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\begin_body

\begin_layout Title
Rešitev šeste domače naloge Linearne Algebre
\end_layout

\begin_layout Author

\noun on
Anton Luka Šijanec
\end_layout

\begin_layout Date
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
today
\end_layout

\end_inset


\end_layout

\begin_layout Abstract
Za boljšo preglednost sem svoje rešitve domače naloge prepisal na računalnik.
 Dokumentu sledi še rokopis.
 Naloge je izdelala asistentka Ajda Lemut.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
newcommand
\backslash
euler{e}
\backslash
newcommand
\backslash
rang{
\backslash
text{rang}}
\end_layout

\end_inset


\end_layout

\begin_layout Enumerate
Naj bosta 
\begin_inset Formula $V,W$
\end_inset

 vektorska prostora.
 Pokaži, da je množica vseh linearnih preslikav 
\begin_inset Formula $\mathcal{L}\left(V,W\right)=\left\{ A:V\to W:A\text{ linearna}\right\} $
\end_inset

 vektorski prostor.
\end_layout

\begin_deeper
\begin_layout Paragraph
Rešitev
\end_layout

\begin_layout Standard
Definirali smo, da za linearno preslikavo velja aditivnost 
\begin_inset Formula $L\left(v_{1}+v_{2}\right)=Lv_{1}+Lv_{2}$
\end_inset

 in homogenost 
\begin_inset Formula $L\alpha v=\alpha Lv$
\end_inset

, skupaj 
\begin_inset Formula $L\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)=\alpha_{1}Lv_{2}+\alpha_{2}Lv_{2}$
\end_inset

.
\end_layout

\begin_layout Standard
Vektorski prostor pa smo definirali kot urejeno trojico 
\begin_inset Formula $\left(V,+,\cdot\right)\ni:$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $\left(V,+\right)$
\end_inset

 je Abelova grupa: komutativnost, asociativnost, inverzi, enota, notranjost
\end_layout

\begin_layout Enumerate
aksiomi množenja s skalarjem iz polja 
\begin_inset Formula $F$
\end_inset

: 
\begin_inset Formula $\forall\alpha,\beta\in F\forall a,b\in V:$
\end_inset


\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\alpha\cdot\left(a+b\right)=\alpha\cdot a+\alpha\cdot b$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $\left(\alpha+\beta\right)\cdot a=\alpha\cdot a+\beta\cdot a$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $\left(\alpha\cdot\beta\right)\cdot a=\alpha\cdot\left(\beta\cdot a\right)$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $1\cdot a=a$
\end_inset

, kjer je 
\begin_inset Formula $1$
\end_inset

 enota 
\begin_inset Formula $F$
\end_inset


\end_layout

\end_deeper
\begin_layout Standard
Da linearne preslikave 
\begin_inset Formula $L:V\to W$
\end_inset

 sploh obstajajo, privzemam, da sta 
\begin_inset Formula $V$
\end_inset

 in 
\begin_inset Formula $W$
\end_inset

 vektorska prostora nad istim poljem.
\end_layout

\begin_layout Standard
Treba je definirati 
\begin_inset Formula $+$
\end_inset

, 
\begin_inset Formula $F$
\end_inset

 in 
\begin_inset Formula $\cdot$
\end_inset

 ter dokazati, da je pri izbranih 
\begin_inset Formula $+$
\end_inset

, 
\begin_inset Formula $F$
\end_inset

 in 
\begin_inset Formula $\cdot$
\end_inset

 
\begin_inset Formula $\left(\mathcal{L},+,\cdot\right)$
\end_inset

 vektorski prostor po tej definiciji.
 Vzemimo za 
\begin_inset Formula $+$
\end_inset

 operacijo 
\begin_inset Formula $+$
\end_inset

 iz vektorskega prostora 
\begin_inset Formula $W$
\end_inset

 in definirajmo operacijo na 
\begin_inset Formula $\mathcal{L}$
\end_inset

: 
\begin_inset Formula $\forall L_{1},L_{2}\in\mathcal{L}:\quad\left(L_{1}+L_{2}\right)v\coloneqq L_{1}v+L_{2}v$
\end_inset

