end{array}\right]$ +\end_inset + +, + torej +\begin_inset Formula +\[ +\left[\begin{array}{cc} +x' & y'\end{array}\right]\left[\begin{array}{cc} +d_{1} & 0\\ +0 & d_{2} +\end{array}\right]\left[\begin{array}{c} +x'\\ +y' +\end{array}\right]=d_{1}\left(x'\right)^{2}+d_{2}\left(y'\right)^{2} +\] + +\end_inset + +Kaj vemo o +\begin_inset Formula $2\times2$ +\end_inset + + ortogonalnih matrikah? + +\begin_inset Formula $P=\left[\begin{array}{cc} +a & b\\ +c & d +\end{array}\right]\Rightarrow P^{T}P=\left[\begin{array}{cc} +a & c\\ +b & d +\end{array}\right]\left[\begin{array}{cc} +a & b\\ +c & d +\end{array}\right]=\left[\begin{array}{cc} +a^{2}+c^{2} & ab+cd\\ +ab+cd & b^{2}+d^{2} +\end{array}\right]$ +\end_inset + +. + Da je +\begin_inset Formula $P^{T}P=I$ +\end_inset + +, + mora veljati +\begin_inset Formula $ab+cd=0$ +\end_inset + + in +\begin_inset Formula $a^{2}+c^{2}=b^{2}+d^{2}=1$ +\end_inset + +, + torej +\begin_inset Formula $a=\cos\varphi$ +\end_inset + +, + +\begin_inset Formula $c=\sin\varphi$ +\end_inset + +, + +\begin_inset Formula $b=\cos\tau$ +\end_inset + +, + +\begin_inset Formula $d=\sin\tau$ +\end_inset + +. + Iz +\begin_inset Formula $\cos\left(\varphi+\tau\right)=0$ +\end_inset + + sledi +\begin_inset Formula $\tau=\varphi\pm\frac{\pi}{2}$ +\end_inset + +, + torej je +\begin_inset Formula $P_{1}=\left[\begin{array}{cc} +\cos\varphi & -\sin\varphi\\ +\sin\varphi & \cos\varphi +\end{array}\right]$ +\end_inset + + (vrtež za +\begin_inset Formula $\varphi$ +\end_inset + +) ali +\begin_inset Formula $P_{2}=\left[\begin{array}{cc} +\cos\varphi & \sin\varphi\\ +\sin\varphi & -\cos\varphi +\end{array}\right]$ +\end_inset + + (zrcaljenje žez +\begin_inset Formula $\varphi/2$ +\end_inset + +). + +\begin_inset Formula $\det P_{1}=1$ +\end_inset + +, + +\begin_inset Formula $\det P_{2}=-1$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Če je +\begin_inset Formula $A=\left[\begin{array}{cc} +v_{1} & v_{2}\end{array}\right]\left[\begin{array}{cc} +d_{1} & 0\\ +0 & d_{2} +\end{array}\right]\left[\begin{array}{cc} +v_{1} & v_{2}\end{array}\right]^{-1}$ +\end_inset + +, + je tudi +\begin_inset Formula $A=\left[\begin{array}{cc} +v_{1} & -v_{2}\end{array}\right]\left[\begin{array}{cc} +d_{1} & 0\\ +0 & d_{2} +\end{array}\right]\left[\begin{array}{cc} +v_{1} & -v_{2}\end{array}\right]^{-1}$ +\end_inset + +. + Če je +\begin_inset Formula $\left[\begin{array}{cc} +v_{1} & v_{2}\end{array}\right]$ +\end_inset + + ortogonalna, + je tudi +\begin_inset Formula $\left[\begin{array}{cc} +v_{1} & -v_{2}\end{array}\right]$ +\end_inset + + ortogonalna. + Če je +\begin_inset Formula $A$ +\end_inset + + +\begin_inset Formula $2\times2$ +\end_inset + + simetrična matrika, + lahko poiščemo tak vrtež +\begin_inset Formula $P=\left[\begin{array}{cc} +\cos\varphi & -\sin\varphi\\ +\sin\varphi & \cos\varphi +\end{array}\right]$ +\end_inset + +, + da je +\begin_inset Formula $A=P\left[\begin{array}{cc} +d_{1} & 