Add command for linear span
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LinAlg2.tex
85
LinAlg2.tex
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@ -81,6 +81,7 @@
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\newcommand\homkv{\Hom_\K(V, V)}
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\newcommand\homk{\Hom_\K}
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\newcommand\linspan[1]{\left\langle #1 \right\rangle}
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\newcommand\inner[2]{\left\langle #1, #2 \right\rangle}
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\newcommand\norm[1]{\left\lVert #1 \right\rVert}
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\newcommand\ontop[2]{\genfrac{}{}{0pt}{0}{#1}{#2}}
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@ -1030,7 +1031,7 @@ $\le\genfrac{}{}{0pt}{0}{\dim(V)}{n}$, da
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\bar1 & \bar4 & \bar0 \\
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\bar0 & \bar0 & \bar0
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\end{array}\right) \\
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& \implies \eig_{\bar2}(A) = \left\langle\begin{pmatrix}\bar1 \\ \bar1\end{pmatrix} \right\rangle \\
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& \implies \eig_{\bar2}(A) = \linspan{\begin{pmatrix}\bar1 \\ \bar1\end{pmatrix}} \\
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& \implies A \mathrlap{\text{ nicht diagonalisierbar [Lemma \ref{theo:2.1.4} (b)]}}
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\end{alignat*}
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@ -1254,15 +1255,15 @@ $A = \begin{pmatrix}
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& \sim \left( \begin{array}{c c c | c} -1 & 1 & 1 & 0 \\ 0 & -3 & 3 & 0 \\ 0 & 3 & -3 & 0 \end{array} \right)
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\sim \left( \begin{array}{c c c | c} 1 & -1 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)
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\sim \left( \begin{array}{c c c | c} 1 & 0 & -2 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \\
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& \implies \eig_A(3) = \left\langle \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \right\rangle
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& \implies \eig_A(3) = \linspan{\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}}
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\end{align*}
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$\lambda = -3$
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\begin{align*}
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& \left( \begin{array}{c c c | c} 1+3 & 2 & 2 & 0 \\ 2 & -2+3 & 1 & 0 \\ 2 & 1 & -2+3 & 0 \end{array} \right)
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= \left( \begin{array}{c c c | c} 4 & 2 & 2 & 0 \\ 2 & 1 & 1 & 0 \\ 2 & 1 & 1 & 0 \end{array} \right) \\
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& \sim \left( \begin{array}{c c c | c} 2 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \\
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& \implies \eig_A(-3) = \left\langle \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} \right\rangle
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= \left( \begin{array}{c c c | c} 4 & 2 & 2 & 0 \\ 2 & 1 & 1 & 0 \\ 2 & 1 & 1 & 0 \end{array} \right) \\
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& \sim \left( \begin{array}{c c c | c} 2 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \\
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& \implies \eig_A(-3) = \linspan {\begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}}
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\end{align*}
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\item \begin{align*}
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@ -1475,10 +1476,10 @@ $\underset{\mathrlap{\text{\dq fast alle Matrizen sind diagonalisierbar\dq}}}
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\lambda_1 & a_{12} & \dots & a_{1n} \\
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0 & \tl & & \\
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\vdots & & \tilde{A} & \\
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0 & & & \br\end{pmatrix} \\
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0 & & & \br\end{pmatrix} \\
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& \text{Sei }\beta: \begin{cases}
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\overbrace{\langle b_2, \dots, b_n\rangle}^{V}
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& \to \langle b_2, \dots, b_n\rangle \\
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\overbrace{\linspan {b_2, \dots, b_n}}^{V}
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& \to \linspan{ b_2, \dots, b_n} \\
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b_i
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& \mapsto \Phi^{-1}_{\tilde{B}}(C\cdot
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{}_{\tilde{B}}v)
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@ -1501,17 +1502,17 @@ $\underset{\mathrlap{\text{\dq fast alle Matrizen sind diagonalisierbar\dq}}}
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Wegen \ref{eq:2.2.22.1} gilt
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\begin{equation}
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\beta(\tilde{b}_i) \in
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\langle \tilde{b}_1, \dots, \tilde{b}_i \rangle
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\linspan{ \tilde{b}_1, \dots, \tilde{b}_i }
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\label{eq:2.2.22.2}
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\end{equation}
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Wir zeigen nun, dass für die Basis $C=(c_1, \dots, c_n)$ mit
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$c_1 = b_1, c_2 = \tilde{b}_2, \dots, c_n = \tilde{b}_i $
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die Matrix ${}_C M(\alpha)_C$ obere Dreiecksgestalt hat.
