Replace ^+ with ^\dagger
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LinAlg2.tex
58
LinAlg2.tex
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@ -3874,7 +3874,7 @@ $\implies \ker(\alpha) = \langle b_{r+1}, \dots, b_n \rangle_V, \im(\alpha) = \l
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\end{align*}
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\subsubsection{Bemerkung:}
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$\alpha$ invertierbar, $V=W \implies n = m = r \implies \alpha^+ = \alpha^{-1}$ \\
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$\alpha$ invertierbar, $V=W \implies n = m = r \implies \alpha^\dagger = \alpha^{-1}$ \\
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Wir haben eine echte Verallgemeinerung.
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\begin{defin}
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@ -3893,7 +3893,7 @@ Wir haben eine echte Verallgemeinerung.
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\]
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Dann heißt die Matrix
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\[
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A^+ = V^* \Sigma^+ U \in \K^{n \times m}, \Sigma^+ = \left( \begin{smallmatrix}
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A^\dagger = V^* \Sigma^\dagger U \in \K^{n \times m}, \Sigma^\dagger = \left( \begin{smallmatrix}
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\frac1{s_1} \\
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& \ddots \\
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& & \frac1{s_r} \\
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@ -3904,8 +3904,8 @@ Wir haben eine echte Verallgemeinerung.
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\]
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(Moore-Penrose) \underline{Pseudoinverse} von $A$.
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\item Sei $\alpha \in \homk(V, W), \dim(V), \dim(W) < \infty$ und $B, B'$ Orthonormalbasen mit
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${}_{B'} M(\alpha)_B = \Sigma$ und $\alpha^+$ so, dass ${}_B M(\alpha^+)_B = \Sigma^+$.
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Dann heißt $\alpha^+$ (Moore-Penrose) \underline{Pseudoinverse} von $\alpha$.
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${}_{B'} M(\alpha)_B = \Sigma$ und $\alpha^\dagger$ so, dass ${}_B M(\alpha^\dagger)_B = \Sigma^\dagger$.
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Dann heißt $\alpha^\dagger$ (Moore-Penrose) \underline{Pseudoinverse} von $\alpha$.
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\end{itemize}
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\end{defin}
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@ -3914,43 +3914,43 @@ Wir haben eine echte Verallgemeinerung.
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Seien $V, W$ endlich dimensional euklidische/unitäre Vektorräume, \\
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$\alpha \in \Hom(V, W)$. Dann gilt:
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\[
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\alpha^+ \text{ ist pseudoinverse} \iff \begin{aligned}
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& \alpha \circ \alpha^+ \circ \alpha = \alpha \\
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& \alpha^+ \circ \alpha \circ \alpha^+ = \alpha^+ \\
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& \alpha \circ \alpha^+ \text{ selbstadjungiert} \\
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& \alpha^+ \circ \alpha \text{ selbstadjungiert} \\
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\alpha^\dagger \text{ ist pseudoinverse} \iff \begin{aligned}
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& \alpha \circ \alpha^\dagger \circ \alpha = \alpha \\
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& \alpha^\dagger \circ \alpha \circ \alpha^\dagger = \alpha^\dagger \\
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& \alpha \circ \alpha^\dagger \text{ selbstadjungiert} \\
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& \alpha^\dagger \circ \alpha \text{ selbstadjungiert} \\
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\end{aligned}
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\]
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\end{satz}
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\begin{proof}
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Beweis über Matrizen, da äquivalent. Weiters die $\implies$ Richtung nur für \R.
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\begin{itemize}
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\item[$\implies$:] $A = U^T \Sigma V, A^+ = V^T \Sigma^+ U$
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\item[$\implies$:] $A = U^T \Sigma V, A^\dagger = V^T \Sigma^\dagger U$
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\[
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A A^+ = U^T \Sigma \underbrace{V V^T}_{=I} \Sigma^+ U = U^T \Sigma \Sigma^+ U =
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A A^\dagger = U^T \Sigma \underbrace{V V^T}_{=I} \Sigma^\dagger U = U^T \Sigma \Sigma^\dagger U =
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U^T \left( \begin{smallmatrix} 1 \\ & \ddots \\ & & 1 \\ & & & 0 \\ & & & & \ddots \\
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& & & & & 0 \end{smallmatrix} \right) U
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\]
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$A^+ A$.
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$A^\dagger A$.