.
 Dokažimo, da je 
\begin_inset Formula $\left(\mathcal{L},+\right)$
\end_inset

 abelova grupa:
\end_layout

\begin_layout Enumerate
Notranjost operacije: Trdimo, da je 
\begin_inset Formula $L_{1}+L_{2}$
\end_inset

 linearna transformacija.
 Dokaz: 
\begin_inset Formula $\forall v\in V:$
\end_inset


\end_layout

\begin_deeper
\begin_layout Enumerate
Aditivnost: 
\begin_inset Formula $\left(L_{1}+L_{2}\right)\left(v_{1}+v_{2}\right)\overset{\text{def}+}{=}L_{1}\left(v_{1}+v_{2}\right)+L_{2}\left(v_{1}+v_{2}\right)\overset{\text{aditivnost}}{=}L_{1}v_{1}+L_{1}v_{2}+L_{2}v_{1}+L_{2}v_{2}\overset{W\text{V.P.}}{=}L_{1}v_{1}+L_{2}v_{1}+L_{1}v_{2}+L_{2}v_{2}\overset{def+}{=}\left(L_{1}+L_{2}\right)v_{1}+\left(L_{1}+L_{2}\right)v_{2}$
\end_inset


\end_layout

\begin_layout Enumerate
Homogenost: 
\begin_inset Formula $\alpha\left(L_{1}+L_{2}\right)v\overset{\text{def}+}{=}\alpha\left(L_{1}v+L_{2}v\right)\overset{W\text{V.P.}}{=}\alpha L_{1}v+\alpha L_{2}v\overset{\text{homogenost}}{=}L_{1}\alpha v+L_{2}\alpha v\overset{\text{def}+}{=}\left(L_{1}+L_{2}\right)\left(\alpha v\right)$
\end_inset


\end_layout

\end_deeper
\begin_layout Enumerate
Enote: Enota naj bo tista linearna preslikava 
\begin_inset Formula $L_{0}$
\end_inset

, ki slika ves 
\begin_inset Formula $V$
\end_inset

 v 
\begin_inset Formula $0\in W$
\end_inset

.
 Dokaz: 
\begin_inset Formula $\forall L\in\mathcal{L}:\quad$
\end_inset


\begin_inset Formula $Lv+L_{0}v\text{\ensuremath{\overset{\text{def}L_{0}}{=}}}Lv+0\ensuremath{\overset{\left(W,+\right)\text{abelova grupa}}{=}}Lv$
\end_inset


\end_layout

\begin_layout Enumerate
Inverzi
\begin_inset CommandInset label
LatexCommand label
name "enu:Inverzi:-Ker-je"

\end_inset

: Ker je 
\begin_inset Formula $W$
\end_inset

 V.
 P., 
\begin_inset Formula $\forall w\in W\exists!-w\in W\ni:w+\left(-w\right)=0$
\end_inset

, zato 
\begin_inset Formula $\forall L\in\mathcal{L}\exists-L\in\mathcal{L}\ni:-L+L=L_{0}$
\end_inset

 s predpisom 
\begin_inset Formula $-L$
\end_inset

 slika element 
\begin_inset Formula $v\in V$
\end_inset

 v tisti 
\begin_inset Formula $w\in W$
\end_inset

, ki je inverz 
\begin_inset Formula $Lv\in W$
\end_inset

.
 
\begin_inset Formula $-L$
\end_inset

 je res 
\begin_inset Formula $\in\mathcal{L}$
\end_inset

.
 Velja 
\begin_inset Formula $-L\coloneqq$
\end_inset


\begin_inset Formula $\left(-1\right)\cdot L$
\end_inset

, kjer je 
\begin_inset Formula $-1$
\end_inset

 inverz enote polja, ki ga izberemo kasneje.
 
