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Diffstat (limited to 'šola/la/teor.lyx')
| -rw-r--r-- | šola/la/teor.lyx | 2409 |
1 files changed, 2405 insertions, 4 deletions
diff --git a/šola/la/teor.lyx b/šola/la/teor.lyx index ae7dcaf..ea3069a 100644 --- a/šola/la/teor.lyx +++ b/šola/la/teor.lyx @@ -5451,6 +5451,12 @@ x_{i}=\frac{\det A_{i}\left(\vec{b}\right)}{\det A} \end_layout \begin_layout Subsubsection +\begin_inset CommandInset label +LatexCommand label +name "subsec:Formula-za-inverz-matrike" + +\end_inset + Formula za inverz matrike \end_layout @@ -5556,9 +5562,9 @@ X=A^{-1}=\left[\begin{array}{ccc} \vdots & & \vdots\\ \frac{\det A_{1n}\cdot\left(-1\right)^{1+n}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A} \end{array}\right]=\frac{1}{\det A}\left[\begin{array}{ccc} -\frac{\det A_{11}\cdot\left(-1\right)^{1+1}}{\det A} & \cdots & \frac{\det A_{1n}\cdot\left(-1\right)^{1+1}}{\det A}\\ +\det A_{11}\cdot\left(-1\right)^{1+1} & \cdots & \det A_{1n}\cdot\left(-1\right)^{1+1}\\ \vdots & & \vdots\\ -\frac{\det A_{n1}\cdot\left(-1\right)^{n+1}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A} +\det A_{n1}\cdot\left(-1\right)^{n+1} & \cdots & \det A_{nn}\cdot\left(-1\right)^{n+n} \end{array}\right]^{T}=\frac{1}{\det A}\tilde{A}^{T}, \] @@ -11157,7 +11163,13 @@ Velja torej . \end_layout -\begin_layout Definition* +\begin_layout Definition +\begin_inset CommandInset label +LatexCommand label +name "def:vsota-je-direktna" + +\end_inset + Pravimo, da je vsota \begin_inset Formula $W_{1}+W_{2}$ @@ -11172,6 +11184,10 @@ Pravimo, \begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim W_{1}+\dim W_{2}$ \end_inset + oziroma ekvivalentno +\begin_inset Formula $\forall w_{1}\in W_{1},w_{2}\in W_{2}:w_{1}+w_{2}=0\Rightarrow w_{1}=w_{2}=0$ +\end_inset + . \end_layout @@ -12452,6 +12468,12 @@ Odtod sledi: \end_deeper \begin_layout Theorem +\begin_inset CommandInset label +LatexCommand label +name "thm:matrika-kompozituma-linearnih" + +\end_inset + matrika kompozituma linearnih preslikav. Posplošitev formule \begin_inset Formula $P_{\mathcal{\mathcal{D}\leftarrow\mathcal{B}}}=P_{\mathcal{D\leftarrow C}}\cdot P_{\mathcal{C}\leftarrow\mathcal{B}}$ @@ -13345,7 +13367,7 @@ status open \begin_layout Plain Layout Isto oznako uporabljamo tudi za podobne matrike, - vendar podobnost ni povesem enako kot ekvivalentnost. + vendar podobnost ni enako kot ekvivalentnost. \end_layout \end_inset @@ -13662,6 +13684,2385 @@ Torej je res \end_layout \end_deeper +\begin_layout Subsubsection +Podobnost matrik +\end_layout + +\begin_layout Definition* +Kvadratni matriki +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $B$ +\end_inset + + sta podobni, + če +\begin_inset Formula $\exists$ +\end_inset + + taka obrnljiva matrika +\begin_inset Formula $P\ni:B=PAP^{-1}$ +\end_inset + +.ž +\end_layout + +\begin_layout Claim* +Podobnost je ekvivalenčna relacija. +\end_layout + +\begin_layout Proof +Dokazujemo, + da je relacija ekvivalenčna, + torej: +\end_layout + +\begin_deeper +\begin_layout Itemize +refleksivna: + +\begin_inset Formula $A=IAI^{-1}=IAI=A$ +\end_inset + + +\end_layout + +\begin_layout Itemize +simetrična: + +\begin_inset Formula $B=PAP^{-1}\Rightarrow P^{-1}BP=A$ +\end_inset + + +\end_layout + +\begin_layout Itemize +tranzitivna: + +\begin_inset Formula $B=PAP^{-1}\wedge C=QBQ^{-1}\Rightarrow C=QPAP^{-1}Q^{-1}=\left(QP\right)A\left(QP\right)^{-1}$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Remark +\begin_inset CommandInset label +LatexCommand label +name "rem:nista-podobni" + +\end_inset + +Očitno velja podobnost +\begin_inset Formula $\Rightarrow$ +\end_inset + + ekvivalentnost, + toda obrat ne velja vedno. + Na primer +\begin_inset Formula $\left[\begin{array}{cc} +1 & 0\\ +0 & 0 +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $\left[\begin{array}{cc} +0 & 1\\ +0 & 0 +\end{array}\right]$ +\end_inset + + sta ekvivalentni (sta enake velikosti in ranga), + toda nista podobni (dokaz kasneje). +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Standard +Od prej vemo, + da je vsaka matrika ekvivalentna matriki +\begin_inset Formula $\left[\begin{array}{cc} +I_{r} & 0\\ +0 & 0 +\end{array}\right]$ +\end_inset + +, + kjer je +\begin_inset Formula $r$ +\end_inset + + njen rang. + A je vsaka kvadratna matrika podobna kakšni lepi matriki? + Ja. + Vsaka matrika je podobna zgornjetrikotni matriki in jordanski kanonični formi (več o tem kasneje). + Toda a je vsaka kvadratna matrika podobna diagonalni matriki? + Ne. +\end_layout + +\begin_layout Definition* +Matrika +\begin_inset Formula $D$ +\end_inset + + je diagonalna +\begin_inset Formula $\sim d_{ij}\not=0\Rightarrow i=j$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Kdaj je matrika +\begin_inset Formula $A$ +\end_inset + + podobna neki diagonalni matriki? + Kdaj +\begin_inset Formula $\exists$ +\end_inset + + diagonalna +\begin_inset Formula $D$ +\end_inset + + in obrnljiva +\begin_inset Formula $P\ni:A=PDP^{-1}$ +\end_inset + +? + Izpeljimo iz nastavka. + +\begin_inset Formula $D=\left[\begin{array}{ccc} +\lambda_{1} & & 0\\ + & \ddots\\ +0 & & \lambda n +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $P=\left[\begin{array}{ccc} +\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]$ +\end_inset + +, + kjer sta +\begin_inset Formula $D$ +\end_inset + + in +\begin_inset Formula $P$ +\end_inset + + neznani. + Ker mora biti +\begin_inset Formula $P$ +\end_inset + + obrnljiva, + so