master > master: Formatierung
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@ -1475,7 +1475,7 @@ durch
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1 &8 &6 &-3\\
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2 &14 &\alpha &-2\\
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\end{matrix}
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&\mathbf{b}_{\beta} &:= &\begin{vector} 4\\ 0\\ \beta\\\end{vector}
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&\mathbf{b}_{\beta} &:= &\begin{vector} 4\\ 0\\ \beta\\\end{vector}
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\end{mathe}
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gegeben sind.
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@ -1574,15 +1574,15 @@ Dies führt zu einem Fallunterschied:
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Zusammengefasst erhalten wir die allgemeine Form der Lösung:
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\begin{mathe}[mc]{rcl}
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\mathbf{x} &= &\begin{svector} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\\end{svector}\\
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&= &\begin{svector} 32 + 26x_{3} + -13x_{4}\\ -4 - 4x_{3} + 2x_{4}\\ x_{3}\\ x_{4}\\\end{svector}\\
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&= &\begin{svector} 32 + 26x_{3} + -13x_{4}\\ -4 - 4x_{3} + 2x_{4}\\ 0 + 1x_{3} + 0x_{4}\\ 0 + 0x_{3} + 1x_{4}\\\end{svector}\\
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&= &\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ \begin{svector} 26x_{3}\\ -4x_{3}\\ 1x_{3}\\ 0x_{3}\\\end{svector}
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+ \begin{svector} -13x_{4}\\ 2x_{4}\\ 1x_{4}\\ 1x_{4}\\\end{svector}\\
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&= &\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ x_{3}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}
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+ x_{4}\cdot\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}\\
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\mathbf{x} &= &\begin{svector} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\\end{svector}\\
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&= &\begin{svector} 32 + 26x_{3} + -13x_{4}\\ -4 - 4x_{3} + 2x_{4}\\ x_{3}\\ x_{4}\\\end{svector}\\
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&= &\begin{svector} 32 + 26x_{3} + -13x_{4}\\ -4 - 4x_{3} + 2x_{4}\\ 0 + 1x_{3} + 0x_{4}\\ 0 + 0x_{3} + 1x_{4}\\\end{svector}\\
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&= &\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ \begin{svector} 26x_{3}\\ -4x_{3}\\ 1x_{3}\\ 0x_{3}\\\end{svector}
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+ \begin{svector} -13x_{4}\\ 2x_{4}\\ 1x_{4}\\ 1x_{4}\\\end{svector}\\
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&= &\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ x_{3}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}
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+ x_{4}\cdot\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}\\
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\end{mathe}
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mit $x_{3}$, $x_{4}$ frei wählbar.
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@ -1591,14 +1591,14 @@ Dies führt zu einem Fallunterschied:
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Also erhalten wird in diesem Falle
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$\boxed{
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L_{\alpha,\beta}=\left\{
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\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ t_{1}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}
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+ t_{2}\cdot\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}
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\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ t_{1}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}
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+ t_{2}\cdot\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}
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\mid t_{1}, t_{2}\in\reell
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\right\}
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}$,
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oder etwas kompakter formuliert,
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${L_{\alpha,\beta}=\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector} + \vectorspacespan\left\{\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}, \begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}\right\}}$.
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${L_{\alpha,\beta}=\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector} + \vectorspacespan\left\{\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}, \begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}\right\}}$.
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\end{enumerate}
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%% FALL 2
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@ -1617,12 +1617,12 @@ Dies führt zu einem Fallunterschied:
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das heißt
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\begin{mathe}[mc]{rcl}
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\mathbf{x} &= &\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ x_{3}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}
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+ x_{4}\cdot\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}\\
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&= &\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ \frac{\beta-8}{\alpha-4}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}
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+ x_{4}\cdot\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector},\\
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\mathbf{x} &= &\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ x_{3}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}
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+ x_{4}\cdot\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}\\
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&= &\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ \frac{\beta-8}{\alpha-4}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}
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+ x_{4}\cdot\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector},\\
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\end{mathe}
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wobei $x_{4}$ frei wählbar ist.
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@ -1631,14 +1631,14 @@ Dies führt zu einem Fallunterschied:
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Also erhalten wird in diesem Falle
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$\boxed{
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L_{\alpha,\beta}=\left\{
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\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ \frac{\beta-8}{\alpha-4}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}
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+ t\cdot\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}
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\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}
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+ \frac{\beta-8}{\alpha-4}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}
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+ t\cdot\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}
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\mid t\in\reell
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\right\}
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}$,
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oder etwas kompakter formuliert,
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${L_{\alpha,\beta}=\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector} + \frac{\beta-8}{\alpha-4}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector} + \vectorspacespan\left\{\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}\right\}}$.
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${L_{\alpha,\beta}=\begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector} + \frac{\beta-8}{\alpha-4}\cdot\begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector} + \vectorspacespan\left\{\begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}\right\}}$.
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\end{enumerate}
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Wir fassen die Lösung für alle Fälle zusammen:
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@ -1653,9 +1653,9 @@ Wir fassen die Lösung für alle Fälle zusammen:
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für alle $\alpha,\beta\in\reell$,
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wobei
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$\mathbf{u} = \begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}$,
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$\mathbf{v} = \begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}$,
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$\mathbf{w} = \begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}$.
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$\mathbf{u} = \begin{svector} 32\\ -4\\ 0\\ 0\\\end{svector}$,
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$\mathbf{v} = \begin{svector} 26\\ -4\\ 1\\ 0\\\end{svector}$,
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$\mathbf{w} = \begin{svector} -13\\ 2\\ 1\\ 1\\\end{svector}$.