\begin_inset Formula $\forall v\in V,L\in\mathcal{L}:\left(\left(-1\right)\cdot L\right)v\overset{\text{def}\cdot\text{sledi}}{=}\left(-1\right)\left(Lv\right)\overset{\text{def\ensuremath{\cdot},homogenost}}{=}L\left(-1v\right)\overset{\text{karakteristika F}\not=0}{L\left(-v\right)}$
\end_inset

.
 Ta dokaz se sklicuje na določitev polja in skalarnega množenja, ki ga podam
 kasneje.
\end_layout

\begin_layout Enumerate
Asociativnost: 
\begin_inset Formula $\forall L_{1},L_{2},L_{3}\in\mathcal{L}:L_{1}+\left(L_{2}+L_{3}\right)=\left(L_{1}+L_{2}\right)+L_{3}$
\end_inset

 velja očitno iz definicije 
\begin_inset Formula $+$
\end_inset

, saj je 
\begin_inset Formula $W$
\end_inset

 vektorski prostor.
 Komutativnost spet iz istih razlogov.
\end_layout

\begin_layout Standard
Določiti moramo še polje in množenje s skalarjem.
 Vzemimo za 
\begin_inset Formula $F$
\end_inset

 polje vektorskega prostora 
\begin_inset Formula $W$
\end_inset

 in množenje s skalarjem definirajmo takole: 
\begin_inset Formula $\forall v\in V,\alpha\in F:\left(\alpha L\right)v\coloneqq\alpha\left(Lv\right)$
\end_inset

.
 Zopet za vsak slučaj dokažimo še linearnost dobljene preslikave 
\begin_inset Formula $\forall\alpha,\beta\in F\forall L\in\mathcal{L}\forall v_{1},v_{2}\in V:$
\end_inset


\end_layout

\begin_layout Enumerate
Aditivnost: 
\begin_inset Formula $\left(\alpha L\right)\left(v_{1}+v_{2}\right)\overset{\text{def}\cdot}{=}\alpha\left(L\left(v_{1}+v_{2}\right)\right)\overset{\text{aditivnost}}{=}\alpha\left(Lv_{1}+Lv_{2}\right)\overset{W\text{V.P.}}{=}\alpha\left(Lv_{1}\right)+\alpha\left(Lv_{2}\right)\overset{\text{def}\cdot}{=}\left(\alpha L\right)v_{1}+\left(\alpha L\right)v_{2}$
\end_inset


\end_layout

\begin_layout Enumerate
Homogenost: 
\begin_inset Formula $\left(\alpha L\right)\left(\beta v\right)\overset{\text{def}\cdot}{=}\alpha\left(L\left(\beta v\right)\right)\overset{\text{homogenost}}{=}\alpha$
\end_inset


\begin_inset Formula $\beta Lv\overset{\text{F\text{polje}}}{=}\beta\alpha\left(Lv\right)\overset{\text{def}\cdot}{=}\beta\left(\alpha L\right)v$
\end_inset


\end_layout

\begin_layout Standard
Iz tega dokaza sledi tudi obstoj inverzov (
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:Inverzi:-Ker-je"
plural "false"
caps "false"
noprefix "false"

\end_inset

).
\end_layout

\begin_layout Standard
Sedaj lahko dokažemo še štiri aksiome vektorskih prostorov za množenje s
 skalarjem.
 
\begin_inset Formula $\forall\alpha,\beta\in F\forall L_{1},L_{2}\in V:$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
udensdash{$
\backslash
alpha
\backslash
left(L_1+L_2
\backslash
right)
\backslash
overset{?}{=}
\backslash
alpha L_1+
\backslash
alpha L_2$}
\end_layout

\end_inset

: 
\begin_inset Formula $\left(\alpha\left(L_{1}+L_{2}\right)\right)v\overset{\text{def}+\cdot}{=}\alpha\left(L_{1}v+L_{2}v\right)\overset{W\text{V.P.}}{=}\alpha L_{1}+\alpha L_{2}$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
udensdash{$
\backslash
left(
\backslash
alpha+
\backslash
beta
\backslash
right)L_1=
\backslash
alpha L_1+
\backslash
beta L_1$}
\end_layout

\end_inset

: Po definiciji našega 
\begin_inset Formula $\cdot$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
udensdash{$
\backslash
alpha
\backslash
left(
\backslash
beta L_1
\backslash
right)=
\backslash
left(
\backslash
alpha
\backslash
beta
\backslash
right)L_1$}
\end_layout

\end_inset

: Po definiciji našega 
\begin_inset Formula $\cdot$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
udensdash{$1
\backslash
cdot L_1=L_1$}
\end_layout

