1 files changed, 3388 insertions, 19 deletions

diff --git a/šola/la/teor.lyx b/šola/la/teor.lyx
index ea3069a..e25c8bf 100644
--- a/šola/la/teor.lyx
+++ b/šola/la/teor.lyx
@@ -2207,7 +2207,7 @@ Sistem je homogen,
  če je vektor desnih strani ničeln.
 \end_layout
 
-\begin_layout Claim*
+\begin_layout Standard
 Vedno ima rešitev 
 \begin_inset Formula $\vec{0}$
 \end_inset
@@ -3817,7 +3817,7 @@ Dokazujemo ekvivalenco.
 \begin_inset Formula $\left(1\Rightarrow2\right)$
 \end_inset
 
- Že dokazano zgoraj.
+ Sledi iz definicije.
 \end_layout
 
 \begin_layout Labeling
@@ -3825,7 +3825,7 @@ Dokazujemo ekvivalenco.
 \begin_inset Formula $\left(1\Rightarrow3\right)$
 \end_inset
 
- Že dokazano zgoraj.
+ Sledi iz definicije.
 \end_layout
 
 \begin_layout Labeling
@@ -4352,7 +4352,9 @@ status open
 
 \begin_layout Plain Layout
 (tegale ne razumem zares dobro,
- niti med predavanji nismo dokazali)
+ niti med predavanji nismo dokazali) mogoče čim ima stopnico,
+ daljšo od 1,
+ ima ničelno vrstico?
 \end_layout
 
 \end_inset
@@ -4645,16 +4647,19 @@ To si lahko zapomnimo s Saurusovim pravilom.
 
 \begin_layout Example*
 Vektorski produkt.
- 
-\begin_inset Formula $\left\langle \left(x,y,z\right),\left(a_{21},a_{22},a_{23}\right)\times\left(a_{31},a_{32},a_{33}\right)\right\rangle =\det\left[\begin{array}{ccc}
+ Velja:
+\begin_inset Formula 
+\[
+\left\langle \left(x,y,z\right),\left(a_{21},a_{22},a_{23}\right)\times\left(a_{31},a_{32},a_{33}\right)\right\rangle =\det\left[\begin{array}{ccc}
 x & y & z\\
 a_{21} & a_{22} & a_{23}\\
 a_{31} & a_{32} & a_{33}
-\end{array}\right]$
+\end{array}\right],
+\]
+
 \end_inset
 
-.
- Torej je 
+torej je 
 \begin_inset Formula $\left[\left(a_{11},a_{12},a_{13}\right),\left(a_{21},a_{22},a_{23}\right),\left(a_{31},a_{32},a_{33}\right)\right]$
 \end_inset
 
@@ -4969,9 +4974,14 @@ Obravnavajmo še splošen primer:
 \end_inset
 
 
+\begin_inset Formula $\det B$
+\end_inset
+
+ zapišimo na dva načina ne levo in desno stran enačbe.
+ 
 \begin_inset Formula 
 \[
-\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det AB=\det B=\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A\det B
+\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det AB=\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A\det B
 \]
 
 \end_inset
@@ -11963,8 +11973,7 @@ mapsto
 \backslash
 left[
 \backslash
-begin{array}{ccc}
-Le_{1} & 
+begin{array}{ccc}Le_{1} & 
 \backslash
 cdots & Le_{n}
 \backslash
@@ -12004,8 +12013,7 @@ udensdash{$
 \backslash
 left[
 \backslash
-begin{array}{ccc}
-L_{A}e_{1} & 
+begin{array}{ccc}L_{A}e_{1} & 
 \backslash
 cdots & L_{A}e_{n}
 \backslash
@@ -15135,7 +15143,14 @@ Algebraično večkratnost
 .
 \end_layout
 
-\begin_layout Claim*
+\begin_layout Claim
+\begin_inset CommandInset label
+LatexCommand label
+name "claim:geom<=alg"
+
+\end_inset
+
+
 \begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$
 \end_inset
 
@@ -15216,7 +15231,13 @@ Dokaza ne razumem.
  Obupam.
 \end_layout
 
-\begin_layout Claim*
+\begin_layout Claim
+\begin_inset CommandInset label
+LatexCommand label
+name "claim:mi=ni=>diag"
+
+\end_inset
+
 Matriko s paroma različnimi lastnimi vrednostmi 
 \begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
 \end_inset
@@ -16051,18 +16072,3366 @@ Dokazujemo ekvivalenco:
 \end_inset
 
 .
- NADALJUJ TULE AAAA 
+ Oglejmo si izraz 
+\begin_inset Formula 
+\[
+\left(D-\lambda_{1}I\right)\cdots\left(D-\lambda_{k}I\right)=\left[\begin{array}{cccc}
+0 &  &  & 0\\
+ & \lambda_{2}-\lambda_{1}\\
+ &  & \ddots\\
+0 &  &  & \lambda_{k}-\lambda_{1}
+\end{array}\right]\cdots\left[\begin{array}{cccc}
+\lambda_{1}-\lambda_{k} &  &  & 0\\
+ & \ddots\\
+ &  & \lambda_{k-1}-\lambda_{k}\\
+0 &  &  & 0
+\end{array}\right]=0
+\]
+
+\end_inset
+
+Sedaj pa še izraz
+\begin_inset Formula 
+\[
+\left(A-\lambda_{1}I\right)\cdots\left(A-\lambda_{k}I\right)=\left(PDP^{-1}-\lambda_{1}I\right)\cdots\left(PDP^{-1}-\lambda_{k}I\right)=\left(PDP^{-1}-\lambda_{1}PP^{-1}\right)\cdots\left(PDP^{-1}-\lambda_{k}PP^{-1}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=P\left(D-\lambda_{1}I\right)\cancel{P^{-1}}\cdots\cancel{P}\left(D-\lambda_{k}I\right)P^{-1}=P\left(D-\lambda_{1}I\right)\cdots\left(D-\lambda_{k}I\right)P^{-1}=P0P^{-1}=0,
+\]
+
+\end_inset
+
+torej ta polinom anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Ker deli 
+\begin_inset Formula $m_{A}\left(x\right)$
+\end_inset
+
+ —
+ vsebuje vse ničle 
+\begin_inset Formula $m_{A}\left(x\right)$
+\end_inset
+
+,
+ je prav to minimalen polinom 
+\begin_inset Formula $A$
+\end_inset
+
+ —
+ ima najmanjšo stopnjo možno.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Potrebujemo nekaj lem:
+\begin_inset CommandInset counter
+LatexCommand set
+counter "theorem"
+value "0"
+lyxonly "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Lemma
+Za vse matrike 
+\begin_inset Formula $A,B$
+\end_inset
+
+ velja 
+\begin_inset Formula $\n\left(AB\right)\leq\n\left(A\right)+\n\left(B\right)\sim\dim\Ker\left(AB\right)\leq\dim\Ker\left(A\right)+\dim\Ker\left(B\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Oglejmo si preslikavo 
+\begin_inset Formula $L:\Ker AB\to\Ker A$
+\end_inset
+
+,
+ ki slika 
+\begin_inset Formula $x\mapsto Bx$
+\end_inset
+
+.
+ Je dobro definirana,
+ kajti 
+\begin_inset Formula $x\in\Ker AB\Rightarrow ABx=0\Rightarrow Bx\in\Ker A$
+\end_inset
+
+.
+ Po osnovnem dimenzijskem izreku za preslikavo 
+\begin_inset Formula $L$
+\end_inset
+
+ velja 
+\begin_inset Formula $\dim\Ker L+\dim\Slika L=\dim\Ker AB$
+\end_inset
+
+.
+ Ker velja 
+\begin_inset Formula $Lx=0\Rightarrow Bx=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\Ker L\subseteq\Ker B$
+\end_inset
+
+ in zato 
+\begin_inset Formula $\dim\Ker L\leq\dim\Ker B$
+\end_inset
+
+.
+ Poleg tega iz definicije velja 
+\begin_inset Formula $\Slika L\subseteq\Ker A$
+\end_inset
+
+ in zato 
+\begin_inset Formula $\dim\Slika L\leq\dim\Ker A$
+\end_inset
+
+.
+ Vstavimo te neenakosti v enačbo iz dimenzijskega izreka:
+\begin_inset Formula 
+\[
+\dim\Ker L+\dim\Slika L=\dim\Ker AB
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\dim\Ker B+\dim\Ker A\geq\dim\Ker AB
+\]
+
+\end_inset
+
+Lemo lahko posplošimo na več faktorkev,
+ torej 
+\begin_inset Formula $\n\left(A_{1}\cdots A_{k}\right)\leq\n A_{1}+\cdots+\n A_{k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Nadaljujmo z dokazom 
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+.
+ Denimo,
+ da 
+\begin_inset Formula $\left(x-\lambda_{1}\right)\cdots\left(x-\lambda_{k}\right)$
+\end_inset
+
+ anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Upoštevamo 
+\begin_inset Formula 
+\[
+\n\left(\left(A-\lambda_{1}\right)\cdots\left(A-\lambda_{k}\right)\right)\leq\n\left(A-\lambda_{1}\right)+\cdots+\n\left(A-\lambda_{k}\right)
+\]
+
+\end_inset
+
+Členi na desni strani so geometrijske večkratnosti,
+ ker pa 
+\begin_inset Formula $A$
+\end_inset
+
+ anhilira polinom po predpostavki,
+ je ta produkt ničelna preslikava in je dimenzija jedra dimenzija celega prostora.
+\begin_inset Formula 
+\[
+\n\left(0\right)=n=n_{1}+\cdots+n_{k}\leq m_{1}+\cdots+m_{k}
+\]
+
+\end_inset
+
+Ker 
+\begin_inset Formula $\forall i:m_{i}\leq n_{i}$
+\end_inset
+
+ (
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "claim:geom<=alg"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+),
+ velja v zgornji neenačbi enakost,
+ torej je matrika po 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "claim:mi=ni=>diag"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ diagonalizabilna.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Subsubsection
+Korenski podprostori
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A\in M_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{r_{1}}\cdots\left(x-\lambda_{k}\right)^{r_{k}}$
+\end_inset
+
+ njen minimalni polinom.
+ 
+\begin_inset Formula $\forall i\in\left\{ 1..k\right\} $
+\end_inset
+
+ označimo z 
+\begin_inset Formula $W_{i}\coloneqq\Ker\left(A-\lambda_{i}I\right)^{r_{i}}$
+\end_inset
+
+ korenski podprostor matrike 
+\begin_inset Formula $A$
+\end_inset
+
+ za lastno vrednost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+.
+ Vpeljimo še oznako 
+\begin_inset Formula $V_{i}\coloneqq\Ker\left(A-\lambda_{i}I\right)^{1}$
+\end_inset
+
+ (tu potenca ni 
+\begin_inset Formula $r_{i}$
+\end_inset
+
+,
+ temveč je 
+\begin_inset Formula $1$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Definition*
+\begin_inset CommandInset counter
+LatexCommand set
+counter "theorem"
+value "0"
+lyxonly "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact
+Očitno je 
+\begin_inset Formula $\Ker\left(A-\lambda_{i}I\right)\subseteq\Ker\left(A-\lambda_{i}\right)^{2}\subseteq\Ker\left(A-\lambda_{i}I\right)^{3}\subseteq\cdots$
+\end_inset
+
+,
+ kajti če 
+\begin_inset Formula $x\in\Ker\left(A-\lambda_{i}I\right)^{m}\Rightarrow\left(A-\lambda_{i}I\right)^{m}x=0\Rightarrow\left(A-\lambda_{i}I\right)\left(A-\lambda_{i}I\right)^{m}x=0\Rightarrow x\in\Ker\left(A-\lambda_{i}I\right)^{m+1}$
+\end_inset
+
+.
+ Izkaže se,
+ da so vse inkluzije do 
+\begin_inset Formula $r_{i}-te$
+\end_inset
+
+ potence stroge,
+ od 
+\begin_inset Formula $r_{i}-$
+\end_inset
+
+te potence dalje pa so vse inkluzije enačaji,
+ torej za 
+\begin_inset Formula $W_{i}=\Ker\left(A-\lambda_{i}I\right)^{r_{i}}$
+\end_inset
+
+ velja 
+\begin_inset Formula 
+\[
+\Ker\left(A-\lambda_{i}I\right)\subset\Ker\left(A-\lambda_{i}I\right)^{2}\subset\cdots\subset\Ker\left(A-\lambda_{i}I\right)^{r_{i}}=\Ker\left(A-\lambda_{i}I\right)^{r_{i}+1}=\cdots
+\]
+
+\end_inset
+
+Poleg tega se izkaže,
+ da je 
+\begin_inset Formula $\dim W_{i}=n_{i}$
+\end_inset
+
+ (algebraična večkratnost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact
+\begin_inset Formula $\dim V_{i}=\dim\Ker\left(A-\lambda_{i}I\right)=m_{1}$
+\end_inset
+
+ (geometrijska večkratnost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Claim
+\begin_inset CommandInset label
+LatexCommand label
+name "claim:vsota-kor-podpr-je-vse"
+
+\end_inset
+
+
+\begin_inset Formula $\mathbb{C}^{n}=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{k}$
+\end_inset
+
+ —
+ vsota vseh korenskih podprostorov je vse in ta vsota je direktna.
+ Tej vsoti pravimo 
 \begin_inset Quotes gld
 \end_inset
 