njeni stolpični vektorji LN. +\begin_inset Formula +\[ +A=PDP^{-1}\Leftrightarrow AP=PD\Leftrightarrow A\left[\begin{array}{ccc} +\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc} +\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]\left[\begin{array}{ccc} +\lambda_{1} & & 0\\ + & \ddots\\ +0 & & \lambda_{n} +\end{array}\right]\text{ in }P\text{ obrnljiva} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +A\vec{v_{1}} & \cdots & A\vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc} +\lambda_{1}\vec{v_{1}} & \cdots & \lambda_{n}\vec{v_{n}}\end{array}\right]\text{ in }v_{i}\text{ so LN} +\] + +\end_inset + + +\begin_inset Formula +\[ +A\vec{v_{1}}=\lambda_{1}\vec{v_{1}},\dots,A\vec{v_{n}}=\lambda_{n}v_{n}\text{ in }\forall i:v_{i}\not=0 +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +Porodi se naloga, + imenovana +\begin_inset Quotes gld +\end_inset + +Lastni problem +\begin_inset Quotes grd +\end_inset + +. + Iščemo pare +\begin_inset Formula $\left(\lambda,\vec{v}\right)$ +\end_inset + +, + ki zadoščajo enačbi +\begin_inset Formula $A\vec{v}=\lambda\vec{v}$ +\end_inset + +. + +\end_layout + +\begin_layout Definition* +Pravimo, + da je +\begin_inset Formula $\lambda$ +\end_inset + + je lastna vrednost matrike +\begin_inset Formula $A$ +\end_inset + +, + če obstaja tak +\begin_inset Formula $\vec{v}\not=0$ +\end_inset + +, + da je +\begin_inset Formula $A\vec{v}=\lambda\vec{v}$ +\end_inset + +. + V tem primeru pravimo, + da je +\begin_inset Formula $\vec{v}$ +\end_inset + + lastni vektor, + ki pripada lastni vrednosti +\begin_inset Formula $\lambda$ +\end_inset + +. + Paru +\begin_inset Formula $\left(\lambda,\vec{v}\right)$ +\end_inset + +, + ki zadošča enačbi, + pravimo lastni par matrike +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Nalogo +\begin_inset Quotes gld +\end_inset + +Lastni problem +\begin_inset Quotes grd +\end_inset + + rešujemo v dveh korakih. + Najprej najdemo vse +\begin_inset Formula $\lambda$ +\end_inset + +, + nato za vsako poiščemo pripadajoče +\begin_inset Formula $\vec{v}$ +\end_inset + +, + ki za lastno vrednost obstajajo po definiciji. +\end_layout + +\begin_layout Standard +Za nek +\begin_inset Formula $v\not=0$ +\end_inset + + pišimo +\begin_inset Formula $Av=\lambda v=\lambda Iv\Leftrightarrow Av-\lambda Iv=0\Leftrightarrow\left(A-\lambda I\right)v=0$ +\end_inset + + za nek +\begin_inset Formula $v\not=0\Leftrightarrow\Ker\left(A-\lambda I\right)\not=\left\{ 0\right\} \overset{\text{K.O.M.}}{\Longleftrightarrow}A-\lambda I$ +\end_inset + + ni obrnljiva +\begin_inset Formula $\Leftrightarrow\det\left(A-\lambda I\right)=0$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Polinom +\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-xI\right)$ +\end_inset + + je karakteristični polinom matrike +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Premislek zgoraj nam pove, + da so lastne vrednosti +\begin_inset Formula $A$ +\end_inset + + ničle +\begin_inset Formula $p_{A}\left(x\right)$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Karakteristični polinom lahko nima nobene ničle: + +\begin_inset Formula $A=\left[\begin{array}{cc} +0 & 1\\ +-1 & 0 +\end{array}\right]$ +\end_inset + +, + +\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\det\left[\begin{array}{cc} +-\lambda & 1\\ +-1 & -\lambda +\end{array}\right]=x^{2}+1$ +\end_inset + +, + katerega ničli sta +\begin_inset Formula $\lambda_{1}=i$ +\end_inset + + in +\begin_inset Formula $\lambda_{2}=-i$ +\end_inset + +, + ki nista realni števili. + V nadaljevanju se zato omejimo na kompleksne matrike in kompleksne lastne vrednosti, + saj ima po Osnovnem izreku Algebre polinom s kompleksnimi koeficienti vedno vsaj kompleksne ničle. +\end_layout + +\begin_layout Standard +Kako pa iščemo lastne vektorje za lastno vrednost +\begin_inset Formula $\lambda$ +\end_inset + +? + Spomnimo se na +\begin_inset Formula $Av=\lambda v\Leftrightarrow v\in\Ker\left(A-\lambda I\right)$ +\end_inset + +. + Rešiti moramo homogen sistem linearnih enačb. + Po definiciji so lastni vektorji neničelni, + zato nas trivialna rešitev ne zanima. +\end_layout + +\begin_layout Definition* +Množici +\begin_inset Formula $\Ker\left(A-\lambda I\right)$ +\end_inset + + pravimo lastni podprostor matrike +\begin_inset Formula $A$ +\end_inset + +, + ki pripada +\begin_inset Formula $\lambda$ +\end_inset + +. + Slednji vsebuje +\begin_inset Formula $\vec{0}$ +\end_inset + + in množico vektorjev, + ki so vsi lastni vektorji +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Exercise* +Izračunaj lastne vrednosti od +\begin_inset Formula $A=\left[\begin{array}{cc} +0 & 1\\ +-1 & 0 +\end{array}\right]$ +\end_inset + +. + Od prej vemo, + da +\begin_inset Formula $\lambda_{1}=i$ +\end_inset + +, + +\begin_inset Formula $\lambda_{2}=-i$ +\end_inset + +. + Izračunajmo +\begin_inset Formula $\Ker\left(A-iI\right)$ +\end_inset + + in +\begin_inset Formula $\Ker\left(A+iI\right)$ +\end_inset + +: +\begin_inset Formula +\[ +\Ker\left(A-iI\right):\quad\left[\begin{array}{cc} +-i & 1\\ +-1 & -i +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right]=0\quad\Longrightarrow\quad-ix+y=0,-x-iy=0\quad\Longrightarrow\quad y=ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c} +1\\ +i +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +\Ker\left(A+iI\right):\quad\left[\begin{array}{cc} +i & 1\\ +-1 & i +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right]=0\quad\Longrightarrow\quad