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Dies ist äquivalent zu
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\[\alpha(c_i)\in \langle c_1, \dots, c_n \rangle \forall i=1, \dots, n \]
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\[\alpha(c_i)\in \linspan{ c_1, \dots, c_n } \forall i=1, \dots, n \]
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\begin{itemize}
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\item [$i=1$:] $\alpha(c_1) = \alpha(b_1) = \lambda_1 b_1
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\in \langle b_1 \rangle = \langle c_1 \rangle$
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\in \linspan{ b_1 } = \linspan{c_1}$
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\item [$i>1$:]
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\begin{align*}
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& \alpha(c_i) = \alpha(\tilde{b}_i) =
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@ -1524,8 +1525,8 @@ $\underset{\mathrlap{\text{\dq fast alle Matrizen sind diagonalisierbar\dq}}}
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& = \sigma_i b_1+ \beta(\sum_{j=2}^n \mu_{ij} b_j)
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= \sigma_i b_1 + \beta(\tilde{b}_i) \\
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& \underbrace{\in}_{\text{\ref{eq:2.2.22.2}}}
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\langle b_1,\tilde{b}_2,\dots,\tilde{b}_i\rangle
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= \langle c_1, \dots, c_i \rangle
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\linspan{ b_1,\tilde{b}_2,\dots,\tilde{b}_i}
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= \linspan{ c_1, \dots, c_i }
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\end{align*}
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\end{itemize}
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\end{itemize}
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@ -1630,7 +1631,7 @@ Wir wollen zeigen, dass $\alpha/A$ genau dann eine Jordan-Normalform besitzt, we
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\begin{enumerate} [label=\arabic*)]
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\item $\alpha(C^m) \subseteq V_{m-1}$
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\item $\alpha(C^m)$ linear unabhängig
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\item $\langle \alpha(C^m) \rangle \cap V_{m-2} = \{0\}$
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\item $\linspan{ \alpha(C^m) } \cap V_{m-2} = \{0\}$
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\end{enumerate}
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\begin{proof}[Zwischenbeweis]
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\begin{itemize}
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@ -1650,16 +1651,16 @@ Wir wollen zeigen, dass $\alpha/A$ genau dann eine Jordan-Normalform besitzt, we
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\end{proof}
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Es folgt, dass
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\[
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\underbrace{V_{m-2} \oplus \langle \alpha(C^m) \rangle \oplus
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\overset{\langle D^{m-1} \rangle}{\langle C^{m-1} \rangle}}_{V_{m-1}} \oplus
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\overset{\langle D^m \rangle}{\langle C^m \rangle} = V
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\underbrace{V_{m-2} \oplus \linspan{ \alpha(C^m) } \oplus
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\overset{\linspan{ D^{m-1} }}{\linspan{C^{m-1}}}}_{V_{m-1}} \oplus
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\overset{\linspan{ D^m }}{\linspan{C^m}} = V
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\]
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Setze $D^m := C^m$ und definiere induktiv für $D^i \subseteq V_i$ die Menge
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$D^{i-1} := \alpha(D^i) \cup C^{i-1} \subseteq V_{i-1}$ sodass mit einer Basis $B^{i-2}$ von
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$V_{i-2}$ die Menge $B^{i-2} \cup D^{i-1}$ Basis von $V_{i-1}$ ist, also
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\[
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V_{i-2} \oplus \underbrace{\langle \alpha(D^i) \rangle \oplus \langle C^{i-1} \rangle}_
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{\langle D^i \rangle}
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V_{i-2} \oplus \underbrace{\linspan{ \alpha(D^i) } \oplus \linspan {C^{i-1}}}_
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{\linspan{ D^i }}
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= V_{i-1} \text{$\leftarrow$ das geht nach obiger Behauptung}
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\]
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Nach Konstruktion ist $(D^1, \dots, D^m)$ Basis von $V$.
