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\begin_body
\begin_layout Title
Rešitev osme 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}
\end_layout
\end_inset
\end_layout
\begin_layout Enumerate
Dokaži, da je
\begin_inset Formula $\left[\left(x,y,z\right),\left(u,v,w\right)\right]=2xu-yu-xv+2yv-zv-yw+zw$
\end_inset
skalarni produkt in ugotovi, ali je
\begin_inset Formula
\[
A=\left[\begin{array}{ccc}
0 & 2 & -2\\
0 & 1 & 0\\
-1 & 2 & -1
\end{array}\right]
\]
\end_inset
normalna preslikava glede na
\begin_inset Formula $\left[\cdot,\cdot\right]$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Paragraph
Rešitev
\end_layout
\begin_layout Standard
Predpostavljam polje
\begin_inset Formula $\mathbb{R}$
\end_inset
in vektorski prostor
\begin_inset Formula $V=\mathbb{R}^{3}$
\end_inset
, saj v kompleksnem to ni skalarni produkt (protiprimer pozitivne definitnosti
je
\begin_inset Formula $\left[\left(1,1,1+i\right),\left(1,1,1+i\right)\right]=2$
\end_inset
).
\begin_inset Formula $\langle\cdot,\cdot\rangle:V\times V\to\mathbb{R}$
\end_inset
je skalarni produkt, če zadošča naslednjim lastnostim.
Dokažimo jih za
\begin_inset Formula $\left[\cdot,\cdot\right]$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\langle v,v\rangle>0$
\end_inset
\begin_inset Formula
\[
\left[\left(x,y,z\right),\left(x,y,z\right)\right]=2x^{2}-2xy+2y^{2}-2yz+z^{2}=2\left(x^{2}-xy+y^{2}\right)-2yz+z^{2}=
\]
\end_inset
\begin_inset Formula
\[
=2\left(\left(x-\frac{y}{2}\right)^{2}+\frac{3y^{2}}{4}\right)-2yz+z^{2}=2\left(x-\frac{y}{2}\right)^{2}-\frac{3y^{2}}{4}-2yz+z^{2}=
\]
\end_inset
\begin_inset Formula
\[
=2\left(x-\frac{y}{2}\right)^{2}+\left(\frac{\sqrt{3}y}{\sqrt{2}}-\frac{z\sqrt{2}}{\sqrt{3}}\right)^{2}+\frac{z^{2}}{3}\geq0
\]
\end_inset
Sedaj poiščimo ničle.
Fiksirajmo poljubna
\begin_inset Formula $y$
\end_inset
,
\begin_inset Formula $z$
\end_inset
in uporabimo obrazec za ničle kvadratne enačbe:
\begin_inset Formula
\[
x_{1,2}=\frac{2y\pm\sqrt{4y^{2}-8\left(2y^{2}-2yz+z^{2}\right)}}{4}
\]
\end_inset
Iščemo pozitivne diskriminante.
\begin_inset Formula
\[
4y^{2}-8\left(2y^{2}-2yz+z^{2}\right)=-12y^{2}+16yz-8z^{2}=4
\]
\end_inset
Fiksirajmo poljuben
\begin_inset Formula $z$
\end_inset
.
Vodilni koeficient kvadratne enačbe je negativen.
Uporabimo obrazec:
\begin_inset Formula
\[
y_{1,2}=\frac{-16z\pm\sqrt{256z^{2}-384z^{2}=-128z^{2}}}{-24}
\]
\end_inset
Diskriminanta je nenegativna
\begin_inset Formula $\Leftrightarrow z=0$
\end_inset
.
Torej
\begin_inset Formula $z=0$
\end_inset
, zato
\begin_inset Formula $y=0$
\end_inset
in tudi
\begin_inset Formula $x=0$
\end_inset
glede na obrazce.