ix+y=0,-x+y=0\quad\Longrightarrow\quad y=-ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c} +1\\ +-i +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +\Ker\left(A-iI\right)=\Lin\left\{ \left[\begin{array}{c} +1\\ +i +\end{array}\right]\right\} ,\quad\Ker\left(A+iI\right)=\Lin\left\{ \left[\begin{array}{c} +1\\ +-i +\end{array}\right]\right\} +\] + +\end_inset + + +\end_layout + +\begin_layout Exercise* +Vstavimo lastna vektorja v +\begin_inset Formula $P$ +\end_inset + + in lastne vrednosti v +\begin_inset Formula $D$ +\end_inset + + na pripadajoči mesti. + Dobimo obrnljivo +\begin_inset Formula $P$ +\end_inset + + in velja +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + + +\begin_inset Formula +\[ +P=\left[\begin{array}{cc} +1 & 1\\ +i & -i +\end{array}\right],\quad D=\left[\begin{array}{cc} +i & 0\\ +0 & -i +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Exercise* +Temu početju pravimo +\begin_inset Quotes gld +\end_inset + +diagonalizacija matrike +\begin_inset Formula $A$ +\end_inset + + +\begin_inset Quotes grd +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Primer matrike, + ki ni diagonalizabilna: + +\begin_inset Formula $A=\left[\begin{array}{cc} +0 & 1\\ +0 & 0 +\end{array}\right]$ +\end_inset + +. + +\begin_inset Formula $\det\left(A-\lambda I\right)=\left[\begin{array}{cc} +-\lambda & 1\\ +0 & -\lambda +\end{array}\right]=\lambda^{2}$ +\end_inset + +. + Ničli/lastni vrednosti sta +\begin_inset Formula $\lambda_{1}=0$ +\end_inset + + in +\begin_inset Formula $\lambda_{2}=0$ +\end_inset + +. + Toda +\begin_inset Formula $\Ker\left(A-0I\right)=\Ker A=\Lin\left\{ \left[\begin{array}{c} +1\\ +0 +\end{array}\right]\right\} $ +\end_inset + + in +\begin_inset Formula $P=\left[\begin{array}{cc} +1 & 1\\ +0 & 0 +\end{array}\right]$ +\end_inset + + ni obrnljiva. + S tem dokažemo trditev v primeru +\begin_inset CommandInset ref +LatexCommand ref +reference "rem:nista-podobni" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. + +\begin_inset Formula $\left[\begin{array}{cc} +1 & 0\\ +0 & 0 +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $\left[\begin{array}{cc} +0 & 1\\ +0 & 0 +\end{array}\right]$ +\end_inset + + nista podobni, + ker je prva diagonalna, + druga pa ni podobna diagonalni matriki (ne da se je diagonalizirati). +\end_layout + +\begin_layout Standard +Lastne vrednosti lahko definiramo tudi za linearne preslikave, + saj so linearne preslikave linearno izomorfne matrikam. +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + vektorski prostor nad +\begin_inset Formula $F=\mathbb{C}$ +\end_inset + + in +\begin_inset Formula $L:V\to V$ +\end_inset + + linearna preslikava. + Število +\begin_inset Formula $\lambda\in F$ +\end_inset + + je lastna vrednost +\begin_inset Formula $L$ +\end_inset + +, + le obstaja tak neničelni +\begin_inset Formula $v\in V$ +\end_inset + +, + da velja +\begin_inset Formula $Lv=\lambda v$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Kako pa rešujemo +\begin_inset Quotes gld +\end_inset + +Lastni problem +\begin_inset Quotes grd +\end_inset + + za linearne preslikave? + +\begin_inset Formula $Lv=\lambda v\Leftrightarrow Lv-\lambda\left(id\right)v=0\Leftrightarrow\left(L-\lambda\left(id\right)\right)v=0\Leftrightarrow v\in\Ker\left(L-\lambda\left(id\right)\right)\overset{v\not=0}{\Longleftrightarrow}\det\left(L-\lambda\left(id\right)\right)=0$ +\end_inset + +. + Toda determinante linearne preslikave nismo definirali. + Lahko pa determinanto izračunamo na matriki, + ki pripada tej linearni preslikavi. + Toda dvem različnim bazam pripadata različni matriki linearne preslikave. + Dokazati je treba, + da sta determinanti dveh matrik, + pripadajočih eni linearni preslikavi, + enaki, + četudi sta matriki v različnih bazah. +\end_layout + +\begin_layout Lemma +\begin_inset CommandInset label +LatexCommand label +name "lem:Podobni-matriki-imata" + +\end_inset + +Podobni matriki imata isto determinanto. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $B=PAP^{-1}$ +\end_inset + + za neko obrnljivo +\begin_inset Formula $P$ +\end_inset + +. + Tedaj +\begin_inset Formula $\det B=\det PAP^{-1}=\det P\det A\det P^{-1}=\det P\det P^{-1}\det A=\det PP^{-1}\det A=\det I\det A=1\cdot\det A=\det A$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Proof +\begin_inset Formula $L:V\to V$ +\end_inset + + naj bo linearna preslikava, + +\begin_inset Formula $V$ +\end_inset + + prostor nad +\begin_inset Formula $F=\mathbb{C}$ +\end_inset + +, + +\begin_inset Formula $\mathcal{B}$ +\end_inset + + in +\begin_inset Formula $\mathcal{C}$ +\end_inset + + pa bazi +\begin_inset Formula $V$ +\end_inset + +. + Priredimo matriki +\begin_inset Formula $L_{\mathcal{B}\leftarrow\mathcal{B}}$ +\end_inset + + in +\begin_inset Formula $L_{\mathcal{C}\leftarrow\mathcal{C}}$ +\end_inset + +. + Spomnimo se izreka +\begin_inset CommandInset ref +LatexCommand vref +reference "thm:matrika-kompozituma-linearnih" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +: + +\begin_inset Formula $\left[KL\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$ +\end_inset + +. + +\begin_inset Formula $L=\left[id\circ L\circ id\right]$ +\end_inset + +, + zato +\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\circ L\circ id\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}\left[id\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{C}}}}=P\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}P^{-1}$ +\end_inset + + za neko obrnljivo +\begin_inset Formula $P$ +\end_inset + +. + Torej sta matriki +\begin_inset Formula $\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}$ +\end_inset + + in +\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}$ +\end_inset + + podobni, + torej imata po lemi +\begin_inset CommandInset ref +LatexCommand vref +reference "lem:Podobni-matriki-imata" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + isto determinanto. