From e0993405941a9f487141336ebc09eb4328555e40 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Anton=20Luka=20=C5=A0ijanec?= <anton@sijanec.eu>
Date: Mon, 20 May 2024 13:08:04 +0200
Subject: dn8

---
 "\305\241ola/la/dn8/dokument.lyx" | 639 +++++++++++++++++++++++++++++++++++++-
 "\305\241ola/la/dn8/sage.png"     | Bin 0 -> 158070 bytes
 2 files changed, 630 insertions(+), 9 deletions(-)
 create mode 100644 "\305\241ola/la/dn8/sage.png"

diff --git "a/\305\241ola/la/dn8/dokument.lyx" "b/\305\241ola/la/dn8/dokument.lyx"
index 3032fdd..c01e7bc 100644
--- "a/\305\241ola/la/dn8/dokument.lyx"
+++ "b/\305\241ola/la/dn8/dokument.lyx"
@@ -53,7 +53,7 @@ theorems-ams
 \spacing single
 \use_hyperref false
 \papersize default
-\use_geometry false
+\use_geometry true
 \use_package amsmath 1
 \use_package amssymb 1
 \use_package cancel 1
@@ -78,12 +78,12 @@ theorems-ams
 \shortcut idx
 \color #008000
 \end_index
-\leftmargin 1cm
-\topmargin 0cm
-\rightmargin 1cm
+\leftmargin 2cm
+\topmargin 2cm
+\rightmargin 2cm
 \bottommargin 2cm
-\headheight 1cm
-\headsep 1cm
+\headheight 2cm
+\headsep 2cm
 \footskip 1cm
 \secnumdepth 3
 \tocdepth 3
@@ -166,7 +166,7 @@ Dokaži, da je
 A=\left[\begin{array}{ccc}
 0 & 2 & -2\\
 0 & 1 & 0\\
-1 & 2 & -1
+-1 & 2 & -1
 \end{array}\right]
 \]
 
@@ -193,7 +193,12 @@ Predpostavljam polje
 \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
@@ -211,6 +216,87 @@ Predpostavljam polje
 \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
@@ -218,6 +304,30 @@ Predpostavljam polje
 \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
@@ -225,6 +335,313 @@ Predpostavljam polje
 \end_inset
 
 
+\begin_inset Formula 
+\[
+\left[\alpha\left(\left(x_{1},y_{1},z_{1}\right)+\left(x_{2},y_{2},z_{2}\right)\right),\left(u,v,w\right)\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=2\alpha\left(x_{1}+x_{2}\right)u-\alpha\left(y_{1}+y_{2}\right)u-\alpha\left(x_{1}+x_{2}\right)v+2\alpha\left(y_{1}+y_{2}\right)v-\alpha\left(z_{1}+z_{2}\right)v-\alpha\left(y_{1}+y_{2}\right)w+\alpha\left(z_{1}+z_{2}\right)w=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\alpha\left(2\left(x_{1}+x_{2}\right)u-\left(y_{1}+y_{2}\right)u-\left(x_{1}+x_{2}\right)v+2\left(y_{1}+y_{2}\right)v-\left(z_{1}+z_{2}\right)v-\left(y_{1}+y_{2}\right)w+\left(z_{1}+z_{2}\right)w\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\alpha\left(2x_{1}u+2x_{2}u-y_{1}u-y_{2}u-x_{1}v-x_{2}v+2y_{1}v+2y_{2}v-z_{1}v-z_{2}v-y_{1}w-y_{2}w+z_{1}w+z_{2}w\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\alpha\left(2x_{1}u-y_{1}u-x_{1}v+2y_{1}v-z_{1}v-y_{1}w+z_{1}w\right)+\alpha\left(2x_{2}u-y_{2}u-x_{2}v+2y_{2}v-z_{2}v-y_{2}w+z_{2}w\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\alpha\left[\left(x_{1},y_{1},z_{1}\right),\left(u,v,w\right)\right]+\alpha\left[\left(x_{2},y_{2},z_{2}\right),\left(u,v,w\right)\right]
+\]
+
+\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 pozitivno definitna matrika 
+\begin_inset Formula $M$
+\end_inset
+
+, da velja 
+\begin_inset Formula $\left[u,v\right]=\langle u,Mv\rangle=u^{*}v$
+\end_inset
+
+, kjer je 
+\begin_inset Formula $\langle\cdot,\cdot\rangle$
+\end_inset
+
+ standardni skalarni produkt.
+\end_layout
+
+\begin_layout Itemize
+Na predavanjih 2024-04-17 smo dokazali, da 
+\begin_inset Formula $\left[L^{*}\right]_{C\leftarrow B}=\left(\left[L\right]_{B\leftarrow C}\right)^{*}$
+\end_inset
+
+, torej 
+\begin_inset Formula $PLP^{-1}=\left(P^{-1}L^{*}P\right)^{*}$
+\end_inset
+
+.
+\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
 
 \end_deeper
@@ -240,6 +657,78 @@ Pokaži
  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 $\Leftrightarrow A^{*}A=A^{*}$
+\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
+
+
+\end_layout
+
+\end_deeper
 \begin_layout Enumerate
 Naj bo 
 \begin_inset Formula $w_{1}=\left(1,1,1,1\right)$
@@ -250,12 +739,106 @@ Naj bo
 \end_inset
 
  in 
-\begin_inset Formula $w_{3}=\left(6,0,2,0\right)$
+\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
+Uporabimo fourierov razvoj 
+\begin_inset Formula $y$
+\end_inset
+
+ po 
+\begin_inset Formula $B$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+y=\sum_{i=1}^{n}\frac{\left\langle y,v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }v_{i}=\left\langle \left(-6,0,2,0\right),\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\right\rangle v_{i}+\left\langle \left(-6,0,2,0\right),\left(\frac{1}{2},\frac{1}{2},\frac{-1}{2},\frac{-1}{2}\right)\right\rangle \left(\frac{1}{2},\frac{1}{2},\frac{-1}{2},\frac{-1}{2}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)+2\left(\frac{1}{2},\frac{1}{2},\frac{-1}{2},\frac{-1}{2}\right)=\left(2,2,2,2\right)+\left(1,1,-1,-1\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
 \begin_layout Enumerate
 Poišči singularni razcep matrike
 \begin_inset Formula 
@@ -509,5 +1092,43 @@ 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/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
+
+
+\begin_inset External
+	template PDFPages
+	filename /mnt/slu/shramba/upload/www/d/4ladn8.jpg
+
+\end_inset
+
+
+\end_layout
+
 \end_body
 \end_document
diff --git "a/\305\241ola/la/dn8/sage.png" "b/\305\241ola/la/dn8/sage.png"
new file mode 100644
index 0000000..1cd41f5
Binary files /dev/null and "b/\305\241ola/la/dn8/sage.png" differ
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