Skalarni produkt je res pozitivno definiten.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall v,u\in V:\langle v,u\rangle=\langle u,v\rangle$
\end_inset
\begin_inset Formula
\[
\left[\left(u,v,w\right),\left(x,y,z\right)\right]=\left[\left(x,y,z\right),\left(u,v,w\right)\right]
\]
\end_inset
\begin_inset Formula
\[
2ux-vx-uy-2vy-wy-vz+wz=2xu-yu-xv+2yv-zv-vz+wz
\]
\end_inset
\begin_inset Formula
\[
0=0
\]
\end_inset
Skalarni produkt je res simetričen.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in\mathbb{C}\forall u_{1},u_{2},v\in V:\langle\alpha_{1}v_{1}+\alpha_{2}v_{2},v\rangle=\alpha_{1}\langle u_{1},v\rangle+\alpha_{2}\langle u_{2},v\rangle$
\end_inset
\begin_inset Formula
\[
\left[\alpha_{1}\left(x_{1},y_{1},z_{1}\right)+\alpha_{2}\left(x_{2},y_{2},z_{2}\right),\left(u,v,w\right)\right]=
\]
\end_inset
\begin_inset Formula
\[
=\left[\left(\alpha_{1}x_{1}+\alpha_{2}x_{2},\alpha_{1}y_{1}+\alpha_{2}y_{2},\alpha_{1}z_{1}+\alpha_{2}z_{2}\right),\left(u,v,w\right)\right]=
\]
\end_inset
\begin_inset Formula
\[
=2\left(\alpha_{1}x_{1}+\alpha_{2}x_{2}\right)u-\left(\alpha_{1}y_{1}+\alpha_{2}y_{2}\right)u-\left(\alpha_{1}x_{1}+\alpha_{2}x_{2}\right)v+
\]
\end_inset
\begin_inset Formula
\[
+2\left(\alpha_{1}y_{1}+\alpha_{2}y_{2}\right)v-\left(\alpha_{1}z_{1}+\alpha_{2}z_{2}\right)v-\left(\alpha_{1}y_{1}+\alpha_{2}y_{2}\right)w+\left(\alpha_{1}z_{1}+\alpha_{2}z_{2}\right)w=
\]
\end_inset
\begin_inset Formula
\[
=2\alpha_{1}x_{1}u+2\alpha_{2}x_{2}u-\alpha_{1}y_{1}u-\alpha_{2}y_{2}u-\alpha_{1}x_{1}v-\alpha_{2}x_{2}v+
\]
\end_inset
\begin_inset Formula
\[
+2\alpha_{1}y_{1}v+2\alpha_{2}y_{2}v-\alpha_{1}z_{1}v-\alpha_{2}z_{2}v-\alpha_{1}y_{1}w-\alpha_{2}y_{2}w+\alpha_{1}z_{1}w+\alpha_{2}z_{2}w=
\]
\end_inset
\begin_inset Formula
\[
=\alpha_{1}\left(2x_{1}u-y_{1}u-x_{1}v+2y_{1}v-z_{1}v-y_{1}w+z_{1}w\right)+\alpha_{2}\left(2x_{2}u-y_{2}u-x_{2}v+2y_{2}v-z_{2}v-y_{2}w+z_{2}w\right)=
\]
\end_inset
\begin_inset Formula
\[
=\alpha_{1}\left[\left(x_{1},y_{1},z_{1}\right),\left(u,v,w\right)\right]+\alpha_{2}\left[\left(x_{2},y_{2},z_{2}\right),\left(u,v,w\right)\right]
\]
\end_inset
Skalarni produkt je res homogen in linearen v prvem faktorju.
\end_layout
\begin_layout Standard
Po definiciji
\begin_inset Formula $A$
\end_inset
normalna
\begin_inset Formula $\Leftrightarrow A^{*}A=AA^{*}$
\end_inset
.
Izračunajmo torej matriko
\begin_inset Formula $A^{*}$
\end_inset
.
\end_layout
\begin_layout Itemize
Na predavanjih 2024-05-08 smo dokazali, da za vsak skalarni produkt
\begin_inset Formula $\left[u,v\right]$
\end_inset
obstaja taka ortogonalna (
\begin_inset Formula $M^{*}=M^{-1}$
\end_inset
) pozitivno definitna matrika
\begin_inset Formula $M$
\end_inset
, da velja
\begin_inset Formula $\left[u,v\right]=\langle u,Mv\rangle$
\end_inset
, kjer je
\begin_inset Formula $\langle\cdot,\cdot\rangle$
\end_inset
standardni skalarni produkt.