-LA1P FMF 2024-03-20
+korenski razcep matrike 
+\begin_inset Formula $A$
+\end_inset
+
+
 \begin_inset Quotes grd
 \end_inset
 
- stran 2.
+.
+\end_layout
+
+\begin_layout Remark*
+Dokazujmo:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $V_{1}+\cdots+V_{k}$
+\end_inset
+
+ je tudi direktna,
+ ampak ni nujno enaka 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\mathbb{C}^{n}=V_{1}+\cdots+V_{k}\Leftrightarrow A$
+\end_inset
+
+ se da diagonalizirati (povedano prej).
+ Dokažimo trditev 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "claim:vsota-kor-podpr-je-vse"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+ Dokažimo najprej,
+ da če 
+\begin_inset Formula $w_{1}\in W_{1},\dots,w_{k}\in W_{k}$
+\end_inset
+
+ zadoščajo 
+\begin_inset Formula $w_{1}+\cdots+w_{k}=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $w_{1}=\cdots=w_{k}=0$
+\end_inset
+
+ (direktna).
+ Delajmo indukcijo po številu členov:
+\end_layout
+
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $w_{1}=0\Rightarrow w_{1}=0$
+\end_inset
+
+ je očitno.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Indukcijska predpostavka:
+ 
+\begin_inset Formula $w_{1}+\cdots+w_{i}=0\Rightarrow w_{1}=\cdots=w_{i}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Naj bodo 
+\begin_inset Formula $w_{1},\dots,w_{i+1}$
+\end_inset
+
+ taki,
+ da 
+\begin_inset Formula 
+\[
+w_{1}+\cdots+w_{i+1}=0\quad\quad\quad\quad/\cdot\left(A-\lambda_{i+1}I\right)^{r_{i+1}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+w_{1}'+\cdots+w_{i}'+0=0,
+\]
+
+\end_inset
+
+kajti 
+\begin_inset Formula $w_{i+1}\in\Ker\left(A-\lambda_{i+1}I\right)^{r_{i+1}}$
+\end_inset
+
+.
+ Ker je vsak korenski prostor 
+\begin_inset Formula $W_{j}$
+\end_inset
+
+ invarianten za 
+\begin_inset Formula $\left(A-\lambda_{i+1}I\right)^{r_{i+1}}$
+\end_inset
+
+,
+ ...
+ Tega dokaza ne najdem,
+ zato 
+\series bold
+tega dokaza ne razumem
+\series default
+.
+ Po definiciji je 
+\begin_inset Formula $V$
+\end_inset
+
+ invarianten za 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ če za vsak 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ velja 
+\begin_inset Formula $Av\in V$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Plain Layout
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+Nadaljuj 
+\begin_inset Quotes gld
+\end_inset
+
+LA1P FMF 2024-03-20.pdf
+\begin_inset Quotes grd
+\end_inset
+
+ na strani 3.
+\end_layout
+
+\end_inset
+
+Če predpostavimo,
+ da je vsota direktna,
+ je lahko dokazati,
+ da je vsota cel prostor.
+ V karakteristični polinom,
+ ki po Caylay-Hamiltonu anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ vstavimo 
+\begin_inset Formula $A$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $0=\left(-1\right)^{n}\left(A-\lambda_{1}I\right)^{r_{1}}\cdots\left(A-\lambda_{k}I\right)^{r_{k}}=A_{1}\cdots A_{k}$
+\end_inset
+
+.
+ Ker je vsota direktna,
+ velja 
+\begin_inset Formula $\n\left(A_{1}\cdots A_{n}\right)=\n\left(0\right)=n=\n A_{1}\cdots\n A_{k}=\dim\left(W_{1}+\cdots+W_{k}\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $W_{1}+\cdots+W_{k}=\mathbb{C}^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Če predpostavimo,
+ da je 
+\begin_inset Formula $W_{i}\cap W_{j}=\left\{ 0\right\} $
+\end_inset
+
+ za 
+\begin_inset Formula $i\not=j$
+\end_inset
+
+,
+ lahko od tod izpeljemo direktnost vsote korenskih podprostorov.
+ Dokaz z indukcijo:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $W_{1}$
+\end_inset
+
+ je direktna vsota.
+ Očitno (
+\begin_inset Formula $\forall w_{1}\in W_{1}:w_{1}=0\Rightarrow w_{1}=0$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Itemize
+Indukcijska predpostavka:
+ 
+\begin_inset Formula $w_{1}+\cdots+w_{i}=0\Rightarrow w_{1}=\cdots=w_{i}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Naj bodo 
+\begin_inset Formula $w_{1},\dots,w_{i+1}$
+\end_inset
+
+ taki,
+ da 
+\begin_inset Formula 
+\[
+w_{1}+\cdots+w_{i+1}=0\quad\quad\quad\quad/\cdot\left(A-\lambda_{i+1}I\right)^{r_{i+1}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{1}+\cdots+\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{i}+0=0
+\]
+
+\end_inset
+
+Ker 
+\begin_inset Formula $\left(A-\lambda_{h}I\right)^{r_{h}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(A-\lambda_{k}I\right)^{r_{k}}$
+\end_inset
+
+ za vsaka 
+\begin_inset Formula $h,k$
+\end_inset
+
+ komutirata (gre namreč za polinom,
+ v katerega je vstavljen 
+\begin_inset Formula $A$
+\end_inset
+
+),
+ velja za vsak 
+\begin_inset Formula $j$
+\end_inset
+
+ iz 
+\begin_inset Formula $\left(A-\lambda_{j}I\right)^{r_{j}}w_{j}=0=\left(A-\lambda_{i+1}I\right)^{r_{i+1}}\left(A-\lambda_{j}I\right)^{r_{j}}w_{j}$
+\end_inset
+
+ tudi
+\begin_inset Formula 
+\[
+\left(A-\lambda_{j}I\right)^{r_{j}}\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{j}=0
+\]
+
+\end_inset
+
+Ker je po I.
+ P.
+ 
+\begin_inset Formula $W_{1}+\cdots+W_{i}$
+\end_inset
+
+ direktna,
+ velja za vsak 
+\begin_inset Formula $j$
+\end_inset
+
+ 
+\begin_inset Formula $w_{j}\in W_{j}$
+\end_inset
+
+,
+ toda zaradi našega množenja tudi 
+\begin_inset Formula $w_{j}\in W_{i+1}$
+\end_inset
+
+.
+ Zaradi predpostavke 
+\begin_inset Formula $m=n\Rightarrow W_{m}\cup W_{n}=\left\{ 0\right\} $
+\end_inset
+
+ velja za vsak
+\begin_inset Formula $j\in\left\{ 1..i\right\} :$
+\end_inset
+
+ 
+\begin_inset Formula $w_{j}=0$
+\end_inset
+
+.
+ V prvi enačbi ostane le še 
+\begin_inset Formula $w_{i+1}=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+Dokazati je treba še 
+\begin_inset Formula $i\not=j\Rightarrow W_{i}\cup W_{j}=\left\{ 0\right\} $
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $W_{i}$
+\end_inset
+
+ invarianten za 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ t.
+ j.
+ 
+\begin_inset Formula $v\in W_{i}\Rightarrow Av\in W_{i}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $v\in W_{i}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula 
+\[
+\left(A-\lambda_{i}I\right)^{r_{i}}v=0\quad\quad\quad\quad/\cdot A
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+A\left(A-\lambda_{i}I\right)^{r_{i}}v=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+Av\in\Ker\left(A-\lambda_{i}I\right)^{r_{i}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+Av\in W_{i}
+\]
+
+\end_inset
+
+Ker so vsi 
+\begin_inset Formula $W_{i}$
+\end_inset
+
+ invariantni za 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ so tudi njihovi preseki invariantni za 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Definirajmo torej linearno preslikavo 
+\begin_inset Formula $L:W_{i}\cap W_{j}\to W_{i}\cap W_{j}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $v\mapsto Av$
+\end_inset
+
+.
+ Vemo,
+ da ima 
+\begin_inset Formula $L$
+\end_inset
+
+ vsaj eno lastno vrednost 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ in pripadajoči lastni vektor 
+\begin_inset Formula $w$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $w\in W_{i}\cap W_{j}$
+\end_inset
+
+ in 
+\begin_inset Formula $Lw=\lambda w$
+\end_inset
+
+,
+ toda 
+\begin_inset Formula $Lw=Aw=\lambda w$
+\end_inset
+
+.
+ Ker velja 
+\begin_inset Formula $Av=\lambda v\Rightarrow A^{q}v=\lambda^{q}v\Rightarrow p\left(A\right)v=p\left(\lambda\right)v$
+\end_inset
+
+ za vsak polinom 
