### A Pluto.jl notebook ### # v0.20.21 using Markdown using InteractiveUtils # ╔═╡ fb99b46e-a0f4-499a-ab30-5f4ac1757689 begin using Printf using PlutoTeachingTools using PlutoUI using LinearAlgebra using CairoMakie ultramarine40 = RGBf(100/255.0, 143/255.0, 255/255.0) indigo50 = RGBf(120/255.0, 94/255.0, 240/255.0) magenta50 = RGBf(220/255.0, 38/255.0, 127/255.0) orange40 = RGBf(254/255.0, 97/255.0, 0/255.0) gold20 = RGBf(255/255.0, 176/255.0, 0/255.0) TableOfContents() end # ╔═╡ fac42b42-35d2-11f1-a407-bdbaa61d4b67 html"""

Höhere Mathematik II

Michael Schlottke-Lakemper, Manuel Torrilhon

Universität Augsburg

""" # ╔═╡ 3b4ec4eb-af39-4f41-90cc-663d1b2eccab md""" # 4 Lineare Algebra II ## 4.1 Determinanten Das lineare Gleichungssystem $A\,x=b$ mit $A\in\mathbb{R}^{n\times n}$ ist eindeutig lösbar, falls $\mathrm{rg}(A)=n$, d.h. alle Spalten von $A$ linear unabhängig sind. Zum Testen von $A$ hätten wir gerne eine Funktion $\psi:\mathbb{R}^{n\times n}\to\mathbb{R},\,A\mapsto\psi(A) = \begin{cases} 0 &\text{Spalten von A lin. abh.}\\\neq 0 &\text{Spalten von A lin. unabh.}\end{cases}$ """ # ╔═╡ c5d2f769-f27b-40e4-9602-8b962d5d7934 md""" ##### Definition: Für eine Permutation $\pi\in S_n$ ($S_n$ ist die symmetrische Gruppe, die Menge aller bijektiven Abbildungen auf $\{1,\ldots,n\}$ mit der Verknüpfung von Funktionen als Gruppenverknüpfung) definieren wir $\mathrm{sign}: S_n \to \{\pm 1\}, \pi\mapsto\mathrm{sign}(\pi) = \begin{cases} +1 & \text{falls }\pi\text{ gerade} \\[2pt] -1 & \text{falls }\pi\text{ ungerade.}\end{cases}$ Un/gerade heißt un/gerade Anzahl von paarweisen Vertauschungen von $\{1,\ldots,n\}$. ##### Beispiel $\pi = (2, 3, 1 ,4) \in S_4$ | $i$ | 1 | 2 | 3 | 4 | |:------:|:-:|:-:|:-:|:-:| |$\pi(i)$| 2 | 3 | 1 | 4 | | (1) | 2 | 1 | 3 | 4 | | (2) | 1 | 2 | 3 | 4 | $\Rightarrow 2 \text{ Vertauschungen} \Rightarrow \pi \text{ gerade} \Rightarrow \mathrm{sign}(\pi) = +1.$ ##### Definition Für $A = (a_{ij}) \in \mathrm{R}^{n\times n}$ definieren wir die *Determinante* von $A$ durch $\det(A) := \sum_{\pi\in S_n} \mathrm{sign}(\pi)\cdot a_{1\pi(1)}\cdot a_{2\pi(2)}\cdot\,\cdots\,\cdot a_{n\pi(n)}$ Es gilt dann + $\det(A) = 0 \Leftrightarrow A$ ist singulär $\Leftrightarrow A\,x=b$ nicht (eindeutig) lösbar + $\det(A) \neq 0 \Leftrightarrow A$ ist regulär $\Leftrightarrow A\,x=b$ eindeutig lösbar ##### Beispiel: 1. $\det\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} = \underset{\pi=(1,2)\text{ gerade}}{+a_{11}\,a_{22}} - \underset{\pi=(2,1)\text{ ungerade}}{a_{12}\,a_{21}}$ 2. $\det\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix} = a_{11}\,a_{22}\,a_{33}-a_{11}\,a_{23}\,a_{32}-a_{12}\,a_{21}\,a_{33}$\ $\phantom{\det\begin{bmatrix}a_{00}&a_{00}&a_{00}\end{bmatrix}=}\!+a_{12}\,a_{23}\,a_{31}+a_{13}\,a_{21}\,a_{32}-a_{13}\,a_{22}\,a_{31}$ ##### Satz (Laplace'sche Entwicklung für det): Sei $A\in\mathbb{R}^{n\times n}$, dann gilt $\det(A) = \sum\limits_{j=1}^n(-1)^{i+j}\,a_{ij}\,\det(\hat{A}_{ij})$, wobei $\hat{A}_{ij}\in\mathbb{R}^{(n-1)\times(n-1)}$ die Matrix ist, die durch Streichung der $i$-ten Zeile und $j$-ten Spalte aus $A$ entsteht ($i$ beliebig, "Entwicklung nach der $i$-ten Zeile"). """ # ╔═╡ def3641a-20ba-4961-b509-442da4ab9710 md""" ##### Beispiel: 1. $\det\begin{bmatrix}1&2&3\\-1&0&1\\1&2&1\end{bmatrix}\overset{i=1}{=} (-1)^{1+1}\cdot 1\cdot\det(\hat{A}_{11})+(-1)^{1+2}\cdot2\cdot\det(\hat{A}_{12})$\ $\phantom{\det\begin{bmatrix}0&-0&0\end{bmatrix}\overset{i=1}{=}}\! +(-1)^{1+3}\cdot3\cdot\det(\hat{A}_{13})$\ $\phantom{\det\begin{bmatrix}0&-0&0\end{bmatrix}\overset{i=1}{=}}\!\!\! = 1\cdot\det\begin{bmatrix}0&1\\2&1\end{bmatrix}-2\cdot\det\begin{bmatrix}-1&1\\1&1\end{bmatrix}+3\cdot\det\begin{bmatrix}-1&0\\1&2\end{bmatrix}$\ $\phantom{\det\begin{bmatrix}0&-0&0\end{bmatrix}\overset{i=1}{=}}\!\!\! =1\cdot(0\cdot1-2\cdot1)-2(-1\cdot1-1\cdot1)+3\cdot(-1\cdot2-1\cdot0)=-4.$ 2. $\det\begin{bmatrix}1&2&3&4\\1&0&1&0\\1&-1&1&-1\\4&3&2&1\end{bmatrix} \overset{i=2}{=} (-1)^{2+1}\cdot1\cdot\det(\hat{A}_{21})+(-1)^{2+2}\cdot0\cdot\det(\hat{A}_{22})$\ $\phantom{\det\begin{bmatrix}1&-1&1&-1\end{bmatrix}\overset{i=2}{=}}\!+(-1)^{2+3}\cdot1\cdot\det(\hat{A}_{23})+(-1)^{2+4}\cdot0\cdot\det(\hat{A}_{24})$\ $\phantom{\det\begin{bmatrix}1&-1&1&-1\end{bmatrix}\overset{i=2}{=}}\!\!\!=-\det\begin{bmatrix}2&3&4\\-1&1&-1\\3&2&1\end{bmatrix}-\det\begin{bmatrix}1&2&4\\1&-1&-1\\4&3&1\end{bmatrix}=\cdots$ """ # ╔═╡ a193fb98-232f-491c-b192-539fc8ab6f5e Foldable("Exkurs (Determinaten als alternierende Linearformen)", md"""##### Definition: Eine Abbildung $\psi:\underbrace{V\times V\times \dots\times V}_{\text{n- mal}}\to W,\,(V_1,V_2,\dots,V_n)\mapsto\psi(V_1,V_2,\dots,V_n)$ mit Vektorräumen $V$ und $W$ heißt _$n$-fach linear_, falls $\psi$ in jedem Argument eine lineare Abbildung ist. Falls $W=\mathbb{R}$, so heißt $\psi$ eine $n$-fache Linearform (Für $n=2$ "Bilinearform") Wir betrachten den Spezialfall $W=\mathbb{R}$ und $\dim(V) = n$. ##### Definition: Eine $n$-fache Linearform $\psi$ heißt alternierend, falls $\psi(\dots,V,\dots,W,\dots)= -\psi(\dots,W,\dots,V,\dots)$ gilt. D.h. Vorzeichenwechsel bei Vertauschung zweier Argumente. ##### Satz: Für eine $n$-fache alternierende Linearform $\psi$ gilt: 1. $\psi(\dots,V,\dots,V,\dots)=0\quad$ (zwei gleiche Argumente) 2. Mit Permutation $\pi\in S_n$ gilt $\psi(V_1,\dots,V_n)=\mathrm{sign}(\pi)\psi(V_{\pi(1)},V_{\pi(2)},\dots,V_{\pi(n)})$ 2. $\{V_1,V_2,\dots,V_n\}$ linear abhängig $\Rightarrow\psi(V_1,V_2,\dots,V_n)=0.