From 5298dbf1e7eb71a11cf7bcd04fb9078cb078ab61 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Anton=20Luka=20=C5=A0ijanec?= <anton@sijanec.eu>
Date: Sat, 29 Jun 2024 22:42:33 +0200
Subject: la/teor.lyx, grem spat

---
 "\305\241ola/la/teor.lyx" | 2409 ++++++++++++++++++++++++++++++++++++++++++++-
 1 file changed, 2405 insertions(+), 4 deletions(-)

(limited to 'šola/la/teor.lyx')

diff --git "a/\305\241ola/la/teor.lyx" "b/\305\241ola/la/teor.lyx"
index ae7dcaf..ea3069a 100644
--- "a/\305\241ola/la/teor.lyx"
+++ "b/\305\241ola/la/teor.lyx"
@@ -5451,6 +5451,12 @@ x_{i}=\frac{\det A_{i}\left(\vec{b}\right)}{\det A}
 \end_layout
 
 \begin_layout Subsubsection
+\begin_inset CommandInset label
+LatexCommand label
+name "subsec:Formula-za-inverz-matrike"
+
+\end_inset
+
 Formula za inverz matrike
 \end_layout
 
@@ -5556,9 +5562,9 @@ X=A^{-1}=\left[\begin{array}{ccc}
 \vdots &  & \vdots\\
 \frac{\det A_{1n}\cdot\left(-1\right)^{1+n}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A}
 \end{array}\right]=\frac{1}{\det A}\left[\begin{array}{ccc}
-\frac{\det A_{11}\cdot\left(-1\right)^{1+1}}{\det A} & \cdots & \frac{\det A_{1n}\cdot\left(-1\right)^{1+1}}{\det A}\\
+\det A_{11}\cdot\left(-1\right)^{1+1} & \cdots & \det A_{1n}\cdot\left(-1\right)^{1+1}\\
 \vdots &  & \vdots\\
-\frac{\det A_{n1}\cdot\left(-1\right)^{n+1}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A}
+\det A_{n1}\cdot\left(-1\right)^{n+1} & \cdots & \det A_{nn}\cdot\left(-1\right)^{n+n}
 \end{array}\right]^{T}=\frac{1}{\det A}\tilde{A}^{T},
 \]
 
@@ -11157,7 +11163,13 @@ Velja torej
 .
 \end_layout
 
-\begin_layout Definition*
+\begin_layout Definition
+\begin_inset CommandInset label
+LatexCommand label
+name "def:vsota-je-direktna"
+
+\end_inset
+
 Pravimo,
  da je vsota 
 \begin_inset Formula $W_{1}+W_{2}$
@@ -11170,6 +11182,10 @@ Pravimo,
 
  oziroma ekvivalentno če je 
 \begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim W_{1}+\dim W_{2}$
+\end_inset
+
+ oziroma ekvivalentno 
+\begin_inset Formula $\forall w_{1}\in W_{1},w_{2}\in W_{2}:w_{1}+w_{2}=0\Rightarrow w_{1}=w_{2}=0$
 \end_inset
 
 .
@@ -12452,6 +12468,12 @@ Odtod sledi:
 
 \end_deeper
 \begin_layout Theorem
+\begin_inset CommandInset label
+LatexCommand label
+name "thm:matrika-kompozituma-linearnih"
+
+\end_inset
+
 matrika kompozituma linearnih preslikav.
  Posplošitev formule 
 \begin_inset Formula $P_{\mathcal{\mathcal{D}\leftarrow\mathcal{B}}}=P_{\mathcal{D\leftarrow C}}\cdot P_{\mathcal{C}\leftarrow\mathcal{B}}$
@@ -13345,7 +13367,7 @@ status open
 
 \begin_layout Plain Layout
 Isto oznako uporabljamo tudi za podobne matrike,
- vendar podobnost ni povesem enako kot ekvivalentnost.
+ vendar podobnost ni enako kot ekvivalentnost.
 \end_layout
 
 \end_inset
@@ -13661,6 +13683,2385 @@ Torej je res
 .
 \end_layout
 
+\end_deeper
+\begin_layout Subsubsection
+Podobnost matrik
+\end_layout
+
+\begin_layout Definition*
+Kvadratni matriki 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $B$
+\end_inset
+
+ sta podobni,
+ če 
+\begin_inset Formula $\exists$
+\end_inset
+
+ taka obrnljiva matrika 
+\begin_inset Formula $P\ni:B=PAP^{-1}$
+\end_inset
+
+.ž
+\end_layout
+
+\begin_layout Claim*
+Podobnost je ekvivalenčna relacija.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo,
+ da je relacija ekvivalenčna,
+ torej:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+refleksivna:
+ 
+\begin_inset Formula $A=IAI^{-1}=IAI=A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+simetrična:
+ 
+\begin_inset Formula $B=PAP^{-1}\Rightarrow P^{-1}BP=A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+tranzitivna:
+ 
+\begin_inset Formula $B=PAP^{-1}\wedge C=QBQ^{-1}\Rightarrow C=QPAP^{-1}Q^{-1}=\left(QP\right)A\left(QP\right)^{-1}$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Remark
+\begin_inset CommandInset label
+LatexCommand label
+name "rem:nista-podobni"
+
+\end_inset
+
+Očitno velja podobnost 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ ekvivalentnost,
+ toda obrat ne velja vedno.
+ Na primer 
+\begin_inset Formula $\left[\begin{array}{cc}
+1 & 0\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[\begin{array}{cc}
+0 & 1\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ sta ekvivalentni (sta enake velikosti in ranga),
+ toda nista podobni (dokaz kasneje).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Od prej vemo,
+ da je vsaka matrika ekvivalentna matriki 
+\begin_inset Formula $\left[\begin{array}{cc}
+I_{r} & 0\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $r$
+\end_inset
+
+ njen rang.
+ A je vsaka kvadratna matrika podobna kakšni lepi matriki?
+ Ja.
+ Vsaka matrika je podobna zgornjetrikotni matriki in jordanski kanonični formi (več o tem kasneje).
+ Toda a je vsaka kvadratna matrika podobna diagonalni matriki?
+ Ne.