\end_layout
\begin_layout Itemize
Izpeljimo predpis za
\begin_inset Formula $A^{*}$
\end_inset
pri podani matriki
\begin_inset Formula $A$
\end_inset
in skalarnem produktu
\begin_inset Formula $\left[\cdot,\cdot\right]$
\end_inset
s pripadajočo matriko
\begin_inset Formula $M$
\end_inset
:
\begin_inset Formula
\[
\left[A^{*}x,y\right]=\left[x,Ay\right]\text{, uporabimo prvo točko:}
\]
\end_inset
\begin_inset Formula
\[
\left\langle A^{*}x,My\right\rangle =\left\langle x,MAy\right\rangle \text{, pišimo \ensuremath{z=My}:}
\]
\end_inset
\begin_inset Formula
\[
\left\langle A^{*}x,z\right\rangle =\left\langle x,MAM^{-1}z\right\rangle \text{, upoštevajmo drugo točko:}
\]
\end_inset
\begin_inset Formula
\[
\left\langle A^{*}x,z\right\rangle =\left\langle M^{-1}A^{\square}Mx,z\right\rangle \text{, kjer je \ensuremath{A^{\square}} adjungacija \ensuremath{A} pri standardnem skalarnem produktu}
\]
\end_inset
\begin_inset Formula
\[
\Rightarrow A^{*}=M^{-1}A^{\square}M=M^{-1}\overline{A}^{T}M\overset{A\in M\left(\mathbb{R}\right)}{=}M^{-1}A^{T}M
\]
\end_inset
\end_layout
\begin_layout Itemize
Potrebujemo še matriko skalarnega produkta.
\begin_inset Formula
\[
\left\langle \left(x,y,z\right),M\left(u,v,w\right)\right\rangle =\left[\begin{array}{ccc}
x & y & z\end{array}\right]\left[\begin{array}{ccc}
m_{11} & m_{12} & m_{13}\\
m_{21} & m_{22} & m_{23}\\
m_{31} & m_{32} & m_{33}
\end{array}\right]\left[\begin{array}{c}
u\\
v\\
w
\end{array}\right]=\left[\begin{array}{ccc}
x & y & z\end{array}\right]\left[\begin{array}{c}
um_{11}+vm_{12}+wm_{13}\\
um_{21}+vm_{22}+wm_{23}\\
um_{31}+vm_{32}+wm_{33}
\end{array}\right]=
\]
\end_inset
\begin_inset Formula
\[
=\begin{array}{ccccc}
& xum_{11} & xvm_{12} & xwm_{13} & +\\
+ & yum_{21} & yvm_{22} & ywm_{23} & +\\
+ & zum_{31} & zvm_{32} & zwm_{33}
\end{array}=\left[\left(x,y,z\right),\left(u,v,w\right)\right]=2xu-yu-xv+2yv-zv-yw+zw\text{, torej}
\]
\end_inset
\begin_inset Formula
\[
M=\left[\begin{array}{ccc}
2 & -1 & 0\\
-1 & 2 & -1\\
0 & -1 & 1
\end{array}\right]\text{, njen inverz pa je }M^{-1}=\left[\begin{array}{ccc}
1 & 1 & 1\\
1 & 2 & 2\\
1 & 2 & 3
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Itemize
Izračunamo
\begin_inset Formula $A^{*}$
\end_inset
po formuli
\begin_inset Formula $A^{*}=M^{-1}A^{T}M$
\end_inset
in preverimo
\begin_inset Formula $A^{*}A=AA^{*}$
\end_inset
:
\begin_inset Formula
\[
A^{*}=\left[\begin{array}{ccc}
1 & 1 & 1\\
1 & 2 & 2\\
1 & 2 & 3
\end{array}\right]\left[\begin{array}{ccc}
0 & 0 & -1\\
2 & 1 & 2\\
-2 & 0 & -1
\end{array}\right]\left[\begin{array}{ccc}
2 & -1 & 0\\
-1 & 2 & -1\\
0 & -1 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-1 & 2 & -1\\
-2 & 3 & -1\\
-6 & 6 & -2
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
A^{*}A=\left[\begin{array}{ccc}
1 & -2 & 3\\
1 & -3 & 5\\
2 & -10 & 14
\end{array}\right]\not=\left[\begin{array}{ccc}
8 & -6 & 2\\
-2 & 3 & -1\\
3 & -2 & 1
\end{array}\right]=AA^{*}
\]
\end_inset
\begin_inset Formula $A$
\end_inset
ni normala matrika.