$ 3. $\{V_1,V_2,\dots,V_n\}$ linear unabhängig $\Rightarrow\psi(V_1,V_2,\dots,V_n)\neq0.$ ##### Beweis: zu 1: $\psi(\dots,V,\dots,V,\dots) = -\psi(\dots,V,\dots,V,\dots) \Rightarrow \psi=0$ zu 2: Symmetrische Gruppe $S_n = \{$bijektive Abb. auf $\{1,\dots,n\}\}$ Definiere $\mathrm{sign}: S_n\to\{\pm1\},\,\pi\mapsto\mathrm{sign}(\pi)=\begin{cases}+1& \pi\text{gerade,}\\-1&\pi\text{ungerade.}\end{cases}$ "Ungerade" heißt ungerade Anzahl von paarweisen Vertauschungen von $\{1,2,\dots,n\}$ $|S_n|=n! = 1,2,3,\dots,n$ zu 3: Annahme: $v_1 = \sum\limits_{k=2}^n\alpha_k\,v_k$ ($v_1$ wird durch $v_2,\dots,v_n$ dargstelllt) $\psi\left(\sum\limits_{k=2}^n\alpha_k\,v_k,v_2,\dots,v_n\right) = \sum\limits_{k=2}^n\alpha_k\psi(v_k,v_2,\dots,v_n)=0$ (wegen 1.) zu 4: Wir wählen irgendwelche $c_1,\dots,c_n\in V$ mit $\psi(c_1,\dots,c_n)\neq 0. \{v_1,\dots,v_n\}$ linear unabhängig, also Basis. $c_i=\sum\limits_{j=1}^n\alpha_{ij}v_j$ Basisdarstellung, einsetzen in $\psi$ $\begin{align*}0&\neq\psi(c_1,\dots,c_n) = \psi\left(\sum\limits_{j=1}^n\alpha_{1j}v_k,\dots,\sum\limits_{j=1}^n\alpha_{nj}v_j\right)\\&= \sum\limits_{j_1=1}^n\alpha_{1j_1}\sum\limits_{j_2=1}^n\alpha_{2j_2} \cdots \sum\limits_{j_n=1}^n\alpha_{nj_n}\psi(v_{j_1},v_{j_2},\dots,v_{j_n})\\&= \psi(v_1,v_2,\dots,v_n)\left(\sum\limits_{\pi\in S_n}\underbrace{\mathrm{sign}(\pi)\alpha_{1\pi(1)}\alpha_{2\pi(2)}\cdots\alpha_{n\pi(n)}}_{n! \text{ Summanden}\neq0}\right)\\&\Rightarrow \psi(v_1,v_2,\dots,v_n)\neq 0\end{align*}$ ##### Bemerkung: Nach Wahl einer Basis $\{v_1,v_2,\dots,v_n\}$ ist $\psi$ einzig durch die Wahl des Wertes $\psi(v_1,\dots,v_n)\in\mathbb{R}$ _festgelegt_. Andere Auswertungen folgen durch Basisdarstellung und obige Formel. ##### Definition: Sei $\{b_1,\dots,b_n\}$ eine Basis von $V$ und $\psi$ eine $n$-fache alternierende Linearform. Für eine lineare Abbildung $\varphi:V\to V$ ist durch $\det(\varphi)=\frac{\psi(\varphi(b_1),\dots,\varphi(b_n))}{\psi(b_1,\dots,b_n)}$ die _Determinante_ von $\varphi$ definiert. ##### Bemerkung: 1. Es lässt sich zeigen, dass die Determinante unabhängig von der Basis-Wahl ist(!) 2. Es gilt wie gewünscht: $\det(\varphi)\neq 0$, falls $\varphi$ invertierbar, d.h. $\{\varphi(b_1),\dots,\varphi(b_n)\}$ lin. unabhängig, $\det(\varphi)=0$, falls $\varphi$ nicht invertierbar. #### Berechnung der Determinante Wir wählen die kanonische Basis $\{e_1,\dots,e_n\}$ und die lineare Abbildung $\varphi$ ist gegeben durch $\varphi(e_j)=\sum\limits_{i=1}^na_{ij}e_i$ mit der Matrix $A=(a_{ij})$ $\begin{align*}\det(A)&=\frac{\psi(\varphi(e_1),\dots,\varphi(e_n))}{\psi(e_1,\dots,e_n)}=\frac{\psi(e_1,\dots,e_n)\left(\sum\limits_{\pi\in S_n}\mathrm{sign}(\pi)a_{1,\pi(1)}a_{2,\pi(2)}\cdots a_{n,\pi(n)}\right)}{\psi(e_1,\dots,e_n)}\\&=\sum\limits_{\pi\in S_n}\mathrm{sign}(\pi)a_{1,\pi(1)}a_{2,\pi(2)}\cdots a_{n,\pi(n)}\end{align*}$ Es gilt $\det(A)=0\Leftrightarrow$ $A$ ist singulär $\Leftrightarrow A\,x=b$ ist _nicht_ eindeutig lösbar oder gar nicht. """) # ╔═╡ 4843a1ee-4425-4fc4-8168-066f15874621 md""" ##### Beweis (Ansatz): Für $i=1$ vergleiche mit $\det(A)=\sum\limits_{\pi\in S_n}\mathrm{sign}(\pi)a_{1,\pi(1)}\cdots a_{n,\pi(n)}$. Das lässt sich schreiben für festes $i\in\{1,\dots,n\}: \det(A)=a_{i1}\,d_{i1}+a_{i2}\,d_{i2}+\dots+a_{in}\,d_{in}$, wobei $d_{ij}$ Summen von Produkten mit $n-1$ Faktoren _ohne_ Element $a_{ij}$ sind. Zu zeigen ist: $d_{ij} = (-1)^{i+j}\det(\tilde{A}_{ij}).$ ##### Bemerkung: 1. Analog funktioniert die "Entwicklung nach der $j$-ten _Spalte_": $\det(A) = \sum\limits_{i=1}^n(-1)^{i+j}\,a_{ij}\,\det(\hat{A}_{ij}),\quad j$ beliebig. 2. Anzahl Multiplikationen wie in alter Formel, aber Nullen können besser ausgenutzt werden. ##### Beispiel: $\begin{align*}\det\begin{bmatrix}5&2&-2&1\\3&0&1&4\\0&0&0&2\\1&0&3&-4\end{bmatrix}&\overset{i=3}{=}(-1)^{3+4}\cdot2\cdot\det\begin{bmatrix}5&2&-2\\3&0&1\\1&0&3\end{bmatrix}\\&\overset{j=2}{=}-2\cdot(-1)^{2+1}\cdot2\cdot\det\begin{bmatrix}3&1\\1&3\end{bmatrix}=4\cdot(9-1)=32\end{align*}$ """ # ╔═╡ cb7df674-832a-4143-a8bf-63f8143cc875 md""" ### 4.1.1 Eigenschaften der Determinante ##### Satz: Für $A,B\in\mathbb{R}^{n\times n}$ gilt 1. $\det(\lambda A)=\lambda^n\det(A)\quad \lambda\in\mathbb{R}$ 2. $\det(A) = \det(A^T)$ 3. $A=\begin{bmatrix}\lambda_1&0&\cdots&0\\0&\lambda_2&&\vdots\\\vdots&&\ddots&\\0&\cdots&&\lambda_n\end{bmatrix} = \mathrm{diag}(\lambda_1,\dots,\lambda_n)$ Diagonalmatrix: $\det(A)=\lambda_1\cdot\,\cdots\,\cdot\lambda_n$ 3. $\det(A\cdot B) = \det(A)\cdot \det(B) = \det(B\cdot A)$ 4. $\det(A^{-1})=\frac{1}{\det(A)}\quad (A$ invertierbar, d.h. $\det(A)\neq0$) """ # ╔═╡ c3698a4a-ed13-4e71-a249-e729b3cdb19d md""" ##### Beweis: Für 1.-3. betrachte $\det(A)=\sum\limits_{\pi\in S_n}\mathrm{sign}(\pi)\,a_{1,\pi(1)}\,\dots\,a_{n,\pi(n)}.$ zu 1: Jedes Element $a_{ij}$ wird mit $\lambda$ multipliziert $\Rightarrow \det(\lambda A)=\sum\limits_{\pi\in S_n}\mathrm{sign}(\pi)\,(\lambda a_{1,\pi(1)})\,(\lambda a_{2,\pi(2)})\,\dots = \lambda^n\det(A).$ zu 2: $\begin{align*}\det(A)&=\sum\limits_{\pi\in S_n}\mathrm{sign}(\pi)\,a_{1,\pi(1)}\,\dots\,a_{n,\pi(n)} = \sum\limits_{\pi\in S_n}\mathrm{sign}(\pi)\,a_{\pi^{-1}(1),1}a_{\pi^{-1}(2),2}\,\dots \\&= \sum\limits_{\pi^{-1}\in S_n}\mathrm{sign}(\pi^{-1})\,a_{\pi^{-1}(1),1}\,\dots\,a_{\pi^{-1}(n),n} = \det(A^T)\end{align*}$ zu 3: Nur die Permutation $a_{11}\,a_{22}\,\dots\,a_{nn}$ ist ungleich Null. zu 4: Seien $\varphi_A,\varphi_B$ die lineare Abbildung zu $A,B.