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%% AUFGABE 1-2
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\let\altsectionname\sectionname
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@ -2391,10 +2391,10 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
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Betrachte die folgenden Vektoren in $\reell^{3}$:
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\begin{mathe}[mc]{rclqrclqrclqrcl}
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\mathbf{v} &= &\begin{svector} 0\\ 0\\ 0\\\end{svector},
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&\mathbf{v}^{\prime} &= &\begin{svector} 1\\ 0\\ 0\\\end{svector},
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&\mathbf{w} &= &\begin{svector} 0\\ 1\\ 0\\\end{svector},
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&\mathbf{w}^{\prime} &= &\begin{svector} 0\\ 1\\ 1\\\end{svector}.\\
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\mathbf{v} &= &\begin{svector} 0\\ 0\\ 0\\\end{svector},
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&\mathbf{v}^{\prime} &= &\begin{svector} 1\\ 0\\ 0\\\end{svector},
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&\mathbf{w} &= &\begin{svector} 0\\ 1\\ 0\\\end{svector},
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&\mathbf{w}^{\prime} &= &\begin{svector} 0\\ 1\\ 1\\\end{svector}.\\
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\end{mathe}
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Bis auf 2-Dimensionalität erfüllen diese die Voraussetzungen in \Cref{satz:main:ueb:2:ex:2a}.
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@ -2474,7 +2474,7 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
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gelten muss, da $\gamma_{1}\neq\gamma_{2}$.
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Eingesetzt in die erste Gleichung oben liefert
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$2x+y=\gamma\cdot 0=0$.
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Darum muss $\begin{svector} x\\ y\\\end{svector}$
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Darum muss $\begin{svector} x\\ y\\\end{svector}$
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das LGS $(A|\mathbf{b})$ lösen, wobei
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\begin{mathe}[mc]{rclqrcl}
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@ -2482,7 +2482,7 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
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1 &-3\\
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2 &1\\
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\end{smatrix},
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&\mathbf{b} &= &\begin{svector} 7\\ 0\\\end{svector}
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&\mathbf{b} &= &\begin{svector} 7\\ 0\\\end{svector}
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\end{mathe}
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\begin{algorithm}[\rtab][\rtab]
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@ -2635,10 +2635,10 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
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Wir arbeiten im Vektorraum $\reell^{3}$ und betrachten die Vektoren
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\begin{mathe}[mc]{rclqrclqrclqrcl}
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\mathbf{v}_{1} &= &\begin{svector} 1\\ 3\\ 1\\\end{svector}
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&\mathbf{v}_{2} &= &\begin{svector} -2\\ 5\\ -2\\\end{svector}
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&\mathbf{w}_{1} &= &\begin{svector} 4\\ -3\\ -3\\\end{svector}
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&\mathbf{w}_{2} &= &\begin{svector} 0\\ 1\\ 1\\\end{svector}\\
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\mathbf{v}_{1} &= &\begin{svector} 1\\ 3\\ 1\\\end{svector}
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&\mathbf{v}_{2} &= &\begin{svector} -2\\ 5\\ -2\\\end{svector}
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&\mathbf{w}_{1} &= &\begin{svector} 4\\ -3\\ -3\\\end{svector}
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&\mathbf{w}_{2} &= &\begin{svector} 0\\ 1\\ 1\\\end{svector}\\
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\end{mathe}
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\textbf{Zu berechnen:}
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@ -2746,7 +2746,7 @@ und daraus die Parameter abzulesen.
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Also ist die homogene Lösung gegeben durch
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\begin{mathe}[mc]{rcl}
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\mathbf{t} &= &\beta\begin{svector} 28\\ -8\\ -11\\ -77\\\end{svector},
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\mathbf{t} &= &\beta\begin{svector} 28\\ -8\\ -11\\ -77\\\end{svector},
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\quad\text{mit $\beta\in\reell$ frei wählbar}.
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\end{mathe}
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\end{algorithm}
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@ -2766,7 +2766,7 @@ Wir können nun \eqcref{eq:0:ueb:3:ex:1} fortsetzen und erhalten
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\mathbf{\xi}=t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
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\,\text{und}\,
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\exists{\beta\in\reell:~}
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\mathbf{t}=\beta\begin{svector} 28\\ -8\\ -11\\ -77\\\end{svector}\\
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\mathbf{t}=\beta\begin{svector} 28\\ -8\\ -11\\ -77\\\end{svector}\\
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&\Longleftrightarrow
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&\exists{\beta\in\reell:~}
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\mathbf{\xi}=\beta\cdot(
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@ -2782,10 +2782,10 @@ Es gilt
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\begin{mathe}[mc]{rcccccl}
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\mathbf{u}
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&= &28\begin{svector} 1\\ 3\\ 1\\\end{svector}
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-8\begin{svector} -2\\ 5\\ -2\\\end{svector}
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&= &\begin{svector} 44\\ 44\\ 44\\\end{svector}
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&= &44\begin{svector} 1\\ 1\\ 1\\\end{svector}.\\
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&= &28\begin{svector} 1\\ 3\\ 1\\\end{svector}
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-8\begin{svector} -2\\ 5\\ -2\\\end{svector}
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&= &\begin{svector} 44\\ 44\\ 44\\\end{svector}
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&= &44\begin{svector} 1\\ 1\\ 1\\\end{svector}.\\
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\end{mathe}
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Aus \eqcref{eq:1:ueb:3:ex:1} ergibt sich der zu berechnende Untervektorraum
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@ -2796,8 +2796,8 @@ als
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\cap\vectorspacespan\{\mathbf{w}_{1},\mathbf{w}_{2}\}
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&= &U
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&= &\vectorspacespan\{\mathbf{u}\}
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&= &\vectorspacespan\{44\begin{svector} 1\\ 1\\ 1\\\end{svector}\}
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&= &\vectorspacespan\{\begin{svector} 1\\ 1\\ 1\\\end{svector}\}.\\
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&= &\vectorspacespan\{44\begin{svector} 1\\ 1\\ 1\\\end{svector}\}
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&= &\vectorspacespan\{\begin{svector} 1\\ 1\\ 1\\\end{svector}\}.\\
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\end{mathe}
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%% AUFGABE 3-2
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@ -4164,7 +4164,7 @@ für alle $A,B\in R$.