\end_inset

: 
\begin_inset Formula $\left(1\cdot L_{1}\right)v\overset{\text{def}\cdot}{=}1\cdot\left(L_{1}v\right)\overset{W\text{V.P.}}{=}L_{1}v$
\end_inset


\end_layout

\end_deeper
\begin_layout Enumerate
Naj bo 
\begin_inset Formula $Z:\mathbb{R}^{3}\to\mathbb{R}^{3}$
\end_inset

 zrcaljenje preko ravnine 
\begin_inset Formula $x+y+z=0$
\end_inset

.
 Določi matriko 
\begin_inset Formula $Z$
\end_inset

 v standardni bazi.
\end_layout

\begin_deeper
\begin_layout Paragraph
Rešitev
\end_layout

\begin_layout Standard
Tri točke na taki ravnini so 
\begin_inset Formula $\left(0,0,0\right)$
\end_inset

, 
\begin_inset Formula $\left(1,0,-1\right)$
\end_inset

 in 
\begin_inset Formula $\left(0,1,-1\right)$
\end_inset

.
 Normala ravnine je 
\begin_inset Formula $\left(1,0,-1\right)\times\left(0,1,-1\right)=\left(1,1,1\right)$
\end_inset

.
 Parametrično to ravnino zapišemo kot 
\begin_inset Formula $\left\{ s\vec{r}+p\vec{q};s,p\in\mathbb{R}\right\} $
\end_inset

, kjer 
\begin_inset Formula $\vec{r}=\left(1,0,-1\right)$
\end_inset

 in 
\begin_inset Formula $\vec{q}=\left(0,1,-1\right)$
\end_inset

.
 Za določitev matrike linearne preslikave 
\begin_inset Formula $Z$
\end_inset

 bomo zrcalili vektorje standardne baze 
\begin_inset Formula $\left(1,0,0\right)$
\end_inset

, 
\begin_inset Formula $\left(0,1,0\right)$
\end_inset

 in 
\begin_inset Formula $\left(0,0,1\right)$
\end_inset

 čez to ravnino.
 Zrcaljenje 
\begin_inset Formula $\vec{t}$
\end_inset

 v 
\begin_inset Formula $Z\vec{t}$
\end_inset

 čez ravnino je opisano z enačbo 
\begin_inset Formula $Z\vec{t}=\vec{t}+2\left(\hat{t}-\vec{t}\right)=2\hat{t}-\vec{t}$
\end_inset

, kjer s 
\begin_inset Formula $\hat{t}$
\end_inset

 označim pravokotno projekcijo točke 
\begin_inset Formula $\vec{t}$
\end_inset

 na ravnino.
 Torej najprej tako projicirajmo standardno bazo na ravnino.
\begin_inset Formula 
\[
\langle\hat{t}-\vec{t},\vec{q}\rangle=0=\langle\hat{t}-\vec{t},\vec{r}\rangle\quad\text{(pravokotna projekcija)}
\]

\end_inset


\begin_inset Formula 
\[
\langle s\vec{r}+p\vec{q}-\vec{t},\vec{q}\rangle=0=\langle s\vec{r}+p\vec{q}-\vec{t},\vec{r}\rangle\quad\text{(parametrični zapis \ensuremath{\hat{t}})}
\]

\end_inset


\begin_inset Formula 
\[
s\langle\vec{r},\vec{q}\rangle+p\langle\text{\ensuremath{\vec{q},\vec{q}\rangle-\langle\vec{t},\vec{q}\rangle=0=s\langle\vec{r},\vec{r}\rangle+p\langle\vec{q},\vec{r}\rangle-\langle\vec{t},\vec{r}\rangle}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{align*}
\end_layout

\begin_layout Plain Layout

s
\backslash
langle
\backslash
vec{r},
\backslash
vec{q}
\backslash
rangle+p
\backslash
langle
\backslash
vec{q},
\backslash
vec{q}
\backslash
rangle&=
\backslash
langle
\backslash
vec{t},
\backslash
vec{q}
\backslash
rangle
\backslash

\backslash

\end_layout

\begin_layout Plain Layout

s
\backslash
langle
\backslash
vec{r},
\backslash
vec{r}
\backslash
rangle+p
\backslash
langle
\backslash
vec{q},
\backslash
vec{r}
\backslash
rangle&=
\backslash
langle
\backslash
vec{t},
\backslash
vec{r}
\backslash
rangle
\end_layout