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@ -1875,16 +1876,16 @@ Angenommen \(\alpha - \lambda \id: V \to V\) nilpotent. Dann besitzt \(\alpha\)
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\item
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\begin{enumerate}[label=\alph*)]
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\item Setze zunächst $v_i^k = b_i^k, i = 1, \dots, r_k$. $D_k := (v_1^k, \dots, v_{r_k}^k)$ \\
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Setze $v_i^{k-1} := (\alpha - \lambda \id)(v_i^k) \in \langle B_{k-1} \rangle,
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Setze $v_i^{k-1} := (\alpha - \lambda \id)(v_i^k) \in \linspan{ B_{k-1} },
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i = 1, \dots, r_k$ \\
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Ergänze gegebenenfalls $(v_1^{k-1}, \dots, v_{r_k}^{k-1},
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v_{r_{k+1}}^{k-1}, \dots, v_{r_{k-1}}^{k-1})=:D_{k-1}$, sodass \\
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$\langle D_{k-1} \rangle = \langle B_{k-1} \rangle$
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$\linspan{ D_{k-1} } = \linspan B_{k-1}$
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\item Führe 3a) iterativ aus. \\
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Setze $v_i^{l-1} := (\alpha - \lambda \id)(v_i^l), i = 1, \dots, r_l$ \\
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Ergänze gegebenenfalls
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$v_1^{l-1}, \dots, v_{r_l}^{l-1}, v_{r_{l+1}}^{l-1}, \dots, v_{r_{l-1}}^{l-1} =:D_{l-1}$,
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sodass $\langle D_{l-1} \rangle = \langle B_{l-1} \rangle$
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sodass $\linspan{ D_{l-1} } = \linspan{B_{l-1}}$
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\end{enumerate}
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\item Sei $B_\lambda = (D_1, \dots, D_k) \implies {}_{B_\lambda} M(\alpha|_{v_\lambda})_{B_\lambda}$ hat
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Jordan-Normalform mit Eigenwert $\lambda$.
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@ -1900,7 +1901,7 @@ Angenommen \(\alpha - \lambda \id: V \to V\) nilpotent. Dann besitzt \(\alpha\)
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0 & 0 & 0 & 1 & -1 \\
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0 & 0 & 0 & 0 & 1
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\end{pmatrix}
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, \chi_A(\lambda) = (\lambda - 1)^5 \\
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, \chi_A(\lambda) = (\lambda - 1)^5 \\
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& (A - 1\cdot I) = \begin{pmatrix}
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0 & 0 & 2 & 3 & 4 \\
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0 & 0 & 0 & -2 & -3 \\
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@ -1908,7 +1909,7 @@ Angenommen \(\alpha - \lambda \id: V \to V\) nilpotent. Dann besitzt \(\alpha\)
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0 & 0 & 0 & 0 & -1 \\
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0 & 0 & 0 & 0 & 0
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\end{pmatrix}
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\implies \ker(A - I) = \langle ( \underbrace{e_1, e_2}_{B_1} ) \rangle \\
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\implies \ker(A - I) = \linspan{ ( \underbrace{e_1, e_2}_{B_1} ) } \\
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& (A-I)^2 = \begin{pmatrix}
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0 & 0 & 0 & 0 & 1 \\
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0 & 0 & 0 & 0 & 2 \\
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@ -1916,9 +1917,9 @@ Angenommen \(\alpha - \lambda \id: V \to V\) nilpotent. Dann besitzt \(\alpha\)
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0
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\end{pmatrix}
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\implies \ker((A-I)^2) = \langle (\underbrace{e_1, e_2}_{B_1}, \underbrace{e_3, e_4}_{B_2}) \rangle \\
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\implies \ker((A-I)^2) = \linspan{ (\underbrace{e_1, e_2}_{B_1}, \underbrace{e_3, e_4}_{B_2}) } \\
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& (A-I)^3 = 0 \implies \ker((A-I)^3) =
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\langle(\underbrace{e_1, e_2}_{B_1}, \underbrace{e_3, e_4}_{B_2}, \underbrace{e_5}_{B_3}) \rangle \\
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\linspan{(\underbrace{e_1, e_2}_{B_1}, \underbrace{e_3, e_4}_{B_2}, \underbrace{e_5}_{B_3}) } \\
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& B_1 = (e_1, e_2), B_2 = (e_3, e_4), B_3 = (e_5)
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\end{align*}
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\begin{align*}
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@ -2312,7 +2313,7 @@ Auch skalare Produkte können eindeutig fortgesetzt werden.
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Sei $(a_1, a_2, \dots) \subseteq V$ linear unabhängig. Dann existiert genau ein Orthonormalsystem
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$(b_1, b_2, \dots)$ mit
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\begin{enumerate}[label=\roman*)]
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\item $\forall k: \langle a_1, \dots, a_k \rangle = \langle b_1, \dots, b_k \rangle =: U_k$
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\item $\forall k: \linspan{ a_1, \dots, a_k } = \linspan{b_1, \dots, b_k} =: U_k$
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\item Die Basistransformationsmatrix $M_k$ zwischen der Basen $(a_1, \dots, a_k)$ und $(b_1, \dots, b_k)$
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von $U_k$ hat positive Determinante.