\end_layout
\begin_layout Itemize
Da preverimo pravilnost matrike
\begin_inset Formula $A^{*}$
\end_inset
, lahko napravimo preizkus:
\begin_inset Float figure
placement H
wide false
sideways false
status open
\begin_layout Plain Layout
\begin_inset Graphics
filename sage.png
width 100col%
\end_inset
\begin_inset Caption Standard
\begin_layout Plain Layout
Preizkus s programom SageMath.
\end_layout
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
Dokazati, da
\begin_inset Formula $A$
\end_inset
ni normalna, je moč še lažje.
Dokažemo lahko namreč, da eden izmed potrebnih pogojev za normalnost matrike
ni izpolnjen.
Na primer:
\begin_inset Formula $AA^{*}=A^{*}A\rightarrow A=PDP^{-1}$
\end_inset
, kjer je
\begin_inset Formula $P$
\end_inset
ortogonalna in
\begin_inset Formula $D$
\end_inset
diagonalna
\begin_inset Formula $\Rightarrow$
\end_inset
lastni vektorji
\begin_inset Formula $A$
\end_inset
tvorijo ortogonalno množico.
\end_layout
\begin_layout Standard
Lastne vrednosti
\begin_inset Formula $A$
\end_inset
so (s kalkulatorjem)
\begin_inset Formula $\left\{ -2,1\right\} $
\end_inset
, kjer ima 1 algebrajsko večkratnost 2.
Lastni vektorji:
\begin_inset Formula
\[
A-\left(-2\right)I=\left[\begin{array}{ccc}
2 & 2 & -2\\
0 & 3 & 0\\
-1 & 2 & 1
\end{array}\right]\sim\left[\begin{array}{ccc}
2 & 2 & -2\\
0 & 3 & 0\\
0 & 3 & 0
\end{array}\right]\sim\left[\begin{array}{ccc}
2 & 2 & -2\\
0 & 3 & 0\\
0 & 0 & 0
\end{array}\right]\sim\left[\begin{array}{ccc}
2 & 0 & -2\\
0 & 3 & 0\\
0 & 0 & 0
\end{array}\right]\Rightarrow x=z,y=0\Rightarrow v_{1}=\left(1,0,1\right)
\]
\end_inset
\begin_inset Formula
\[
A-1I=\left[\begin{array}{ccc}
-1 & 2 & -2\\
0 & 0 & 0\\
-1 & 2 & -2
\end{array}\right]\sim\left[\begin{array}{ccc}
-1 & 2 & -2\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right]\Rightarrow x=2y-2z\Rightarrow v_{2}=\left(2,1,0\right),\quad v_{3}=\left(-2,0,1\right)
\]
\end_inset
\begin_inset Formula
\[
\left[v_{1},v_{2}\right]=\left[\left(1,0,1\right),\left(2,1,0\right)\right]=4-0-1+0-1-0+0=2\not=0\Rightarrow v_{1}\not\perp v_{2}\Rightarrow A\text{ ni normalna}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
Pokaži
\begin_inset Formula $A:V\to V$
\end_inset
je normalna
\begin_inset Formula $\Leftrightarrow AA^{*}-A^{*}A$
\end_inset
je pozitivno semidefinitna.
