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| -rw-r--r-- | šola/la/teor.lyx | 3407 |
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 |