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Proof +Alternativen dokaz, + da imata podobni matriki iste lastne vrednosti: + +\begin_inset Formula $A$ +\end_inset + + podobna +\begin_inset Formula +\[ +B\Rightarrow B=PAP^{-1}\Rightarrow B-xI=P\left(A-xI\right)P^{-1}\Rightarrow\det\left(B-xI\right)=\det\left(A-xI\right)\Rightarrow p_{A}=p_{B}, +\] + +\end_inset + +torej so lastne vrednosti enake. + Kaj pa lastni vektorji? + Naj bo +\begin_inset Formula $v$ +\end_inset + + lastni vektor +\begin_inset Formula $A$ +\end_inset + +, + torej +\begin_inset Formula +\[ +Av=\lambda v\Rightarrow PAv=\lambda Pv\Rightarrow PAP^{-1}Pv=\lambda Pv\Rightarrow BPv=\lambda Pv, +\] + +\end_inset + +torej za +\begin_inset Formula $v$ +\end_inset + + lastni vektor +\begin_inset Formula $A$ +\end_inset + + sledi, + da je +\begin_inset Formula $Pv$ +\end_inset + + lastni vektor +\begin_inset Formula $B$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Linearni transformaciji torej priredimo tako matriko, + ki ima v začetnem in končnem prostoru isto bazo. + Tedaj lahko izračunamo lastne pare na tej matriki. +\end_layout + +\begin_layout Theorem* +Schurov izrek. + Vsaka kompleksna kvadratna matrika je podobna zgornjetrikotni matriki. +\end_layout + +\begin_layout Proof +Indukcija po velikosti matrike. +\end_layout + +\begin_deeper +\begin_layout Itemize +Baza: + +\begin_inset Formula $A_{1\times1}$ +\end_inset + + je zgornjetrikotna. +\end_layout + +\begin_layout Itemize +Korak: + Po I. + P. + trdimo, + da je vsaka +\begin_inset Formula $A_{\left(n-1\right)\times\left(n-1\right)}$ +\end_inset + + podobna kaki zgornjetrikotni matriki. + Dokažimo še za poljubno +\begin_inset Formula $A_{n\times n}$ +\end_inset + +. + Naj bo +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $v_{1}$ +\end_inset + + pripadajoči lastni vektor ter +\begin_inset Formula $v_{2},\dots,v_{n}$ +\end_inset + + dopolnitev +\begin_inset Formula $v_{1}$ +\end_inset + + do baze +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Potem je matrika +\begin_inset Formula $P=\left[\begin{array}{ccc} +v_{1} & \cdots & v_{n}\end{array}\right]$ +\end_inset + + obrnljiva. +\begin_inset Formula +\[ +AP=\left[\begin{array}{ccc} +Av_{1} & \cdots & Av_{n}\end{array}\right]=\left[\begin{array}{ccc} +v_{1} & \cdots & v_{n}\end{array}\right]\left[\begin{array}{cccc} +\lambda & a_{1,2} & \cdots & a_{1,n}\\ +0 & \vdots & & \vdots\\ +\vdots & \vdots & & \vdots\\ +0 & a_{m,n} & \cdots & a_{m.n} +\end{array}\right]=P\left[\begin{array}{cc} +\lambda & B\\ +0 & C +\end{array}\right] +\] + +\end_inset + +Po I. + P. + obstaja taka zgornjetrikotna +\begin_inset Formula $T$ +\end_inset + + in obrnljiva +\begin_inset Formula $Q$ +\end_inset + +, + da +\begin_inset Formula $C=QTQ^{-1}$ +\end_inset + +. +\begin_inset Formula +\[ +\left[\begin{array}{cc} +1 & 0\\ +0 & Q +\end{array}\right]^{-1}P^{-1}AP\left[\begin{array}{cc} +1 & 0\\ +0 & Q +\end{array}\right]=\left[\begin{array}{cc} +1 & 0\\ +0 & Q +\end{array}\right]^{-1}\left[\begin{array}{cc} +\lambda & B\\ +0 & C +\end{array}\right]\left[\begin{array}{cc} +1 & 0\\ +0 & Q +\end{array}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left[\begin{array}{cc} +\lambda & C\\ +0 & Q^{-1}B +\end{array}\right]\left[\begin{array}{cc} +1 & 0\\ +0 & Q +\end{array}\right]=\left[\begin{array}{cc} +\lambda & CQ\\ +0 & B +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula $A$ +\end_inset + + je torej podobna +\begin_inset Formula $\left[\begin{array}{cc} +\lambda & CQ\\ +0 & B +\end{array}\right]$ +\end_inset + +, + ki je zgornjetrikotna. +\end_layout + +\end_deeper +\begin_layout Proof +\begin_inset Note Note +status open + +\begin_layout Plain Layout +TODO karakterizacija linearnih preslikav +\begin_inset Quotes gld +\end_inset + +LA1V FMF 2024-03-12 +\begin_inset Quotes grd +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Zadosten pogoj za diagonalizabilnost +\end_layout + +\begin_layout Theorem +\begin_inset CommandInset label +LatexCommand label +name "thm:lave-razl-lavr-so-LN" + +\end_inset + +Lastni vektorji, + ki pripadajo različnim lastnim vrednostim, + so linearno neodvisni. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A_{n\times n}$ +\end_inset + + matrika, + +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + njene lastne vrednosti in +\begin_inset Formula $v_{1},\dots,v_{k}$ +\end_inset + + njim pripadajoči lastni vektorji. + Dokazujemo +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + paroma različni +\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$ +\end_inset + + LN. + Dokaz z indukcijo po +\begin_inset Formula $k$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Itemize +Baza +\begin_inset Formula $k=1$ +\end_inset + +: + Elementi +\begin_inset Formula $\left\{ \lambda_{1}\right\} $ +\end_inset + + so trivialno paroma različni in +\begin_inset Formula $v_{1}$ +\end_inset + + je kot neničen vektor LN. +\end_layout + +\begin_layout Itemize +Korak: + Dokazujemo +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k+1}$ +\end_inset + + so paroma različne +\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$ +\end_inset + + so LN, + vedoč I. + P. + Denimo, + da +\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$ +\end_inset + +. + Množimo z +\begin_inset Formula $A$ +\end_inset + +: +\begin_inset Formula +\[ +A\left(\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}\right)=\alpha_{1}Av_{1}+\cdots+\alpha_{k+1}Av_{k+1}=\alpha_{1}\lambda_{1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0 +\] + +\end_inset + +Množimo začetno enačbo z +\begin_inset Formula $\lambda_{k+1}$ +\end_inset + + (namesto z +\begin_inset Formula $A$ +\end_inset + +, + kot smo to storili zgoraj): +\begin_inset Formula +\[ +\alpha_{1}\lambda_{k+1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0 +\] + +\end_inset + +Odštejmo eno enačbo od druge, + dobiti moramo 0, + saj odštevamo 0 od 0: +\begin_inset Formula +\[ +\alpha_{1}\left(\lambda_{1}-\lambda_{k+1}\right)v_{1}+\cdots+\alpha_{k}\left(\lambda_{k}-\lambda_{k+1}\right)v_{k}+\cancel{\alpha_{k+1}\left(\lambda_{k+1}-\lambda_{k+1}\right)v_{k+1}}=0 +\] + +\end_inset + +Ker so lastne vrednosti paroma različne ( +\begin_inset Formula $\lambda_{i}=\lambda_{j}\Rightarrow i=j$ +\end_inset + +), + so njihove razlike neničelne. + Ker so +\begin_inset Formula $v_{1},\dots,v_{k}$ +\end_inset + + po predpostavki LN, + sledi +\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=0$ +\end_inset + +. + Vstavimo te konstante v +\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$ +\end_inset + + in dobimo +\begin_inset Formula $\alpha_{k+1}v_{k+1}=0$ +\end_inset + +. + Ker je +\begin_inset Formula $v_{k+1}$ +\end_inset + + neničeln (je namreč lastni vektor), + sledi +\begin_inset Formula $\alpha_{k+1}=0$ +\end_inset + +, + torej +\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=\alpha_{k+1}=0$ +\end_inset + +, + zatorej so +\begin_inset Formula $v_{1},\dots,v_{k+1}$ +\end_inset + + res LN. +\end_layout + +\end_deeper +\begin_layout Corollary +\begin_inset CommandInset label +LatexCommand label +name "cor:vsota-lastnih-podpr-direktna" + +\end_inset + +Vsota vseh lastnih podprostorov matrike je direktna (definicija +\begin_inset CommandInset ref +LatexCommand vref +reference "def:vsota-je-direktna" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +). +\end_layout + +\begin_layout Proof +Naj bodo +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + vse paroma različne lastne vrednosti matrike +\begin_inset Formula $A\in M_{n}\left(\mathbb{C}\right)$ +\end_inset + +. + Pripadajoči lastni podprostori so torej +\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :V_{i}=\Ker\left(A-\lambda_{i}I\right)$ +\end_inset + +. + Trdimo, + da je vsota teh podprostorov direktna, + torej +\begin_inset Formula $\forall v_{1}\in V_{1},\dots,v_{k}\in V_{k}:v_{1}+\cdots+v_{k}=0\Rightarrow v_{1}=\cdots=v_{k}=0$ +\end_inset + +. + To sledi iz izreka +\begin_inset CommandInset ref +LatexCommand vref +reference "thm:lave-razl-lavr-so-LN" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. +\end_layout + +\begin_layout Corollary* +Če ima +\begin_inset Formula $n\times n$ +\end_inset + + matrika +\begin_inset Formula $n$ +\end_inset + + paroma različnih lastnih vrednosti, + je podobna diagonalni matriki. +\end_layout + +\begin_layout Proof +Po posledici +\begin_inset CommandInset ref +LatexCommand vref +reference "cor:vsota-lastnih-podpr-direktna" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + je vsota lastnih podprostorov matrike +\begin_inset Formula $A_{n\times n}$ +\end_inset + + direktna. + Če je torej lastnih podprostorov +\begin_inset Formula $n$ +\end_inset + +, + je njihova vsota cel prostor +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Matriko se da diagonalizirati, + kadar je vsota vseh lastnih podprostorov enaka podprostoru +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + + (tedaj so namreč stolpci matrike +\begin_inset Formula $P$ +\end_inset + + linearno neodvisni, + zato je +\begin_inset Formula $P$ +\end_inset + + obrnljiva). +\end_layout + +\begin_layout Subsubsection +Algebraične in geometrijske vekčratnosti +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $A_{n\times n}$ +\end_inset + + matrika. + +\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\left(-1\right)^{n}\left(\lambda-\lambda_{1}\right)^{n_{1}}\cdots\left(\lambda-\lambda_{k}\right)^{n_{k}}$ +\end_inset + +, + kjer so +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + vse paroma različne lastne vrednosti +\begin_inset Formula $A$ +\end_inset + +. + Stopnji ničle — + +\begin_inset Formula $n_{i}$ +\end_inset + + — + rečemo algebraična večkratnost lastne vrednosti +\begin_inset Formula $\lambda_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Geometrijska večkratnost lastne vrednosti +\begin_inset Formula $\lambda_{i}$ +\end_inset + + je +\begin_inset Formula $\dim\Ker\left(A-\lambda_{i}I\right)=\n$ +\end_inset + + +\begin_inset Formula $\left(A-\lambda_{i}I\right)=m_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Algebraično večkratnost +\begin_inset Formula $\lambda_{i}$ +\end_inset + + označimo z +\begin_inset Formula $n_{i}$ +\end_inset + + in je večkratnost ničle +\begin_inset Formula $\lambda_{i}$ +\end_inset + + v +\begin_inset Formula $p_{A}\left(\lambda\right)$ +\end_inset + + (karakterističnem polinomu). + Geometrijsko večkratnost +\begin_inset Formula $\lambda_{i}$ +\end_inset + + pa označimo z +\begin_inset Formula $m_{i}$ +\end_inset + + in je dimenzija lastnega podprostora za +\begin_inset Formula $\lambda_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$ +\end_inset + + — + geometrijska večkratnost lastne vrednosti je kvečjemu tolikšna, + kot je algebraična večkratnost te lastne vrednosti. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $v_{1},\dots,v_{m_{i}}$ +\end_inset + + baza za lastni podprostor +\begin_inset Formula $V_{i}=\Ker\left(A-\lambda_{i}I\right)$ +\end_inset + + in naj bo +\begin_inset Formula $v_{m_{i}+1},\dots,v_{n}$ +\end_inset + + njena dopolnitev do baze +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Tedaj velja: + +\begin_inset Formula $Av_{1}=\lambda_{1}v_{1}$ +\end_inset + +, + ..., + +\begin_inset Formula $Av_{m_{i}}=\lambda_{m_{i}}v_{m_{i}}$ +\end_inset + +, + +\begin_inset Formula $Av_{m_{i}+1}=$ +\end_inset + + linearna kombinacija +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + +, + ..., + +\begin_inset Formula $Av_{n}=$ +\end_inset + + linearna kombinacija +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + +. + Naj bo +\begin_inset Formula $P=\left[\begin{array}{cccccc} +v_{1} & \cdots & v_{m_{i}} & v_{m_{i}+1} & \cdots & v_{n}\end{array}\right]$ +\end_inset + +, + ki je obrnljiva. +\begin_inset Formula +\[ +P^{-1}AP=\cdots=\left[\begin{array}{cc} +\lambda_{i}I_{m_{i}} & B\\ +0 & C +\end{array}\right] +\] + +\end_inset + + +\series bold +Dokaza ne razumem. + Obupam. +\end_layout + +\begin_layout Claim* +Matriko s paroma različnimi lastnimi vrednostmi +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + je moč diagonalizirati +\begin_inset Formula $\Leftrightarrow\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $V_{i}$ +\end_inset + + lastno podprostor lastne vrednosti +\begin_inset Formula $\lambda_{i}$ +\end_inset + +. + Vemo, + da se da +\begin_inset Formula $A_{n\times n}$ +\end_inset + + diagonalizirati +\begin_inset Formula $\Leftrightarrow A$ +\end_inset + + ima +\begin_inset Formula $n$ +\end_inset + + LN stolpičnih vektorjev +\begin_inset Formula $\Leftrightarrow\Ker\left(A-\lambda_{1}I\right)+\cdots+\Ker\left(A-\lambda_{k}I\right)=\mathbb{C}^{n}\Leftrightarrow\dim\left(V_{i}+\cdots+V_{k}\right)=\dim V_{i}+\cdots+\dim V_{k}\Leftrightarrow$ +\end_inset + + vsota lastnih podprostorov je direktna +\begin_inset Formula $\Leftrightarrow\dim\left(V_{1}+\cdots+V_{n}\right)=n\Leftrightarrow\dim V_{1}+\cdots+\dim V_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n_{1}+\cdots+n_{m}$ +\end_inset + +. + Toda ker po prejšnjem izreku +\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$ +\end_inset + +, + mora veljati +\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Minimalni polinom matrike +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+\cdots+c_{n}x^{n}\in\mathbb{C}\left[x\right]$ +\end_inset + + polinom in +\begin_inset Formula $A$ +\end_inset + + matrika. + +\begin_inset Formula $p\left(A\right)\coloneqq c_{0}A^{0}+\cdots+c_{n}A^{n}=c_{0}I+\cdots+c_{n}A^{n}$ +\end_inset + +. + Če je +\begin_inset Formula $p\left(A\right)=0$ +\end_inset + + (ničelna matrika), + pravimo, + da polinom +\begin_inset Formula $p$ +\end_inset + + anhilira/uniči matriko +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Fact* +\begin_inset Formula $p\left(A\right)=0\Rightarrow p\left(P^{-1}AP\right)=0$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Izkaže se, + da karakteristični polimom anhilira matriko — + +\begin_inset Formula $p_{A}\left(A\right)=0$ +\end_inset + +. + Dokaz kasneje. +\end_layout + +\begin_layout Definition* +Polinom +\begin_inset Formula $m\left(x\right)$ +\end_inset + + je minimalen polinom +\begin_inset Formula $A$ +\end_inset + +, + če velja: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $m\left(A\right)=0$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $m$ +\end_inset + + ima vodilni koeficient 1 +\end_layout + +\begin_layout Enumerate +med vsemi polinomi, + ki zadoščajo prvi in drugi zahtevi, + ima +\begin_inset Formula $m$ +\end_inset + + najnižjo stopnjo +\end_layout + +\end_deeper +\begin_layout Claim* +eksistenca minimalnega polinoma — + Minimalni polinom obstaja. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A_{n\times n}$ +\end_inset + + matrika. + Očitno je +\begin_inset Formula $M_{n\times n}\left(\mathbb{C}\right)$ +\end_inset + + vektorski prostor dimenzije +\begin_inset Formula $n^{2}$ +\end_inset + +. + Matrike +\begin_inset Formula $\left\{ I,A,A^{2},\dots,A^{n^{2}}\right\} $ +\end_inset + + so linearno odvisne, + ker je moč te množice za 1 večja od moči vektorskega prostora. + Torej +\begin_inset Formula $\exists c_{0},\cdots,c_{n^{2}}\in\mathbb{C}$ +\end_inset + +, + ki niso vse 0 +\begin_inset Formula $\ni:c_{0}I+c_{1}A+c_{2}A^{2}+\cdots+c_{n^{2}}A^{n^{2}}=0$ +\end_inset + +. + Torej polinom +\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+c_{1}x^{1}+c_{2}x^{2}+\cdots+c_{n^{2}}x^{n^{2}}$ +\end_inset + + anhilira +\begin_inset Formula $A$ +\end_inset + +. + Če ta polinom delimo z njegovim vodilnim koeficientom, + dobimo polinom, + ki ustreza prvima dvema zahevama za minimalni polinom. + Če med vsemi takimi izberemo takega z najnižjo stopnjo, + le-ta ustreza še tretji zahtevi. +\end_layout + +\begin_layout Theorem* +Če je +\begin_inset Formula $m\left(x\right)$ +\end_inset + + minimalni polinom za +\begin_inset Formula $A$ +\end_inset + + in če +\begin_inset Formula $p\left(x\right)$ +\end_inset + + anhilira +\begin_inset Formula $A$ +\end_inset + +, + potem +\begin_inset Formula $m\left(x\right)\vert p\left(x\right)$ +\end_inset + + ( +\begin_inset Formula $m\left(x\right)$ +\end_inset + + deli +\begin_inset Formula $p\left(x\right)$ +\end_inset + +). +\end_layout + +\begin_layout Proof +Delimo +\begin_inset Formula $p$ +\end_inset + + z +\begin_inset Formula $m$ +\end_inset + +: + +\begin_inset Formula $\exists k\left(x\right),r\left(x\right)\ni:p\left(x\right)=k\left(x\right)m\left(x\right)+r\left(x\right)\wedge\deg r\left(x\right)<\deg m\left(x\right)$ +\end_inset + +. + Vstavimo +\begin_inset Formula $A$ +\end_inset + + na obe strani: +\begin_inset Formula +\[ +0=p\left(A\right)=k\left(A\right)m\left(A\right)+r\left(A\right)=k\left(A\right)\cdot0+r\left(A\right)=0+r\left(A\right)=r\left(A\right)=0 +\] + +\end_inset + +Sledi +\begin_inset Formula $r\left(x\right)=0$ +\end_inset + +, + kajti če +\begin_inset Formula $r$ +\end_inset + + ne bi bil ničeln polinom, + bi ga lahko delili z vodilnim koeficientom in po predpostavki +\begin_inset Formula $\deg r\left(x\right)<\deg m\left(x\right)$ +\end_inset + + bi imel manjšo stopnjo kot +\begin_inset Formula $m\left(x\right)$ +\end_inset + +, + torej bi ustrezal zahtevam 1 in 2 za minimalni polinom in bi imel manjšo stopnjo od +\begin_inset Formula $m$ +\end_inset + +, + torej +\begin_inset Formula $m$ +\end_inset + + ne bi bil minimalni polinom, + kar bi vodilo v protislovje. +\end_layout + +\begin_layout Corollary* +enoličnost minimalnega polinoma. + Naj bosta +\begin_inset Formula $m_{1}$ +\end_inset + + in +\begin_inset Formula $m_{2}$ +\end_inset + + minimalna polinoma matrike +\begin_inset Formula $A$ +\end_inset + +. + Ker +\begin_inset Formula $m$ +\end_inset + + po definiciji anhilira +\begin_inset Formula $A$ +\end_inset + +, + iz prejšnje trditve sledi, + če vstavimo +\begin_inset Formula $m=m_{1}$ +\end_inset + + in +\begin_inset Formula $p=m_{2}$ +\end_inset + +, + +\begin_inset Formula $m_{1}\vert m_{2}$ +\end_inset + +. + Toda če vstavimo +\begin_inset Formula $m=m_{2}$ +\end_inset + + in +\begin_inset Formula $p=m_{1}$ +\end_inset + +, + +\begin_inset Formula $m_{2}\vert m_{1}$ +\end_inset + +. + Iz +\begin_inset Formula $m_{1}\vert m_{2}\wedge m_{2}\vert m_{1}$ +\end_inset + + sledi, + da se +\begin_inset Formula $m_{1}$ +\end_inset + + in +\begin_inset Formula $m_{2}$ +\end_inset + + razlikujeta le za konstanten faktor, + ki pa je po definiciji minimalnega polinoma 1, + torej +\begin_inset Formula $m_{1}=m_{2}$ +\end_inset + +. + Zaradi enoličnosti lahko označimo minimalni polinom +\begin_inset Formula $A$ +\end_inset + + z +\begin_inset Formula $m_{A}\left(x\right)$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Ničle minimalnega polinoma +\end_layout + +\begin_layout Claim* +\begin_inset Formula $m_{A}\left(x\right)$ +\end_inset + + in +\begin_inset Formula $p_{A}\left(x\right)$ +\end_inset + + imata iste ničle +\begin_inset Formula $\sim$ +\end_inset + + ničle +\begin_inset Formula $m_{A}\left(x\right)$ +\end_inset + + so lastne vrednosti +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Ker je +\begin_inset Formula $p_{A}\left(x\right)$ +\end_inset + + (dokaz kasneje), + velja po trditvi v dokazu enoličnosti, + da +\begin_inset Formula $m_{A}\vert p_{A}$ +\end_inset + +, + torej je vsaka ničla +\begin_inset Formula $m_{A}$ +\end_inset + + tudi ničla +\begin_inset Formula $p_{A}$ +\end_inset + +. + Treba je dokazati še, + da je vsaka ničla +\begin_inset Formula $p_{A}$ +\end_inset + + tudi ničla +\begin_inset Formula $m_{A}$ +\end_inset + +, + natančneje: + Treba je dokazati, + da če je +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost matrike +\begin_inset Formula $A$ +\end_inset + +, + je +\begin_inset Formula $m_{A}\left(\lambda\right)=0$ +\end_inset + +. + Naj bo +\begin_inset Formula $v\not=0$ +\end_inset + + lastni vektor za +\begin_inset Formula $\lambda$ +\end_inset + +. + Tedaj +\begin_inset Formula $Av=\lambda v$ +\end_inset + +. + Potem velja +\begin_inset Formula $A^{2}v=AAv=A\lambda v=\lambda Av=\lambda\lambda v=\lambda^{2}v$ +\end_inset + + in splošneje +\begin_inset Formula $A^{n}v=\lambda^{n}v$ +\end_inset + +. + Sedaj recimo, + da je +\begin_inset Formula $m_{A}\left(x\right)=d_{0}x^{0}+\cdots+d_{r}x^{r}$ +\end_inset + +. + Potem je, + ker minimalni polinom anhilira +\begin_inset Formula $A$ +\end_inset + +, + +\begin_inset Formula +\[ +m_{A}\left(\lambda\right)v=\left(d_{0}+d_{1}\lambda+d_{2}\lambda^{2}+\cdots+d_{r}\lambda^{r}\right)v=d_{0}v+d_{1}\lambda v+d_{2}\lambda^{2}v+\cdots+d_{r}\lambda^{r}v= +\] + +\end_inset + + +\begin_inset Formula +\[ +=d_{0}v+d_{1}Av+d_{2}A^{2}v+\cdots+d_{r}A^{r}v=\left(d_{0}+d_{1}A+d_{2}A^{2}+\cdots+d_{r}A^{r}\right)v=m_{A}\left(A\right)v=0v=0 +\] + +\end_inset + +Ker +\begin_inset Formula $m_{A}\left(\lambda\right)v=0$ +\end_inset + + in +\begin_inset Formula $v\not=0$ +\end_inset + + (je namreč lastni vektor), + velja +\begin_inset Formula $m_{A}\left(\lambda\right)=0$ +\end_inset + +. +\end_layout + +\begin_layout Paragraph* +Lastnosti +\end_layout + +\begin_layout Standard +Ker je +\begin_inset Formula $p_{A}\left(x\right)=\left(-1\right)^{n}\left(x-\lambda_{1}\right)^{n_{1}}\cdots\left(x-\lambda_{k}\right)^{n_{k}}$ +\end_inset + + in +\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{r_{1}}\cdots\left(x-\lambda_{k}\right)^{r_{k}}$ +\end_inset + +, + sledi iz +\begin_inset Formula $m_{A}\vert