\end_layout
\begin_deeper
\begin_layout Paragraph
Rešitev
\end_layout
\begin_layout Itemize
Definiciji:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $A:V\to V$
\end_inset
je normalna
\begin_inset Formula $\Leftrightarrow A^{*}A=AA^{*}$
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $A:V\to V$
\end_inset
je pozitivno semidefinitna
\begin_inset Formula $\Leftrightarrow A=A^{*}\wedge\forall v\in V:\left\langle Av,v\right\rangle \geq0$
\end_inset
\end_layout
\end_deeper
\begin_layout Itemize
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Po predpostavki velja
\begin_inset Formula $AA^{*}=A^{*}A\Rightarrow AA^{*}-A^{*}A=0$
\end_inset
.
\begin_inset Formula
\[
AA^{*}-A^{*}A\overset{?}{=}\left(AA^{*}-A^{*}A\right)^{*}\Leftrightarrow0=0^{*}
\]
\end_inset
\begin_inset Formula
\[
\left\langle \left(AA^{*}-A^{*}A\right)v,v\right\rangle =\left\langle 0v,v\right\rangle \overset{\text{homogenost}}{=}0\left\langle v,v\right\rangle =0\geq0
\]
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
Po predpostavki velja
\begin_inset Formula $\left(AA^{*}-A^{*}A\right)^{*}=AA^{*}-A^{*}A$
\end_inset
in
\begin_inset Formula $\forall v\in V:\left\langle \left(AA^{*}-A^{*}A\right)v,v\right\rangle \geq0$
\end_inset
.
\begin_inset Formula
\[
\sled\left(AA^{*}-A^{*}A\right)=\sled\left(AA^{*}\right)-\sled\left(A^{*}A\right)\overset{\text{lastnost sledi}}{=}\sled\left(AA^{*}\right)-\sled\left(A^{*}A\right)=0
\]
\end_inset
Sled
\begin_inset Formula $M$
\end_inset
je vsota lastnih vrednosti
\begin_inset Formula $M$
\end_inset
, torej je vsota lastnih vrednosti
\begin_inset Formula $\left(AA^{*}-A^{*}A\right)=0$
\end_inset
.
\begin_inset Formula $AA^{*}-A^{*}A\geq0\Rightarrow$
\end_inset
vse lastne vrednosti so nenegativne.
Iz teh dveh trditev sledi, da je vsaka lastna vrednost
\begin_inset Formula $AA^{*}-A^{*}A=0$
\end_inset
.
\begin_inset Formula $AA^{*}-A^{*}A\geq0\Rightarrow AA^{*}-A^{*}A$
\end_inset
normalna.
Normalne matrike je moč diagonalizirati v ortonormirani bazi:
\begin_inset Formula
\[
AA^{*}-A^{*}A=PDP^{-1}\overset{\text{diagonalci so lastne vrednosti}}{=}P0P^{-1}=0
\]
\end_inset
\begin_inset Formula
\[
AA^{*}=A^{*}A\Rightarrow A\text{ je normalna}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
Naj bo
\begin_inset Formula $w_{1}=\left(1,1,1,1\right)$
\end_inset
,
\begin_inset Formula $w_{2}=\left(3,3,-1,-1\right)$
\end_inset
in
\begin_inset Formula $y=\left(6,0,2,0\right)$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Enumerate
Poišči ortonormirano bazo za
\begin_inset Formula $W=\Lin\left\{ w_{1},w_{2}\right\} $
\end_inset
glede na standardni skalarni produkt.
\end_layout
\begin_layout Enumerate
Izrazi
\begin_inset Formula $y$
\end_inset
kot vsoto vektorja iz
\begin_inset Formula $W$
\end_inset
in vektorja iz
\begin_inset Formula $W^{\perp}$
\end_inset
.