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\end{enumerate}
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@ -2330,10 +2331,10 @@ Auch skalare Produkte können eindeutig fortgesetzt werden.
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& \forall i \in [n]: \inner{c_{n+1}}{b_i} = \inner{a_{n+1}}{b_i} -
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\sum\limits_{j=1}^n \inner{a_{n+1}}{b_j} \underbrace{\inner{b_j}{b_i}}_{\delta_{ij}} \\
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& = \inner{a_{n+1}}{b_i}
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- \inner{a_{n+1}}{b_i} = 0 \implies c_{n+1} \bot \langle b_1, \dots, b_n \rangle
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- \inner{a_{n+1}}{b_i} = 0 \implies c_{n+1} \bot \linspan{ b_1, \dots, b_n }
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\end{align*}
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$b_{n+1} = \dfrac{c_{n+1}}{\norm{c_{n+1}}} \implies (b_1, \dots, b_{n+1})$ Orthonormalsystem mit \\
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$\langle b_1, \dots, b_n \rangle = \langle a_1, \dots, a_n \rangle$
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$\linspan{ b_1, \dots, b_n } = \linspan{a_1, \dots, a_n}$
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\begin{align*}
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& b_1 = \mu_{11} a_1 \\
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& b_2 = \mu_{21} a_1 + \mu_{22} a_2 \\
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@ -2447,13 +2448,13 @@ $M^\bot$ ist immer Unterraum von $V$, selbst wenn $M$ kein Unterraum ist.
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\item $(b_1, \dots, b_r)$ Orthonormalbasis von $U$.\\
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$(b_1, \dots, b_r, b_{r+1}, \dots, b_n)$ Orthonormalbasis von $V$. \newline
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[die existiert laut Satz \ref{theo:3.1.17}] \\
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Behauptung: $U^\bot = \langle b_{r+1}, \dots, b_n \rangle$ \\
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Behauptung: $U^\bot = \linspan{ b_{r+1}, \dots, b_n }$ \\
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Beweis: $\subseteq$: Sei $v\in U^\bot, v = \sum_{i=1}^n \lambda_i b_i$
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\[
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\forall i \in [r]: 0 = \inner v{b_i} = \inner{\sum_{j=1}^n \lambda_j b_j}{b_i} =
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\sum_{j=1}^n \lambda_j \underbrace{\inner{b_j}{b_i}}_{\delta_{ij}} = \lambda_i
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\]
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$\supseteq: v\in \langle b_{r+1}, \dots, b_n \rangle \overset{!}{\implies} v \in U^\bot$ \\
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$\supseteq: v\in \linspan{ b_{r+1}, \dots, b_n } \overset{!}{\implies} v \in U^\bot$ \\
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$\implies \sum_{j=r+1}^n \lambda_j b_j, u = \sum_{i=1}^r \mu_i b_i \in U$ \\
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$\implies \inner vu = \sum_{j=r+1}^n \lambda_j \sum_{i=1}^r \underbrace{\inner{b_j}{b_i}}_{=0}=0$ \\
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$\implies$ a)
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@ -2704,12 +2705,12 @@ Behauptung: $\inner fp > 0$
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\item[$n=1$:] $\exists$ Eigenvektor $e_1 \in V \setminus\{0\}$ mit $\alpha(e_1) = \lambda e_1$.\\
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o.B.d.A.: $\norm{e_1} = 1 \implies v$ ist Orthonormalbasis aus Eigenvektoren
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\item[$n-1 \to n$:] $\exists$ Eigenvektor $e_1 \in V \setminus\{0\}$ mit $\alpha(e_1) = \lambda e_1$.\\
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o.B.d.A.: $\norm{e_1} = 1 \; U= \langle e_1 \rangle ^\bot$
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o.B.d.A.: $\norm{e_1} = 1 \; U= \linspan{ e_1 } ^\bot$
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\begin{itemize}
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\item $V = \langle e_1 \rangle \oplus U, \alpha(U) \overset{\text{!}}{\subseteq} U,
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\alpha(\langle e_1 \rangle) \subseteq \overset{\checkmark}{\langle}e_1 \rangle$
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\item $V = \linspan{ e_1 } \oplus U, \alpha(U) \overset{\text{!}}{\subseteq} U,
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\alpha(\linspan{e_1}) \overset{\checkmark}{\subseteq} \linspan{e_1 }$
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\end{itemize}
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$\implies \alpha = \alpha|_{\langle e_1 \rangle} \oplus \alpha|_U$ \\
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$\implies \alpha = \alpha|_{\linspan{ e_1 }} \oplus \alpha|_U$ \\
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Sei $v \in U\implies 0 = \inner{v}{e_1}
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\;\;\; [e_1 \in \eig_\alpha(\lambda) \iff e_1 \in \eig_{\alpha^*}(\overline\lambda)]$
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\begin{align*}
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@ -3031,8 +3032,8 @@ Das sind genau die Längen- und Winkelerhaltenden Abbildungen.