$ ($A,B$ regulär, also $\det\neq0$). $\begin{align*}\det(A\,B) &= \det(\varphi_A\circ\varphi_B) = \frac{\psi(\varphi_A(\varphi_B(e_1)),\varphi_A(\varphi_B(e_2)),\dots,\varphi_A(\varphi_B(e_n)))}{\psi(e_1\dots e_n)}\\&=\frac{\psi(\varphi_A(\varphi_B(e_1)),\dots\varphi_A(\varphi_B(e_n)))}{\psi(\varphi_B(e_1),\dots,\varphi_B(e_n))}\frac{\psi(\varphi_B(e_1),\dots,\varphi_B(e_n))}{\psi(e_1\dots e_n)}\\&=\det(\varphi_A)\cdot\det(\varphi_B)=\det(A)\cdot\varphi(B)\end{align*}$ Für $A$ oder $B$ singulär, ist $A\cdot B$ auch singulär ($\det(A\,B)=0$). zu 5: $1=\det(I)=\det(A\cdot A^{-1})=\det(A)\det(A^{-1})\Rightarrow \det(A^{-1})=\frac{1}{\det(A)}$. """ # ╔═╡ 1058cc8a-3d7c-4c68-ba03-c408c608bcae md""" ##### Bemerkung: 1. Für $A$ Diagonalmatrix gilt $\det(A)=\prod\limits_{i=1}^n a_{ii}$, für $A$ obere/untere Dreiecksmatrix auch. 2. Die erlaubten Operationen im Gauß-Algorithmus haben folgende Effekte auf die Determinante: + Skalieren einer Zeile mit Faktor $\lambda$ skaliert Determinante um Faktor $\lambda$ + Vertauschen zweier Zeilen ändert das Vorzeichen der Determinante + Addieren eines beliebigen Vielfaches einer Zeiler zu einer anderen lässt die Determinante unverändert 3. Für $\det(A+B)$ gibt es keine Vereinfachung ($\neq \det(A)+\det(B)!!!$) """ # ╔═╡ e2891539-b1c0-495e-a2ee-7eebcccd9ff0 md""" ## 4.2 Eigenwerte und Eigenvektoren ##### Motivation: Eine lineare Abbildung $\varphi:V\to V$ wirkt auf die meisten Vektoren "kompliziert". Ideal wäre eine Basis, in der die zu $\varphi$ gehörende Matrix in Diagonalform ist. Die Leitfrage hier ist: gibt es Vektoren $v\in V$, die von $\varphi$ nur gestreckt oder gestaucht werden, d.h. für die $\varphi(v) = \lambda\,v$ ($\lambda\in\mathbb{R}$) gilt? """ # ╔═╡ c41cf772-d812-4268-8d6d-6da462cfab90 md""" ##### Beispiel: Gegeben eine lineare Abbildung $\varphi$ bzgl. der kanonischen Basis $\{e_1,\dots,e_n\}$ Beispielsweise für $n=2$, mit $\varphi(e_1)=\begin{bmatrix}1\\2\end{bmatrix},\, \varphi(e_2)=\begin{bmatrix}2\\-2\end{bmatrix},\, $also $A=\begin{bmatrix}1&\phantom{-}2\\2&-2\end{bmatrix}$ Wir stellen fest: $\varphi(e_1-2e_2)\mathrel{\widehat{=}} A\begin{bmatrix}\phantom{-}1\\-2\end{bmatrix} = \begin{bmatrix}-3\\\phantom{-}6\end{bmatrix}=-3\begin{bmatrix}\phantom{-}1\\-2\end{bmatrix}\mathrel{\widehat{=}}-3(e_1-2e_2)$ $\phantom{Wir stellen fe:\!}\varphi(2e_1+e_2)\mathrel{\widehat{=}}A\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}4\\2\end{bmatrix}=2\begin{bmatrix}2\\1\end{bmatrix}\mathrel{\widehat{=}}2(e_1+e_2)$ Wir wählen als neue Basis $\{v_1,v_2\}=\{e_1-2e_2,2e_1+e_2\}$ Offenbar ist $\varphi(v_1)=-3v_1,\,\varphi(v_2)=2v_2$, also $A = \begin{bmatrix}-3&0\\\phantom{-}0&2\end{bmatrix}$ bzgl. $\{v_1,v_2\}$. Genauer: Koordinatentransformation bei Basiswechsel! Seien $\{v_1,v_2,\dots,v_n\}$ und $\{w_1,w_2,\dots,w_n\}$ 2 Basen von $V$. Für $u\in V$ gilt $u=\sum\limits_{i=1}^nv_i\,\alpha_i=\sum\limits_{i=1}^nw_i\,\beta_i$. Wie hängen $\alpha_i,\beta_i$ zusammen? Dafür zunächst: wie hängen $\{v_i\}$ und $\{w_i\}$ zusammen? Wir haben $\begin{align*} v_1 &= w_1\,t_{11}+w_2\,t_{21}+\dots + w_n\,t_{n1}\\v_2&=w_1\,t_{12}+w_2\,t_{22}+\dots+w_n\,t_{n2}\\&\;\;\vdots\end{align*}$ also für jeden Basisvektor die Darstellung $v_i=\sum\limits_{j=1}^nw_j\,t_{ji},$ mit $T=(t_{ij})\in\mathbb{R}^{n\times n}$ regulär (invertierbar). Einsetzen: $u=\sum\limits_{i=1}^nv_i\,\alpha_i=\underbrace{\sum\limits_{i=1}^n\sum\limits_{j=1}^nw_j\,t_{ji}\,\alpha_i}_{=\sum\limits_{j=1}^nw_j\,\left(\sum\limits_{i=1}^nt_{ji}\,\alpha_i\right)} \overset{!}{=}\sum\limits_{j=1}^nw_j\,\beta_j\Rightarrow \sum\limits_{j=1}^nw_j\,\left(\sum\limits_{i=1}^nt_{ji}\,\alpha_i-\beta_j\right)=0$. Also $\beta_j=\sum\limits_{i=1}^nt_{ji}\,\alpha_i$, bzw. $\tilde x = T\,x$ wobei $\tilde x$ der Koordinatenvektor bzgl. $\{w_i\}$ und $x$ der Koordinatenvektor bzgl. $\{v_i\}$ ist. Jetzt: $\varphi:V\to V$ lineare Abbildung.\ $y=\varphi(x)$ in Basis $\{v_1,\dots,v_n\}\;\,:\, y=A\,x,\quad x,y$ Koordinaten bzgl. $\{v_i\}$, $\kern39pt$ in Basis $\{w_1,\dots,w_n\}:\tilde{y} = \tilde{A}\,\tilde{x},\quad \tilde{x},\tilde{y}$ Koordinaten bzgl. $\{w_i\}$. $\Rightarrow \tilde{y}=T\,y=T\,A\,x=\underbrace{T\,A\,T^{-1}}_{\tilde{A}}\overbrace{T\,x}^{\tilde{x}}=T\,A\,T^{-1}\tilde{x}=\tilde{A}\,\tilde{x}.$ Damit erhält man mit $\tilde A = T\,A\,T^{-1}$ die Koordinatendarstellung von $\varphi$ bzgl. $\{w_i\}$. """ # ╔═╡ b08e67ac-2759-4041-8367-ad72b989c4fb md""" ##### Definition: Zwei Matrizen $A,B\in\mathbb{R}^{n\times n}$ heißen _ähnlich_, falls ein reguläres $T\in\mathbb{R}^{n\times n}$ existiert, sodass $B=T\,A\,T^{-1}$. ##### Bemerkung: 1. Ähnlichkeit erzeugt Äquivalenzklassen auf $\mathbb{R}^{n\times n}$, d.h. der Menge der Matrizen. 2. Ähnliche $A$ und $B$ haben bestimmte Dinge gemeinsam, sogenannte "Invarianten" (bzgl. Basiswechsel). Zum Beispiel: $\det(B) = \det(T\,A\,T^{-1}) = \det(T)\,\det(A)\,\det(T^{-1})= \det(A).$ 3. Je nach Definition des Basiswechsels gilt auch $S^{-1}\,A\,S$ mit $S=T^{-1}$. 4. Ideal wäre eine einfache Form von $B$, z.B. $D=\begin{bmatrix}\lambda_1&0&\cdots&0\\0&\lambda_2&&\vdots\\\vdots&&\ddots&\\0&\cdots&&\lambda_n\end{bmatrix}=\mathrm{diag}(\lambda_1,\dots,\lambda_n),$ d.h. $S^{-1}\,A\,S=D$, bzw $A\,S=S\,D$. Sei $S=\underbrace{\begin{bmatrix}|&|&&|\\s_1&s_2&\cdots&s_n\\|&|&&|\end{bmatrix}}_{\text{Spalten}}$, dann $SD = \begin{bmatrix}|&|&&|\\\lambda_1\,s_1&\lambda_2\,s_2&\cdots&\lambda_n\,s_n\\|&|&&|\end{bmatrix},\quad AS= \begin{bmatrix}|&|&&|\\A\,s_1&A\,s_2&\cdots& A\,s_n\\|&|&&|\end{bmatrix},$ also $A\,s_i\overset{!