\begin_layout Plain Layout


\backslash
end{align*}
\end_layout

\end_inset

Dobimo sistem enačb z neznankama 
\begin_inset Formula $s$
\end_inset

 in 
\begin_inset Formula $p$
\end_inset

, parametroma projekcije.
 Vstavimo 
\begin_inset Formula $\vec{r}=\left(1,0,-1\right)$
\end_inset

 in 
\begin_inset Formula $\vec{q}=\left(0,1,-1\right)$
\end_inset

 ter za 
\begin_inset Formula $\vec{t}$
\end_inset

 posamično vse tri točke standardne baze in izračunajmo njihove projekcije.
\begin_inset Formula 
\[
s\cdot1+p\cdot2=\langle\vec{t},\left(0,1,-1\right)\rangle
\]

\end_inset


\begin_inset Formula 
\[
s\cdot2+p\cdot1=\langle\vec{t},\left(1,0,-1\right)\rangle
\]

\end_inset


\end_layout

\begin_layout Standard
Nato izračunamo še njihovo zrcaljenje iz projekcij po enačbi za zrcaljenje
 zgoraj.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
setlength{
\backslash
columnseprule}{0.2pt}
\backslash
begin{multicols}{3}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\vec{b_{1}}=\left(1,0,0\right)$
\end_inset


\begin_inset Formula 
\[
s+2p=0
\]

\end_inset


\begin_inset Formula 
\[
2s+p=1
\]

\end_inset


\begin_inset Formula 
\[
s=-2p
\]

\end_inset


\begin_inset Formula 
\[
p-4p=1
\]

\end_inset


\begin_inset Formula 
\[
p=-\frac{1}{3},\quad s=\frac{2}{3}
\]

\end_inset


\begin_inset Formula 
\[
\hat{t}=\left(\frac{2}{3},-\frac{1}{3},-\frac{1}{3}\right)
\]

\end_inset


\begin_inset Formula 
\[
Z\vec{b_{1}}=\left(\frac{1}{3},-\frac{2}{3},-\frac{2}{3}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\vec{b_{2}}=\left(0,1,0\right)$
\end_inset


\begin_inset Formula 
\[
s+2p=1
\]

\end_inset


\begin_inset Formula 
\[
2s+p=0
\]

\end_inset


\begin_inset Formula 
\[
p=-2s
\]

\end_inset


\begin_inset Formula 
\[
s-4s=1
\]

\end_inset


\begin_inset Formula 
\[
p=\frac{2}{3},\quad s=-\frac{1}{3}
\]

\end_inset


\begin_inset Formula 
\[
\hat{b_{2}}=\left(-\frac{1}{3},\frac{2}{3},-\frac{1}{3}\right)
\]

\end_inset


\begin_inset Formula 
\[
Z\vec{b_{2}}=\left(-\frac{2}{3},\frac{1}{3},-\frac{2}{3}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\vec{b_{3}}=\left(0,0,1\right)$
\end_inset


\begin_inset Formula 
\[
s+2p=-1
\]

\end_inset


\begin_inset Formula 
\[
2s+p=-1
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
s=-2p-1
\]

\end_inset


\begin_inset Formula 
\[
2\left(-2p-1\right)+p=-1
\]

\end_inset


\begin_inset Formula 
\[
p=-\frac{1}{3},\quad s=-\frac{1}{3}
\]

\end_inset


\begin_inset Formula 
\[
\hat{b_{2}}=\left(-\frac{1}{3},-\frac{1}{3},\frac{2}{3}\right)
\]

\end_inset


\begin_inset Formula 
\[
Z\vec{b_{3}}=\left(-\frac{2}{3},-\frac{2}{3},\frac{1}{3}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
end{multicols}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Dobljene z 
\begin_inset Formula $Z$
\end_inset

 preslikane (čez ravnino zrcaljene) vektorje po stolpcih zložimo v matriko
 
\begin_inset Formula $Z$
\end_inset

:
\begin_inset Formula 
\[
Z=\left[\begin{array}{ccc}
1/3 & -2/3 & -2/3\\
-2/3 & 1/3 & -2/3\\
-2/3 & -2/3 & 1/3
\end{array}\right]
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Enumerate
Določi rang matrike
\begin_inset Formula 
\[
B=\left[\begin{array}{cccc}
-2-t & 4 & 5+t & 4\\
1 & -1 & -2 & 1\\
-t & 3 & 1+t & 4+t
\end{array}\right]
\]