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Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
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\begin{mathe}[mc]{rcl}
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z\in\kmplx &\mapsto &\begin{svector} \ReTeil(z)\\ \ImTeil(z)\\\end{svector}\in\reell^{2},\\
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z\in\kmplx &\mapsto &\begin{svector} \ReTeil(z)\\ \ImTeil(z)\\\end{svector}\in\reell^{2},\\
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\mathbf{x}\in\reell^{2} &\mapsto &x_{1}+\imageinh x_{2}\in\kmplx.\\
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\end{mathe}
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@ -4173,12 +4173,12 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
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\item
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\begin{claim*}
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Für alle $z\in\kmplx\ohne\{0\}$ existieren eindeutige Werte $r\in(0,\infty)$ und $\alpha\in[0,2\pi)$,
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dann $z=r\cdot\begin{svector} \cos(\alpha)\\ \sin(\alpha)\\\end{svector}$ (unter der o.\,s. Identifizierung).
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dann $z=r\cdot\begin{svector} \cos(\alpha)\\ \sin(\alpha)\\\end{svector}$ (unter der o.\,s. Identifizierung).
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\end{claim*}
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\begin{proof}
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Unter der Identifizierung können wir $z=\begin{svector} x\\ y\\\end{svector}$ schreiben, wobei $x,y\in\reell$.
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Da $z\neq 0=\begin{svector} 0\\ 0\\\end{svector}$, muss entweder $x\neq 0$ oder $y\neq 0$ gelten.
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Unter der Identifizierung können wir $z=\begin{svector} x\\ y\\\end{svector}$ schreiben, wobei $x,y\in\reell$.
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Da $z\neq 0=\begin{svector} 0\\ 0\\\end{svector}$, muss entweder $x\neq 0$ oder $y\neq 0$ gelten.
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Zur {\bfseries Existenz}:
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Sei $r:=\sqrt{x^{2}+y^{2}}$. Dann $r>0$ weil $(x,y)\neq (0,0)$.\\
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@ -4226,15 +4226,15 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
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\end{kompaktenum}
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Darum gilt in allen Fällen
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$r\cdot\begin{svector} \cos(\alpha)\\ \sin(\alpha)\\\end{svector}
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=\begin{svector} r\cos(\alpha)\\ r\sin(\alpha)\\\end{svector}
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=\begin{svector} x\\ y\\\end{svector}
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$r\cdot\begin{svector} \cos(\alpha)\\ \sin(\alpha)\\\end{svector}
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=\begin{svector} r\cos(\alpha)\\ r\sin(\alpha)\\\end{svector}
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=\begin{svector} x\\ y\\\end{svector}
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=z$.
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Zur {\bfseries Eindeutigkeit}:
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Seien $r_{i}\in(0,\infty)$, $\alpha_{i}\in[0,2\pi)$
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mit
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$r_{i}\cdot\begin{svector} \cos(\alpha_{i})\\ \sin(\alpha_{i})\\\end{svector}=z$
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$r_{i}\cdot\begin{svector} \cos(\alpha_{i})\\ \sin(\alpha_{i})\\\end{svector}=z$
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für $i\in\{1,2\}$.
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\textbf{Zu zeigen:} $r_{1}=r_{2}$ und $\alpha_{1}=\alpha_{2}$.
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Es gilt
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@ -4294,10 +4294,10 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
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\item
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\begin{claim*}
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Seien $z_{1},z_{2}\in\kmplx\ohne\{0\}$ mit Darstellungen
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$z_{i}=r_{i}\cdot\begin{svector} \cos(\alpha_{i})\\ \sin(\alpha_{i})\\\end{svector}$
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$z_{i}=r_{i}\cdot\begin{svector} \cos(\alpha_{i})\\ \sin(\alpha_{i})\\\end{svector}$
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für $i\in\{1,2\}$.
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Dann gilt die Rechenregel
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$z_{1}z_{2}=r_{1}r_{2}\cdot\begin{svector} \cos(\alpha_{1}+\alpha_{2})\\ \sin(\alpha_{1}+\alpha_{2})\\\end{svector}$.
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$z_{1}z_{2}=r_{1}r_{2}\cdot\begin{svector} \cos(\alpha_{1}+\alpha_{2})\\ \sin(\alpha_{1}+\alpha_{2})\\\end{svector}$.