}{=}\lambda_i\,s_i,\quad i=1,\dots,n$. Diagonalisierung ist aber nicht immer möglich. """ # ╔═╡ e3f03b99-68e3-45f7-b3e0-97f45e1da6b9 md""" ##### Definition: Sei $A\in \mathbb{R}^{n\times n}$. Falls für ein $v\in\mathbb{R}^n\setminus\{0\}$ und $\lambda\in\mathbb{R}$ gilt $A\,v=\lambda\,v$, so heißt $\lambda$ _Eigenwert_ und $v$ _Eigenvektor_ von $A$. Die Menge der Eigenvektoren zum Eigenwert $\lambda$ heißt _Eigenraum_ $\mathrm{ER}(\lambda)=\{v\in\mathbb{R}^n \mid A\,v=\lambda\,v\}$. Die Menge aller Eigenwerte von $A$ heißt _Spektrum_ $\sigma(A) = \{\lambda\in \mathbb{R} \mid \exists v\in\mathbb{R}^n, A\,v=\lambda\,v\}$. ##### Bemerkung: 1. $\lambda$ ist invariant, denn für $B=T\,A\,T^{-1}$ folgt aus \ $T\cdot\mid\, A\,v=\lambda\,v \Leftrightarrow T\,A\,T^{-1}\,T\,v = \lambda T\,v \Leftrightarrow B\,\tilde{v}=\lambda\,\tilde{v},\quad$ mit $\tilde{v}=Tv$ Eigenvektor. 2. Später werden wir $\lambda\in\mathbb{C}$ und $v\in\mathbb{C}^n$ zulassen, selbst wenn $A\in\mathbb{R}^{n\times n}.$ """ # ╔═╡ aa1e42de-dc2f-40e1-ad09-a3fb9f869c3f md""" ##### Beispiel: 1. Oben $A=\begin{bmatrix} 1&\phantom{-}2\\2&-2\end{bmatrix},\, \lambda_1=-3,\,v_1=\begin{bmatrix}\phantom{-}1\\-2\end{bmatrix},\,\lambda_2=2,\, v_2=\begin{bmatrix}2\\1\end{bmatrix}$ $\mathrm{ER}(-3)=\left\{\alpha\begin{bmatrix}\phantom{-}1\\-2\end{bmatrix}:\alpha\in\mathbb{R}\right\},\, \mathrm{ER}(2)=\left\{\alpha\begin{bmatrix}2\\1\end{bmatrix}:\alpha\in\mathbb{R}\right\}$ 2. $A=I_n,\, \lambda = 1,\, \mathrm{ER}(1) = \mathbb{R}^n$ 3. $A = \begin{bmatrix}0\end{bmatrix}\in\mathbb{R}^{n\times n},\, \lambda = 0, \mathrm{ER}(0) = \mathbb{R}^n$ 4. $A = \begin{bmatrix} 2&0&0\\0&2&0\\0&0&3\end{bmatrix},\, \lambda_1=2,\, v_1=\begin{bmatrix}1\\0\\0\end{bmatrix},\, \lambda_2 = 2,\, v_2=\begin{bmatrix}0\\1\\0\end{bmatrix},\, \lambda_3=3,\, v_3=\begin{bmatrix}0\\0\\1\end{bmatrix}$ $\mathrm{ER}(2) = \left\{\alpha\begin{bmatrix}1\\0\\0\end{bmatrix}+\beta\begin{bmatrix}0\\1\\0\end{bmatrix}: \alpha,\beta\in\mathbb{R}\right\},\quad \mathrm{ER}(3)=\left\{\alpha\begin{bmatrix}0\\0\\1\end{bmatrix}: \alpha\in\mathbb{R}\right\}$ """ # ╔═╡ 9faaa5df-21f9-432e-ae98-dee5c4043a57 md""" ##### Satz: Existieren zu $A\in\mathbb{R}^{n\times n}$ $n$ linear unabhängige Eigenvektoren $v_i (i=1,\dots,n)$, genau dann ist $A$ diagonalisierbar, d.h. $A$ ist ähnlich zu einer Diagonalmatrix. ##### Beweis: (Diagonalisierbar $\Rightarrow$ $n$ Eigenvektoren): Klar, denn $\mathrm{diag}(\lambda_1,\dots,\lambda_n)$ hat $n$ Eigenvektoren (siehe oben). ($n$ Eigenvektoren $\Rightarrow$ diagonalisierbar): Seien $v_i\in\mathbb{R}^n$ die Eigenvektoren zu Eigenwerten $\lambda_i\in\mathbb{R}$. Setze $S=\begin{bmatrix}|&|&&|\\v_1&v_2&\cdots&v_n\\|&|&&|\end{bmatrix}$, dann gilt $A\,S = \begin{bmatrix}|&&|\\\lambda_1\,s_1&\cdots&\lambda_n\,s_n\\|&&|\end{bmatrix}=S\,D$ mit $D=\mathrm{diag}(\lambda_1,\dots,\lambda_n)$, also $D=S^{-1}\,A\,S$. $A$ ist _diagonal-ähnlich_. """ # ╔═╡ f8998cfc-8838-45b6-82d3-bcdad20a60ee md""" Wir benötigen Eigenwerte und Eigenvektoren von $A$. ### 4.2.1 Berechnung der Eigenwerte Es gilt $A\,v = \lambda\,v$, also $(\lambda\,I_n-A)\,v = 0$, d.h. die Matrix $\lambda \,I_n-A$ besitzt einen nicht-trivialen Nullraum $\Leftrightarrow$ das lineare Gleichungssystem $(\lambda\,I_n-A)\,v = 0$ ist mehrdeutig lösbar $\Leftrightarrow$ die Matrix $\lambda\,I_n-A$ ist _nicht_ regulär $\Leftrightarrow \det(\lambda\,I_n-A)=0\Leftrightarrow \mathrm{Ker}(\lambda\,I_n-A)\neq\{0\}$. Das ist eine Gleichung für $\lambda$! ##### Satz und Definition: Für $A\in\mathbb{R}^{n\times n}$ gilt $\lambda\in\mathbb{R}$ Eigenwert von $A \Leftrightarrow \det(\lambda\,I_n-A)=0$ Die Funktion $p_A(t):= \det(t\,I_n-A)$ heißt _charakteristisches Polynom_. """ # ╔═╡ 16ad9117-7b0a-4e96-be3a-44bb0dea2ab8 md""" ##### Beispiel: 1. $A = \begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix},\quad \lambda\,\begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} \lambda - 1 & -2 \\ -1 & \lambda - 2\end{bmatrix},$ $p_A(t) = \det(t\,I - A) = \det\left(\begin{bmatrix} t - 1 & -2 \\ -1 & t - 2 \end{bmatrix}\right) = (t-1)(t-2) - 2 = t^2 - 3t,$ $p_A(t) \overset{!}{=} 0 \Rightarrow t^2 - 3t = 0 \Rightarrow t_1 = 0, t_2 = 3.$ 2. $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix},\quad \lambda\,\begin{bmatrix} 1&0&0\\ 0 & 1&0 \\ 0&0&1 \end{bmatrix} - \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix} = \begin{bmatrix} \lambda & -1 & 0 \\ -1 & \lambda-1 & -1 \\ -1 & -2 & \lambda-2 \end{bmatrix},$ $p_A(t) = \det(t\,I - A) = \det\left(\begin{bmatrix} t & -1 & 0 \\ -1 & t-1 & -1 \\ -1 & -2 & t-2 \end{bmatrix}\right)$ $\phantom{p_A(t) = \det(t\,I - A) }= t((t-1)(t-2) - 2) - (t - 2) - 1 = t^3 - 3t^2 -t + 1.$ """ # ╔═╡ adb1b0c5-3441-486f-b677-b47f54bc68c3 md""" ##### Satz: Für $A\in\mathbb{R}^{n\times n}$ hat das charakteristische Polynom die Form $p_A(t) = t^n-\mathrm{sp}(A)\,t^{n-1}+\dots+(-1)^n\,\det(A),$ wobei $\mathrm{sp}(A)=\sum\limits_{i=1}^na_{ii}$ die "Spur" von $A$ ist. ##### Beweis: $\begin{align*}\det(\overbrace{\lambda I_n-A}^{=:(b_{ij)})})&=\det\begin{bmatrix}\lambda-a_{11}&-a_{12}&\cdots&\cdots\\-a_{21}&\lambda-a_{22}\\\vdots&&\ddots&\\\vdots&&&\lambda-a_{nn}\end{bmatrix}\\&=\sum\limits_{\pi\in S_n}\mathrm{sign}(\pi)\,b_{1,\pi(1)}\,\cdots\,b_{n,\pi(n)}\\&=(\lambda-a_{11})\,(\lambda-a_{22})\,\cdots\,(\lambda-a_{nn})+\overbrace{\sum\limits_{\pi\in S_n\setminus\{\mathrm{id}\}}\mathrm{sign}(\pi)\,b_{1,\pi(1)}\,\cdots\,b_{n,\pi(n)}}^{\text{max. }n-2\text{ Diagonalelemente}}\\&=\lambda^n-(\underbrace{a_{11}+a_{22}+\dots+a_{nn}}_{\mathrm{sp}(A)})\,\lambda^{n-1}+\dots\lambda^{n-2}+\dots\end{align*}$ Konstanter Koeffizient von $p_A(\lambda)$ ist $p_A(0)=\det(-A)=(-1)^n\,\det(A)$. """ # ╔═╡ 59d1416f-4a17-41f7-91ef-61b619b69af0 md""" ##### Bemerkung: 1. $p_A(\lambda)$ ist invariant unter Basiswechsel, denn $\begin{align*}B&=T\,A\,T^{-1},\, p_B(\lambda)=\det(\lambda\,I_n -B)=\det(\lambda\,T\,T^{-1} -T\,A\,T^{-1})\\&=\det(T\,(\lambda\,I_n-A)\,T^{-1})=\det(T)\,p_A(\lambda)\,\det(T^{-1})=p_A(\lambda).