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