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\end{claim*}
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\begin{proof}
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@ -4305,10 +4305,10 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
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\begin{longmathe}[mc]{RCL}
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z_{1}z_{2}
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&= &\begin{svector} \ReTeil(z_{1})\ReTeil(z_{2})-\ImTeil(z_{1})\ImTeil(z_{2})\\ \ReTeil(z_{1})\ImTeil(z_{2})+\ReTeil(z_{1})\ImTeil(z_{2})\\\end{svector}\\
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&= &\begin{svector} r_{1}\cos(\alpha_{1})\cdot r_{2}\cos(\alpha_{2})-r_{1}\sin(\alpha_{1})\cdot r_{2}\sin(\alpha_{2})\\ r_{1}\cos(\alpha_{1})\cdot r_{2}\sin(\alpha_{2})+r_{1}\cos(\alpha_{1})\cdot r_{2}\sin(\alpha_{2})\\\end{svector}\\
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&= &r_{1}r_{2}\begin{svector} \cos(\alpha_{1})\cos(\alpha_{2})-\sin(\alpha_{1})\sin(\alpha_{2})\\ \cos(\alpha_{1})\sin(\alpha_{2})+\cos(\alpha_{1})\sin(\alpha_{2})\\\end{svector}\\
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&= &r_{1}r_{2}\begin{svector} \cos(\alpha_{1}+\alpha_{2})\\ \sin(\alpha_{1}+\alpha_{2})\\\end{svector}.\\
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&= &\begin{svector} \ReTeil(z_{1})\ReTeil(z_{2})-\ImTeil(z_{1})\ImTeil(z_{2})\\ \ReTeil(z_{1})\ImTeil(z_{2})+\ReTeil(z_{1})\ImTeil(z_{2})\\\end{svector}\\
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&= &\begin{svector} r_{1}\cos(\alpha_{1})\cdot r_{2}\cos(\alpha_{2})-r_{1}\sin(\alpha_{1})\cdot r_{2}\sin(\alpha_{2})\\ r_{1}\cos(\alpha_{1})\cdot r_{2}\sin(\alpha_{2})+r_{1}\cos(\alpha_{1})\cdot r_{2}\sin(\alpha_{2})\\\end{svector}\\
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&= &r_{1}r_{2}\begin{svector} \cos(\alpha_{1})\cos(\alpha_{2})-\sin(\alpha_{1})\sin(\alpha_{2})\\ \cos(\alpha_{1})\sin(\alpha_{2})+\cos(\alpha_{1})\sin(\alpha_{2})\\\end{svector}\\
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&= &r_{1}r_{2}\begin{svector} \cos(\alpha_{1}+\alpha_{2})\\ \sin(\alpha_{1}+\alpha_{2})\\\end{svector}.\\
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\end{longmathe}
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Die letzte Vereinfachung folgt aus der trigonometrischen Additionsregel.
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@ -4317,9 +4317,9 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
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\item
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\begin{claim*}[de Moivre]
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Sei $z\in\kmplx\ohne\{0\}$ mit Darstellungen
|
||||
$z=r\cdot\begin{svector} \cos(\alpha)\\ \sin(\alpha)\\\end{svector}$.
|
||||
$z=r\cdot\begin{svector} \cos(\alpha)\\ \sin(\alpha)\\\end{svector}$.
|
||||
Dann gilt die Potenzregel
|
||||
$z^{n}=r^{n}\cdot\begin{svector} \cos(n\alpha)\\ \sin(n\alpha)\\\end{svector}$
|
||||
$z^{n}=r^{n}\cdot\begin{svector} \cos(n\alpha)\\ \sin(n\alpha)\\\end{svector}$
|
||||
für alle $n\in\ntrlpos$.
|
||||
\end{claim*}
|
||||
|
||||
@ -4331,10 +4331,10 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
|
||||
Die Gleichung gilt offensichtlich für $n=1$.
|
||||
|
||||
\item[\uwave{{\bfseries Induktionsvoraussetzung:}}]
|
||||
Sei $n>1$. Angenommen, $z^{n-1}=r^{n-1}\cdot\begin{svector} \cos((n-1)\alpha)\\ \sin((n-1)\alpha)\\\end{svector}$.
|
||||
Sei $n>1$. Angenommen, $z^{n-1}=r^{n-1}\cdot\begin{svector} \cos((n-1)\alpha)\\ \sin((n-1)\alpha)\\\end{svector}$.
|
||||
|
||||
\item[\uwave{{\bfseries Induktionsschritt:}}]
|
||||
\textbf{Zu zeigen:} $z^{n}=r^{n}\cdot\begin{svector} \cos(n\alpha)\\ \sin(n\alpha)\\\end{svector}$.\\
|
||||
\textbf{Zu zeigen:} $z^{n}=r^{n}\cdot\begin{svector} \cos(n\alpha)\\ \sin(n\alpha)\\\end{svector}$.\\
|
||||
Per rekursive Definition vom Potenzieren
|
||||
gilt zunächst $z^{n}=z^{n-1}\cdot z$ (Multiplikation innerhalb der Algebra $\kmplx$).
|
||||
Aufgabe 6-2(b) zur Folge gilt somit
|
||||
@ -4342,11 +4342,11 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
|
||||
\begin{mathe}[mc]{rcl}
|
||||
z^{n}=z^{n-1}\cdot z
|
||||
&\textoverset{IV}{=}
|
||||
&r^{n-1}\cdot\begin{svector} \cos((n-1)\alpha)\\ \sin((n-1)\alpha)\\\end{svector}
|
||||
\cdot r\cdot\begin{svector} \cos(\alpha)\\ \sin(\alpha)\\\end{svector}\\
|
||||
&r^{n-1}\cdot\begin{svector} \cos((n-1)\alpha)\\ \sin((n-1)\alpha)\\\end{svector}
|
||||
\cdot r\cdot\begin{svector} \cos(\alpha)\\ \sin(\alpha)\\\end{svector}\\
|
||||
&\textoverset{(2b)}{=}
|
||||
&r^{n-1}r\cdot\begin{svector} \cos((n-1)\alpha+\alpha)\\ \sin((n-1)\alpha+\alpha)\\\end{svector}\\
|
||||
&= &r^{n}\cdot\begin{svector} \cos(n\alpha)\\ \sin(n\alpha)\\\end{svector}.\\
|
||||
&r^{n-1}r\cdot\begin{svector} \cos((n-1)\alpha+\alpha)\\ \sin((n-1)\alpha+\alpha)\\\end{svector}\\
|
||||
&= &r^{n}\cdot\begin{svector} \cos(n\alpha)\\ \sin(n\alpha)\\\end{svector}.\\
|
||||
\end{mathe}
|
||||
|
||||
Darum gilt die Gleichung für $n$.