\end{align*}$ Insbesondere ist $\mathrm{sp}(A)$ invariant. 2. $\lambda = 0 \Leftrightarrow 0\in\sigma(A)\Leftrightarrow p_A(0)=0\Leftrightarrow \det(A)=0$, also $A$ singulär. 3. $\mathrm{Ker}(\lambda\,I_n-A) = \mathrm{ER}(\lambda) \Rightarrow$ Eigenvektoren befinden sich im Kern. """ # ╔═╡ 2ca178fb-0d09-49af-88bd-37aa02d229ba md""" ### 4.2.2 Vorgehen bei Diagonalisierung von $A$ 1. Berechnung des charakteristischen Polynoms $p_A(t)=\det(t\,I_n -A)$ 2. Finden der Nullstellen $p_A(\lambda_i)=0,\quad i=1,2,\dots,n$ 3. Lösen des LGS $(\lambda_i\,I_n - A)v=0,\quad i=1,\dots,n$ 4. Bilden der Transformationsmatrix $S=\begin{bmatrix}|&|&&|\\v_1&v_2&\cdots&v_n\\|&|&&|\end{bmatrix}$ mit $n$ _linear unabhängigen_ Eigenvektoren. $\Rightarrow S^{-1}\,A\,S =\mathrm{diag}(\lambda_1,\dots,\lambda_n)$ mit $\lambda_i$ Eigenwerte zu $v_i.$ _Achtung_: Lassen sich keine $n$ linear unabhängige Eigenvektoren finden, ist $A$ _nicht_ diagonalisierbar. """ # ╔═╡ 36bbca21-c2a7-4e91-b0e4-9bc08c45a042 # ╔═╡ 029fd333-ad3f-4867-8250-7285bb483602 md""" ##### Satz: Sei $\lambda$ Eigenwert zu $A\in\mathbb{R}^{n\times n}$ und $\mathrm{dim}(\mathrm{ER}(\lambda))=\mathrm{dim}(\mathrm{Ker}(\lambda I_n-A))$ die Anzahl linear unabhängiger Eigenvektoren zu $\lambda$. Dann gilt $1\leq \mathrm{dim}(\mathrm{ER}(\lambda))\leq K_\lambda,$ wobei $K_\lambda$ die Vielfachheit der Nullstelle $\lambda$ von $p_A(t)$ ist. ##### Bemerkung: 1. Das heißt zu einem Eigenwert $\lambda$ gibt es mindestens einen linear unabhängigen Eigenvektor und maximal $K_\lambda$ linear unabhängige Eigenvektoren. 2. Erinnere: z.B. $x^3-x^2-8x+12 = (x+3)(x-4x+4)=(x+3)(x-2)^2$, wobei $-3$ einfache Nullstelle und $2$ doppelte Nullstelle. 3. Sprechweise: $\;\,K_\lambda \kern38pt=:$ "algebraische Vielfachheit" von $\lambda$,\ $\phantom{Sprechweise}\mathrm{dim}(\mathrm{ER}(\lambda))=:$ "geometrische Vielfachheit" von $\lambda$. """ # ╔═╡ 73a5b4bc-ff97-433f-bb50-7a2a3e9e55ad md""" ##### Beispiel: 1. $A = \begin{bmatrix} 3 & -1 \\ 1 & \phantom{-}1 \end{bmatrix}$, $\det\begin{bmatrix} \lambda-3 & 1 \\ -1 & \lambda-1 \end{bmatrix} = (\lambda-3)\,(\lambda-1)+1 = \lambda^2 - 4\lambda - 4 = (\lambda - 2)^2$ $p_A(\lambda) = (\lambda - 2)^2$, $\lambda_{1,2}=2$, alg. Vielfachheit $2$. $\lambda_1 = 2: (\lambda\,I_2 - A)\,v = \begin{bmatrix} -1 & 1 \\ -1 & 1 \end{bmatrix}\,v = 0$ $\mathrm{ER}(2) = \left\{\begin{bmatrix} \alpha\\\alpha\end{bmatrix}, \alpha\in\mathbb{R}\right\}$, nicht diagonalisierbar, da geometrische Vielfachheit $1$! 2. $A = \begin{bmatrix} -3 & 2 & \phantom{-}1 \\\phantom{-}2 & 0 & \phantom{-}2\\\phantom{-}1 & 2 & -3\end{bmatrix}$, $\det\begin{bmatrix} \lambda+3 & -2 & -1 \\-2 & \lambda & -2\\-1 & -2 & \lambda+3\end{bmatrix}=\lambda^3 + 6\lambda^2 - 32$ $p_A(\lambda) = \lambda^3 + 6\lambda^2 - 32, \lambda_1 = 2, \Rightarrow p_A(\lambda) = (\lambda-2)\,(\lambda+4)^2$ $\lambda_1 = 2$, alg. Vielfachheit $1$. $\mathrm{ER}(2) = \left\{\alpha\begin{bmatrix} \phantom{-}9 \\ -2 \\\phantom{-}1\end{bmatrix} : \alpha \in \mathbb{R}\right\}$ $\lambda_2 = -4$, alg. Vielfachheit $2$. $\mathrm{ER}(-4) = \left\{\begin{bmatrix} -2\beta - \alpha \\ \beta \\ \alpha \end{bmatrix}: \alpha,\beta\in\mathbb{R} \right\}$, z.B. $v_2 = \begin{bmatrix} -1\\0\\1\end{bmatrix}, v_3 = \begin{bmatrix} -2\\1\\0\end{bmatrix}$. Damit diagonalisierbar. """ # ╔═╡ e69cb10f-709a-4a4f-be30-be962d74ff22 md""" ##### Beweis: $\dim(\mathrm{ER}) \geq 1$, klar per Konstruktion, denn $\det(\lambda\,I_n-A) = 0 \Rightarrow \mathrm{Ker}(\lambda\,I_n - A) \neq \{0\}$. Sei $\{v_1,\dots,v_m\}$ Basis von $\mathrm{ER}(\lambda_0)$, d.h. $\mathrm{dim}(\mathrm{ER})=m$. Wir basteln eine Basis $\{v_1,\dots,v_m,w_{m+1},\dots,w_n\}$ des gesamten Raums. Bezüglich dieser Basis hat die lineare Abbildung $\varphi_A$ die Matrix-Koordinaten $B = \begin{bmatrix} \lambda_0 & \cdots & 0 \\ \vdots & \ddots & \vdots & \star \\ 0 & \cdots & \lambda_0\\ & 0 & & \tilde{A} \end{bmatrix}$ Dieses $B$ hat dasselbe charakteristische Polynom wie $A$. $p_B(t)=p_A(t)=\det\begin{bmatrix} t-\lambda_0 & \cdots & 0 \\ \vdots & \ddots & \vdots & \star \\ 0 & \cdots & t-\lambda_0\\ & 0 & & t\,I_{n-m}-\tilde{A} \end{bmatrix}$ $\phantom{p_B(t)=p_A(t)}=(t-\lambda_0)^m\det(t\,I_{n-m}-\tilde{A})\Rightarrow m = \mathrm{dim}(\mathrm{ER}(\lambda_v))\leq K_{\lambda_0}.$ """ # ╔═╡ 79b9e65d-84af-480e-ba32-8fa09367c48c md""" ##### Bemerkung: 1. Hat $p_A(t)$ $n$ Nullstellen $\lambda_i,\,i=1,\dots,n$, mit Vielfachheit gezählt _und_ ist die algebraische gleich der geometrischen Vielfachheit, so ist die Matrix $A$ diagonalisierbar. 2. Hat $p_A(t)$ $n$ _verschiedene_ Nullstellen, so ist $A$ diagonalisierbar. 