+\begin_inset Formula $p$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $p\left(A\right)w=p\left(\lambda\right)w$
+\end_inset
+
+ za vsak polinom 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ Uporabimo polinom 
+\begin_inset Formula $p\left(x\right)=\left(x-\lambda_{i}\right)^{r_{i}}$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $\left(A-\lambda_{i}\right)^{r_{i}}w=\left(\lambda-\lambda_{i}\right)^{r_{i}}w$
+\end_inset
+
+.
+ Leva stran enačbe je 0,
+ PDDRAA 
+\begin_inset Formula $w$
+\end_inset
+
+ ni 0,
+ torej 
+\begin_inset Formula $\left(\lambda-\lambda_{i}\right)^{r_{i}}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lambda=\lambda_{i}$
+\end_inset
+
+.
+ Vendar lahko namesto tistega polimoma uporabimo polinom 
+\begin_inset Formula $p\left(x\right)=\left(x-\lambda_{j}\right)^{r_{j}}$
+\end_inset
+
+,
+ kar pokaže 
+\begin_inset Formula $\lambda=\lambda_{j}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lambda_{j}=\lambda_{i}$
+\end_inset
+
+,
+ kar je v 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+ s tem,
+ da so lastne vrednosti 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ paroma različne.
+ Torej 
+\begin_inset Formula $w=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Jordanska kanonična forma
+\end_layout
+
+\begin_layout Standard
+Vsaka kvadratna matrika je podobna posebni zgornjetrikotni matriki,
+ ki ji pravimo JKF.
+ To je bločno diagonalna matrika,
+ ki ima za diagonalne bloke t.
+ i.
+ 
+\begin_inset Quotes gld
+\end_inset
+
+jordanske kletke
+\begin_inset Quotes grd
+\end_inset
+
+,
+ to so matrike oblike:
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccccc}
+\lambda & 1\\
+ & \lambda & 1\\
+ &  & \ddots & \ddots\\
+ &  &  & \lambda & 1\\
+ &  &  &  & \lambda
+\end{array}\right].
+\]
+
+\end_inset
+
+Jordanska matrika je sestavljena iz jordanskih kletk po diagonali (
+\begin_inset Formula $J_{i}$
+\end_inset
+
+ so jordanske kletke):
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccc}
+J_{1} &  & 0\\
+ & \ddots\\
+0 &  & J_{m}
+\end{array}\right].
+\]
+
+\end_inset
+
+Običajno zahtevamo še,
+ da so JK,
+ ki imajo isto lastno vrednost,
+ skupaj,
+ ter da so JK padajoče urejene po lastni vrednosti od največje do najmanjše.
+\end_layout
+
+\begin_layout Theorem*
+Za vsako kvadratno kompleksno matriko 
+\begin_inset Formula $A\in M_{n\times n}\left(\mathbb{C}\right)$
+\end_inset
+
+ obstaja taka jordanska matrika 
+\begin_inset Formula $J$
+\end_inset
+
+ in taka obrnljiva matrika 
+\begin_inset Formula $P$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $A=PJP^{-1}$
+\end_inset
+
+.
+ ZDB vsaka 
+\begin_inset Formula $A\in M_{n\times n}\left(\mathbb{C}\right)$
+\end_inset
+
+ je podobna neki Jordanski matriki.
+\end_layout
+
+\begin_layout Standard
+Procesu iskanja jordanske matrike pravimo 
+\begin_inset Quotes gld
+\end_inset
+
+jordanifikacija
+\begin_inset Quotes grd
+\end_inset
+
+.
+ Kako konstruiramo 
+\begin_inset Formula $J$
+\end_inset
+
+ in 
+\begin_inset Formula $P$
+\end_inset
+
+?
+ Izračunamo lastne vrednosti in pripadajoče korenske podprostore.
+ 
+\end_layout
+
+\begin_layout Itemize
+Naj bo 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ lastna vrednost 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Za preglednost pišimo 
+\begin_inset Formula $N\coloneqq A-\lambda I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Izračunamo lastne vektorje in lastni podprostor 
+\begin_inset Formula $\Ker N^{r}$
+\end_inset
+
+ ter ga izrazimo z njegovo bazo,
+ recimo ji 
+\begin_inset Formula $B_{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Nato izberemo 
+\begin_inset Quotes gld
+\end_inset
+
+pomožne baze
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $\mathcal{B}_{1},\dots,\mathcal{B}_{r}$
+\end_inset
+
+,
+ ki pripadajo prostorom 
+\begin_inset Formula $\Ker N^{1},\dots,\Ker N^{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Pomožno bazo 
+\begin_inset Formula $\mathcal{B}_{r-1}$
+\end_inset
+
+ dopolnimo do baze 
+\begin_inset Formula $\mathcal{B}_{r}$
+\end_inset
+
+ z elementi 
+\begin_inset Formula $\mathcal{B}_{r}$
+\end_inset
+
+ 
+\begin_inset Formula $u_{1},\dots,u_{k_{1}}$
+\end_inset
+
+.
+ Potem je 
+\begin_inset Formula $\mathcal{B}_{r-1}\cup$
+\end_inset
+
+
+\begin_inset Formula $\left\{ u_{1},\dots,u_{k_{1}}\right\} $
+\end_inset
+
+ 
+\begin_inset Quotes gld
+\end_inset
+
+popravek pomožne baze 
+\begin_inset Formula $\mathcal{B}_{r}$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Vektorje 
+\begin_inset Formula $\left\{ u_{1},\dots,u_{k_{1}}\right\} \in\Ker N^{r}$
+\end_inset
+
+ pomnožimo z matriko 
+\begin_inset Formula $N$
+\end_inset
+
+,
+ dobljeni 
+\begin_inset Formula $Nu_{1},\dots,Nu_{k_{1}}$
+\end_inset
+
+ ležijo v 
+\begin_inset Formula $\Ker N^{r-1}$
+\end_inset
+
+.
+ Množica 
+\begin_inset Formula $\mathcal{B}_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{k_{1}}\right\} $
+\end_inset
+
+ je linearno neodvisna.
+ Izberemo take 
+\begin_inset Formula $v_{1},\dots,v_{k_{2}}\in B_{r-1}$
+\end_inset
+
+,
+ ki dopolnijo LN 
+\begin_inset Formula $B_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{2}\right\} $
+\end_inset
+
+ do baze 
+\begin_inset Formula $\Ker N^{r-1}$
+\end_inset
+
+.
+ Potem je 
+\begin_inset Formula $\mathcal{B}_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{k_{1}}\right\} \cup\left\{ v_{1},\dots,v_{k_{2}}\right\} $
+\end_inset
+
+ popravek pomožne baze 
+\begin_inset Formula $\mathcal{B}_{r-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Izberemo take 
+\begin_inset Formula $w_{1},\dots,w_{k_{3}}\in\mathcal{B}_{r-2}$
+\end_inset
+
+,
+ ki 
+\begin_inset Formula $\mathcal{B}_{r-3}\cup\left\{ N^{2}u_{1},\dots,N^{2}u_{k_{1}}Nv_{1},\dots,Nv_{k_{2}}\right\} $
+\end_inset
+
+ dopolnijo do baze 
+\begin_inset Formula $\Ker N^{r-2}$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $\mathcal{B}_{r-3}\cup\left\{ N^{2}u_{1},\cdots,N^{2}u_{k_{1}},Nv_{1},\dots,Nv_{k_{2}},w_{1},\dots,w_{k_{3}}\right\} $
+\end_inset
+
+ popravek pomožne baze 
+\begin_inset Formula $B_{r-2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Postopek ponavljamo,
+ dokler ne popravimo vseh možnih baz.
+\end_layout
+
+\begin_layout Standard
+Dobimo t.
+ i.
+ 
+\begin_inset Quotes gld
+\end_inset
+
+jordanske verige
+\begin_inset Quotes grd
+\end_inset
+
+.
+ Ena jordanska veriga je 
+\begin_inset Formula $\left(u,Nu,N^{2}u,\dots,N^{x}u\right)$
+\end_inset
+
+,
+ torej preslikanje elementa 
+\begin_inset Formula $u$
+\end_inset
+
+,
+ ki začne kot dopolnitev baze korenskega podprostora 
+\begin_inset Formula $\Ker N^{x+1}$
+\end_inset
+
+ in je na koncu 
+\begin_inset Formula $x-$
+\end_inset
+
+krat preslikan z 
+\begin_inset Formula $N$
+\end_inset
+
+,
+ torej konča v korenskem podprostoru 
+\begin_inset Formula $\Ker N$
+\end_inset
+
+.
+ Nekatere verige se začno v največjem korenskem podprostoru 
+\begin_inset Formula $\Ker N^{r}$
+\end_inset
+
+,
+ nekatere šele kasneje,
+ v 
+\begin_inset Formula $\Ker N^{1}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\Ker N^{2}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\Ker N^{3}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Imamo torej 
+\begin_inset Formula $k_{1}$
+\end_inset
+
+ jordanskih verig dolžine 
+\begin_inset Formula $r$
+\end_inset
+
+,
+ 
