Merge branch 'lorenz'
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23
LinAlg2.tex
23
LinAlg2.tex
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@ -528,10 +528,13 @@ Es gibt also zu jedem $\K$-VR V mit $\dim(V)=n$ eine nicht ausgeartete alternier
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\begin{align*}
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\det(\alpha\beta) & = \frac{\varphi(\alpha(\beta(a_1)), \dots, \alpha(\beta(a_n)))}
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{\varphi(a_1, \dots, a_n)} \\
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& =\frac{\varphi(\alpha(\beta(a_1)), \dots, \alpha(\beta(a_n)))}
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{\varphi(\beta(a_1), \dots, \beta(a_n))}\cdot \cdots \\
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& \cdots \frac{\varphi(\beta(a_1), \dots, \beta(a_n))}
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{\varphi(a_1, \dots, a_n)}\underbrace{=}_{\text{Satz \ref{theo:1.3.2}}}
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& \begin{multlined}
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=\frac{\varphi(\alpha(\beta(a_1)), \dots, \alpha(\beta(a_n)))}
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{\varphi(\beta(a_1), \dots, \beta(a_n))}\cdot
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\frac{\varphi(\beta(a_1), \dots, \beta(a_n))}
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{\varphi(a_1, \dots, a_n)}
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\end{multlined} \\
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& \underbrace{=}_{\mathclap{\text{Satz \ref{theo:1.3.2}}}}
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\det(\alpha)\det(\beta)
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\end{align*}
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\end{enumerate}
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@ -580,11 +583,11 @@ Schreibweise für $A=(a_{ij})$:
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Sei $A=(a_1, \dots, a_n)\in\K^{n\times n}$. Dann gilt
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\begin{enumerate}[label=\alph*)]
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\item $\det(A)=\det(A^T)$
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\item $\forall i, j\in[n]: i<n:
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\det((a_1, \dots, \underbrace{a_j}_{i}, \dots, \underbrace{a_i}_{j}, \dots, a_n))=\det(A)$
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\item $\forall i, j\in[n]: i<j:
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\det((a_1, \dots, \underbrace{a_j}_{i}, \dots, \underbrace{a_i}_{j}, \dots, a_n))=-\det(A)$
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\item $\forall i\in[n]: \lambda_1, \dots, \lambda_n\in\K: \det((a_1, \dots, a_i+
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\sum\limits_{\substack{j=1\\j\neq i}}^n\lambda_ja_j, \dots, a_n))=\det(A)$
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\item $\forall i\in[n]: \lambda\in\K: \det((a_1, \dots, \lambda a_i, \dots, a_n)) = \det(A)$
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\item $\forall i\in[n]: \lambda\in\K: \det((a_1, \dots, \lambda a_i, \dots, a_n)) = \lambda \det(A)$
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\item $\exists i, j\in[n]: i\neq j\land a_i=a_j \implies \det(A)=0$
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\item $\forall \lambda \in \K: \det(\lambda A)=\lambda^n \det(A)$
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\item $A$ invertierbar $\implies \det(A^{-1})=\det(A)^{-1}$
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@ -632,7 +635,7 @@ Vielfachen einer Zeile zu einer anderen durch. Raus kommt eine obere Dreiecksmat
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0 & \dots & \dots & b_{nn}
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\end{pmatrix}
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\end{equation}
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Operationen 2) ändern die Determinante nicht, Operationen 1) ändern das Vorzeichen.
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Die Operationen von 2) ändern die Determinante gar nicht, die Operationen von 1) ändern das Vorzeichen.
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\begin{satz}
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@ -857,7 +860,7 @@ da obige Matrix aus $M_{ij}$ durch Spaltenadditionen hervorgeht.
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Dann gilt
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\begin{align*}
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{}_B M(\alpha)_B & = {}_B M(\id)_C \cdot {}_C M(\alpha)_C \cdot {}_C M(\id)_B \\
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& = {}_C M(\id)_{B^{-1}} \cdot {}_C M(\alpha)_C \cdot {}_C M(\id)_B
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& = {{}_C M(\id)_{B}}^{-1} \cdot {}_C M(\alpha)_C \cdot {}_C M(\id)_B
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\end{align*}
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Also ist ${}_C M(\alpha)_C$ diagonalisierbar.
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\item[$\impliedby$:] Sei ${}_C M(\alpha)_C$ diagonalisierbar und $P$ invertierbar mit
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@ -1075,7 +1078,7 @@ $\le\genfrac{}{}{0pt}{0}{\dim(V)}{n}$, da
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\begin{itemize}
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\item[$r=1$:] $v_1$ ist linear unabhängig.
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\item[$r-1\mapsto r$:] \begin{equation}\label{eq:2.2.10.1}
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\mu_1 v_1 + \cdots + \mu_1 v_1 = 0 \end{equation}
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\mu_1 v_1 + \cdots + \mu_r v_r = 0 \end{equation}
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\[ \implies A(\mu_1 v_1 + \cdots + \mu_r v_r) = 0 \]
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\begin{equation}\label{eq:2.2.10.2}
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\implies \lambda_1\mu_1 v_1 + \cdots \lambda_r \mu_r v_r = 0
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