\end_layout
\begin_layout Paragraph
Rešitev
\end_layout
\begin_layout Enumerate
Uporabimo Gram-Schmidtov postopek in sproti normiramo bazne vektorje:
\begin_inset Formula
\[
v_{1}=\left(1,1,1,1\right)/2=\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)
\]
\end_inset
\begin_inset Formula
\[
\tilde{v_{2}}=\left(3,3,-1,-1\right)-\left\langle \left(3,3,-1,-1\right),v_{1}\right\rangle v_{1}=\left(3,3,-1,-1\right)-\left\langle \left(3,3,-1,-1\right),\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\right\rangle \left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)=
\]
\end_inset
\begin_inset Formula
\[
=\left(3,3,-1,-1\right)-2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)=\left(2,2,-2,-2\right),\quad\quad\quad v_{2}=\tilde{v_{2}}/\left|\left|\tilde{v_{2}}\right|\right|=\text{\ensuremath{\tilde{v_{2}}/4}}=\left(\frac{1}{2},\frac{1}{2},\frac{-1}{2},\frac{-1}{2}\right)
\]
\end_inset
Baza za
\begin_inset Formula $W$
\end_inset
je
\begin_inset Formula $B=\left\{ v_{1},v_{2}\right\} =\left\{ \left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2},\frac{-1}{2},\frac{-1}{2}\right)\right\} $
\end_inset
.
\end_layout
\begin_layout Enumerate
Dopolnimo
\begin_inset Formula $B$
\end_inset
do baze
\begin_inset Formula $\mathbb{R}^{4}$
\end_inset
.
Dopolnitev
\begin_inset Formula $\left\{ u_{1},u_{2}\right\} $
\end_inset
bo ortonormirana baza za
\begin_inset Formula $W^{\perp}$
\end_inset
, nato uporabimo Fourierov razvoj po dopolnjeni bazi.
Bazo podprostora dopolnimo tako, da rešimo sistem enačb.
\begin_inset Formula
\[
\left\langle \left(x_{1},y_{1},z_{1},w_{1}\right),\left(3,3,-1,-1\right)\right\rangle =0\quad\quad\quad\left\langle \left(x_{2},y_{2},z_{2},w_{2}\right),\left(1,1,1,1\right)\right\rangle =0
\]
\end_inset
\begin_inset Formula
\[
\left[\begin{array}{cccc}
1 & 1 & 1 & 1\\
3 & 3 & -1 & -1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & 1 & 0 & 0\\
0 & 0 & 1 & 1
\end{array}\right]\Rightarrow x=-y,\quad z=-w\Rightarrow\tilde{u_{1}}=\left(1,-1,0,0\right),\quad\tilde{u_{2}}=\left(0,0,1,-1\right)
\]
\end_inset
\begin_inset Formula
\[
u_{1}=\tilde{u_{1}}/\left|\left|\tilde{u_{1}}\right|\right|=\left(\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0,0\right)\quad\quad\quad u_{2}=\tilde{u_{2}}/\left|\left|\tilde{u}_{2}\right|\right|=\left(0,0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}}\right)
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
y=\sum_{i=1}^{n}\frac{\left\langle y,v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }v_{i}=\left\langle \left(6,0,2,0\right),\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\right\rangle v_{1}+\left\langle \left(6,0,2,0\right),\left(\frac{1}{2},\frac{1}{2},\frac{-1}{2},\frac{-1}{2}\right)\right\rangle v_{2}+
\]
\end_inset
\begin_inset Formula
\[
+\left\langle \left(6,0,2,0\right),\left(\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0,0\right)\right\rangle u_{1}+\left\langle \left(6,0,2,0\right),\left(0,0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},\right)\right\rangle u_{2}=4v_{1}+2v_{2}+\frac{6}{\sqrt{2}}u_{1}+\frac{2}{\sqrt{2}}u_{2}=
\]
\end_inset
\begin_inset Formula
\[
=\left(2,2,2,2\right)+\left(1,1,-1,-1\right)+\left(3,-3,0,0\right)+\left(0,0,1,-1\right)=\left(3,3,1,1\right)\in W+\left(3,-3,1,-1\right)\in W^{\perp}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Enumerate
Poišči singularni razcep matrike
\begin_inset Formula
\[
A=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & -2 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right]\text{.}
\]
\end_inset
\end_layout
\begin_deeper
\begin_layout Paragraph
Rešitev
\end_layout
\begin_layout Itemize
Iščemo
\begin_inset Formula $U$
\end_inset
,
\begin_inset Formula $\Sigma$
\end_inset
in
\begin_inset Formula $V$
\end_inset
, da velja
\begin_inset Formula $A=U\Sigma V^{*}$
\end_inset
.