+\begin_inset Formula $k_{2}$
+\end_inset
+
+ jordanskih verig dolžine 
+\begin_inset Formula $r-1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $k_{3}$
+\end_inset
+
+ jordanskih verig dolžine 
+\begin_inset Formula $r-2$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $k_{r}$
+\end_inset
+
+ jordanskih verig dolžine 1.
+ Skupaj je jordanskih verig 
+\begin_inset Formula $k_{1}+\cdots+k_{r}=\dim\Ker N$
+\end_inset
+
+.
+ Jordanskih verig za lastno vrednost 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ je torej toliko,
+ kot je njena geometrijska večkratnost.
+\end_layout
+
+\begin_layout Standard
+Vsaki jordanski verigi dolžine 
+\begin_inset Formula $k$
+\end_inset
+
+ pripada ena jordanska kletka velikosti 
+\begin_inset Formula $k\times k$
+\end_inset
+
+.
+ 
+\begin_inset Formula $k-$
+\end_inset
+
+vektorjev iz verige zložimo v 
+\begin_inset Formula $P$
+\end_inset
+
+ tako,
+ da je vektor z začetka verige (torej tisti iz popravljene baze večjega prostora) na levi strani v matriki.
+\end_layout
+
+\begin_layout Example*
+Poišči jordansko kanonično formo matrike
+\begin_inset Formula 
+\[
+A=\left[\begin{array}{cccc}
+0 & 1 & -1 & 2\\
+0 & 2 & 2 & 2\\
+0 & 0 & 2 & 0\\
+0 & 0 & 0 & 2
+\end{array}\right].
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Najprej izračunamo karakteristični polinom:
+ 
+\begin_inset Formula $\det\left(A-\lambda I\right)=x\left(x-2\right)^{3}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\lambda_{1}=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $n_{1}=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\lambda_{2}=2$
+\end_inset
+
+,
+ 
+\begin_inset Formula $n_{2}=3$
+\end_inset
+
+.
+ Lastni vektorji:
+ 
+\begin_inset Formula $\Ker\left(A-0I\right)=\Lin\left\{ \left(1,0,0,0\right)\right\} $
+\end_inset
+
+,
+ 
+\begin_inset Formula $\Ker\left(A-2I\right)=\Lin\left\{ \left(3,0,-2,2\right),\left(1,2,0,0\right)\right\} $
+\end_inset
+
+.
+ Če bi dobili 4 lastne vektorje,
+ bi lahko matriko diagonalizirali.
+ Tako je ne moremo.
+ Ker 
+\begin_inset Formula $n_{1}=1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $r_{1}$
+\end_inset
+
+ največ 
+\begin_inset Formula $1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\Ker\left(A-0I\right)=\Ker\left(A-0I\right)^{2}=\cdots$
+\end_inset
+
+.
+ Izračunamo korenske podprostore
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+za lastno vrednost 0:
+ 
+\begin_inset Formula $\Ker\left(A-0I\right)=\Ker\left(A\right)=\Ker\left(A^{2}\right)$
+\end_inset
+
+.
+ Dobimo eno verigo 
+\begin_inset Formula $\left(\left(1,0,0,0\right)\right)$
+\end_inset
+
+ dolžine 
+\begin_inset Formula $1$
+\end_inset
+
+ za lastno vrednost 
+\begin_inset Formula $0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+za lastno vrednost 2:
+ 
+\begin_inset Formula $\Ker\left(A-2I\right)^{2}=\Lin\left\{ \left(1,0,0,2\right),\left(-1,0,1,0\right),\left(1,2,0,0\right)\right\} $
+\end_inset
+
+,
+ 
+\begin_inset Formula $\Ker\left(A-2I\right)^{3}=\Ker\left(A-2I\right)^{2}$
+\end_inset
+
+.
+ Opazimo,
+ da je 
+\begin_inset Formula $\left(1,0,0,2\right)$
+\end_inset
+
+ dopolnitev baze 
+\begin_inset Formula $\Ker\left(A-2I\right)$
+\end_inset
+
+ do baze 
+\begin_inset Formula $\Ker\left(A-2I\right)^{2}$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $\left\{ \left(1,0,0,2\right)\right\} $
+\end_inset
+
+ 
+\begin_inset Quotes gld
+\end_inset
+
+popravljena baza
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $N^{2}$
+\end_inset
+
+.
+ Preslikamo 
+\begin_inset Formula $\left(A-2I\right)\left(1,0,0,2\right)=\left(2,4,0,0\right)$
+\end_inset
+
+,
+ kar tvori verigo dolžine 2 
+\begin_inset Formula $\left(\left(1,0,0,2\right),\left(2,4,0,0\right)\right)$
+\end_inset
+
+.
+ Edini linearno neodvisen od 
+\begin_inset Formula $\left(2,4,0,0\right)$
+\end_inset
+
+ v 
+\begin_inset Formula $\mathcal{B}_{1}$
+\end_inset
+
+ je 
+\begin_inset Formula $\left(3,0,-2,2\right)$
+\end_inset
+
+,
+ zato je slednji začetek zadnje tretje verige dolžine 1 
+\begin_inset Formula $\left(\left(3,0,-2,2\right)\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Tri verige,
+ ki jih dobimo,
+ so 
+\begin_inset Formula $\left(\left(1,0,0,0\right)\right)$
+\end_inset
+
+ za lastno vrednost 
+\begin_inset Formula $0$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(\left(1,0,0,2\right),\left(2,4,0,0\right)\right)$
+\end_inset
+
+ ter 
+\begin_inset Formula $\left(\left(3,0,-2,2\right)\right)$
+\end_inset
+
+ obe za lastno vrednost 2.
+ Zložimo jih v matriko 
+\begin_inset Formula $P$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+P=\left[\begin{array}{cccc}
+1 & 1 & 2 & 3\\
+0 & 0 & 4 & 0\\
+0 & 0 & 0 & -2\\
+0 & 2 & 0 & 2
+\end{array}\right]
+\]
+
+\end_inset
+
+V matriko 
+\begin_inset Formula $J$
+\end_inset
+
+ pa zložimo kletke pripadajočih velikosti:
+\begin_inset Formula 
+\[
+J=\left[\begin{array}{cccc}
+0 &  &  & 0\\
+ & 2 & 1\\
+ &  & 2\\
+0 &  &  & 2
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+In velja 
+\begin_inset Formula $A=PJP^{-1}$
+\end_inset
+
+ (
+\begin_inset Formula $P^{-1}$
+\end_inset
+
+ izračunamo z Gaussom).
+\end_layout
+
+\begin_layout Subsubsection
+Funkcije matrik
+\end_layout
+
+\begin_layout Standard
+Če poznamo razcep 
+\begin_inset Formula $A=PJP^{-1}$
+\end_inset
+
+,
+ prevedemo računanje potenc 
+\begin_inset Formula $A$
+\end_inset
+
+ na računanje potenc matrike 
+\begin_inset Formula $J$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula 
+\[
+A^{n}=\left(PJP^{-1}\right)\left(PJP^{-1}\right)\cdots\left(PJP^{-1}\right)=PJP^{-1}PJP^{-1}\cdots PJP^{-1}=PJ^{n}P^{-1}.
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $J$
+\end_inset
+
+ bločno diagonalna matrika,
+ sestavljena iz jordanskih kletk,
+ se potenciranje 
+\begin_inset Formula $J$
+\end_inset
+
+ prevede na potenciranje kletk,
+ kajti 
+\begin_inset Formula 
+\[
+J^{n}=\left[\begin{array}{ccc}
+J_{1}^{n} &  & 0\\
+ & \ddots\\
+0 &  & J_{m}^{n}
+\end{array}\right].
+\]
+
+\end_inset
+
+Potenciranje jordanske kletke:
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccccc}
+\lambda & 1 &  &  & 0\\
+ & \lambda & 1\\
+ &  & \ddots & \ddots\\
+ &  &  & \lambda & 1\\
+0 &  &  &  & \lambda
+\end{array}\right]^{n}=\left(\lambda I+\left[\begin{array}{ccc}
+1 &  & 0\\
+ & \ddots\\
+0 &  & 1
+\end{array}\right]\right)^{n}=\left(\lambda I+N\right)^{n}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\binom{n}{0}\left(\lambda I\right)^{n}N^{0}+\binom{n}{1}\left(\lambda N\right)^{n-1}N^{1}+\cdots+\binom{n}{n}\left(\lambda N\right)^{0}N^{n}=\binom{n}{0}\lambda^{n}+\binom{n}{1}\lambda^{n-1}N^{1}+\cdots+\binom{n}{n}N^{n}
+\]
+
+\end_inset
+
+Poraja se vprašanje,
+ kako potencirati 
+\begin_inset Formula $N=\left[\begin{array}{ccccc}
+0 & 1 &  &  & 0\\
+ & \ddots & \ddots\\
+ &  &  & \ddots\\
+ &  &  & \ddots & 1\\
+0 &  &  &  & 0
+\end{array}\right]$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $N^{2}=\left[\begin{array}{ccccc}
+0 & 0 & 1 &  & 0\\
+ & \ddots & \ddots & \ddots\\
+ &  &  & \ddots & 1\\
+ &  &  & \ddots & 0\\
+0 &  &  &  & 0
+\end{array}\right]$
+\end_inset
+
+ in tako dalje (
+\begin_inset Quotes gld
+\end_inset
+
+diagonalo
+\begin_inset Quotes grd
+\end_inset
+
+ enic pomikamo gor in desno).
+ Za 
+\begin_inset Formula $r\times r$
+\end_inset
+