|
||||
@ -4362,21 +4362,21 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
|
||||
$%
|
||||
z^{0}
|
||||
=1+\imageinh 0
|
||||
=\begin{svector} 1\\ 0\\\end{svector}
|
||||
=r^{0}\cdot\begin{svector} \cos(0\alpha)\\ \sin(0\alpha)\\\end{svector}%
|
||||
=\begin{svector} 1\\ 0\\\end{svector}
|
||||
=r^{0}\cdot\begin{svector} \cos(0\alpha)\\ \sin(0\alpha)\\\end{svector}%
|
||||
$.
|
||||
Für $n=-1$ liefert uns die Rechenregel für Multiplikation innerhalb $\kmplx$,
|
||||
dass
|
||||
$r^{-1}\cdot\begin{svector} \cos(-\alpha)\\ \sin(-\alpha)\\\end{svector}$
|
||||
$r^{-1}\cdot\begin{svector} \cos(-\alpha)\\ \sin(-\alpha)\\\end{svector}$
|
||||
eine hinreichende Konstruktion für ein Inverses von $z$ ist,
|
||||
und darum ist dies wegen Eindeutigkeit des Inverses gleich $z^{-1}$.
|
||||
Für $n<0$ allgemein wenden wir schließlich
|
||||
$%
|
||||
z^{n}
|
||||
=(z^{-1})^{|n|}
|
||||
=(r^{-1}\cdot\begin{svector} \cos(-\alpha)\\ \sin(-\alpha)\\\end{svector})^{|n|}
|
||||
=(r^{-1})^{|n|}\cdot\begin{svector} \cos(|n|\cdot-\alpha)\\ \sin(|n|\cdot-\alpha)\\\end{svector}
|
||||
=r^{n}\cdot\begin{svector} \cos(n\alpha)\\ \sin(n\alpha)\\\end{svector}%
|
||||
=(r^{-1}\cdot\begin{svector} \cos(-\alpha)\\ \sin(-\alpha)\\\end{svector})^{|n|}
|
||||
=(r^{-1})^{|n|}\cdot\begin{svector} \cos(|n|\cdot-\alpha)\\ \sin(|n|\cdot-\alpha)\\\end{svector}
|
||||
=r^{n}\cdot\begin{svector} \cos(n\alpha)\\ \sin(n\alpha)\\\end{svector}%
|
||||
$
|
||||
an.
|
||||
\end{rem*}
|
||||
@ -4713,9 +4713,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
|
||||
\item
|
||||
\begin{claim*}
|
||||
Sei $K=\rtnl$. Dann sind
|
||||
${\mathbf{v}_{1}=\begin{svector} 1\\ 2\\ 2\\\end{svector}}$,
|
||||
${\mathbf{v}_{2}=\begin{svector} 3\\ 2\\ 1\\\end{svector}}$, und
|
||||
${\mathbf{v}_{3}=\begin{svector} 2\\ 1\\ -1\\\end{svector}}$
|
||||
${\mathbf{v}_{1}=\begin{svector} 1\\ 2\\ 2\\\end{svector}}$,
|
||||
${\mathbf{v}_{2}=\begin{svector} 3\\ 2\\ 1\\\end{svector}}$, und
|
||||
${\mathbf{v}_{3}=\begin{svector} 2\\ 1\\ -1\\\end{svector}}$
|
||||
über $K$ \fbox{linear unabhängig}.
|
||||
\end{claim*}
|
||||
|
||||
@ -4771,9 +4771,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
|
||||
\item
|
||||
\begin{claim*}
|
||||
Sei $K=\mathbb{F}_{5}$. Dann sind
|
||||
${\mathbf{v}_{1}=\begin{svector} 1\\ 2\\ 2\\\end{svector}}$,
|
||||
${\mathbf{v}_{2}=\begin{svector} 3\\ 2\\ 1\\\end{svector}}$, und
|
||||
${\mathbf{v}_{3}=\begin{svector} 2\\ 1\\ 4\\\end{svector}}$
|
||||
${\mathbf{v}_{1}=\begin{svector} 1\\ 2\\ 2\\\end{svector}}$,
|
||||
${\mathbf{v}_{2}=\begin{svector} 3\\ 2\\ 1\\\end{svector}}$, und
|
||||
${\mathbf{v}_{3}=\begin{svector} 2\\ 1\\ 4\\\end{svector}}$
|
||||
über $K$ \fbox{nicht linear unabhängig}.