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Sei $l := \frac{v}{\norm v} \overset{\text{d)}}{\implies} \alpha(l)$ ist ONS
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$\implies \norm{\alpha(l)} = 1 \implies \norm{\alpha(v)} = \norm v$. \\
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Es folgt $\inner{\alpha(v)}{\alpha(w)} = \inner vw \checkmark$.
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\item $v, w$ linear unabhängig. Sei $(e_1, e_2)$ ONS mit $\langle\{e_1, e_2\}\rangle
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= \langle\{ v, w \}\rangle$.
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\item $v, w$ linear unabhängig. Sei $(e_1, e_2)$ ONS mit $\linspan{\{e_1, e_2\}}
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= \linspan{\{ v, w \}}$.
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(Gram-Schmidt liefert Existenz)
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\begin{align}
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\implies & (\alpha(e_1), \alpha(e_2)) \text{ ist ONS} \nonumber \\ \nonumber
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@ -3857,8 +3858,8 @@ Sei ${}_B M(\alpha)_B = \begin{pmatrix}s_1 \\
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& & & 0 \\
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& & & & \ddots \\
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& & & & & 0\end{pmatrix}$ \\
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$\implies \ker(\alpha) = \langle b_{r+1}, \dots, b_n \rangle_V, \im(\alpha) = \langle b'_1, \dots b_r' \rangle_W,
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\ker(\alpha)^\bot = \langle b_1, \dots, b_r \rangle_V$
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$\implies \ker(\alpha) = \linspan{ b_{r+1}, \dots, b_n }_V, \im(\alpha) = \linspan{b'_1, \dots b_r'},
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\ker(\alpha)^\bot = \linspan{ b_1, \dots, b_r }_V$
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\begin{align*}
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\alpha: V & \to
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@ -3971,7 +3972,7 @@ Wir haben eine echte Verallgemeinerung.
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& = \inner{\nu(v) - \nu(v)}{w} = 0
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\end{align*}
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\end{enumerate}
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$\implies (b_1, \dots, b_n)$ ONB mit $\langle b_{r+1}, \dots, b_n \rangle_V = \ker(\nu) =
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$\implies (b_1, \dots, b_n)$ ONB mit $\linspan{ b_{r+1}, \dots, b_n }_V = \ker(\nu) =
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\ker(\alpha)$
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\begin{align*}
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& \sum_{i=1}^n \lambda_i b_i & & \overset{\nu}{\mapsto} \sum_{i=1}^r \lambda_i b_i
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@ -4052,7 +4053,7 @@ Wir haben eine echte Verallgemeinerung.
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\end{pmatrix}
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\end{align*}
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Es gilt $\ker(\alpha^\dagger) = \ker(\nu') = \im(\alpha)^\bot
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= \langle b_{r+1}', \dots, b_r' \rangle \implies a^\dagger_{i\_} = 0 \forall i > r$
|
||||
= \linspan{ b_{r+1}', \dots, b_r' } \implies a^\dagger_{i\_} = 0 \forall i > r$
|
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\[
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||||
\implies {}_B M(\alpha^\dagger)_{B'} =
|
||||
\begin{pmatrix}
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|
|
@ -4102,7 +4103,7 @@ Sei $Ax = b$ Lineares Gleichungssystem mit $L(A,b) = \emptyset$. Versuche ein $x
|
|||
$\norm{Ax-b}_{\K^m}$ minimal, $\norm{\alpha(v) - w}$ minimal.
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Sei $b_1, \dots, b_n$ Orthonormalbasis von $V$, \\
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$b_1', \dots, b_m'$ ONB von $W$.
|
||||
$\langle b_1, \dots, b_r \rangle = \ker(\alpha)^\bot, \langle b_1', \dots b_r'\rangle = \im(\alpha)$
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||||
$\linspan{ b_1, \dots, b_r } = \ker(\alpha)^\bot, \linspan {b_1', \dots b_r'} = \im(\alpha)$
|
||||
$v = \sum_{i=1}^n \lambda_i b_i \implies \alpha(v) = \sum_{i=1}^r s_i \lambda_i b_i'$
|
||||
$w = \sum_{i=1}^n \mu_i b_i'$
|
||||
|
||||
|
|
|
|||
Loading…
Reference in New Issue