+ jordansko kletko,
+ kadar 
+\begin_inset Formula $n\geq r$
+\end_inset
+
+ (sicer dobimo le prvih nekaj naddiagonal),
+ sledi
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccccc}
+\lambda & 1 &  &  & 0\\
+ & \lambda & 1\\
+ &  & \ddots & \ddots\\
+ &  &  & \lambda & 1\\
+0 &  &  &  & \lambda
+\end{array}\right]^{n}=\left[\begin{array}{ccccc}
+\lambda^{n} & n\lambda^{n-1} & \cdots & \binom{n}{r-2}\lambda^{n-r+2} & \binom{n}{r-1}\lambda^{n-r+1}\\
+ & \lambda^{n} & n\lambda^{n-1} & \ddots & \binom{n}{r-2}\lambda^{n-r+2}\\
+ &  & \ddots & \ddots & \vdots\\
+ &  &  & \lambda^{n} & n\lambda^{n-1}\\
+0 &  &  &  & \lambda^{n}
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Za računanje poljubne funkcije jordanske kletke pa velja predpis
+\begin_inset Formula 
+\[
+f\left(\left[\begin{array}{ccccc}
+\lambda & 1 &  &  & 0\\
+ & \lambda & 1\\
+ &  & \ddots & \ddots\\
+ &  &  & \lambda & 1\\
+0 &  &  &  & \lambda
+\end{array}\right]\right)=\left[\begin{array}{ccccc}
+f\left(\lambda\right) & f'\left(\lambda\right) & \frac{f''\left(\lambda\right)}{2} & \cdots & \frac{f^{\left(k-1\right)\left(\lambda\right)}}{\left(k-1\right)!}\\
+ & f\left(\lambda\right) & f'\left(\lambda\right) & \ddots & \cdots\\
+ &  & \ddots & \ddots & \frac{f''\left(\lambda\right)}{2}\\
+ &  &  & f\left(\lambda\right) & f'\left(\lambda\right)\\
+0 &  &  &  & f\left(\lambda\right)
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+In torej za računanje poljubne funkcije poljubne matrike 
+\begin_inset Formula $f\left(A\right)=f\left(PJP^{-1}\right)=Pf\left(J\right)P^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Vektorski prostori s skalarnim produktom
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor nad poljem 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ nenujno končno razsežen.
+ Preslikavi 
+\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle :V\times V\to\mathbb{R}$
+\end_inset
+
+ pravimo skalarni produtkt,
+ če zadošča naslednjim lastnostim:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+pozitivna definitnost:
+ 
+\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle >0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+simetričnost:
+ 
+\begin_inset Formula $\forall u,v\in V:\left\langle v,u\right\rangle =\left\langle u,v\right\rangle $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+linearnost v prvem faktorju:
+ 
+\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in\mathbb{R},v_{1},v_{2}\in V:\left\langle \alpha_{1}v_{1}+\alpha_{2}v_{2},v\right\rangle =\alpha_{1}\left\langle v_{1},v\right\rangle +\alpha_{2}\left\langle v_{2},v\right\rangle $
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Corollary*
+linearnost v drugem faktorju.
+ 
+\begin_inset Formula $\left\langle u,\beta_{1}v_{1}+\beta_{2}v_{2}\right\rangle =\left\langle \beta_{1}v_{1}+\beta_{2}v_{2},v\right\rangle =\beta_{1}\left\langle v_{1},v\right\rangle +\beta_{2}\left\langle v_{2},v\right\rangle =\beta_{1}\left\langle v,v_{1}\right\rangle +\beta_{2}\left\langle v,v_{2}\right\rangle $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Skalarni produkt z 0:
+ 
+\begin_inset Formula $\left\langle 0,v\right\rangle =\left\langle 0\cdot v+0\cdot v,v\right\rangle =0\left\langle v,v\right\rangle +0\left\langle v,v\right\rangle =0\Rightarrow\left\langle v,0\right\rangle =0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Alternativna formulacija 1:
+ 
+\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$
+\end_inset
+
+.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Dokazujemo ekvivalenco:
+ alternativna formulacija 1 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ originalna definicija 1.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Predpostavimo 
+\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle \geq0$
+\end_inset
+
+ in izjavo negirajmo:
+ 
+\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \leq0\Rightarrow v=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Predpostavimo 
+\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$
+\end_inset
+
+ 
+\end_layout
+
+\end_deeper
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Primeri vektorskih prostorov s skalarnim produktom:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ s standardnim skalarnim produktom:
+ 
+\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\alpha_{1}\beta_{1}+\cdots+\alpha_{n}\beta_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ z nestandardnim skalarnim produktom:
+ Za pojubne 
+\begin_inset Formula $\gamma_{1}>0,\dots,\gamma_{n}>0$
+\end_inset
+
+ definirajmo 
+\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\gamma_{1}\alpha_{1}\beta_{1}+\cdots+\gamma_{n}\alpha_{n}\beta_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+neskončno razsežen primer s standardnim skalarnim produktom:
+ 
+\begin_inset Formula $V=C\left[a,b\right]\sim$
+\end_inset
+
+ zvezne 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $\forall f,g\in V:\left\langle f,g\right\rangle =\int_{a}^{b}f\left(x\right)g\left(x\right)dx$
+\end_inset
+
+.
+ Zveznost je potrebna za dokaz aksioma 1,
+ sicer za neznano neničelno funkcijo 
+\begin_inset Formula $f\left(x\right)=\begin{cases}
+1 & ;x=0\\
+0 & ;\text{drugače}
+\end{cases}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)g\left(x\right)dx=0$
+\end_inset
+
+.
+ Temu pravimo standardni skalarni produkt v 
+\begin_inset Formula $C\left[a,b\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+neskončno razsežen primer z nestandardnim skalarnim produktom:
+ Naj bo 
+\begin_inset Formula $w:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna,
+ ki zadošča 
+\begin_inset Formula $\forall x\in\left[a,b\right]:w\left(x\right)>0$
+\end_inset
+
+.
+ Ostalo kot prej.
+ 
+\begin_inset Formula $\forall f,g\in V:\left\langle f,g\right\rangle _{w}=\int_{a}^{b}f\left(x\right)g\left(x\right)w\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Vektorski prostor s skalarnnim produktom je tak par 
+\begin_inset Formula $\left(V,\left\langle \cdot,\cdot\right\rangle \right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle $
+\end_inset
+
+ skalarni produkt na 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ To je torej vektorski prostor,
+ za katerega izberemo in fiksiramo skalarni produkt.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor nad poljem 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ nenujno končno razsežen.
+ Preslikavi 
+\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle :V\times V\to\mathbb{C}$
+\end_inset
+
+ pravimo skalarni produtkt,
+ če zadošča naslednjim lastnostim:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+pozitivna definitnost:
+ 
+\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle \in\mathbb{R}\wedge\left\langle v,v\right\rangle >0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+konjugirana simetričnost:
+ 
+\begin_inset Formula $\forall u,v\in V:\left\langle v,u\right\rangle =\overline{\left\langle u,v\right\rangle }$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+linearnost v prvem faktorju:
+ 
+\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in\mathbb{R},v_{1},v_{2}\in V:\left\langle \alpha_{1}v_{1}+\alpha_{2}v_{2},v\right\rangle =\alpha_{1}\left\langle v_{1},v\right\rangle +\alpha_{2}\left\langle v_{2},v\right\rangle $
+\end_inset
+
+
 \end_layout
 
