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f4c36cb129 | ||
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32a9e0e08a |
@@ -15,20 +15,17 @@
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\begin{proof}
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\begin{proof}
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Sei $x \in \mathbb{K}^{n}$. Dann ist
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Sei $x \in \mathbb{K}^{n}$. Dann ist
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\begin{salign}
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\begin{salign*}
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\Vert (\mathbb{I} + B) x \Vert
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\Vert (\mathbb{I} + B) x \Vert
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&= \Vert x + B x\Vert \\
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&= \Vert x + B x\Vert \\
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&\stackrel{\text{Dreiecksungl.}}{\ge } \Vert x \Vert - \Vert Bx \Vert \\
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&\stackrel{\text{Dreiecksungl.}}{\ge } \Vert x \Vert - \Vert Bx \Vert \\
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&\stackrel{\Vert Bx \Vert \le \Vert B \Vert \Vert x \Vert}{\ge }
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&\stackrel{\Vert Bx \Vert \le \Vert B \Vert \Vert x \Vert}{\ge }
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\Vert x \Vert - \Vert B \Vert \cdot \Vert x \Vert \\
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\Vert x \Vert - \Vert B \Vert \cdot \Vert x \Vert \\
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&= ( \underbrace{1 - \Vert B \Vert}_{> 0}) \Vert x \Vert
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&= ( \underbrace{1 - \Vert B \Vert}_{> 0}) \Vert x \Vert
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.\end{salign}
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\intertext{Also hat die Gleichung $(\mathbb{I} + B) x = 0$ nur die Lösung $x = 0$, also
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Also hat die Gleichung $(\mathbb{I} + B) x = 0$ nur die Lösung $x = 0$, also
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ist $(\mathbb{I} + B)$ injektiv und mit \ref{lemma:linabb} regulär.
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ist $(\mathbb{I} + B)$ injektiv und mit \ref{lemma:linabb} regulär.
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Bleibt zu zeigen: $\Vert (\mathbb{I} + B)^{-1} \Vert \le \frac{1}{1 - \Vert B \Vert}$.
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Es gilt}
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Bleibt zu zeigen: $\Vert (\mathbb{I} + B)^{-1} \Vert \le \frac{1}{1 - \Vert B \Vert}$.
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Es gilt
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\begin{salign}
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1 &= \Vert \mathbb{I}\Vert \\
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1 &= \Vert \mathbb{I}\Vert \\
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&= \Vert (\mathbb{I} + B) (\mathbb{I} + B)^{-1} \Vert \\
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&= \Vert (\mathbb{I} + B) (\mathbb{I} + B)^{-1} \Vert \\
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&= \Vert (\mathbb{I} + B)^{-1} + B (\mathbb{I} + B)^{-1} \Vert \\
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&= \Vert (\mathbb{I} + B)^{-1} + B (\mathbb{I} + B)^{-1} \Vert \\
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@@ -36,7 +33,7 @@
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- \Vert B (\mathbb{I} + B)^{-1} \Vert \\
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- \Vert B (\mathbb{I} + B)^{-1} \Vert \\
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&\ge \Vert (\mathbb{I} + B)^{-1} \Vert - \Vert B \Vert \cdot \Vert (\mathbb{I} + B)^{-1} \Vert \\
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&\ge \Vert (\mathbb{I} + B)^{-1} \Vert - \Vert B \Vert \cdot \Vert (\mathbb{I} + B)^{-1} \Vert \\
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&= (1 - \Vert B \Vert) \Vert (\mathbb{I} + B)^{-1} \Vert
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&= (1 - \Vert B \Vert) \Vert (\mathbb{I} + B)^{-1} \Vert
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.\end{salign}
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.\end{salign*}
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Damit folgt die Behauptung.
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Damit folgt die Behauptung.
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\end{proof}
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\end{proof}
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+9
-2
@@ -162,10 +162,17 @@
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% uses regular expressions to calculate the widest stackrel
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% uses regular expressions to calculate the widest stackrel
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% to put additional padding on both sides of relation symbols
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% to put additional padding on both sides of relation symbols
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\NewEnviron{salign}
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\NewEnviron{salign}
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{
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\begin{align}
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\lec_insert_padding:V \BODY
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\end{align}
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}
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% starred version that does no equation numbering
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\NewEnviron{salign*}
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{
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{
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\begin{align*}
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\begin{align*}
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\lec_insert_padding:V \BODY
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\lec_insert_padding:V \BODY
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.\end{align*}
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\end{align*}
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}
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}
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% some helper variables
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% some helper variables
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@@ -210,7 +217,7 @@
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}
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}
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% replace all relations with align characters (&) and add the needed padding
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% replace all relations with align characters (&) and add the needed padding
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\regex_replace_all:nnN
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\regex_replace_all:nnN
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{ (&=|&\c{le}|&\c{ge}|&\c{stackrel}{.*?}{.*?}|&\c{neq}) }
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{ (\c{approx}&|&\c{approx}|\c{equiv}&|&\c{equiv}|=&|&=|\c{le}&|&\c{le}|\c{ge}&|&\c{ge}|&\c{stackrel}{.*?}{.*?}|\c{stackrel}{.*?}{.*?}&|&\c{neq}|\c{neq}&) }
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{ \c{kern} \u{l_tmp_dim_needed} \1 \c{kern} \u{l_tmp_dim_needed} }
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{ \c{kern} \u{l_tmp_dim_needed} \1 \c{kern} \u{l_tmp_dim_needed} }
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\l__lec_text_tl
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\l__lec_text_tl
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\l__lec_text_tl
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\l__lec_text_tl
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