+\end_layout
+
+\begin_layout Definition*
+Matrika 
+\begin_inset Formula $D$
+\end_inset
+
+ je diagonalna 
+\begin_inset Formula $\sim d_{ij}\not=0\Rightarrow i=j$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Kdaj je matrika 
+\begin_inset Formula $A$
+\end_inset
+
+ podobna neki diagonalni matriki?
+ Kdaj 
+\begin_inset Formula $\exists$
+\end_inset
+
+ diagonalna 
+\begin_inset Formula $D$
+\end_inset
+
+ in obrnljiva 
+\begin_inset Formula $P\ni:A=PDP^{-1}$
+\end_inset
+
+?
+ Izpeljimo iz nastavka.
+ 
+\begin_inset Formula $D=\left[\begin{array}{ccc}
+\lambda_{1} &  & 0\\
+ & \ddots\\
+0 &  & \lambda n
+\end{array}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $P=\left[\begin{array}{ccc}
+\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]$
+\end_inset
+
+,
+ kjer sta 
+\begin_inset Formula $D$
+\end_inset
+
+ in 
+\begin_inset Formula $P$
+\end_inset
+
+ neznani.
+ Ker mora biti 
+\begin_inset Formula $P$
+\end_inset
+
+ obrnljiva,
+ so njeni stolpični vektorji LN.
+\begin_inset Formula 
+\[
+A=PDP^{-1}\Leftrightarrow AP=PD\Leftrightarrow A\left[\begin{array}{ccc}
+\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc}
+\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]\left[\begin{array}{ccc}
+\lambda_{1} &  & 0\\
+ & \ddots\\
+0 &  & \lambda_{n}
+\end{array}\right]\text{ in }P\text{ obrnljiva}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccc}
+A\vec{v_{1}} & \cdots & A\vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc}
+\lambda_{1}\vec{v_{1}} & \cdots & \lambda_{n}\vec{v_{n}}\end{array}\right]\text{ in }v_{i}\text{ so LN}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+A\vec{v_{1}}=\lambda_{1}\vec{v_{1}},\dots,A\vec{v_{n}}=\lambda_{n}v_{n}\text{ in }\forall i:v_{i}\not=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Porodi se naloga,
+ imenovana 
+\begin_inset Quotes gld
+\end_inset
+
+Lastni problem
+\begin_inset Quotes grd
+\end_inset
+
+.
+ Iščemo pare 
+\begin_inset Formula $\left(\lambda,\vec{v}\right)$
+\end_inset
+
+,
+ ki zadoščajo enačbi 
+\begin_inset Formula $A\vec{v}=\lambda\vec{v}$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da je 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ je lastna vrednost matrike 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ če obstaja tak 
+\begin_inset Formula $\vec{v}\not=0$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $A\vec{v}=\lambda\vec{v}$
+\end_inset
+
+.
+ V tem primeru pravimo,
+ da je 
+\begin_inset Formula $\vec{v}$
+\end_inset
+
+ lastni vektor,
+ ki pripada lastni vrednosti 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+ Paru 
+\begin_inset Formula $\left(\lambda,\vec{v}\right)$
+\end_inset
+
+,
+ ki zadošča enačbi,
+ pravimo lastni par matrike 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Nalogo 
+\begin_inset Quotes gld
+\end_inset
+
+Lastni problem
+\begin_inset Quotes grd
+\end_inset
+
+ rešujemo v dveh korakih.
+ Najprej najdemo vse 
+\begin_inset Formula $\lambda$
+\end_inset
+
+,
+ nato za vsako poiščemo pripadajoče 
+\begin_inset Formula $\vec{v}$
+\end_inset
+
+,
+ ki za lastno vrednost obstajajo po definiciji.
+\end_layout
+
+\begin_layout Standard
+Za nek 
+\begin_inset Formula $v\not=0$
+\end_inset
+
+ pišimo 
+\begin_inset Formula $Av=\lambda v=\lambda Iv\Leftrightarrow Av-\lambda Iv=0\Leftrightarrow\left(A-\lambda I\right)v=0$
+\end_inset
+
+ za nek 
+\begin_inset Formula $v\not=0\Leftrightarrow\Ker\left(A-\lambda I\right)\not=\left\{ 0\right\} \overset{\text{K.O.M.}}{\Longleftrightarrow}A-\lambda I$
+\end_inset
+
+ ni obrnljiva 
+\begin_inset Formula $\Leftrightarrow\det\left(A-\lambda I\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Polinom 
+\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-xI\right)$
+\end_inset
+
+ je karakteristični polinom matrike 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Premislek zgoraj nam pove,
+ da so lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+ ničle 
+\begin_inset Formula $p_{A}\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Karakteristični polinom lahko nima nobene ničle:
+ 
+\begin_inset Formula $A=\left[\begin{array}{cc}
+0 & 1\\
+-1 & 0
+\end{array}\right]$
+\end_inset
+
+,
+ 
+\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\det\left[\begin{array}{cc}
+-\lambda & 1\\
+-1 & -\lambda
+\end{array}\right]=x^{2}+1$
+\end_inset
+
+,
+ katerega ničli sta 
+\begin_inset Formula $\lambda_{1}=i$
+\end_inset
+
+ in 
+\begin_inset Formula $\lambda_{2}=-i$
+\end_inset
+
+,
+ ki nista realni števili.
+ V nadaljevanju se zato omejimo na kompleksne matrike in kompleksne lastne vrednosti,
+ saj ima po Osnovnem izreku Algebre polinom s kompleksnimi koeficienti vedno vsaj kompleksne ničle.
+\end_layout
+
+\begin_layout Standard
+Kako pa iščemo lastne vektorje za lastno vrednost 
+\begin_inset Formula $\lambda$
+\end_inset
+
+?
+ Spomnimo se na 
+\begin_inset Formula $Av=\lambda v\Leftrightarrow v\in\Ker\left(A-\lambda I\right)$
+\end_inset
+
+.
+ Rešiti moramo homogen sistem linearnih enačb.
+ Po definiciji so lastni vektorji neničelni,
+ zato nas trivialna rešitev ne zanima.