 \end_deeper
+\begin_layout Corollary*
+konjugirana linearnost v drugem faktorju.
+ 
+\begin_inset Formula $\left\langle u,\beta_{1}v_{1}+\beta_{2}v_{2}\right\rangle =\overline{\left\langle \beta_{1}v_{1}+\beta_{2}v_{2},v\right\rangle }=\overline{\beta_{1}\left\langle v_{1},v\right\rangle +\beta_{2}\left\langle v_{2},v\right\rangle }=\overline{\beta_{1}}\overline{\left\langle v_{1},v\right\rangle }+\overline{\beta_{2}}\overline{\left\langle v_{2},v\right\rangle }=\overline{\beta_{1}}\left\langle v,v_{1}\right\rangle +\overline{\beta_{2}}\left\langle v,v_{2}\right\rangle $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Skalarni produkt z 0:
+ 
+\begin_inset Formula $\left\langle 0,v\right\rangle =\left\langle 0\cdot v+0\cdot v,v\right\rangle =0\left\langle v,v\right\rangle +0\left\langle v,v\right\rangle =0\Rightarrow\left\langle v,0\right\rangle =0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Alternativna formulacija 1:
+ 
+\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \in\mathbb{R}\wedge\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri vektorskih prostorov s skalarnim produktom:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+standardni skalarni produkt na 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\alpha_{1}\overline{\beta_{1}}+\cdots+\alpha_{n}\overline{\beta_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+nestandardni skalarni produkt na 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+:
+ Za neke 
+\begin_inset Formula $\gamma_{1}\in\mathbb{\mathbb{R}}^{+},\dots,\gamma_{n}\in\mathbb{R}^{+}$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\gamma_{1}\alpha_{1}\overline{\beta_{1}}+\cdots+\gamma_{n}\alpha_{n}\overline{\beta_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+neskončno razsežen vektorski prostor na 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+ s standardnim skalarnim produktom:
+ Naj bo 
+\begin_inset Formula $V=C\left(\left[a,b\right],\mathbb{C}\right)$
+\end_inset
+
+ —
+ 
+\begin_inset Formula $f=g+ih$
+\end_inset
+
+ za 
+\begin_inset Formula $g,h\in C\left[a,b\right]$
+\end_inset
+
+ (zvezni funkciji iz 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+).
+ Definiramo 
+\begin_inset Formula $\left\langle f_{1},f_{2}\right\rangle =\int_{a}^{b}f_{1}\left(x\right)\overline{f_{2}\left(x\right)}dx=\int_{a}^{b}\left(g_{1}+ih_{1}\right)\left(x\right)\left(g_{2}-ih_{2}\right)\left(x\right)dx=\int_{a}^{b}\left(g_{1}g_{2}+g_{1}g_{2}\right)\left(x\right)dx+i\int_{a}^{b}\left(h_{1}g_{2}-g_{1}h_{2}\right)xdx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+neskončno razsežen vektorski prostor na 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+ z nestandardnim skalarnim produktom:
+ Isto kot zgoraj,
+ le da spet množimo z nekimi funkcijami,
+ kot pri realnem skalarnem produktu.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Norma
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor s skalarnim produktom.
+ 
+\begin_inset Formula $\forall v\in V:\left|\left|v\right|\right|=\sqrt{\left\langle v,v\right\rangle }$
+\end_inset
+
+ je norma 
+\begin_inset Formula $v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph
+Osnovne lastnosti norme
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(\left|\left|v\right|\right|>0\Leftrightarrow v\not=0\right)\wedge\left|\left|0\right|\right|=0$
+\end_inset
+
+ sledi iz prvega aksioma skalarnega produkta
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall\alpha\in F,v\in V:\left|\left|\alpha v\right|\right|=\left|\alpha\right|\left|\left|v\right|\right|$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+trikotniška neenakost:
+ 
+\begin_inset Formula $\forall u,v\in V:\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$
+\end_inset
+
+ sledi iz Cauchy-Schwarzove neenakosti na običajen način.
+\end_layout
+
+\begin_layout Claim*
+Cauchy-Schwarz.
+ Za 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor s skalarnim produktom velja 
+\begin_inset Formula $\forall v\in V:\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Za 
+\begin_inset Formula $v=0$
+\end_inset
+
+ očitno velja 
+\begin_inset Formula $0=0$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $v\not=0$
+\end_inset
+
+ definirajmo 
+\begin_inset Formula 
+\[
+w=u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v
+\]
+
+\end_inset
+
+po prvi lastnosti velja
+\begin_inset Formula 
+\[
+0\leq\left\langle w,w\right\rangle =\left\langle w,u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v\right\rangle =\left\langle w,u\right\rangle -\frac{\overline{\left\langle u,v\right\rangle }}{\left\langle v,v\right\rangle }\left\langle w,v\right\rangle 
+\]
+
+\end_inset
+
+Oglejmo si
+\begin_inset Formula 
+\[
+\left\langle w,v\right\rangle =\left\langle u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,v\right\rangle =\left\langle u,v\right\rangle -\left\langle \frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,v\right\rangle =\cancel{\left\langle u,v\right\rangle }-\frac{\cancel{\left\langle u,v\right\rangle }}{\cancel{\left\langle v,v\right\rangle }}\cancel{\left\langle v,v\right\rangle }=0
+\]
+
+\end_inset
+
+In se vrnimo k prejšnji enačbi:
+\begin_inset Formula 
+\[
+0\leq\left\langle w,w\right\rangle =\left\langle w,u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v\right\rangle =\left\langle w,u\right\rangle -\frac{\overline{\left\langle u,v\right\rangle }}{\left\langle v,v\right\rangle }\left\langle w,v\right\rangle =\left\langle w,u\right\rangle -0=\left\langle w,u\right\rangle =
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left\langle u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,u\right\rangle =\left\langle u,u\right\rangle -\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }\left\langle v,u\right\rangle =\left|\left|u\right|\right|^{2}-\frac{\left\langle u,v\right\rangle \overline{\left\langle u,v\right\rangle }}{\left|\left|v\right|\right|^{2}}=\left|\left|u\right|\right|^{2}-\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0\leq\left|\left|u\right|\right|^{2}-\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}}\leq\left|\left|u\right|\right|^{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left\langle u,v\right\rangle \right|^{2}\leq\left|\left|u\right|\right|^{2}\left|\left|v\right|\right|^{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Z normo lahko izrazimo skalarni produkt:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+V 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left\langle u,v\right\rangle =\frac{1}{4}\left(\left|\left|u+v\right|\right|^{2}-\left|\left|u-v\right|\right|^{2}\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+V 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left\langle u,v\right\rangle =\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokaz v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+.
+ Oglejmo si 
+\begin_inset Formula 
+\[
+\left|\left|u+i^{k}v\right|\right|^{2}=\left\langle u+i^{k}v,u+i^{k}v\right\rangle =\left\langle u,u+i^{k}v\right\rangle +i^{k}\left\langle v,u+i^{k}v\right\rangle =\overline{\left\langle u+i^{k}v,u\right\rangle }+i^{k}\overline{\left\langle u+i^{k}v,v\right\rangle }=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\overline{\left\langle u,u\right\rangle }+\overline{\left\langle i^{k}v,u\right\rangle }+i^{k}\overline{\left\langle u,v\right\rangle }+i^{k}\overline{\left\langle i^{k}v,v\right\rangle }=\left\langle u,u\right\rangle +\left\langle u,i^{k}v\right\rangle +i^{k}\left\langle v,u\right\rangle +i^{k}\left\langle v,i^{k}v\right\rangle =
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +i^{k}\left(-\left(i^{k}\right)\right)\left\langle v,v\right\rangle =
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +1\left\langle v,v\right\rangle 
+\]
+
+\end_inset
+
+Dodajmo vsoto:
+\begin_inset Formula 
+\[
+\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}=\sum_{k=0}^{3}i^{k}\left(\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +1\left\langle v,v\right\rangle \right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\sum_{k=0}^{3}i^{k}\left\langle u,u\right\rangle +\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle +\sum_{k=0}^{3}i^{k}i^{k}\left\langle v,u\right\rangle +\sum_{k=0}^{3}i^{k}\left\langle v,v\right\rangle =0+\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle +0+0,
+\]
+
+\end_inset
+
+kajti 
+\begin_inset Formula $\sum_{k=0}^{3}i^{k}=1+i+\left(-1\right)+\left(-i\right)=0$
+\end_inset
+
+ in 
+\begin_inset Formula $\sum_{k=0}^{3}i^{2k}=1+\left(-1\right)+1+\left(-1\right)=0$
+\end_inset
+
+.
+ Nadaljujmo:
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+=\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle =\sum_{k=0}^{3}1\left\langle u,v\right\rangle =4\left\langle u,v\right\rangle 
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Ortogonalne množice in ortogonalne baze
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ VPSSP 
+\begin_inset Formula $\forall u,v\in V:u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+trivialne opombe.
+ 
+\begin_inset Formula $\forall v\in V:v\perp\vec{0}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\forall v\in V:v\not=0\Leftrightarrow v\not\perp v$
+\end_inset
+
+ (prvi aksiom skalarnega produkta),
+ 
+\begin_inset Formula $\forall v\in V:u\perp v\Leftrightarrow v\perp u$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ VPSSP in 
+\begin_inset Formula $v_{1},\dots,v_{k}\in V$