|
||||
\end{claim*}
|
||||
|
||||
@ -4814,9 +4814,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
|
||||
\item
|
||||
\begin{claim*}
|
||||
Sei $K=\kmplx$. Dann sind
|
||||
${\mathbf{v}_{1}=\begin{svector} 1\\ \imageinh\\ 0\\\end{svector}}$,
|
||||
${\mathbf{v}_{2}=\begin{svector} 1+\imageinh\\ -\imageinh\\ 1-2\imageinh\\\end{svector}}$, und
|
||||
${\mathbf{v}_{3}=\begin{svector} \imageinh\\ 1-\imageinh\\ 2-\imageinh\\\end{svector}}$
|
||||
${\mathbf{v}_{1}=\begin{svector} 1\\ \imageinh\\ 0\\\end{svector}}$,
|
||||
${\mathbf{v}_{2}=\begin{svector} 1+\imageinh\\ -\imageinh\\ 1-2\imageinh\\\end{svector}}$, und
|
||||
${\mathbf{v}_{3}=\begin{svector} \imageinh\\ 1-\imageinh\\ 2-\imageinh\\\end{svector}}$
|
||||
über $K$ \fbox{nicht linear unabhängig}.
|
||||
\end{claim*}
|
||||
|
||||
@ -4871,9 +4871,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
|
||||
|
||||
\begin{claim*}
|
||||
Sei $K=\reell$. Dann sind
|
||||
${\mathbf{v}_{1}=\begin{svector} 1\\ 0\\ 0\\ 1\\ 0\\ 0\\\end{svector}}$,
|
||||
${\mathbf{v}_{2}=\begin{svector} 1\\ 1\\ 0\\ -1\\ 1\\ -2\\\end{svector}}$, und
|
||||
${\mathbf{v}_{3}=\begin{svector} 0\\ 1\\ 1\\ -1\\ 2\\ -1\\\end{svector}}$
|
||||
${\mathbf{v}_{1}=\begin{svector} 1\\ 0\\ 0\\ 1\\ 0\\ 0\\\end{svector}}$,
|
||||
${\mathbf{v}_{2}=\begin{svector} 1\\ 1\\ 0\\ -1\\ 1\\ -2\\\end{svector}}$, und
|
||||
${\mathbf{v}_{3}=\begin{svector} 0\\ 1\\ 1\\ -1\\ 2\\ -1\\\end{svector}}$
|
||||
über $K$ \fbox{linear unabhängig}.
|
||||
\end{claim*}
|
||||
|
||||
@ -5045,14 +5045,14 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
und betrachte die Vektoren
|
||||
|
||||
\begin{mathe}[mc]{rclqrcl}
|
||||
\mathbf{v}_{1}=\mathbf{v}_{2}=\ldots=\mathbf{v}_{n-1} &:= &\begin{svector} 1\\ 0\\\end{svector}
|
||||
&\mathbf{v}_{n} &:= &\begin{svector} 0\\ 1\\\end{svector}\\
|
||||
\mathbf{v}_{1}=\mathbf{v}_{2}=\ldots=\mathbf{v}_{n-1} &:= &\begin{svector} 1\\ 0\\\end{svector}
|
||||
&\mathbf{v}_{n} &:= &\begin{svector} 0\\ 1\\\end{svector}\\
|
||||
\end{mathe}
|
||||
|
||||
Dann gilt $\mathbf{v}_{n}\notin\vectorspacespan(\mathbf{v}_{1},\mathbf{v}_{2},\ldots,\mathbf{v}_{n-1})$,
|
||||
weil $\vectorspacespan(\mathbf{v}_{1},\mathbf{v}_{2},\ldots,\mathbf{v}_{n-1})
|
||||
=\vectorspacespan(\mathbf{v}_{1})
|
||||
=\{\begin{svector} t\\ 0\\\end{svector}\mid t\in K\}\notni \begin{svector} 0\\ 1\\\end{svector}$.
|
||||
=\{\begin{svector} t\\ 0\\\end{svector}\mid t\in K\}\notni \begin{svector} 0\\ 1\\\end{svector}$.
|
||||
Andererseits sind die $n-1\geq 2$ Vektoren,
|
||||
$\mathbf{v}_{1},\mathbf{v}_{2},\ldots,\mathbf{v}_{n}$
|
||||
per Wahl nicht linear unabhängig (weil die alle gleich sind).