+\end_layout
+
+\begin_layout Definition*
+Množici 
+\begin_inset Formula $\Ker\left(A-\lambda I\right)$
+\end_inset
+
+ pravimo lastni podprostor matrike 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ ki pripada 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+ Slednji vsebuje 
+\begin_inset Formula $\vec{0}$
+\end_inset
+
+ in množico vektorjev,
+ ki so vsi lastni vektorji 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Exercise*
+Izračunaj lastne vrednosti od 
+\begin_inset Formula $A=\left[\begin{array}{cc}
+0 & 1\\
+-1 & 0
+\end{array}\right]$
+\end_inset
+
+.
+ Od prej vemo,
+ da 
+\begin_inset Formula $\lambda_{1}=i$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\lambda_{2}=-i$
+\end_inset
+
+.
+ Izračunajmo 
+\begin_inset Formula $\Ker\left(A-iI\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\Ker\left(A+iI\right)$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\Ker\left(A-iI\right):\quad\left[\begin{array}{cc}
+-i & 1\\
+-1 & -i
+\end{array}\right]\left[\begin{array}{c}
+x\\
+y
+\end{array}\right]=0\quad\Longrightarrow\quad-ix+y=0,-x-iy=0\quad\Longrightarrow\quad y=ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c}
+1\\
+i
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\Ker\left(A+iI\right):\quad\left[\begin{array}{cc}
+i & 1\\
+-1 & i
+\end{array}\right]\left[\begin{array}{c}
+x\\
+y
+\end{array}\right]=0\quad\Longrightarrow\quad ix+y=0,-x+y=0\quad\Longrightarrow\quad y=-ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c}
+1\\
+-i
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\Ker\left(A-iI\right)=\Lin\left\{ \left[\begin{array}{c}
+1\\
+i
+\end{array}\right]\right\} ,\quad\Ker\left(A+iI\right)=\Lin\left\{ \left[\begin{array}{c}
+1\\
+-i
+\end{array}\right]\right\} 
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Exercise*
+Vstavimo lastna vektorja v 
+\begin_inset Formula $P$
+\end_inset
+
+ in lastne vrednosti v 
+\begin_inset Formula $D$
+\end_inset
+
+ na pripadajoči mesti.
+ Dobimo obrnljivo 
+\begin_inset Formula $P$
+\end_inset
+
+ in velja 
+\begin_inset Formula $A=PDP^{-1}$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+P=\left[\begin{array}{cc}
+1 & 1\\
+i & -i
+\end{array}\right],\quad D=\left[\begin{array}{cc}
+i & 0\\
+0 & -i
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Exercise*
+Temu početju pravimo 
+\begin_inset Quotes gld
+\end_inset
+
+diagonalizacija matrike 
+\begin_inset Formula $A$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Primer matrike,
+ ki ni diagonalizabilna:
+ 
+\begin_inset Formula $A=\left[\begin{array}{cc}
+0 & 1\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\det\left(A-\lambda I\right)=\left[\begin{array}{cc}
+-\lambda & 1\\
+0 & -\lambda
+\end{array}\right]=\lambda^{2}$
+\end_inset
+
+.
+ Ničli/lastni vrednosti sta 
+\begin_inset Formula $\lambda_{1}=0$
+\end_inset
+
+ in 
+\begin_inset Formula $\lambda_{2}=0$
+\end_inset
+
+.
+ Toda 
+\begin_inset Formula $\Ker\left(A-0I\right)=\Ker A=\Lin\left\{ \left[\begin{array}{c}
+1\\
+0
+\end{array}\right]\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $P=\left[\begin{array}{cc}
+1 & 1\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ ni obrnljiva.
+ S tem dokažemo trditev v primeru 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "rem:nista-podobni"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+ 
+\begin_inset Formula $\left[\begin{array}{cc}
+1 & 0\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[\begin{array}{cc}
+0 & 1\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ nista podobni,
+ ker je prva diagonalna,
+ druga pa ni podobna diagonalni matriki (ne da se je diagonalizirati).
+\end_layout
+
+\begin_layout Standard
+Lastne vrednosti lahko definiramo tudi za linearne preslikave,
+ saj so linearne preslikave linearno izomorfne matrikam.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor nad 
+\begin_inset Formula $F=\mathbb{C}$
+\end_inset
+
+ in 
+\begin_inset Formula $L:V\to V$
+\end_inset
+
+ linearna preslikava.
+ Število 
+\begin_inset Formula $\lambda\in F$
+\end_inset
+
+ je lastna vrednost 
+\begin_inset Formula $L$
+\end_inset
+
+,
+ le obstaja tak neničelni 
+\begin_inset Formula $v\in V$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $Lv=\lambda v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Kako pa rešujemo 
+\begin_inset Quotes gld
+\end_inset
+
+Lastni problem
+\begin_inset Quotes grd
+\end_inset
+
+ za linearne preslikave?
+ 
+\begin_inset Formula $Lv=\lambda v\Leftrightarrow Lv-\lambda\left(id\right)v=0\Leftrightarrow\left(L-\lambda\left(id\right)\right)v=0\Leftrightarrow v\in\Ker\left(L-\lambda\left(id\right)\right)\overset{v\not=0}{\Longleftrightarrow}\det\left(L-\lambda\left(id\right)\right)=0$
+\end_inset
+
+.
+ Toda determinante linearne preslikave nismo definirali.
+ Lahko pa determinanto izračunamo na matriki,
+ ki pripada tej linearni preslikavi.
+ Toda dvem različnim bazam pripadata različni matriki linearne preslikave.
+ Dokazati je treba,
+ da sta determinanti dveh matrik,
+ pripadajočih eni linearni preslikavi,
+ enaki,
+ četudi sta matriki v različnih bazah.
+\end_layout
+
+\begin_layout Lemma
+\begin_inset CommandInset label
+LatexCommand label
+name "lem:Podobni-matriki-imata"
+
+\end_inset
+
+Podobni matriki imata isto determinanto.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $B=PAP^{-1}$
+\end_inset
+
+ za neko obrnljivo 
+\begin_inset Formula $P$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\det B=\det PAP^{-1}=\det P\det A\det P^{-1}=\det P\det P^{-1}\det A=\det PP^{-1}\det A=\det I\det A=1\cdot\det A=\det A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $L:V\to V$
+\end_inset
+
+ naj bo linearna preslikava,
+ 
+\begin_inset Formula $V$
+\end_inset
+
+ prostor nad 
+\begin_inset Formula $F=\mathbb{C}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathcal{C}$
+\end_inset
+
+ pa bazi 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Priredimo matriki 
+\begin_inset Formula $L_{\mathcal{B}\leftarrow\mathcal{B}}$
+\end_inset
+
+ in 
+\begin_inset Formula $L_{\mathcal{C}\leftarrow\mathcal{C}}$
+\end_inset
+
+.
