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update la5

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christian
2019-11-20 20:22:12 +01:00
parent 952bb0985f
commit 902c7802fa
2 changed files with 8 additions and 7 deletions
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ws2019/la/uebungen/la5.pdf
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ws2019/la/uebungen/la5.tex
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@@ -2,6 +2,7 @@
\usepackage{enumerate} \usepackage{enumerate}
\usepackage{array} \usepackage{array}
\usepackage{mathtools}
\title{Übungsblatt Nr. 5} \title{Übungsblatt Nr. 5}
\author{Christian Merten, Mert Biyikli} \author{Christian Merten, Mert Biyikli}
@@ -202,10 +203,10 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
Bleibt zu zeigen: char$K \not\in \{2, \ldots, n+1\} \iff Bleibt zu zeigen: char$K \not\in \{2, \ldots, n+1\} \iff
k+1 \neq 0$ $\forall k \in \{0, \ldots, n\} $. k+1 \neq 0$ $\forall k \in \{0, \ldots, n\} $.
\begin{align*} \begin{align*}
&k + 1 \neq 0 \\ &k + 1 \neq 0 \\[-2mm]
\stackrel{k \ge 0}{\iff} & k + 1 \neq \text{char}K \\ \stackrel{\mathclap{\strut k \ge 0}}{\qquad \iff \qquad} &k + 1 \neq \text{char}K \\[-2mm]
\stackrel{1 \le k + 1 \le n + 1}\iff & \text{char}K = 0 \lor \text{char}K > n + 1 \\ \stackrel{\mathclap{\strut 1 \le k + 1 \le n + 1}}{\qquad \iff \qquad} &\text{char}K = 0 \lor \text{char}K > n + 1 \\[1mm]
\iff & \text{char}K \not\in \{2, \ldots, n+1\} \qquad \iff \qquad & \text{char}K \not\in \{2, \ldots, n+1\}
.\end{align*} .\end{align*}
\end{proof} \end{proof}
\item Bestimmen Sie $\psi(\text{ker }K) \subset K^{n+2}$. \item Bestimmen Sie $\psi(\text{ker }K) \subset K^{n+2}$.
@@ -239,7 +240,7 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
Damit folgt: Damit folgt:
\[ \[
\psi(\text{ker }\partial) = \psi(\text{ker }\partial) =
\{(a, \underbrace{0, \ldots, 0}_{n+1\text{-mal}}) \mid c \in K\} \{(a, \underbrace{0, \ldots, 0}_{n+1\text{-mal}}) \mid a \in K\}
.\] .\]
\item $\text{char }K \in \{2, \ldots, n+1\} $. Dann gilt für $k = \text{char }K-1$: \item $\text{char }K \in \{2, \ldots, n+1\} $. Dann gilt für $k = \text{char }K-1$:
\[ \[
@@ -321,12 +322,12 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
\begin{align*} \begin{align*}
(*(f_1 + f_2))(\varphi) &= ((f_1 + f_2)^{*})(\varphi) (*(f_1 + f_2))(\varphi) &= ((f_1 + f_2)^{*})(\varphi)
= \varphi \circ (f_1 + f_2) = \varphi \circ (f_1 + f_2)
= \varphi \circ f_1 + \varphi \circ f_2 \stackrel{\varphi \text{ linear}}{=} \varphi \circ f_1 + \varphi \circ f_2
= (*(f_1))(\varphi) + (*(f_2))(\varphi) \\ = (*(f_1))(\varphi) + (*(f_2))(\varphi) \\
(*(a f_1))(\varphi) (*(a f_1))(\varphi)
&= ((a f_1)^{*})(\varphi) &= ((a f_1)^{*})(\varphi)
= \varphi \circ (a f_1) = \varphi \circ (a f_1)
= a (\varphi \circ f_1) \stackrel{\varphi \text{ linear}}{=} a \cdot (\varphi \circ f_1)
= a\cdot (*(f_1))(\varphi) = a\cdot (*(f_1))(\varphi)
.\end{align*} .\end{align*}
\end{proof} \end{proof}