|
||||
@ -5079,9 +5079,9 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
Sei $V=K^{2}$ und betrachte
|
||||
|
||||
\begin{mathe}[mc]{rclqrclqrcl}
|
||||
\mathbf{v}_{1} &:= &\begin{svector} 0\\ 1\\\end{svector},
|
||||
&\mathbf{v}_{2} &:= &\begin{svector} 1\\ 0\\\end{svector},
|
||||
&\mathbf{v}_{3} &:= &\begin{svector} 1\\ 1\\\end{svector}.\\
|
||||
\mathbf{v}_{1} &:= &\begin{svector} 0\\ 1\\\end{svector},
|
||||
&\mathbf{v}_{2} &:= &\begin{svector} 1\\ 0\\\end{svector},
|
||||
&\mathbf{v}_{3} &:= &\begin{svector} 1\\ 1\\\end{svector}.\\
|
||||
\end{mathe}
|
||||
|
||||
Es ist einfach zu sehen, dass die echten Teilsysteme
|
||||
@ -5269,13 +5269,13 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
\begin{mathe}[mc]{rcl}
|
||||
x_{2}:=1,\,x_{3}:=0,\,x_{4}:=0
|
||||
&\Longrightarrow
|
||||
&\mathbf{x}=\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},\\
|
||||
&\mathbf{x}=\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},\\
|
||||
x_{2}:=0,\,x_{3}:=1,\,x_{4}:=0
|
||||
&\Longrightarrow
|
||||
&\mathbf{x}=\begin{svector} 2\\ 0\\ 1\\ 0\\\end{svector},\\
|
||||
&\mathbf{x}=\begin{svector} 2\\ 0\\ 1\\ 0\\\end{svector},\\
|
||||
x_{2}:=0,\,x_{3}:=0,\,x_{4}:=1
|
||||
&\Longrightarrow
|
||||
&\mathbf{x}=\begin{svector} -2\\ 0\\ 0\\ 1\\\end{svector}.\\
|
||||
&\mathbf{x}=\begin{svector} -2\\ 0\\ 0\\ 1\\\end{svector}.\\
|
||||
\end{mathe}
|
||||
\end{algorithm}
|
||||
|
||||
@ -5286,9 +5286,9 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
&= &\{\mathbf{x}\in V\mid A_{1}\mathbf{x}=\zerovector\}
|
||||
&= &\vectorspacespan\underbrace{
|
||||
\left\{
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 2\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} -2\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 2\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} -2\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\right\}
|
||||
}_{=:B_{1}}\\
|
||||
\end{mathe}
|
||||
@ -5317,13 +5317,13 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
\begin{mathe}[mc]{rcl}
|
||||
x_{2}:=1,\,x_{3}:=0,\,x_{4}:=0
|
||||
&\Longrightarrow
|
||||
&\mathbf{x}=\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},\\
|
||||
&\mathbf{x}=\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},\\
|
||||
x_{2}:=0,\,x_{3}:=1,\,x_{4}:=0
|
||||
&\Longrightarrow
|
||||
&\mathbf{x}=\begin{svector} 1\\ 0\\ 1\\ 0\\\end{svector},\\
|
||||
&\mathbf{x}=\begin{svector} 1\\ 0\\ 1\\ 0\\\end{svector},\\
|
||||
x_{2}:=0,\,x_{3}:=0,\,x_{4}:=1
|
||||
&\Longrightarrow
|
||||
&\mathbf{x}=\begin{svector} 1\\ 0\\ 0\\ 1\\\end{svector}.\\
|
||||
&\mathbf{x}=\begin{svector} 1\\ 0\\ 0\\ 1\\\end{svector}.\\
|
||||
\end{mathe}
|
||||
\end{algorithm}
|
||||
|
||||
@ -5334,9 +5334,9 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
&= &\{\mathbf{x}\in V\mid A_{2}\mathbf{x}=\zerovector\}
|
||||
&= &\vectorspacespan\underbrace{
|
||||
\left\{
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 1\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} 1\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 1\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} 1\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\right\}
|
||||
}_{=:B_{2}}\\
|
||||
\end{mathe}
|
||||
@ -5374,10 +5374,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
\begin{mathe}[mc]{rcl}
|
||||
x_{2}:=1,\,x_{4}:=0
|
||||
&\Longrightarrow
|
||||
&\mathbf{x}=\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},\\
|
||||
&\mathbf{x}=\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},\\
|
||||
x_{2}:=0,\,x_{4}:=1
|
||||
&\Longrightarrow
|
||||
&\mathbf{x}=\begin{svector} 4\\ 0\\ 3\\ 1\\\end{svector}.\\
|
||||
&\mathbf{x}=\begin{svector} 4\\ 0\\ 3\\ 1\\\end{svector}.\\
|
||||
\end{mathe}
|
||||
|
||||
gegeben.
|
||||
@ -5390,8 +5390,8 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
&= &\{\mathbf{x}\in V\mid A_{3}\mathbf{x}=\zerovector\}
|
||||
&= &\vectorspacespan\underbrace{
|
||||
\left\{
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 4\\ 0\\ 3\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 4\\ 0\\ 3\\ 1\\\end{svector}
|
||||
\right\}
|
||||
}_{=:B_{3}}\\
|
||||
\end{mathe}
|
||||
@ -5406,21 +5406,21 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
\begin{mathe}[mc]{rcl}
|
||||
U_{1}+U_{2}
|
||||
&= &\vectorspacespan\big\{
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 2\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} -2\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 2\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} -2\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\big\}
|
||||
+\vectorspacespan\big\{
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 1\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} 1\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 1\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} 1\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\big\}\\
|
||||
&= &\vectorspacespan\big\{
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 2\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} -2\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} 1\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 2\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} -2\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} 1\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\big\}.\\
|
||||
\end{mathe}
|
||||
|
||||
@ -5492,10 +5492,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
|
||||
\begin{mathe}[mc]{c}
|
||||
\left\{
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 2\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} -2\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 0\\ 1\\ 0\\\end{svector}
|
||||
\begin{svector} 1\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 2\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} -2\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 0\\ 1\\ 0\\\end{svector}
|
||||
\right\}\\
|
||||
\end{mathe}
|
||||
|
||||
@ -5520,10 +5520,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
|
||||
\begin{mathe}[mc]{c}
|
||||
\left\{
|
||||
\begin{svector} 1\\ 0\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 0\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 0\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} 0\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\begin{svector} 1\\ 0\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 0\\ 1\\ 0\\ 0\\\end{svector},
|
||||
\begin{svector} 0\\ 0\\ 1\\ 0\\\end{svector},
|
||||
\begin{svector} 0\\ 0\\ 0\\ 1\\\end{svector}
|
||||
\right\}\\
|
||||
\end{mathe}
|
||||
|
||||
@ -5853,8 +5853,8 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
mit der kanonischen Basis
|
||||
|
||||
\begin{mathe}[mc]{cqc}
|
||||
\mathbf{v}_{1}:=\begin{svector} 1\\ 0\\\end{svector},
|
||||
&\mathbf{v}_{2}:=\begin{svector} 0\\ 1\\\end{svector}.\\
|
||||
\mathbf{v}_{1}:=\begin{svector} 1\\ 0\\\end{svector},
|
||||
&\mathbf{v}_{2}:=\begin{svector} 0\\ 1\\\end{svector}.\\
|
||||
\end{mathe}
|
||||
|
||||
\Cref{claim:2-3:ueb:8:ex:3} zufolge ist
|
||||
@ -5862,8 +5862,8 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
\begin{mathe}[mc]{cqcqcqc}
|
||||
\mathbf{v}_{1},
|
||||
&\mathbf{v}_{2},
|
||||
&\mathbf{v}_{3}:=\imageinh\cdot\mathbf{v}_{1} = \begin{svector} \imageinh\\ 0\\\end{svector},
|
||||
&\mathbf{v}_{4}:=\imageinh\cdot\mathbf{v}_{2} = \begin{svector} 0\\ \imageinh\\\end{svector}.\\
|
||||
&\mathbf{v}_{3}:=\imageinh\cdot\mathbf{v}_{1} = \begin{svector} \imageinh\\ 0\\\end{svector},
|
||||
&\mathbf{v}_{4}:=\imageinh\cdot\mathbf{v}_{2} = \begin{svector} 0\\ \imageinh\\\end{svector}.\\
|
||||
\end{mathe}
|
||||
|
||||
eine Basis für $W:=\kmplx^{2}$, wenn dies als $\reell$-Vektorraum betrachtet wird.