+ Spomnimo se izreka 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "thm:matrika-kompozituma-linearnih"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left[KL\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $L=\left[id\circ L\circ id\right]$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\circ L\circ id\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}\left[id\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{C}}}}=P\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}P^{-1}$
+\end_inset
+
+ za neko obrnljivo 
+\begin_inset Formula $P$
+\end_inset
+
+.
+ Torej sta matriki 
+\begin_inset Formula $\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}$
+\end_inset
+
+ podobni,
+ torej imata po lemi 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "lem:Podobni-matriki-imata"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ isto determinanto.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Alternativen dokaz,
+ da imata podobni matriki iste lastne vrednosti:
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ podobna 
+\begin_inset Formula 
+\[
+B\Rightarrow B=PAP^{-1}\Rightarrow B-xI=P\left(A-xI\right)P^{-1}\Rightarrow\det\left(B-xI\right)=\det\left(A-xI\right)\Rightarrow p_{A}=p_{B},
+\]
+
+\end_inset
+
+torej so lastne vrednosti enake.
+ Kaj pa lastni vektorji?
+ Naj bo 
+\begin_inset Formula $v$
+\end_inset
+
+ lastni vektor 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ torej
+\begin_inset Formula 
+\[
+Av=\lambda v\Rightarrow PAv=\lambda Pv\Rightarrow PAP^{-1}Pv=\lambda Pv\Rightarrow BPv=\lambda Pv,
+\]
+
+\end_inset
+
+torej za 
+\begin_inset Formula $v$
+\end_inset
+
+ lastni vektor 
+\begin_inset Formula $A$
+\end_inset
+
+ sledi,
+ da je 
+\begin_inset Formula $Pv$
+\end_inset
+
+ lastni vektor 
+\begin_inset Formula $B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Linearni transformaciji torej priredimo tako matriko,
+ ki ima v začetnem in končnem prostoru isto bazo.
+ Tedaj lahko izračunamo lastne pare na tej matriki.
+\end_layout
+
+\begin_layout Theorem*
+Schurov izrek.
+ Vsaka kompleksna kvadratna matrika je podobna zgornjetrikotni matriki.
+\end_layout
+
+\begin_layout Proof
+Indukcija po velikosti matrike.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $A_{1\times1}$
+\end_inset
+
+ je zgornjetrikotna.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Po I.
+ P.
+ trdimo,
+ da je vsaka 
+\begin_inset Formula $A_{\left(n-1\right)\times\left(n-1\right)}$
+\end_inset
+
+ podobna kaki zgornjetrikotni matriki.
+ Dokažimo še za poljubno 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ lastna vrednost 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $v_{1}$
+\end_inset
+
+ pripadajoči lastni vektor ter 
+\begin_inset Formula $v_{2},\dots,v_{n}$
+\end_inset
+
+ dopolnitev 
+\begin_inset Formula $v_{1}$
+\end_inset
+
+ do baze 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ Potem je matrika 
+\begin_inset Formula $P=\left[\begin{array}{ccc}
+v_{1} & \cdots & v_{n}\end{array}\right]$
+\end_inset
+
+ obrnljiva.
+\begin_inset Formula 
+\[
+AP=\left[\begin{array}{ccc}
+Av_{1} & \cdots & Av_{n}\end{array}\right]=\left[\begin{array}{ccc}
+v_{1} & \cdots & v_{n}\end{array}\right]\left[\begin{array}{cccc}
+\lambda & a_{1,2} & \cdots & a_{1,n}\\
+0 & \vdots &  & \vdots\\
+\vdots & \vdots &  & \vdots\\
+0 & a_{m,n} & \cdots & a_{m.n}
+\end{array}\right]=P\left[\begin{array}{cc}
+\lambda & B\\
+0 & C
+\end{array}\right]
+\]
+
+\end_inset
+
+Po I.
+ P.
+ obstaja taka zgornjetrikotna 
+\begin_inset Formula $T$
+\end_inset
+
+ in obrnljiva 
+\begin_inset Formula $Q$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $C=QTQ^{-1}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left[\begin{array}{cc}
+1 & 0\\
+0 & Q
+\end{array}\right]^{-1}P^{-1}AP\left[\begin{array}{cc}
+1 & 0\\
+0 & Q
+\end{array}\right]=\left[\begin{array}{cc}
+1 & 0\\
+0 & Q
+\end{array}\right]^{-1}\left[\begin{array}{cc}
+\lambda & B\\
+0 & C
+\end{array}\right]\left[\begin{array}{cc}
+1 & 0\\
+0 & Q
+\end{array}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left[\begin{array}{cc}
+\lambda & C\\
+0 & Q^{-1}B
+\end{array}\right]\left[\begin{array}{cc}
+1 & 0\\
+0 & Q
+\end{array}\right]=\left[\begin{array}{cc}
+\lambda & CQ\\
+0 & B
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula $A$
+\end_inset
+
+ je torej podobna 
+\begin_inset Formula $\left[\begin{array}{cc}
+\lambda & CQ\\
+0 & B
+\end{array}\right]$
+\end_inset
+
+,
+ ki je zgornjetrikotna.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO karakterizacija linearnih preslikav 
+\begin_inset Quotes gld
+\end_inset
+
+LA1V FMF 2024-03-12
+\begin_inset Quotes grd
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Zadosten pogoj za diagonalizabilnost
+\end_layout
+
+\begin_layout Theorem
+\begin_inset CommandInset label
+LatexCommand label
+name "thm:lave-razl-lavr-so-LN"
+
+\end_inset
+
+Lastni vektorji,
+ ki pripadajo različnim lastnim vrednostim,
+ so linearno neodvisni.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ matrika,
+ 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ njene lastne vrednosti in 
+\begin_inset Formula $v_{1},\dots,v_{k}$
+\end_inset
+
+ njim pripadajoči lastni vektorji.