3. Die Bedingung von $n$ Nullstellen für $p_A(t)$ mit Grad $n$ legt nahe, $\lambda_i\in\mathbb{C}$ zuzulassen. """ # ╔═╡ eceefaa1-1bcc-4515-ab40-4ee58825ab04 md""" ##### Beispiel: $A=\begin{bmatrix}\phantom{-}2&1\\-2&4\end{bmatrix},\, \det\begin{bmatrix}\lambda-2&-1\\2&\lambda-4\end{bmatrix}=(\lambda-2)(\lambda-4)+2=\lambda^2-6\lambda + 10$ $\lambda_{1,2}=3\pm\sqrt{9-10}=3\pm i$ Einsetzen von $\lambda_1$ liefert:\ $\begin{bmatrix}1+i&-1\\2&-1+i\end{bmatrix}\,x=0\Rightarrow \begin{bmatrix}2&-1+i\\1+i&-1\end{bmatrix}\,x=0\Rightarrow\begin{bmatrix}2&-1+i\\0&0\end{bmatrix}\,x=0$\ $x_2 = \alpha,\quad x_1 = \frac12(1-i)\alpha$ $\mathrm{ER}(3+1)=\left\{\begin{bmatrix}1-i\\2\end{bmatrix}\alpha:\alpha\in\mathbb{R}\right\},\quad \mathrm{ER}(3-i)=\left\{\begin{bmatrix}1+i\\2\end{bmatrix}\alpha:\alpha\in\mathbb{R}\right\}$\ $\Rightarrow$ diagonalisierbar in $\mathbb{C}^{n\times n}$ (nicht in $\mathbb{R}^{n\times n}$!) $\begin{bmatrix}3+i&0\\0&3-i\end{bmatrix}=S^{-1}\,A\,S$ mit $S=\begin{bmatrix}1-i&1+i\\2&2\end{bmatrix}$ """ # ╔═╡ 030934c2-c218-4036-80f1-264ff86f8e9f md""" ## 4.3 Spektralsatz """ # ╔═╡ a837ff7b-e137-4e37-b27b-27b7bcbc7b2b md""" ##### Satz (Spektralsatz): Erfüllt eine Matrix $A\in\mathbb{R}^{n\times n}$ die Eigenschaft $A^T\,A=A\,A^T$, sowie $\sigma(A) \subset \mathbb{R}$, genau dann ist $A$ diagonalisierbar _mit orthonormalen_ Eigenvektoren. ##### Bemerkung: 1. $\{A\in\mathbb{R}^{n\times n} : A^T\,A = A\,A^T \} \subset \{A\in\mathbb{R}^{n\times n} : A \text{ diagonalisierbar}\} \subset \mathbb{R}^{n\times n}$ 2. Die Matrix $A$ mit $A^TA=AA^T$ heißt "normal". z.B. $A=\begin{bmatrix}1&2\\-2&1\end{bmatrix}$ oder $A=\begin{bmatrix}2&3\\3&-1\end{bmatrix}$ 3. Beweis folgt später """ # ╔═╡ 8f585b55-b23f-4007-a677-11ff85b038b4 md""" ##### Definition: Sei $\varphi: V\to V$ eine lineare Abbildung in $V$ und $V$ habe ein Skalarprodukt $\langle \cdot,\cdot\rangle$. Die _adjungierte Abbildung_ $\varphi^\star$ ist definiert durch $\langle \varphi(v),w\rangle = \langle v,\varphi^\star(w)\rangle\quad \forall v,w\in V$ ##### Satz: Eigenschaften von $\varphi^*$ und der Adjunktion 1. $\varphi^*$ ist eindeutig und existiert. 2. Die Adjunktion $(\cdot)^*$ ist linear, d.h. $(\alpha\varphi+\beta\psi)^*=\alpha\varphi^*+\beta\psi^*\quad \alpha,\beta\in\mathbb{R},\,\varphi,\psi$ lineare Abbildungen 3. $(\varphi\circ\psi)^*=\psi^*\circ\varphi^*\quad \varphi,\psi\in V$ 4. $(\varphi^*)^*=\varphi$ 5. Falls $A$ die Matrix zu $\varphi$ ist (bzgl. orthonormaler Basis), so ist $A^T$ die Matrix zu $\varphi^*$. ##### Beweis: Zu 3: $\langle \underbrace{\varphi(\psi(v))}_{\varphi\circ\psi},w\rangle = \langle \psi(v),\varphi^*(w)\rangle = \langle v,\underbrace{\psi^*(\varphi^*(w))}_{(\varphi\circ\psi)^*}\rangle$ Zu 5: $\langle \cdot,\cdot\rangle$ Euklidische Skalarprodukt, d.h. $\langle v,w\rangle = \sum\limits_{i=1}^n v_i\,w_i$ mit $v=\sum\limits_{i=1}^nv_i\,b_i,\,w=\sum\limits_{i=1}^nw_i\,b_i,\,\{b_i\}$ orthonormale Basis $u=\varphi(v),\, u_i=\sum\limits_{j=1}^na_{ij}\,v_j.$ $\langle \varphi(v),w\rangle = \sum\limits_{i=1}^n\sum\limits_{j=1}^n a_{ij}\,v_j\,w_i=\sum\limits_{j=1}^nv_j\,\sum\limits_{i=1}^na_{ij}\,w_i$ $y = \varphi^*(w),\, y_j=\sum\limits_{i=1}^na_{ji}^*\,w_i.$ $\langle\varphi(v),w\rangle = \sum\limits_{j=1}^nv_j\,\underbrace{\sum\limits_{i=1}^n a_{ij}\,w_i}_{\overset{!}{=}\sum\limits_{i=1}^na_{ji}^*\,w_i} = \langle v,\varphi^*(w)\rangle \Rightarrow a_{ji^*}=a_{ij},\quad$ d.h. $A^*=A^T$. """ # ╔═╡ ffc7bc79-1275-455c-9977-44271d674874 md""" ##### Bemerkung: 1. $\langle x,y\rangle$ kann in $\mathbb{R}^n$ als $x^T\,y$ geschreiben werden. Damit gilt $\langle A\,v,w\rangle = (A\,v)^T\,w = v^T\,A^T\,w=v^T(\underbrace{A^T\,w}_{\varphi^*(w)})\widehat{=}\langle v,A^T\,w\rangle$ 2. Wegen $(\varphi\circ\psi)^*=\psi^*\circ\varphi^*$ gilt $(A\,B)^T=B^T\,A^T$ ##### Definition: Eine Abbildung $\varphi:V\to V$ heißt _selbst-adjungiert_, falls gilt $\varphi=\varphi^*$. ##### Bemerkung: 1. Für die Matrizen gilt dann $A^T=A$, die Matrix ist _symmetrisch_. 2. Symmetrische Matrizen sind "normal", d.h. $A^T\,A=A\,A^T=A^2.$ """ # ╔═╡ d8fbdb5a-83ff-4ba1-80ce-20e8cf404695 md""" ##### Definition: Eine Abbildung $\varphi:V\to V$ heißt _orthogonal_, falls gilt $\langle \varphi(x),\varphi(y)\rangle = \langle x,y\rangle\quad\forall x,y\in V.$ ##### Satz: Für $\varphi$ orthogonal gilt $\varphi^*=\varphi^{-1}$ ##### Beweis: $\langle \varphi(x),\varphi(y)\rangle=\langle x,\varphi^*(\varphi(y))\rangle = \langle x,y\rangle \Leftrightarrow \langle x,\underbrace{\varphi^*(\varphi(y))-y}_{=0}\rangle = 0\quad \forall x,y\in V$\ Also gilt $\varphi^*(\varphi(y))=y.$ """ # ╔═╡ bb223846-09ef-410d-837e-d04c6407ed4c md""" ##### Bemerkung; 1. Orthogonale Matrizen sind normale Matrizen: $A^T=A^{-1}\Rightarrow A^T\,A=A^{-1}\,A=I_n=A\,A^{-1}=A\,A^T$ 2. Orthogonale Abbildungen erhalten die Norm ("Länge") $\lVert \varphi(x)\rVert^2 = \langle \varphi(x),\varphi(x)\rangle =\langle x,x\rangle = \lVert x \rVert^2\quad \forall x\in V$ (Drehung/Spiegelung) ##### Beispiel: $A=\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&-\frac{1}{\sqrt{3}}\\0&\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\end{bmatrix},\, A^T=\begin{bmatrix}\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{6}}&\frac{2}{\sqrt{6}}&-\frac{1}{\sqrt{6}}\\-\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}\end{bmatrix}$ $A^T\,A = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = A\,A^T$ """ # ╔═╡ 0bcf3c7c-de4b-4b98-b429-1cd5002b1fe4 md""" ##### Bemerkung: 1. Wird $B=S^{-1}\,A\,S=T\,A\,T^{-1}$ bezüglich einer orthonormierten Basis gemacht, so gilt $S^{-1}=S^T$ also $B=S^T\,A\,S$, d.h. $S$ orthogonal (bzw. $T=S^T$ und $B=T\,A\,T^{-1})$. 2. Für normale Abbildungen $\varphi\circ\varphi^*=\varphi^*\circ\varphi$ gilt (bzgl orthonormaler Basis) $A\,A^T=A^T\,A$. Dies ist invariant: Für $B=S^T\,A\,S$ gilt: $\begin{align*}B\,B^T&=S^T\,A\,S(S^T\,A\,S)^T = S^T\,A\underbrace{S\,S^T}_{I}A^T\,S\\&=S^T\underbrace{A\,A^T}_{=A^T\,A}S=S^T\,A^T\,S\,S^T\,A\,S\\&=\underbrace{S^T\,A^T\,S}_{(S^T\,A\,S)^T}\,S^T\,A\,S=B^T\,B.