+\end_inset
+
+.
+ Množica 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
+\end_inset
+
+ je:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+ortogonalna,
+ če 
+\begin_inset Formula $v_{1}\not=0\wedge\cdots\wedge v_{k}\not=0$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall i,j\in\left\{ 1..k\right\} :i\not=j\Rightarrow v_{i}\perp v_{j}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+normirana,
+ če 
+\begin_inset Formula $\forall v\in\left\{ v_{1},\dots,v_{k}\right\} :\left|\left|v\right|\right|=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+ortonormirana,
+ če je ortogonalna in ortonormirana hkrati.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Iz (ortogonalne) množice 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
+\end_inset
+
+ dobimo (orto)normirano tako,
+ da vsak element delimo z njegovo normo.
+ 
+\begin_inset Formula $\left\{ \frac{v_{1}}{\left|\left|v_{1}\right|\right|},\dots,\frac{v_{k}}{\left|\left|v_{k}\right|\right|}\right\} $
+\end_inset
+
+ je vedno normirana.
+\end_layout
+
+\begin_layout Claim*
+Vsaka ortogonalna množica je linearno neodvisna.
+\end_layout
+
+\begin_layout Proof
+Denimo,
+ da je 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
+\end_inset
+
+ ortogonalna.
+ Vzemimo take 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{k}\ni:\alpha_{1}v_{1}+\cdots+\alpha_{k}v_{k}=0$
+\end_inset
+
+.
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{$
+\backslash
+alpha_1=
+\backslash
+cdots=
+\backslash
+alpha_k=0$}
+\end_layout
+
+\end_inset
+
+.
+ 
+\begin_inset Formula 
+\[
+\forall i\in\left\{ 1..k\right\} :0=\left\langle 0,v_{i}\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{k}v_{k},v\right\rangle =\alpha_{1}\left\langle v_{1},v_{i}\right\rangle +\cdots+\alpha_{i}\left\langle v_{i},v_{i}\right\rangle +\cdots+\alpha_{k}\left\langle v_{k},v_{i}\right\rangle =\cdots
+\]
+
+\end_inset
+
+Ker je množica ortogonalna,
+ je 
+\begin_inset Formula $\left\langle v_{l},v_{k}\right\rangle =0\Leftrightarrow l\not=k$
+\end_inset
+
+.
+ Nadaljujmo ...
+\begin_inset Formula 
+\[
+\cdots=\alpha_{i}\left\langle v_{i},v_{i}\right\rangle 
+\]
+
+\end_inset
+
+Ker 
+\begin_inset Formula $\left\langle v_{i},v_{i}\right\rangle $
+\end_inset
+
+ ni 0,
+ ker je 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+ neničeln (da,
+ tudi to je del definicije ortogonalnosti),
+ je 
+\begin_inset Formula $\alpha_{i}=0$
+\end_inset
+
+.
+ In to za vsak 
+\begin_inset Formula $i$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ni pa vsaka ortogonalna množica ogrodje.
+ Ortogonalni množici,
+ ki je ogrodje,
+ rečemo ortogonalna baza (LN sledi iz ortogonalnost).
+\end_layout
+
+\begin_layout Subsubsection
+Fourierov razvoj
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP,
+ 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} =\mathcal{B}$
+\end_inset
+
+ ortogonalna baza za 
+\begin_inset Formula $V$
+\end_inset
+
+ in 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ poljuben element.
+ Kako razvijemo 
+\begin_inset Formula $v$
+\end_inset
+
+ po 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+,
+ vedoč,
+ da je ta baza ortogonalna?
+ Postopek imenujemo Fourierov razvoj.
+\end_layout
+
+\begin_layout Standard
+Ker je 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+ ogrodje,
+ 
+\begin_inset Formula $\exists\alpha_{1},\dots,\alpha_{n}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
+\end_inset
+
+.
+ Množimo skalarno z 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}\quad\quad\quad\quad/\cdot v_{i}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle v,v_{i}\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n},v_{i}\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle v,v_{i}\right\rangle =\cancel{\alpha_{1}\left\langle v_{1},v_{i}\right\rangle }+\cancel{\cdots}+\alpha_{i}\left\langle v_{i},v_{i}\right\rangle +\cancel{\cdots}+\cancel{\alpha_{n}\left\langle v_{n},v_{i}\right\rangle }=\alpha_{i}\left\langle v_{i},v_{i}\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{\left\langle v,v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }=\alpha_{i}
+\]
+
+\end_inset
+
+Torej 
+\begin_inset Formula $\forall v\in V$
+\end_inset
+
+ velja 
+\begin_inset Formula $v=\sum_{i=1}^{n}\frac{\left\langle v,v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }v_{i}$
+\end_inset
+
+.
+ Koeficientu 
+\begin_inset Formula $\frac{\left\langle v,v_{i}\right\rangle }{\left|\left|v_{i}\right|\right|^{2}}$
+\end_inset
+
+ pravimo Fourierov koeficient.
+ Če je baza ortonormirana,
+ je Fourierov koeficient 
+\begin_inset Formula $\frac{\left\langle v,v_{i}\right\rangle }{\cancel{\left|\left|v_{i}\right|\right|^{2}}}=\left\langle v,v_{i}\right\rangle $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Parsevalova identiteta
+\end_layout
+
+\begin_layout Theorem*
+Parsevalova identiteta.
+ Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ VPSSP in 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
+\end_inset
+
+ njegova ortogonalna baza.
+ Tedaj 
+\begin_inset Formula $\forall v\in V:$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left|v\right|\right|^{2}=\sum_{i=1}^{n}\frac{\left|\left\langle v,v_{i}\right\rangle \right|^{2}}{\left\langle v_{i},v_{i}\right\rangle }.
+\]
+
+\end_inset
+
+Če je baza ortonormirana,
+ se enačba očitno poenostavi v
+\begin_inset Formula 
+\[
+\left|\left|v\right|\right|^{2}=\sum_{i=1}^{n}\left|\left\langle v,v_{i}\right\rangle \right|^{2}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\left|\left|v\right|\right|^{2}=\left\langle v,v\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n},\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}\right\rangle =$
+\end_inset
+
+ (uporabimo linearnost v 1.
+ in konjugirano linearnost v 2.
+ faktorju)
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+= & \alpha_{1}\overline{\alpha_{1}}\left\langle v_{1},v_{1}\right\rangle  & + & \cancel{\cdots} & + & \cancel{\alpha_{1}\overline{\alpha_{n}}\left\langle v_{1},v_{n}\right\rangle } & +\\
+ & \vdots &  &  &  & \vdots\\
++ & \cancel{\alpha_{n}\overline{\alpha_{1}}\left\langle v_{n},v_{1}\right\rangle } & + & \cancel{\cdots} & + & \alpha_{n}\overline{\alpha_{n}}\left\langle v_{n},v_{n}\right\rangle  & =
+\end{array}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\alpha_{1}\overline{\alpha_{1}}\left\langle v_{1},v_{1}\right\rangle +\cdots+\alpha_{n}\overline{\alpha_{n}}\left\langle v_{n},v_{n}\right\rangle =\left|\alpha_{1}\right|^{2}\left|\left|v_{1}\right|\right|^{2}+\cdots+\left|\alpha_{n}\right|^{2}\left|\left|v_{n}\right|\right|^{2}=
+\]
+
+\end_inset
+
+Vstavimo formule za koeficiente po Fourierjevem razvoju:
+\begin_inset Formula 
+\[
+=\left|\frac{\left\langle v,v_{1}\right\rangle }{\left|\left|v_{1}\right|\right|^{\cancel{2}}}\right|^{2}\cancel{\left|\left|v_{1}\right|\right|^{2}}+\cdots+\left|\frac{\left\langle v,v_{n}\right\rangle }{\left|\left|v_{n}\right|\right|^{\cancel{2}}}\right|^{2}\cancel{\left|\left|v_{n}\right|\right|^{2}}=\frac{\left|\left\langle v,v_{1}\right\rangle \right|^{2}}{\left|\left|v_{1}\right|\right|}+\cdots+\frac{\left|\left\langle v,v_{n}\right\rangle \right|^{2}}{\left|\left|v_{n}\right|\right|}=\sum_{i=1}^{n}\frac{\left|\left\langle v_{i},v\right\rangle \right|^{2}}{\left\langle v_{i},v_{i}\right\rangle }
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Projekcija na podprostor
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP in 
+\begin_inset Formula $W$
+\end_inset
+
+ podprostor 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Za vsak 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ želimo izračunati njegovo ortogonalno projekcijo na 
+\begin_inset Formula $W$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Vektor 
+\begin_inset Formula $v'\in W$
+\end_inset
+
+ je ortogonalna projekcija vektorja 
+\begin_inset Formula $v\in V$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall w\in W:\left|\left|v-v'\right|\right|\leq\left|\left|v-w\right|\right|$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $v'$
+\end_inset
+
+ je najbližje 
+\begin_inset Formula $v$
+\end_inset
+
+ izmed vseh elementov 
+\begin_inset Formula $W$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Zadošča preveriti,
+ da je 
+\begin_inset Formula $v-v'$
+\end_inset
+
+ ortogonalen na vse elemente 
+\begin_inset Formula $W$
+\end_inset
+
+ (pitagorov izrek),
+ kajti v tem primeru (če predpostavimo 
+\begin_inset Formula $\left(v'-w\right)\perp\left(v-v'\right)$
+\end_inset
+
+) velja
+\begin_inset Formula 
+\[
+\left|\left|v-w\right|\right|^{2}=\left|\left|v-v'+v'-w\right|\right|=\left|\left|v-v'\right|\right|^{2}+\left|\left|v'-w\right|\right|^{2}\geq\left|\left|v-v'\right|\right|^{2}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Dokaz pitagovorega izreka:
+ 
+\begin_inset Formula $\left|\left|a+b\right|\right|^{2}=\left\langle a+b,a+b\right\rangle =\left\langle a,a\right\rangle +\cancel{\left\langle a,b\right\rangle +\left\langle b,a\right\rangle }+\left\langle b,b\right\rangle =\left|\left|a\right|\right|^{2}+\left|\left|b\right|\right|^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $\left\{ w_{1},\dots,w_{k}\right\} $
+\end_inset
+
+ ortogonalna baza za 
+\begin_inset Formula $W$
+\end_inset
+
+.
+ Formula za ortogonalno projekcijo se glasi:
+\begin_inset Formula 
+\[