+ Dokazujemo 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ paroma različni 
+\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$
+\end_inset
+
+ LN.
+ Dokaz z indukcijo po 
+\begin_inset Formula $k$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza 
+\begin_inset Formula $k=1$
+\end_inset
+
+:
+ Elementi 
+\begin_inset Formula $\left\{ \lambda_{1}\right\} $
+\end_inset
+
+ so trivialno paroma različni in 
+\begin_inset Formula $v_{1}$
+\end_inset
+
+ je kot neničen vektor LN.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Dokazujemo 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k+1}$
+\end_inset
+
+ so paroma različne 
+\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$
+\end_inset
+
+ so LN,
+ vedoč I.
+ P.
+ Denimo,
+ da 
+\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$
+\end_inset
+
+.
+ Množimo z 
+\begin_inset Formula $A$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+A\left(\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}\right)=\alpha_{1}Av_{1}+\cdots+\alpha_{k+1}Av_{k+1}=\alpha_{1}\lambda_{1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0
+\]
+
+\end_inset
+
+Množimo začetno enačbo z 
+\begin_inset Formula $\lambda_{k+1}$
+\end_inset
+
+ (namesto z 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ kot smo to storili zgoraj):
+\begin_inset Formula 
+\[
+\alpha_{1}\lambda_{k+1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0
+\]
+
+\end_inset
+
+Odštejmo eno enačbo od druge,
+ dobiti moramo 0,
+ saj odštevamo 0 od 0:
+\begin_inset Formula 
+\[
+\alpha_{1}\left(\lambda_{1}-\lambda_{k+1}\right)v_{1}+\cdots+\alpha_{k}\left(\lambda_{k}-\lambda_{k+1}\right)v_{k}+\cancel{\alpha_{k+1}\left(\lambda_{k+1}-\lambda_{k+1}\right)v_{k+1}}=0
+\]
+
+\end_inset
+
+Ker so lastne vrednosti paroma različne (
+\begin_inset Formula $\lambda_{i}=\lambda_{j}\Rightarrow i=j$
+\end_inset
+
+),
+ so njihove razlike neničelne.
+ Ker so 
+\begin_inset Formula $v_{1},\dots,v_{k}$
+\end_inset
+
+ po predpostavki LN,
+ sledi 
+\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=0$
+\end_inset
+
+.
+ Vstavimo te konstante v 
+\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $\alpha_{k+1}v_{k+1}=0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $v_{k+1}$
+\end_inset
+
+ neničeln (je namreč lastni vektor),
+ sledi 
+\begin_inset Formula $\alpha_{k+1}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=\alpha_{k+1}=0$
+\end_inset
+
+,
+ zatorej so 
+\begin_inset Formula $v_{1},\dots,v_{k+1}$
+\end_inset
+
+ res LN.
+\end_layout
+
+\end_deeper
+\begin_layout Corollary
+\begin_inset CommandInset label
+LatexCommand label
+name "cor:vsota-lastnih-podpr-direktna"
+
+\end_inset
+
+Vsota vseh lastnih podprostorov matrike je direktna (definicija 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "def:vsota-je-direktna"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Naj bodo 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ vse paroma različne lastne vrednosti matrike 
+\begin_inset Formula $A\in M_{n}\left(\mathbb{C}\right)$
+\end_inset
+
+.
+ Pripadajoči lastni podprostori so torej 
+\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :V_{i}=\Ker\left(A-\lambda_{i}I\right)$
+\end_inset
+
+.
+ Trdimo,
+ da je vsota teh podprostorov direktna,
+ torej 
+\begin_inset Formula $\forall v_{1}\in V_{1},\dots,v_{k}\in V_{k}:v_{1}+\cdots+v_{k}=0\Rightarrow v_{1}=\cdots=v_{k}=0$
+\end_inset
+
+.
+ To sledi iz izreka 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "thm:lave-razl-lavr-so-LN"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Če ima 
+\begin_inset Formula $n\times n$
+\end_inset
+
+ matrika 
+\begin_inset Formula $n$
+\end_inset
+
+ paroma različnih lastnih vrednosti,
+ je podobna diagonalni matriki.
+\end_layout
+
+\begin_layout Proof
+Po posledici 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "cor:vsota-lastnih-podpr-direktna"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ je vsota lastnih podprostorov matrike 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ direktna.
+ Če je torej lastnih podprostorov 
+\begin_inset Formula $n$
+\end_inset
+
+,
+ je njihova vsota cel prostor 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ Matriko se da diagonalizirati,
+ kadar je vsota vseh lastnih podprostorov enaka podprostoru 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+ (tedaj so namreč stolpci matrike 
+\begin_inset Formula $P$
+\end_inset
+
+ linearno neodvisni,
+ zato je 
+\begin_inset Formula $P$
+\end_inset
+
+ obrnljiva).
+\end_layout
+
+\begin_layout Subsubsection
+Algebraične in geometrijske vekčratnosti
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ matrika.
+ 
+\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\left(-1\right)^{n}\left(\lambda-\lambda_{1}\right)^{n_{1}}\cdots\left(\lambda-\lambda_{k}\right)^{n_{k}}$
+\end_inset
+
+,
+ kjer so 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ vse paroma različne lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Stopnji ničle —
+ 
+\begin_inset Formula $n_{i}$
+\end_inset
+
+ —
+ rečemo algebraična večkratnost lastne vrednosti 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Geometrijska večkratnost lastne vrednosti 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ je 
+\begin_inset Formula $\dim\Ker\left(A-\lambda_{i}I\right)=\n$
+\end_inset
+
+
+\begin_inset Formula $\left(A-\lambda_{i}I\right)=m_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Algebraično večkratnost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $n_{i}$
+\end_inset
+
+ in je večkratnost ničle 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ v 
+\begin_inset Formula $p_{A}\left(\lambda\right)$
+\end_inset
+
+ (karakterističnem polinomu).