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\[
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A A^+ A = U^T \Sigma \underbrace{V V^T}_I \Sigma^+ \underbrace{U U^T}_I \Sigma V
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= U^T \underbrace{\Sigma \Sigma^+ \Sigma}_\Sigma V = U^T \Sigma V = A
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A A^\dagger A = U^T \Sigma \underbrace{V V^T}_I \Sigma^\dagger \underbrace{U U^T}_I \Sigma V
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= U^T \underbrace{\Sigma \Sigma^\dagger \Sigma}_\Sigma V = U^T \Sigma V = A
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\]
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\item[$\impliedby$:] Steht noch aus
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% \begin{itemize}
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% \item \begin{equation} \label{eq:3.6.3.1}
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% \begin{aligned}
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% \ker(\alpha) = \ker(\alpha^+ \circ \alpha) && \im(\alpha) = \im(\alpha \circ \alpha^+) \\
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% \ker(\alpha^+) = \ker(\alpha \circ \alpha^+) && \im(\alpha^+) = \im(\alpha^+ \circ \alpha)
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% \ker(\alpha) = \ker(\alpha^\dagger \circ \alpha) && \im(\alpha) = \im(\alpha \circ \alpha^\dagger) \\
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% \ker(\alpha^\dagger) = \ker(\alpha \circ \alpha^\dagger) && \im(\alpha^\dagger) = \im(\alpha^\dagger \circ \alpha)
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% \end{aligned}
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% \end{equation}
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% \tl UE\br\,:
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% $\ker(\alpha) \subseteq \ker(\alpha^+ \circ \alpha) \subseteq \ker(\alpha \circ \alpha^+
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% \circ \alpha) = \ker(\alpha) \implies \ker(\alpha) = \ker(\alpha \circ \alpha^+ \circ \alpha)$
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% \item $\nu := \alpha^+ \circ \alpha$ ist Orthogonalprojektion auf $\ker(\alpha)^\bot$
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% $\ker(\alpha) \subseteq \ker(\alpha^\dagger \circ \alpha) \subseteq \ker(\alpha \circ \alpha^\dagger
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% \circ \alpha) = \ker(\alpha) \implies \ker(\alpha) = \ker(\alpha \circ \alpha^\dagger \circ \alpha)$
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% \item $\nu := \alpha^\dagger \circ \alpha$ ist Orthogonalprojektion auf $\ker(\alpha)^\bot$
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% \item $\nu$ selbstadjungiert $\implies \ker(v) \bot \im(v)$
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% \item $\nu \circ \nu = \alpha^+ \circ \underbrace{\alpha \circ \alpha^+ \circ \alpha}_\alpha
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% = \alpha^+ \circ \alpha = \nu$
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% \item $\nu \circ \nu = \alpha^\dagger \circ \underbrace{\alpha \circ \alpha^\dagger \circ \alpha}_\alpha
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% = \alpha^\dagger \circ \alpha = \nu$
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% \item $\forall u \in \im(\nu), v \in V$: \[
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% \inner{\nu(v) - v}u = \inner{\nu(v) - v}{\nu(w)} =
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% \inner{\nu^2(v) - \nu(v)}{w} = \inner{0}{w} = 0
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@ -3962,8 +3962,8 @@ Wir haben eine echte Verallgemeinerung.
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\begin{satz}
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Sei $\alpha \in \Hom(V, W)$.
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\begin{itemize}
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\item $\alpha$ injektiv $\implies \alpha^+ = (\alpha^* \circ \alpha)^{{}^{-1}} \circ \alpha^*$
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\item $\alpha$ surjektiv $\alpha^+ = \alpha^* \circ (\alpha \circ \alpha^*)^{{}^{-1}}$
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\item $\alpha$ injektiv $\implies \alpha^\dagger = (\alpha^* \circ \alpha)^{{}^{-1}} \circ \alpha^*$
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\item $\alpha$ surjektiv $\alpha^\dagger = \alpha^* \circ (\alpha \circ \alpha^*)^{{}^{-1}}$
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\end{itemize}
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\end{satz}
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\begin{proof}
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@ -3984,7 +3984,7 @@ Wir haben eine echte Verallgemeinerung.
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\beta = \beta$
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\item $\beta \circ \alpha, \alpha \circ \beta$ sind selbstadjungiert.
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\end{itemize}
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$\underset{\text{Satz \ref{theo:3.6.3}}}{\implies} \beta = \alpha^+$
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$\underset{\text{Satz \ref{theo:3.6.3}}}{\implies} \beta = \alpha^\dagger$
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\end{proof}
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\subsubsection{Anwendung: Methode der kleinsten Quadrate}
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@ -3999,13 +3999,13 @@ $w = \sum_{i=1}^n \mu_i b_i'$
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\norm{\alpha(v) - w}^2 = \norm{\sum_{i=1}^r s_i \lambda_i b_i' - \sum_{i=1}^m \mu_i b_i'}^2 =
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\sum_{i=1}^r (s_i \lambda_i - \mu_i)^2 + \sum_{i = r+1}^m \mu_i^2
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\]
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Wird minimal wenn $\lambda_i = \frac{\mu_i}{s_i}, i \in [r]$, insbesondere für $v = \alpha^+(w)$
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Wird minimal wenn $\lambda_i = \frac{\mu_i}{s_i}, i \in [r]$, insbesondere für $v = \alpha^\dagger(w)$
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\begin{satz}
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Sei $\alpha \in \homk(V, W), \K \in \{\R,\C\}, V, W$ endlich dimensional.
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Sei $w \in W$. Dann gilt mit $v^+ = \alpha^+(w)$ dass
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Sei $w \in W$. Dann gilt mit $v^\dagger = \alpha^\dagger(w)$ dass
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\[
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\norm{\alpha(v^+)-w} = \min_{v\in V} \norm{\alpha(v) - w}
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\norm{\alpha(v^\dagger)-w} = \min_{v\in V} \norm{\alpha(v) - w}
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\]
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Alle Vektoren mit dieser Eigenschaft erfüllen die\\ \underline{Normalgleichungen}
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$\ontop{\alpha^* \alpha(v) = \alpha^*(w)}{A^* A x = A^* b}$
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@ -4013,9 +4013,9 @@ Wird minimal wenn $\lambda_i = \frac{\mu_i}{s_i}, i \in [r]$, insbesondere für
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\begin{satz}
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Sei $\alpha \in \Hom(V, W), w \in \im(\alpha)$.
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Dann gilt mit $v^+ = \alpha^+ (w)$:
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Dann gilt mit $v^\dagger = \alpha^\dagger (w)$:
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\[
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\norm{v^+} = \min\{\norm v: \alpha(v) = w \}
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\norm{v^\dagger} = \min\{\norm v: \alpha(v) = w \}
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\]
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\end{satz}
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