\end{align*}$ $\Rightarrow B$ ist normal. """ # ╔═╡ 9f3be73c-80b6-4460-b256-b7cdd56803bc md""" ##### Beweis (Spektralsatz): Zu zeigen: $\varphi\circ\varphi^*=\varphi^*\circ\varphi\Rightarrow$ diagonalisierbar mit orthonormalen Eigenvektoren. Induktion über $n=\mathrm{dim}(V)$: Induktionsverankerung $n=1$: $A\in\mathbb{R}^{1\times1}(a_{11})$ Induktionsschrit ($n-1\to n$) Voraussetzung: $\tilde{A}\in\mathbb{R}^{(n-1)\times(n-1)\$ mit $\tilde{A}\,\tilde{A}^T=\tilde{A}^T\,\tilde{A}$ ist diagonalisierbar mit orthogonalen Eigenvektoren. Zu zeigen: $A\in\mathbb{R}^{n\times n}$ mit $A\,A^T=A^T\,A$ auch diagonalisierbar. Sei $\lambda$ Eigenwert von $A$ (eventuell $\lambda\in\mathbb{C}$) mit $A\,v=\lambda\,v$ ($v$ ist Eigenvektor). 1. Wir zeigen $\lambda$ Eigenwert von $A^T$ mit $A^T\,v=\lambda\,v$ (gleicher Eigenvektor!) Betrachte dazu Skalarprodukt: $\begin{align*}(A^T\,v-\lambda\,v)^T\,(A^T\,v-\lambda\,v) &= v^T\,A\,A^T\,v-v^T\,A\,\lambda\,v-\lambda\,v^T\,A^T\,v+\lambda^2 v^T\,v\\&=v^T\,(A\,A^T-A\,\lambda-\lambda\,A^T+\lambda^2\,I)\,v\\&=(A\,v-\lambda\,v)^T\,(A\,v-\lambda\,v)=0,\end{align*}$ Denn $A\,v = \lambda\,v$. 2. Sei $\{v,w_1,\dots,w_{m-1}\}$ eine Orthonormalbasis. Bezüglich dieser Basis hat $\varphi_A$ die Matrix $B = \begin{bmatrix} \lambda & \alpha_1 & \cdots &\alpha_{n-1} \\ 0 \\ \vdots & & \tilde{A} \\0 \end{bmatrix}$, mit $\varphi(v) = \lambda\,v$ und $\varphi(w_i)=\alpha_i\,v + \sum_{j=1}^{n-1}\tilde{a}_{ij}\,w_j$. $\varphi_A^\star$ hat demnach die folgende Matrixdarstellung: $B = \begin{bmatrix} \lambda & 0 & \cdots & 0 \\ \alpha_1 \\ \vdots & & \tilde{A} \\ \alpha_{n-1} \end{bmatrix}$, aber auch $\varphi_A^\star(v) = \lambda\,v$, d.h. $B^T\,v = \lambda\,v$. $\Rightarrow \alpha_i=0$. $B$ erfüllt $B^T\,B=B\,B^T$ ($\varphi$ normal, invariant für orthonormale Basis), also $B = \begin{bmatrix} \lambda^2 & 0 & \cdots & 0 \\ 0 \\ \vdots & & \tilde{A}^T\,\tilde{A} \\0 \end{bmatrix} = \begin{bmatrix} \lambda^2 & 0 & \cdots & 0 \\ 0 \\ \vdots & & \tilde{A}\,\tilde{A}^T \\0 \end{bmatrix}$, da $\tilde{A}$ normal ist. Nach Induktionsvoraussetzung ist $\tilde{A}$ diagonalisierbar mit orthogonalen Eigenvektoren. """ # ╔═╡ 910abed2-28f6-4078-8841-7ff4f9a75bd4 md""" ### 4.3.1 Anwendungen ##### Satz: $A$ orthogonal $\Rightarrow A$ diagonalisierbar mit orthonormaler Basis. Für Eigenwert $\lambda\in\mathbb{C}$ gilt $|\lambda|=1.$ ##### Beweis: $A\,A^T=A^T\,A=I$, d.h. $A$ normal. $A\,v=\lambda v,\, \lVert v\rVert^2 = \lVert A\,v\rVert^2=\lVert \lambda v\rVert^2=|\lambda|\Rightarrow \lambda = 1$ """ # ╔═╡ ac374485-d23a-4371-bbeb-fbc9f58be638 md""" ##### Beispiel: Drehung in 2D mit $A=\begin{bmatrix}\cos(\alpha)&-\sin(\alpha)\\\sin(\alpha)&\cos(\alpha)\end{bmatrix}.$ $A\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\cos(\alpha)-y\sin(\alpha)\\x\sin(\alpha)+y\cos(\alpha)\end{bmatrix}$ $\Rightarrow\det(\lambda I-A)=\begin{bmatrix}\lambda-\cos(\alpha)&\sin(\alpha)\\-\sin(\alpha)&\lambda-\cos(\alpha)\end{bmatrix}=\lambda^2-2\cos(\alpha)\lambda+1$ $\lambda_{1,2}=\cos(\alpha)\pm\underbrace{\sqrt{\cos^2(\alpha)-1}}_{-\sin^2(\alpha)}=\cos(\alpha)\pm i\sin(\alpha)\Rightarrow |\lambda|^2 = \cos^2(\alpha)+\sin^2(\alpha)=1$ Zum Beispiel: $\alpha=\frac{\pi}{2},\, A=\begin{bmatrix}0&-1\\1&0\end{bmatrix},\ \lambda_{1,2}=\pm i,\, v_{1,2}=\begin{bmatrix}\pm i\\1\end{bmatrix}$ (Drehung um 90°) """ # ╔═╡ 95b1c356-83dc-4450-bbc0-429152ef066f md""" ##### Satz: Für Matrix $A$ symmetrisch ($A=A^T$) ist $A$ diagonalisierbar mit orthonormaler Basis und alle Eigenwerte sind reell. Die Eigenvektoren in $\mathbb{R}^n$ heißen _Hauptachsen_ von $A$. ##### Beweis: $A^TA=A^2=A\,A^T,\,$ also $A$ normal. Annahme: $\lambda,\overline{\lambda}$ sind komplex konjugierte Eigenwerte und $v,\overline{v}$ sind Eigenwerte. Es gilt $\begin{align*}\lambda v^T\,\overline{v} &=(\lambda v)^T\,\overline{v}\\&=(A\,v)^T\,\overline{v}=v^T\,\underset{=A}{A^T}\,\overline{v}\\&=v^T\,(A\,\overline{v}) = v^T\,(\overline{\lambda}\,\overline{v})\\&=\overline{\lambda}\,v^T\,v\Rightarrow \lambda=\overline{\lambda}.\end{align*}$ """ # ╔═╡ ae076f7b-4b9d-4beb-8352-66eee44a30be md""" ##### Beispiel: $A = \begin{bmatrix} 3&2&4\\2&0&2\\4&2&3\end{bmatrix}$ $\det(\lambda I-A)=\begin{bmatrix}\lambda-3&-2&-4\\-2&\lambda&-2\\-4&-2&\lambda-3\end{bmatrix}=\lambda^3-6\lambda^2-15\lambda-8=(\lambda-8)(\lambda+1)^2$ Eigenwert $\lambda_1 = 8$, algebraische Vielfachheit $1\Rightarrow 1$ linear unabängiger Eigenvektor. $\begin{bmatrix}5&-2&-4\\-2&8&-2\\-4&-2&5\end{bmatrix}v=0\Rightarrow v=\alpha\begin{bmatrix}2\\1\\2\end{bmatrix},\,\alpha\in\mathbb{R},$ z.B. $v_1=\begin{bmatrix}2\\1\\2\end{bmatrix}$ Eigenwert $\lambda_{2,3} = -1$, algebraische Vielfachheit $2\Rightarrow$ mindestens $1$ linear unabhängiger Eigenvektor, höchstens $2$. Aber: $A$ ist symmetrisch, also diagonalisierbar, d.h. es müssen auf jeden Fall $2$ linear unabhängige Eigenvektoren entstehen. $\begin{bmatrix} -4&-2&-4\\-2&-1&-2\\-4&-2&-4\end{bmatrix}v=0\Rightarrow v=\alpha\begin{bmatrix}1\\-4\\1\end{bmatrix}+\beta\begin{bmatrix}1\\-2\\0\end{bmatrix},\,$ z.B. $v_2=\begin{bmatrix}1\\-4\\1\end{bmatrix}, v_3=\begin{bmatrix}1\\-2\\0\end{bmatrix}$ Beachte $v_1$ ist orthogonal zu $v_2$ und $v_3$. Wir können zusätzlich $v_2$ orthogonal zu $v_3$ wählen, z.B. $v_2=\begin{bmatrix}1\\-4\\1\end{bmatrix},\,v_3=\begin{bmatrix}-1\\0\\1\end{bmatrix}$ Orthonormierte Basis aus Eigenvektoren $\left\{\begin{bmatrix}\frac{2}{3}\\\frac{1}{3}\\\frac{2}{3}\end{bmatrix},\begin{bmatrix}\frac{1}{\sqrt{18}}\\-\frac{4}{\sqrt{18}}\\\frac{1}{\sqrt{18}}\end{bmatrix},\begin{bmatrix}-\frac{1}{\sqrt{2}}\\0\\\frac{1}{\sqrt{2}}\end{bmatrix}\right\}$ (Spalten von $S$) $S^{-1}\,A\,S=S^T\,A\,S=\begin{bmatrix}8&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$ """ # ╔═╡ c1d1ff69-73c8-476c-a7ab-bd3c4ce699a8 md""" ##### Bemerkung: 1. Der Spannungstensor $T\in\mathbb{R}^{3\times 3}$ in einem Material, liefert Spannungen $s\in\mathbb{R}^3$ auf einer Schnittfläche mit Normalenrichtung $n$, $s = T\,n$, ist immer symmetrisch, d.h. es gibt immer $3$ orthogonale Schnittflächen auf denen nur Druck- oder Zugkräfte herrschen. 