\end_inset

v odvisnosti od parametra 
\begin_inset Formula $t$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Paragraph
Rešitev
\end_layout

\begin_layout Standard
Za 
\begin_inset Formula $A:V\to U$
\end_inset

 smo definirali 
\begin_inset Formula $\rang A\coloneqq\dim\text{Im}A$
\end_inset

, kjer je 
\begin_inset Formula $\text{Im}A\coloneqq\left\{ Av;v\in V\right\} $
\end_inset

.
 Dokazali smo, da je rang matrike enak številu linearno neodvisnih vrstic
 matrike in da velja 
\begin_inset Formula $\rang A=\rang A^{T}$
\end_inset

.
\begin_inset Formula 
\[
\rang\left[\begin{array}{cccc}
-2-t & 4 & 5+t & 4\\
1 & -1 & -2 & 1\\
-t & 3 & 1+t & 4+t
\end{array}\right]=\rang\left[\begin{array}{ccc}
-2-t & 1 & -t\\
4 & -1 & 3\\
5+t & -2 & 1+t\\
4 & 1 & 4+t
\end{array}\right]=
\]

\end_inset


\begin_inset Formula 
\[
=\rang\left[\begin{array}{ccc}
4 & 1 & -3\\
-2-t & 1 & -t\\
5+t & -2 & 1+t\\
4 & 1 & 4+t
\end{array}\right]=\rang\left[\begin{array}{ccc}
4 & 1 & -3\\
3 & -1 & 1\\
5+t & -2 & 1+t\\
4 & 1 & 4+t
\end{array}\right]=
\]

\end_inset


\begin_inset Formula 
\[
=\rang\left[\begin{array}{ccc}
4 & 1 & -3\\
3 & -1 & 1\\
1+t & -3 & -3\\
4 & 1 & 4+t
\end{array}\right]=\rang\left[\begin{array}{ccc}
1 & 2 & -4\\
3 & -1 & 1\\
1+t & -3 & -3\\
4 & 1 & 4+t
\end{array}\right]=
\]

\end_inset


\begin_inset Formula 
\[
=\rang\left[\begin{array}{ccc}
1 & 2 & -4\\
0 & -7 & 13\\
0 & -5-t & 1+4t\\
0 & -7 & 20+t
\end{array}\right]=\rang\left[\begin{array}{ccc}
1 & 2 & -4\\
0 & -7 & 13\\
0 & 0 & \frac{-58+15t}{7}\\
0 & 0 & 7+t
\end{array}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
Rang je vsaj 2, ker sta 
\begin_inset Formula $\left(1,2,-4\right)$
\end_inset

 in 
\begin_inset Formula $\left(0,-7,13\right)$
\end_inset

 linearno neodvisna.
 Rang je kvečjemu 3, ker je manjša izmed stranic matrike dolžine 3.
 Rang ne more biti 2, ker sistem 
\begin_inset Formula $\frac{-58+15t}{7}=7+t=0$
\end_inset

 nima rešitve.
 
\begin_inset Formula $\rang B=3$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Enumerate
Poišči karakteristični in minimalni polinom matrike
\begin_inset Formula 
\[
A=\left[\begin{array}{ccc}
4 & -5 & 3\\
2 & -3 & 2\\
-1 & 1 & 0
\end{array}\right]
\]

\end_inset

in s pomočjo Cayley-Hamiltonovega izreka določi njen inverz.
\end_layout

\begin_deeper
\begin_layout Standard
\begin_inset Formula 
\[
\nabla_{P}\left(\lambda\right)=\det\left(A-\lambda I\right)=\left|\begin{array}{ccc}
4-\lambda & -5 & 3\\
2 & -3-\lambda & 2\\
-1 & 1 & -\lambda
\end{array}\right|=
\]