p_{A}\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\leq n_{i}$ +\end_inset + +. + Poleg tega, + ker +\begin_inset Formula $m_{A}\left(\lambda_{1}\right)=0\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\geq1$ +\end_inset + +. + Toda pozor: + +\series bold +Ni +\series default + res, + da +\begin_inset Formula $r_{i}=m_{i}$ +\end_inset + + (stropnja lastnega podprostora). +\end_layout + +\begin_layout Theorem* +Cayley-Hamilton. + +\begin_inset Formula $p_{A}\left(A\right)=0$ +\end_inset + + — + karakteristični polinom matrike +\begin_inset Formula $A$ +\end_inset + + anhilira matriko +\begin_inset Formula $A$ +\end_inset + + +\end_layout + +\begin_layout Proof +Spomnimo se eksplicitne formule za celico inverza matrike (razdelek +\begin_inset CommandInset ref +LatexCommand vref +reference "subsec:Formula-za-inverz-matrike" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +), + ki pravi +\begin_inset Formula $B^{-1}=\frac{1}{\det B}\tilde{B}^{T}$ +\end_inset + +. + Računajmo in naposled vstavimo +\begin_inset Formula $B=A-xI$ +\end_inset + +: +\begin_inset Formula +\[ +B^{-1}=\frac{1}{\det B}\tilde{B}^{T}\quad\quad\quad\quad/\cdot\left(\det B\right)B +\] + +\end_inset + + +\begin_inset Formula +\[ +\det B\cdot I=B\tilde{B}^{T} +\] + +\end_inset + + +\begin_inset Formula +\[ +\det\left(A-xI\right)\cdot I=p_{A}\left(x\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T} +\] + +\end_inset + +Glede na definicijo +\begin_inset Formula $\tilde{A}$ +\end_inset + + je +\begin_inset Formula $\tilde{\left(A-xI\right)}^{T}$ +\end_inset + + matrika velikosti +\begin_inset Formula $n\times n$ +\end_inset + +, + ki vsebuje polinome stopnje +\begin_inset Formula $<n$ +\end_inset + +, + kajti vsebuje determinante matrik, + katerih elementi so polinomi stopnje +\begin_inset Formula $\leq1$ +\end_inset + +, + torej takele oblike: +\begin_inset Formula +\[ +\tilde{\left(A-xI\right)}^{T}=B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1} +\] + +\end_inset + + +\begin_inset Foot +status open + +\begin_layout Plain Layout +Ne razumem, + zakaj so tu matrike +\begin_inset Formula $B$ +\end_inset + + in ne skalarji. +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-\lambda I\right)=c_{0}+c_{1}x+\cdots+c_{n}x^{n}$ +\end_inset + +. + Kot v enačbi množimo to z +\begin_inset Formula $I$ +\end_inset + +: + +\begin_inset Formula $\det\left(A-\lambda I\right)\cdot I=c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}$ +\end_inset + +. + Oglejmo si še desno stran enačbe: +\begin_inset Formula +\[ +\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}=\left(A-xI\right)\left(B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}\right)=AB_{0}+AB_{1}x+\cdots+AB_{n-1}x^{n-1}-B_{0}x-B_{1}x^{2}-\cdots-B_{n-1}x^{n}= +\] + +\end_inset + + +\begin_inset Formula +\[ +=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}x^{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n} +\] + +\end_inset + +In primerjajmo koeficiente v polinomih pred istoležnimi spremenljivkami na obeh straneh tele enačbe: +\begin_inset Formula +\[ +\det\left(A-xI\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T} +\] + +\end_inset + + +\begin_inset Formula +\[ +c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n} +\] + +\end_inset + + +\begin_inset Formula +\[ +\begin{array}{cccc} +1: & c_{0}I & = & AB_{0}\\ +x: & c_{1}I & = & AB_{1}-B_{0}\\ +x^{2}: & c_{2}I & = & AB_{2}x^{2}-B_{1}\\ +\vdots\\ +x^{n-1}: & c_{n-1}I & = & AB_{n-1}-B_{n-2}\\ +x^{n}: & c_{n}I & = & -B_{n-1} +\end{array} +\] + +\end_inset + +Vstavimo sedaj +\begin_inset Formula $A$ +\end_inset + + v enačbo namesto +\begin_inset Formula $x$ +\end_inset + +: +\begin_inset Formula +\[ +p_{A}\left(A\right)=c_{0}I+c_{1}IA+\cdots+c_{n}IA^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)A+\left(AB_{2}-B_{1}\right)A^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)A^{n-1}-B_{n-1}A^{n}= +\] + +\end_inset + + +\begin_inset Formula +\[ +=AB_{0}+A^{2}B_{1}-AB_{0}+A^{2}B^{2}-B_{1}A^{2}+\cdots+A^{n}B_{n-1}-A^{n-1}B_{n-2}-B_{n-1}A^{n}=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +p_{A}\left(A\right)=0 +\] + +\end_inset + + +\end_layout + +\begin_layout Theorem* +Matriko +\begin_inset Formula $A$ +\end_inset + + se da diagonalizirati +\begin_inset Formula $\Leftrightarrow m_{A}\left(x\right)$ +\end_inset + + ima samo enostavne ničle (nima večkratnih — + potence so vse 1). + Torej +\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{1}\cdots\left(x-\lambda_{k}\right)^{1}$ +\end_inset + + za +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + vse paroma različne lastne vrednosti +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokazujemo ekvivalenco: +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Rightarrow\right)$ +\end_inset + + Po predpostavki je +\begin_inset Formula $A$ +\end_inset + + podobna diagonalni matriki — + +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + + za diagonalno +\begin_inset Formula $D$ +\end_inset + + in obrnljivo +\begin_inset Formula $P$ +\end_inset + +. + BSŠ naj bo +\begin_inset Formula $D=\left[\begin{array}{ccc} +\lambda_{1} & 0 & 0\\ +0 & \cdots & 0\\ +0 & 0 & \lambda_{k} +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $\lambda_{1}\leq\cdots\leq\lambda_{k}$ +\end_inset + +. + NADALJUJ TULE AAAA +\begin_inset Quotes gld +\end_inset + +LA1P FMF 2024-03-20 +\begin_inset Quotes grd +\end_inset + + stran 2. +\end_layout + +\end_deeper \begin_layout Part Vaja za ustni izpit \end_layout |