0\\ +0 & d_{2} +\end{array}\right]P^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Povzetek: + +\begin_inset Formula $ax^{2}+bxy+cy^{2}\overset{\text{vrtež}}{\longrightarrow}d_{1}x^{2}+d_{2}y^{2}$ +\end_inset + + +\end_layout + +\begin_layout Example* +Nariši krivuljo +\begin_inset Formula $4x^{2}+4xy+7y^{2}=1$ +\end_inset + +. + Pripadajoča kvadratna forma: + +\begin_inset Formula +\[ +\left[\begin{array}{cc} +x & y\end{array}\right]\left[\begin{array}{cc} +4 & 2\\ +2 & 7 +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right]=1=\cdots +\] + +\end_inset + +Radi bi se znebili mešanega člena: +\begin_inset Formula +\[ +\cdots=\left[\begin{array}{cc} +x' & y'\end{array}\right]P^{T}\left[\begin{array}{cc} +4 & 2\\ +2 & 7 +\end{array}\right]P\left[\begin{array}{c} +x'\\ +y' +\end{array}\right]=1=\cdots +\] + +\end_inset + +Iščemo tak vrtež +\begin_inset Formula $P$ +\end_inset + +, + da bo +\begin_inset Formula $P^{T}AP$ +\end_inset + + diagonalna. + Izračunamo lastne vrednosti +\begin_inset Formula $A=\left[\begin{array}{cc} +4 & 2\\ +2 & 7 +\end{array}\right]$ +\end_inset + +. + Lastni vrednosti sta +\begin_inset Formula $\left\{ 3,8\right\} $ +\end_inset + +. + Izračunamo lastna vektorja: + +\begin_inset Formula $\left\{ \left[\begin{array}{c} +-2\\ +1 +\end{array}\right],\left[\begin{array}{c} +1\\ +2 +\end{array}\right]\right\} $ +\end_inset + +. + Sta že ortogonalna, + treba ju je še normirati: + +\begin_inset Formula $\left\{ \left[\begin{array}{c} +-\frac{2}{\sqrt{5}}\\ +\frac{1}{\sqrt{5}} +\end{array}\right],\left[\begin{array}{c} +\frac{1}{\sqrt{5}}\\ +\frac{2}{\sqrt{5}} +\end{array}\right]\right\} $ +\end_inset + +. + Izdelamo vrzež: + +\begin_inset Formula $P=\left[\begin{array}{cc} +\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}}\\ +\frac{2}{\sqrt{5}} & \frac{2}{\sqrt{5}} +\end{array}\right]$ +\end_inset + +. + Izračunamo kot vrteža: + +\begin_inset Formula $\frac{\sin\varphi}{\cos\varphi}=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}=2$ +\end_inset + +, + +\begin_inset Formula $\arctan2\approx63,4^{\circ}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Ogledamo si torej kvadratno formo +\begin_inset Formula $\left[\begin{array}{cc} +x' & y'\end{array}\right]\left[\begin{array}{cc} +8 & 0\\ +0 & 3 +\end{array}\right]\left[\begin{array}{c} +x'\\ +y' +\end{array}\right]=8x'^{2}+3y'^{2}=1$ +\end_inset + +, + kar je elipsa ( +\begin_inset Formula $\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$ +\end_inset + +) s polosema +\begin_inset Formula $\frac{1}{\sqrt{8}}$ +\end_inset + + in +\begin_inset Formula $\frac{1}{\sqrt{3}}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Elipso narišemo in jo v koordinatnem sistemu zavrtimo v negativno smer za +\begin_inset Formula $63,4^{\circ}$ +\end_inset + +. + Po zasuku je risba te krivulje risba naše prvotne kvadratne forme. +\end_layout + \begin_layout Part Vaja za ustni izpit \end_layout |