+ Geometrijsko večkratnost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ pa označimo z 
+\begin_inset Formula $m_{i}$
+\end_inset
+
+ in je dimenzija lastnega podprostora za 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$
+\end_inset
+
+ —
+ geometrijska večkratnost lastne vrednosti je kvečjemu tolikšna,
+ kot je algebraična večkratnost te lastne vrednosti.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $v_{1},\dots,v_{m_{i}}$
+\end_inset
+
+ baza za lastni podprostor 
+\begin_inset Formula $V_{i}=\Ker\left(A-\lambda_{i}I\right)$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $v_{m_{i}+1},\dots,v_{n}$
+\end_inset
+
+ njena dopolnitev do baze 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ Tedaj velja:
+ 
+\begin_inset Formula $Av_{1}=\lambda_{1}v_{1}$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $Av_{m_{i}}=\lambda_{m_{i}}v_{m_{i}}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $Av_{m_{i}+1}=$
+\end_inset
+
+ linearna kombinacija 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $Av_{n}=$
+\end_inset
+
+ linearna kombinacija 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $P=\left[\begin{array}{cccccc}
+v_{1} & \cdots & v_{m_{i}} & v_{m_{i}+1} & \cdots & v_{n}\end{array}\right]$
+\end_inset
+
+,
+ ki je obrnljiva.
+\begin_inset Formula 
+\[
+P^{-1}AP=\cdots=\left[\begin{array}{cc}
+\lambda_{i}I_{m_{i}} & B\\
+0 & C
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\series bold
+Dokaza ne razumem.
+ Obupam.
+\end_layout
+
+\begin_layout Claim*
+Matriko s paroma različnimi lastnimi vrednostmi 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ je moč diagonalizirati 
+\begin_inset Formula $\Leftrightarrow\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $V_{i}$
+\end_inset
+
+ lastno podprostor lastne vrednosti 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+.
+ Vemo,
+ da se da 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ diagonalizirati 
+\begin_inset Formula $\Leftrightarrow A$
+\end_inset
+
+ ima 
+\begin_inset Formula $n$
+\end_inset
+
+ LN stolpičnih vektorjev 
+\begin_inset Formula $\Leftrightarrow\Ker\left(A-\lambda_{1}I\right)+\cdots+\Ker\left(A-\lambda_{k}I\right)=\mathbb{C}^{n}\Leftrightarrow\dim\left(V_{i}+\cdots+V_{k}\right)=\dim V_{i}+\cdots+\dim V_{k}\Leftrightarrow$
+\end_inset
+
+ vsota lastnih podprostorov je direktna 
+\begin_inset Formula $\Leftrightarrow\dim\left(V_{1}+\cdots+V_{n}\right)=n\Leftrightarrow\dim V_{1}+\cdots+\dim V_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n_{1}+\cdots+n_{m}$
+\end_inset
+
+.
+ Toda ker po prejšnjem izreku 
+\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$
+\end_inset
+
+,
+ mora veljati 
+\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Minimalni polinom matrike
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+\cdots+c_{n}x^{n}\in\mathbb{C}\left[x\right]$
+\end_inset
+
+ polinom in 
+\begin_inset Formula $A$
+\end_inset
+
+ matrika.
+ 
+\begin_inset Formula $p\left(A\right)\coloneqq c_{0}A^{0}+\cdots+c_{n}A^{n}=c_{0}I+\cdots+c_{n}A^{n}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $p\left(A\right)=0$
+\end_inset
+
+ (ničelna matrika),
+ pravimo,
+ da polinom 
+\begin_inset Formula $p$
+\end_inset
+
+ anhilira/uniči matriko 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Fact*
+\begin_inset Formula $p\left(A\right)=0\Rightarrow p\left(P^{-1}AP\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Izkaže se,
+ da karakteristični polimom anhilira matriko —
+ 
+\begin_inset Formula $p_{A}\left(A\right)=0$
+\end_inset
+
+.
+ Dokaz kasneje.
+\end_layout
+
+\begin_layout Definition*
+Polinom 
+\begin_inset Formula $m\left(x\right)$
+\end_inset
+
+ je minimalen polinom 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ če velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $m\left(A\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $m$
+\end_inset
+
+ ima vodilni koeficient 1
+\end_layout
+
+\begin_layout Enumerate
+med vsemi polinomi,
+ ki zadoščajo prvi in drugi zahtevi,
+ ima 
+\begin_inset Formula $m$
+\end_inset
+
+ najnižjo stopnjo
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+eksistenca minimalnega polinoma —
+ Minimalni polinom obstaja.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ matrika.
+ Očitno je 
+\begin_inset Formula $M_{n\times n}\left(\mathbb{C}\right)$
+\end_inset
+
+ vektorski prostor dimenzije 
+\begin_inset Formula $n^{2}$
+\end_inset
+
+.
+ Matrike 
+\begin_inset Formula $\left\{ I,A,A^{2},\dots,A^{n^{2}}\right\} $
+\end_inset
+
+ so linearno odvisne,
+ ker je moč te množice za 1 večja od moči vektorskega prostora.
+ Torej 
+\begin_inset Formula $\exists c_{0},\cdots,c_{n^{2}}\in\mathbb{C}$
+\end_inset
+
+,
+ ki niso vse 0 
+\begin_inset Formula $\ni:c_{0}I+c_{1}A+c_{2}A^{2}+\cdots+c_{n^{2}}A^{n^{2}}=0$
+\end_inset
+
+.
+ Torej polinom 
+\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+c_{1}x^{1}+c_{2}x^{2}+\cdots+c_{n^{2}}x^{n^{2}}$
+\end_inset
+
+ anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Če ta polinom delimo z njegovim vodilnim koeficientom,
+ dobimo polinom,
+ ki ustreza prvima dvema zahevama za minimalni polinom.
+ Če med vsemi takimi izberemo takega z najnižjo stopnjo,
+ le-ta ustreza še tretji zahtevi.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $m\left(x\right)$
+\end_inset
+
+ minimalni polinom za 
+\begin_inset Formula $A$
+\end_inset
+
+ in če 
+\begin_inset Formula $p\left(x\right)$
+\end_inset
+
+ anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ potem 
+\begin_inset Formula $m\left(x\right)\vert p\left(x\right)$
+\end_inset
+
+ (
+\begin_inset Formula $m\left(x\right)$
+\end_inset
+
+ deli 
+\begin_inset Formula $p\left(x\right)$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Delimo 
+\begin_inset Formula $p$
+\end_inset
+
+ z 
+\begin_inset Formula $m$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\exists k\left(x\right),r\left(x\right)\ni:p\left(x\right)=k\left(x\right)m\left(x\right)+r\left(x\right)\wedge\deg r\left(x\right)<\deg m\left(x\right)$
+\end_inset
+
+.