2. Gilt algebraische Vielfachheit $\neq$ geometrische Vielfachheit, so ist die Matrix $A$ _nicht_ diagonalisierbar. Es gibt jedoch eine Basis-Transformation $S$ so, dass $A$ _Jordan-Form_ bekommt: $S^{-1}\,A\,S = \begin{bmatrix} J_1 & 0 & \cdots & 0 \\ 0 & J_2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & J_q \end{bmatrix} \in \mathbb{C}^{n\times n}$ $\sigma(A) = \{\lambda_1, \ldots, \lambda_q\}$, die $q$ Eigenwerte von $A$ sind echt voneinander verschieden, mit jeweils $K_q$ als die algebraischen Vielfachheiten und $\ell_q$ als die geometrischen Vielfachheiten der Eigenwerte. Jordanblöcke sind $J_i = \begin{bmatrix} \lambda_i & 1 & 0 & \cdots & 0 \\ 0 & \lambda_i & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & 1 & 0 \\ 0 & & 0 & \lambda_{K_i -\ell_i} & 0 \\ 0 & \cdots & \cdots & 0 & \lambda_{K_i}\end{bmatrix}\in\mathbb{C}^{K_i\times K_i}$ """ # ╔═╡ 01bbdd70-bcce-4201-8843-5e7ca611e245 md""" ### 4.3.2 Ausblick Wir lösen das lineare Gleichungssystem $A\,x=b$ mit Eigenwerten und Eigenvektoren! Zum Beispiel $f''=g$ als Funktion mit linearem Operator (der zweiten Ableitung). Annahme: Wir haben eine orthonormale Basis aus Eigenvektoren $\{v_1,\dots,v_n\}$ des Operators in einem geeigneten Funktionenraum. Ansatz für die Lösung: $x=\sum\limits_{i=1}^n\alpha_i\,v_i\Rightarrow v_i$ bekannt, $\alpha_i\in\mathbb{R} =?$ Idee: Schreibe $b=\sum\limits_{i=1}^n\beta_i\,v_i$, dann sind die $\beta_i,v_i$ bekannt, z.B. $\beta_i=\langle v_i,b\rangle$ Einsetzen: $A\,x = A\left(\sum\limits_{i=1}^n\alpha_i\,v_i\right)=\sum\limits_{i=1}^n\alpha_i\,A\,v_i=\sum\limits_{i=1}^n\alpha_i\,\lambda_i\,v_i\overset{!}{=}b=\sum\limits_{i=1}^n\beta_i\,v_i$ Lösung: $\alpha_i=\frac{1}{\lambda_i}\,\beta_i$ """ # ╔═╡ 00000000-0000-0000-0000-000000000001 PLUTO_PROJECT_TOML_CONTENTS = """ [deps] CairoMakie = "13f3f980-e62b-5c42-98c6-ff1f3baf88f0" LinearAlgebra = "37e2e46d-f89d-539d-b4ee-838fcccc9c8e" PlutoTeachingTools = "661c6b06-c737-4d37-b85c-46df65de6f69" PlutoUI = "7f904dfe-b85e-4ff6-b463-dae2292396a8" Printf = "de0858da-6303-5e67-8744-51eddeeeb8d7" [compat] CairoMakie = "~0.15.9" PlutoTeachingTools = "~0.4.7" PlutoUI = "~0.7.80" """ # ╔═╡ 00000000-0000-0000-0000-000000000002 PLUTO_MANIFEST_TOML_CONTENTS = """ # This 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["Artifacts", "JLLWrappers", "Libdl", "Zlib_jll"] git-tree-sha1 = "e51150d5ab85cee6fc36726850f0e627ad2e4aba" uuid = "b53b4c65-9356-5827-b1ea-8c7a1a84506f" version = "1.6.58+0" [[deps.libsixel_jll]] deps = ["Artifacts", "JLLWrappers", "JpegTurbo_jll", "Libdl", "libpng_jll"] git-tree-sha1 = "c1733e347283df07689d71d61e14be986e49e47a" uuid = "075b6546-f08a-558a-be8f-8157d0f608a5" version = "1.10.5+0" [[deps.libva_jll]] deps = ["Artifacts", "JLLWrappers", "Libdl", "Xorg_libX11_jll", "Xorg_libXext_jll", "Xorg_libXfixes_jll", "libdrm_jll"] git-tree-sha1 = "7dbf96baae3310fe2fa0df0ccbb3c6288d5816c9" uuid = "9a156e7d-b971-5f62-b2c9-67348b8fb97c" version = "2.23.0+0" [[deps.libvorbis_jll]] deps = ["Artifacts", "JLLWrappers", "Libdl", "Ogg_jll"] git-tree-sha1 = "11e1772e7f3cc987e9d3de991dd4f6b2602663a5" uuid = "f27f6e37-5d2b-51aa-960f-b287f2bc3b7a" version = "1.3.8+0" [[deps.libwebp_jll]] deps = ["Artifacts", "Giflib_jll", "JLLWrappers", "JpegTurbo_jll", "Libdl", "Libglvnd_jll", "Libtiff_jll", "libpng_jll"] git-tree-sha1 = "4e4282c4d846e11dce56d74fa8040130b7a95cb3" uuid = "c5f90fcd-3b7e-5836-afba-fc50a0988cb2" version = "1.6.0+0" [[deps.nghttp2_jll]] deps = ["Artifacts", "Libdl"] uuid = "8e850ede-7688-5339-a07c-302acd2aaf8d" version = "1.64.0+1" [[deps.p7zip_jll]] deps = ["Artifacts", "CompilerSupportLibraries_jll", "Libdl"] uuid = "3f19e933-33d8-53b3-aaab-bd5110c3b7a0" version = "17.7.0+0" [[deps.x264_jll]] deps = ["Artifacts", "JLLWrappers", "Libdl"] git-tree-sha1 = "14cc7083fc6dff3cc44f2bc435ee96d06ed79aa7" uuid = "1270edf5-f2f9-52d2-97e9-ab00b5d0237a" version = "10164.0.1+0" [[deps.x265_jll]] deps = ["Artifacts", "JLLWrappers", "Libdl"] git-tree-sha1 = "e7b67590c14d487e734dcb925924c5dc43ec85f3" uuid = "dfaa095f-4041-5dcd-9319-2fabd8486b76" version = "4.1.0+0" """ # ╔═╡ Cell order: # ╟─fac42b42-35d2-11f1-a407-bdbaa61d4b67 # ╟─fb99b46e-a0f4-499a-ab30-5f4ac1757689 # ╟─3b4ec4eb-af39-4f41-90cc-663d1b2eccab # ╟─c5d2f769-f27b-40e4-9602-8b962d5d7934 # 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╟─a837ff7b-e137-4e37-b27b-27b7bcbc7b2b # ╟─8f585b55-b23f-4007-a677-11ff85b038b4 # ╟─ffc7bc79-1275-455c-9977-44271d674874 # ╟─d8fbdb5a-83ff-4ba1-80ce-20e8cf404695 # ╟─bb223846-09ef-410d-837e-d04c6407ed4c # ╟─0bcf3c7c-de4b-4b98-b429-1cd5002b1fe4 # ╟─9f3be73c-80b6-4460-b256-b7cdd56803bc # ╟─910abed2-28f6-4078-8841-7ff4f9a75bd4 # ╟─ac374485-d23a-4371-bbeb-fbc9f58be638 # ╟─95b1c356-83dc-4450-bbc0-429152ef066f # ╟─ae076f7b-4b9d-4beb-8352-66eee44a30be # ╟─c1d1ff69-73c8-476c-a7ab-bd3c4ce699a8 # ╟─01bbdd70-bcce-4201-8843-5e7ca611e245 # ╟─00000000-0000-0000-0000-000000000001 # ╟─00000000-0000-0000-0000-000000000002