\end_inset


\begin_inset Formula 
\[
=-3\left(3+\lambda\right)-2\left(4-\lambda\right)-10\lambda+\lambda\left(3+\lambda\right)\left(4-\lambda\right)+10+6=
\]

\end_inset


\begin_inset Formula 
\[
-9-3\lambda-8+2\lambda-10\lambda+\left(3\lambda+\lambda^{2}\right)\left(4-\lambda\right)+16=
\]

\end_inset


\begin_inset Formula 
\[
=-17+16-11\lambda+12\lambda-3\lambda^{2}+4\lambda^{2}-\lambda^{3}=-\lambda^{3}+\lambda^{2}+\lambda-1
\]

\end_inset


\end_layout

\begin_layout Standard
Eno ničlo uganemo (
\begin_inset Formula $\lambda_{1}=1$
\end_inset

), nato 
\begin_inset Formula $-\lambda^{3}+\lambda^{2}+\lambda-1:\lambda-1=-\lambda^{2}+1=\left(1+\lambda\right)\left(1-\lambda\right)$
\end_inset

.
 1 je torej dvojna ničla, 
\begin_inset Formula $\lambda_{2}=-1$
\end_inset

 pa enojna.
 Ker 
\begin_inset Formula $m_{A}\left(\lambda\right)|\nabla_{A}\left(\lambda\right)$
\end_inset

, je kandidat za 
\begin_inset Formula $m_{A}\left(\lambda\right)$
\end_inset

 poleg 
\begin_inset Formula $-\nabla_{A}\left(\lambda\right)$
\end_inset

 še 
\begin_inset Formula $p\left(\lambda\right)=\left(\lambda-x\right)\left(\lambda+1\right)=1-\lambda^{2}$
\end_inset

.
 Po Cayley-Hamiltonovem izreku 
\begin_inset Formula $m_{A}\left(A\right)=0=\nabla_{A}\left(A\right)$
\end_inset

.
 Toda ker 
\begin_inset Formula $I-A^{2}\not=0$
\end_inset

, je 
\begin_inset Formula $m_{A}\left(\lambda\right)=-\nabla_{A}\left(\lambda\right)=\lambda^{3}-\lambda^{2}-\lambda+1$
\end_inset

.
 Izračunajmo inverz:
\begin_inset Formula 
\[
m_{A}\left(A\right)=A^{3}-A^{2}-A+I=0\quad\quad/-I
\]

\end_inset


\begin_inset Formula 
\[
A^{3}-A^{2}-A=-I\quad\quad/\cdot A^{-1}
\]

\end_inset


\begin_inset Formula 
\[
A^{3}A^{-1}-A^{2}A^{-1}-AA^{-1}=-IA^{-1}\quad\quad/\cdot\left(-I\right)
\]

\end_inset


\begin_inset Formula 
\[
-A^{2}+A+I=A^{1}=
\]

\end_inset


\begin_inset Formula 
\[
=-\left[\begin{array}{ccc}
4 & -5 & 3\\
2 & -3 & 2\\
-1 & 1 & 0
\end{array}\right]\left[\begin{array}{ccc}
4 & -5 & 3\\
2 & -3 & 2\\
-1 & 1 & 0
\end{array}\right]+\left[\begin{array}{ccc}
4 & -5 & 3\\
2 & -3 & 2\\
-1 & 1 & 0
\end{array}\right]+\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]=
\]

\end_inset


\begin_inset Formula 
\[
=\left[\begin{array}{ccc}
2 & -3 & 1\\
2 & -3 & 2\\
1 & -1 & 2
\end{array}\right]\text{, kar je res \ensuremath{A^{-1}.}}
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
Rokopisi, ki sledijo, naj služijo le kot dokaz samostojnega reševanja.
 Zavedam se namreč njihovega neličnega izgleda.
\end_layout

\begin_layout Standard
\begin_inset External
	template PDFPages
	filename /mnt/slu/shramba/upload/www/d/LADN6FMF1.pdf
	extra LaTeX "pages=-"

\end_inset


\begin_inset External
	template PDFPages
	filename /mnt/slu/shramba/upload/www/d/LADN6FMF2.pdf
	extra LaTeX "pages=-"

\end_inset


\end_layout

\end_body
\end_document