+ Vstavimo 
+\begin_inset Formula $A$
+\end_inset
+
+ na obe strani:
+\begin_inset Formula 
+\[
+0=p\left(A\right)=k\left(A\right)m\left(A\right)+r\left(A\right)=k\left(A\right)\cdot0+r\left(A\right)=0+r\left(A\right)=r\left(A\right)=0
+\]
+
+\end_inset
+
+Sledi 
+\begin_inset Formula $r\left(x\right)=0$
+\end_inset
+
+,
+ kajti če 
+\begin_inset Formula $r$
+\end_inset
+
+ ne bi bil ničeln polinom,
+ bi ga lahko delili z vodilnim koeficientom in po predpostavki 
+\begin_inset Formula $\deg r\left(x\right)<\deg m\left(x\right)$
+\end_inset
+
+ bi imel manjšo stopnjo kot 
+\begin_inset Formula $m\left(x\right)$
+\end_inset
+
+,
+ torej bi ustrezal zahtevam 1 in 2 za minimalni polinom in bi imel manjšo stopnjo od 
+\begin_inset Formula $m$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $m$
+\end_inset
+
+ ne bi bil minimalni polinom,
+ kar bi vodilo v protislovje.
+\end_layout
+
+\begin_layout Corollary*
+enoličnost minimalnega polinoma.
+ Naj bosta 
+\begin_inset Formula $m_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $m_{2}$
+\end_inset
+
+ minimalna polinoma matrike 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $m$
+\end_inset
+
+ po definiciji anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ iz prejšnje trditve sledi,
+ če vstavimo 
+\begin_inset Formula $m=m_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $p=m_{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $m_{1}\vert m_{2}$
+\end_inset
+
+.
+ Toda če vstavimo 
+\begin_inset Formula $m=m_{2}$
+\end_inset
+
+ in 
+\begin_inset Formula $p=m_{1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $m_{2}\vert m_{1}$
+\end_inset
+
+.
+ Iz 
+\begin_inset Formula $m_{1}\vert m_{2}\wedge m_{2}\vert m_{1}$
+\end_inset
+
+ sledi,
+ da se 
+\begin_inset Formula $m_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $m_{2}$
+\end_inset
+
+ razlikujeta le za konstanten faktor,
+ ki pa je po definiciji minimalnega polinoma 1,
+ torej 
+\begin_inset Formula $m_{1}=m_{2}$
+\end_inset
+
+.
+ Zaradi enoličnosti lahko označimo minimalni polinom 
+\begin_inset Formula $A$
+\end_inset
+
+ z 
+\begin_inset Formula $m_{A}\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Ničle minimalnega polinoma
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $m_{A}\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $p_{A}\left(x\right)$
+\end_inset
+
+ imata iste ničle 
+\begin_inset Formula $\sim$
+\end_inset
+
+ ničle 
+\begin_inset Formula $m_{A}\left(x\right)$
+\end_inset
+
+ so lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $p_{A}\left(x\right)$
+\end_inset
+
+ (dokaz kasneje),
+ velja po trditvi v dokazu enoličnosti,
+ da 
+\begin_inset Formula $m_{A}\vert p_{A}$
+\end_inset
+
+,
+ torej je vsaka ničla 
+\begin_inset Formula $m_{A}$
+\end_inset
+
+ tudi ničla 
+\begin_inset Formula $p_{A}$
+\end_inset
+
+.
+ Treba je dokazati še,
+ da je vsaka ničla 
+\begin_inset Formula $p_{A}$
+\end_inset
+
+ tudi ničla 
+\begin_inset Formula $m_{A}$
+\end_inset
+
+,
+ natančneje:
+ Treba je dokazati,
+ da če je 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ lastna vrednost matrike 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $m_{A}\left(\lambda\right)=0$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $v\not=0$
+\end_inset
+
+ lastni vektor za 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $Av=\lambda v$
+\end_inset
+
+.
+ Potem velja 
+\begin_inset Formula $A^{2}v=AAv=A\lambda v=\lambda Av=\lambda\lambda v=\lambda^{2}v$
+\end_inset
+
+ in splošneje 
+\begin_inset Formula $A^{n}v=\lambda^{n}v$
+\end_inset
+
+.
+ Sedaj recimo,
+ da je 
+\begin_inset Formula $m_{A}\left(x\right)=d_{0}x^{0}+\cdots+d_{r}x^{r}$
+\end_inset
+
+.
+ Potem je,
+ ker minimalni polinom anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ 
+\begin_inset Formula 
+\[
+m_{A}\left(\lambda\right)v=\left(d_{0}+d_{1}\lambda+d_{2}\lambda^{2}+\cdots+d_{r}\lambda^{r}\right)v=d_{0}v+d_{1}\lambda v+d_{2}\lambda^{2}v+\cdots+d_{r}\lambda^{r}v=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=d_{0}v+d_{1}Av+d_{2}A^{2}v+\cdots+d_{r}A^{r}v=\left(d_{0}+d_{1}A+d_{2}A^{2}+\cdots+d_{r}A^{r}\right)v=m_{A}\left(A\right)v=0v=0
+\]
+
+\end_inset
+
+Ker 
+\begin_inset Formula $m_{A}\left(\lambda\right)v=0$
+\end_inset
+
+ in 
+\begin_inset Formula $v\not=0$
+\end_inset
+
+ (je namreč lastni vektor),
+ velja 
+\begin_inset Formula $m_{A}\left(\lambda\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph*
+Lastnosti
+\end_layout
+
+\begin_layout Standard
+Ker je 
+\begin_inset Formula $p_{A}\left(x\right)=\left(-1\right)^{n}\left(x-\lambda_{1}\right)^{n_{1}}\cdots\left(x-\lambda_{k}\right)^{n_{k}}$
+\end_inset
+
+ in 
+\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{r_{1}}\cdots\left(x-\lambda_{k}\right)^{r_{k}}$
+\end_inset
+
+,
+ sledi iz 
+\begin_inset Formula $m_{A}\vert p_{A}\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\leq n_{i}$
+\end_inset
+
+.