+v'=\frac{\left\langle v,w_{1}\right\rangle }{\left\langle w_{1},w_{1}\right\rangle }w_{1}+\cdots+\frac{\left\langle v,w_{k}\right\rangle }{\left\langle w_{k},w_{k}\right\rangle }=\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Dokažimo,
+ da je 
+\begin_inset Formula $v-\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i}$
+\end_inset
+
+ pravokoten na vse elemente 
+\begin_inset Formula $W$
+\end_inset
+
+.
+ Zaradi linearnosti skalarnega produkta zadošča preveriti,
+ da je pravokoten na bazo 
+\begin_inset Formula $W$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\forall j\in\left\{ 1..k\right\} $
+\end_inset
+
+ velja (spomnimo se,
+ da je 
+\begin_inset Formula $\left\langle w_{i},w_{j}\right\rangle =0\Leftrightarrow i\not=j$
+\end_inset
+
+,
+ zato po drugem enačaju ostane le še en člen vsote):
+\begin_inset Formula 
+\[
+\left\langle v-\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i},w_{j}\right\rangle =\left\langle v,w_{j}\right\rangle -\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }\left\langle w_{i},w_{j}\right\rangle =\left\langle v,w_{j}\right\rangle -\frac{\left\langle v,w_{j}\right\rangle }{\cancel{\left\langle w_{j},w_{j}\right\rangle }}\cancel{\left\langle w_{j},w_{j}\right\rangle }=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left\langle v,w_{j}\right\rangle -\left\langle v,w_{j}\right\rangle =0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Obstoj ortogonalne baze —
+ Gram-Schmidtova ortogonalizacija
+\end_layout
+
+\begin_layout Standard
+Radi bi dokazali,
+ da ima vsak KRVPSSP ortogonalno bazo in da je moč vsako ortogonalno množico dopolniti do ortogonalne baze.
+ Konstruktiven dokaz 
+\begin_inset Formula $\ddot{\smile}!$
+\end_inset
+
+ —
+ postopek,
+ imenovan Gram-Schmidtova ortogonalizacija,
+ iz poljubne baze naredi ortogonalno.
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP in 
+\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} $
+\end_inset
+
+ njegova poljubna baza.
+ Naj bo 
+\begin_inset Formula $v_{1}\coloneqq u_{1}$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+v_{2}\coloneqq u_{2}-\frac{\left\langle u_{2},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}=u_{2}-u_{2}'
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{3}\coloneqq u_{3}-\frac{\left\langle u_{3},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}-\frac{\left\langle u_{3},v_{2}\right\rangle }{\left\langle v_{2},v_{2}\right\rangle }v_{2}=u_{3}-u_{3}'
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\cdots
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{n}\coloneqq u_{n}-\sum_{i=1}^{n-1}\frac{\left\langle u_{n},v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }v_{i}=u_{n}-u_{n}'
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Trdimo,
+ da je 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ ortogonalna baza za 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Opazimo,
+ da je 
+\begin_inset Formula $u_{2}'$
+\end_inset
+
+ ortogonalna projekcija 
+\begin_inset Formula $u_{2}$
+\end_inset
+
+ na 
+\begin_inset Formula $\Lin\left\{ v_{1}\right\} $
+\end_inset
+
+,
+ 
+\begin_inset Formula $u_{3}'$
+\end_inset
+
+ ortogonalna projekcija 
+\begin_inset Formula $u_{3}$
+\end_inset
+
+ na 
+\begin_inset Formula $\Lin\left\{ v_{1},v_{2}\right\} $
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $u_{n}'$
+\end_inset
+
+ pa ortogonalna projekcija na 
+\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n-1}\right\} $
+\end_inset
+
+,
+ torej
+\begin_inset Formula 
+\[
+v_{2}=u_{2}-u_{2}'\perp\Lin\left\{ v_{1}\right\} \text{, torej }v_{2}\perp v_{1}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{3}=u_{3}-u_{3}'\perp\Lin\left\{ v_{1},v_{2}\right\} \text{, torej }v_{3}\perp v_{1},v_{3}\perp v_{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\cdots
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{n}=u_{n}-u_{n}'\perp\Lin\left\{ v_{1},\dots,v_{n-1}\right\} \text{, torej }v_{n}\perp v_{1},\dots,v_{n}\perp v_{n-1},
+\]
+
+\end_inset
+
+kar pomeni,
+ da so 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ paroma ortogonalni.
+ Toda vprašanje je,
+ ali so neničelni,
+ kajti to je,
+ ne boste verjeli,
+ prav tako pogoj za ortogonalno množico.
+ 
+\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :$
+\end_inset
+
+ dokažimo neničelnost 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+:-)
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $v_{1}$
+\end_inset
+
+ je neničeln,
+ ker je enak 
+\begin_inset Formula $u_{1}$
+\end_inset
+
+,
+ ki je element baze 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ 
+\begin_inset Formula $v_{2}$
+\end_inset
+
+ je neničeln,
+ ker je 
+\begin_inset Formula $v_{2}=u_{2}-\alpha v_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $u_{2}\not=\alpha v_{1}$
+\end_inset
+
+,
+ ker sta linearno neodvisna,
+ ker tvorita ortogonalno množico.
+ 
+\begin_inset Formula $v_{3}$
+\end_inset
+
+ je neničeln,
+ ker 
+\begin_inset Formula $v_{3}=u_{3}-\left(\beta v_{1}+\gamma v_{2}\right)$
+\end_inset
+
+ in ker so 
+\begin_inset Formula $v_{1},v_{2},u_{3}$
+\end_inset
+
+ LN,
+ 
+\begin_inset Formula $u_{3}\not=\left(\beta v_{1}+\gamma v_{2}\right)$
+\end_inset
+
+.
+ In tako dalje.
+\end_layout
+
+\begin_layout Paragraph*
+Dopolnitev ortogonalne množice do baze
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $\left\{ u_{1},\dots,u_{k}\right\} $
+\end_inset
+
+ ortogonalna množica,
+ torej je linearno neodvisna,
+ torej jo lahko dopolnimo do baze.
+ 
+\begin_inset Formula $\left\{ u_{k+1},\dots,u_{n}\right\} $
+\end_inset
+
+ je dopolnitev do baze.
+ Toda slednja še ni ortogonalna.
+ A nič ne de,
+ uporabimo lahko Gram-Schmidtovo ortogonalizacijo na 
+\begin_inset Formula $\left\{ u_{1},\dots,u_{k},u_{k+1},\dots,u_{n}\right\} $
+\end_inset
+
+ in dobimo ortogonalno bazo 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+.
+ Opazimo,
+ da ker so po predpostavki 
+\begin_inset Formula $u_{1},\dots,u_{k}$
+\end_inset
+
+ ortogonalni,
+ velja 
+\begin_inset Formula $v_{1}=u_{1},\dots,v_{k}=u_{k}$
+\end_inset
+
+ (po GS).
+\end_layout
+
+\begin_layout Example*
+primer GS ortogonalizacije iz analize.
+ Naj bo 
+\begin_inset Formula $V=\mathbb{R}\left[x\right]_{\leq3}$
+\end_inset
+
+.
+ Baza:
+ 
+\begin_inset Formula $u_{1}=1,u_{2}=x,u_{3}=x^{2},u_{4}=x^{3}$
+\end_inset
+
+,
+ skalarni produkt naj bo 
+\begin_inset Formula $\left\langle p,q\right\rangle =\int_{-1}^{1}p\left(x\right)q\left(x\right)dx$
+\end_inset
+
+.
+ Konstruirajmo pripadajočo ortogonalno bazo 
+\begin_inset Formula $v_{1},\dots,v_{4}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+v_{1}\coloneqq u_{1}=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{2}\coloneqq u_{2}-\frac{\left\langle u_{2},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}=x-\frac{\int_{-1}^{1}xdx}{\int_{-1}^{1}dx}=x-0=x=u_{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{3}\coloneqq u_{3}-\frac{\left\langle u_{3},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}-\frac{\left\langle u_{3},v_{2}\right\rangle }{\left\langle v_{2},v_{2}\right\rangle }=x^{2}-\frac{\int_{-1}^{1}x^{2}dx}{\int_{-1}^{1}dx}-\frac{\int_{-1}^{1}x^{3}dx}{\int_{-1}^{1}x^{2}dx}x=\cdots=x²-\frac{1}{3}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{4}\coloneqq\cdots=x^{2}-\frac{3}{5}x
+\]
+
+\end_inset
+
+Sklep:
+ 
+\begin_inset Formula $\left\{ 1,x,x^{2}-\frac{1}{3},x^{2}-\frac{3}{5}x\right\} $
+\end_inset
+
+ je ortogonalna baza za ta vektorski prostor s tem skalarnim produktom.
+ Normirajmo jo!
+ Norme teh baznih vektorjev po vrsti so 
+\begin_inset Formula $\sqrt{2},\sqrt{\frac{2}{3}},\sqrt{\frac{8}{45}},\sqrt{\frac{8}{175}}$
+\end_inset
+
+.
+ Pripadajoča ortonormirana baza je torej 
+\begin_inset Formula $\left\{ \frac{1}{\sqrt{2}},\frac{x}{\sqrt{\frac{2}{3}}},\frac{x^{2}-\frac{1}{3}}{\sqrt{\frac{8}{45}}},\frac{x^{2}-\frac{3}{5}x}{\sqrt{\frac{8}{175}}}\right\} .$
+\end_inset
+
+ Normiranje bi sicer prineslo lepše formule,
+ vendar bi v račune prineslo te objektivno grde konstante.
+\end_layout
+
+\begin_layout Subsubsection
+Ortogonalni komplement
+\end_layout
+
+\begin_layout Definition
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP nad 
+\begin_inset Formula $F$
+\end_inset
+
+ in 
+\begin_inset Formula $S\subseteq V$
+\end_inset
+
+.
+ Ortogonalni komplement 
+\begin_inset Formula $S$
+\end_inset
+
+ je množica 
+\begin_inset Formula $S^{\perp}$
+\end_inset
+
+.
+ Vsebuje vse tiste vektorje iz 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ ki so ortogonalni na 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $S^{\perp}\coloneqq\left\{ v\in V;\forall s\in S:v\perp s\right\} =\left\{ v\in V;v\perp S\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\forall S\subseteq V:S^{\perp}$
+\end_inset
+
+ je podprostor 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazati je treba 
+\begin_inset Formula $\forall u_{1},u_{2}\in S^{\perp},\alpha_{1},\alpha_{2}\in F:\alpha_{1}v_{1}+\alpha_{2}v_{2}\in S^{\perp}$
+\end_inset
+
+.
+ Po definiciji ortogonalnega komplementa velja 
+\begin_inset Formula 
+\[
+\forall s\in S:\left\langle u_{1},s\right\rangle =0\wedge\left\langle u_{2},s\right\rangle =0\Longrightarrow0=\alpha_{1}\left\langle u_{1},s\right\rangle +\alpha_{2}\left\langle u_{2},s\right\rangle =\left\langle \alpha_{1}u_{1}+\alpha_{2}u_{2},s\right\rangle \Longrightarrow\alpha_{1}u_{1}+\alpha_{2}u_{2}\in S^{\perp}
+\]
+
+\end_inset
+
+
+\end_layout
+
 \begin_layout Part
 Vaja za ustni izpit
 \end_layout