\end_layout
\begin_layout Itemize
Diagonalci
\begin_inset Formula $\Sigma$
\end_inset
so singularne vrednosti
\begin_inset Formula $A$
\end_inset
.
Singularne vrednosti
\begin_inset Formula $A$
\end_inset
so koreni lastnih vrednosti
\begin_inset Formula $A^{*}A$
\end_inset
, torej
\begin_inset Formula $\sigma_{1}=2$
\end_inset
,
\begin_inset Formula $\sigma_{2}=1$
\end_inset
,
\begin_inset Formula $\sigma_{3}=0$
\end_inset
.
\begin_inset Formula
\[
A^{*}A=\left[\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & -2 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & -2 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 4 & 0\\
0 & 0 & 0
\end{array}\right]
\]
\end_inset
\begin_inset Formula
\[
\Sigma=\left[\begin{array}{ccc}
\sigma_{1} & 0 & 0\\
0 & \sigma_{2} & 0\\
0 & 0 & \sigma_{3}\\
0 & 0 & 0
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Itemize
Stolpci
\begin_inset Formula $V$
\end_inset
so ortonormirana baza jedra
\begin_inset Formula $A^{*}A-\sigma^{2}I$
\end_inset
za vse singularne vrednosti
\begin_inset Formula $\sigma$
\end_inset
.
\begin_inset Formula
\[
A^{*}A-4I=\left[\begin{array}{ccc}
-3 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & -4
\end{array}\right]\Rightarrow x=z=0\Rightarrow v_{1}=\left(0,1,0\right)
\]
\end_inset
\begin_inset Formula
\[
A^{*}A-1I=\left[\begin{array}{ccc}
0 & 0 & 0\\
0 & 3 & 0\\
0 & 0 & -1
\end{array}\right]\Rightarrow y=z=0\Rightarrow v_{2}=\left(1,0,0\right)
\]
\end_inset
\begin_inset Formula
\[
A^{*}A-0I=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 4 & 0\\
0 & 0 & 0
\end{array}\right]\Rightarrow x=y=0\Rightarrow v_{3}=\left(0,0,1\right)
\]
\end_inset
\begin_inset Formula
\[
V=\left[\begin{array}{ccc}
v_{1} & v_{2} & v_{3}\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Itemize
Stolpci
\begin_inset Formula $U$
\end_inset
so ortonormirana baza in velja
\begin_inset Formula $\forall i\in\left\{ 1..\rang A\right\} :u_{i}=\sigma_{i}^{-1}Av_{i}$
\end_inset
.
Stolpične vektorje
\begin_inset Formula $v_{\rang A+1},\dots,v_{m}$
\end_inset
najdemo tako, da dopolnimo
\begin_inset Formula $v_{1},\dots,v_{\rang A}$
\end_inset
do ONB.
\begin_inset Formula
\[
U=\left[\begin{array}{cccc}
0 & 1 & 0 & 0\\
-1 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0
\end{array}\right]
\]
\end_inset
\end_layout
\begin_layout Itemize
Dobljene matrike zmnožimo, s čimer potrdimo veljavnost singularnega razcepa:
\begin_inset Formula
\[
U\Sigma V^{*}=\left[\begin{array}{cccc}
0 & 1 & 0 & 0\\
-1 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right]\left[\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & -2 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right]=A
\]
\end_inset
\end_layout
\end_deeper
\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/1ladn8a.jpg
\end_inset
\begin_inset External
template PDFPages
filename /mnt/slu/shramba/upload/www/d/1ladn8aq.jpg
\end_inset
\begin_inset External
template PDFPages
filename /mnt/slu/shramba/upload/www/d/1ladn8b.jpg
\end_inset
\begin_inset External
template PDFPages
filename /mnt/slu/shramba/upload/www/d/1ladn8c.jpg
\end_inset
\begin_inset External
template PDFPages
filename /mnt/slu/shramba/upload/www/d/3ladn8.jpg
\end_inset
\end_layout
\begin_layout Standard
\begin_inset External
template PDFPages
filename /mnt/slu/shramba/upload/www/d/4ladn8.jpg
\end_inset
\end_layout
\end_body
\end_document