+ Poleg tega,
+ ker 
+\begin_inset Formula $m_{A}\left(\lambda_{1}\right)=0\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\geq1$
+\end_inset
+
+.
+ Toda pozor:
+ 
+\series bold
+Ni
+\series default
+ res,
+ da 
+\begin_inset Formula $r_{i}=m_{i}$
+\end_inset
+
+ (stropnja lastnega podprostora).
+\end_layout
+
+\begin_layout Theorem*
+Cayley-Hamilton.
+ 
+\begin_inset Formula $p_{A}\left(A\right)=0$
+\end_inset
+
+ —
+ karakteristični polinom matrike 
+\begin_inset Formula $A$
+\end_inset
+
+ anhilira matriko 
+\begin_inset Formula $A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Spomnimo se eksplicitne formule za celico inverza matrike (razdelek 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "subsec:Formula-za-inverz-matrike"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+),
+ ki pravi 
+\begin_inset Formula $B^{-1}=\frac{1}{\det B}\tilde{B}^{T}$
+\end_inset
+
+.
+ Računajmo in naposled vstavimo 
+\begin_inset Formula $B=A-xI$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+B^{-1}=\frac{1}{\det B}\tilde{B}^{T}\quad\quad\quad\quad/\cdot\left(\det B\right)B
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det B\cdot I=B\tilde{B}^{T}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det\left(A-xI\right)\cdot I=p_{A}\left(x\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}
+\]
+
+\end_inset
+
+Glede na definicijo 
+\begin_inset Formula $\tilde{A}$
+\end_inset
+
+ je 
+\begin_inset Formula $\tilde{\left(A-xI\right)}^{T}$
+\end_inset
+
+ matrika velikosti 
+\begin_inset Formula $n\times n$
+\end_inset
+
+,
+ ki vsebuje polinome stopnje 
+\begin_inset Formula $<n$
+\end_inset
+
+,
+ kajti vsebuje determinante matrik,
+ katerih elementi so polinomi stopnje 
+\begin_inset Formula $\leq1$
+\end_inset
+
+,
+ torej takele oblike:
+\begin_inset Formula 
+\[
+\tilde{\left(A-xI\right)}^{T}=B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}
+\]
+
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Ne razumem,
+ zakaj so tu matrike 
+\begin_inset Formula $B$
+\end_inset
+
+ in ne skalarji.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-\lambda I\right)=c_{0}+c_{1}x+\cdots+c_{n}x^{n}$
+\end_inset
+
+.
+ Kot v enačbi množimo to z 
+\begin_inset Formula $I$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\det\left(A-\lambda I\right)\cdot I=c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}$
+\end_inset
+
+.
+ Oglejmo si še desno stran enačbe:
+\begin_inset Formula 
+\[
+\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}=\left(A-xI\right)\left(B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}\right)=AB_{0}+AB_{1}x+\cdots+AB_{n-1}x^{n-1}-B_{0}x-B_{1}x^{2}-\cdots-B_{n-1}x^{n}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}x^{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n}
+\]
+
+\end_inset
+
+In primerjajmo koeficiente v polinomih pred istoležnimi spremenljivkami na obeh straneh tele enačbe:
+\begin_inset Formula 
+\[
+\det\left(A-xI\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\begin{array}{cccc}
+1: & c_{0}I & = & AB_{0}\\
+x: & c_{1}I & = & AB_{1}-B_{0}\\
+x^{2}: & c_{2}I & = & AB_{2}x^{2}-B_{1}\\
+\vdots\\
+x^{n-1}: & c_{n-1}I & = & AB_{n-1}-B_{n-2}\\
+x^{n}: & c_{n}I & = & -B_{n-1}
+\end{array}
+\]
+
+\end_inset
+
+Vstavimo sedaj 
+\begin_inset Formula $A$
+\end_inset
+
+ v enačbo namesto 
+\begin_inset Formula $x$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+p_{A}\left(A\right)=c_{0}I+c_{1}IA+\cdots+c_{n}IA^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)A+\left(AB_{2}-B_{1}\right)A^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)A^{n-1}-B_{n-1}A^{n}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=AB_{0}+A^{2}B_{1}-AB_{0}+A^{2}B^{2}-B_{1}A^{2}+\cdots+A^{n}B_{n-1}-A^{n-1}B_{n-2}-B_{n-1}A^{n}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+p_{A}\left(A\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Matriko 
+\begin_inset Formula $A$
+\end_inset
+
+ se da diagonalizirati 
+\begin_inset Formula $\Leftrightarrow m_{A}\left(x\right)$
+\end_inset
+
+ ima samo enostavne ničle (nima večkratnih —
+ potence so vse 1).
+ Torej 
+\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{1}\cdots\left(x-\lambda_{k}\right)^{1}$
+\end_inset
+
+ za 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ vse paroma različne lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Po predpostavki je 
+\begin_inset Formula $A$
+\end_inset
+
+ podobna diagonalni matriki —
+ 
+\begin_inset Formula $A=PDP^{-1}$
+\end_inset
+
+ za diagonalno 
+\begin_inset Formula $D$
+\end_inset
+
+ in obrnljivo 
+\begin_inset Formula $P$
+\end_inset
+
+.
+ BSŠ naj bo 
+\begin_inset Formula $D=\left[\begin{array}{ccc}
+\lambda_{1} & 0 & 0\\
+0 & \cdots & 0\\
+0 & 0 & \lambda_{k}
+\end{array}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\lambda_{1}\leq\cdots\leq\lambda_{k}$
+\end_inset
+
+.
+ NADALJUJ TULE AAAA 
+\begin_inset Quotes gld
+\end_inset
+
+LA1P FMF 2024-03-20
+\begin_inset Quotes grd
+\end_inset
+
+ stran 2.
+\end_layout
+
 \end_deeper
 \begin_layout Part
 Vaja za ustni izpit
-- 
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