|
||||
@ -5887,11 +5887,11 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
Seien
|
||||
|
||||
\begin{mathe}[mc]{cqcqcqcqc}
|
||||
u_{1} := \begin{svector} 1\\ 2\\ -1\\ 1\\\end{svector},
|
||||
&u_{2} := \begin{svector} -1\\ -2\\ 1\\ 2\\\end{svector},
|
||||
&v_{1} := \begin{svector} 1\\ 2\\ -1\\ -2\\\end{svector},
|
||||
&v_{2} := \begin{svector} -1\\ 3\\ 0\\ -2\\\end{svector},
|
||||
&v_{3} := \begin{svector} 2\\ -1\\ -1\\ 1\\\end{svector}.\\
|
||||
u_{1} := \begin{svector} 1\\ 2\\ -1\\ 1\\\end{svector},
|
||||
&u_{2} := \begin{svector} -1\\ -2\\ 1\\ 2\\\end{svector},
|
||||
&v_{1} := \begin{svector} 1\\ 2\\ -1\\ -2\\\end{svector},
|
||||
&v_{2} := \begin{svector} -1\\ 3\\ 0\\ -2\\\end{svector},
|
||||
&v_{3} := \begin{svector} 2\\ -1\\ -1\\ 1\\\end{svector}.\\
|
||||
\end{mathe}
|
||||
|
||||
Vektoren in $\reell^{4}$ und setze
|
||||
@ -6136,9 +6136,9 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
|
||||
Seien
|
||||
|
||||
\begin{mathe}[mc]{cqcqcqcqc}
|
||||
\mathbf{v}_{1} := \begin{svector} 1\\ -2\\ 3\\ 1\\\end{svector},
|
||||
&\mathbf{v}_{2} := \begin{svector} 2\\ -5\\ 7\\ 0\\\end{svector},
|
||||
&\mathbf{v}_{3} := \begin{svector} -2\\ 6\\ -9\\ -3\\\end{svector}.\\
|
||||
\mathbf{v}_{1} := \begin{svector} 1\\ -2\\ 3\\ 1\\\end{svector},
|
||||
&\mathbf{v}_{2} := \begin{svector} 2\\ -5\\ 7\\ 0\\\end{svector},
|
||||
&\mathbf{v}_{3} := \begin{svector} -2\\ 6\\ -9\\ -3\\\end{svector}.\\
|
||||
\end{mathe}
|
||||
|
||||
Vektoren in $\reell^{4}$ und sei $\phi:\reell^{3}\to\reell^{4}$
|
||||
@ -6538,11 +6538,11 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
|
||||
Seien
|
||||
|
||||
\begin{mathe}[mc]{ccccc}
|
||||
v_{1} = \begin{vector} 2\\ 1\\\end{vector},
|
||||
&v_{2} = \begin{vector} -1\\ 1\\\end{vector},
|
||||
&w_{1} = \begin{vector} 1\\ 1\\ 0\\\end{vector},
|
||||
&w_{2} = \begin{vector} 1\\ -1\\ 2\\\end{vector},
|
||||
&w_{3} = \begin{vector} 0\\ 3\\ -1\\\end{vector}.
|
||||
v_{1} = \begin{vector} 2\\ 1\\\end{vector},
|
||||
&v_{2} = \begin{vector} -1\\ 1\\\end{vector},
|
||||
&w_{1} = \begin{vector} 1\\ 1\\ 0\\\end{vector},
|
||||
&w_{2} = \begin{vector} 1\\ -1\\ 2\\\end{vector},
|
||||
&w_{3} = \begin{vector} 0\\ 3\\ -1\\\end{vector}.
|
||||
\end{mathe}
|
||||
|
||||
Zuerst beoachten wir dass
|
||||
@ -6614,9 +6614,9 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
|
||||
|
||||
\begin{mathe}[mc]{rcccccl}
|
||||
\phi(x)
|
||||
&= &\begin{vector} 2x_{1}\\ -x_{2}\\\end{vector}
|
||||
&= &\begin{vector} 2\\ 0\\\end{vector}x_{1}
|
||||
+ \begin{vector} 0\\ -1\\\end{vector}x_{2}
|
||||
&= &\begin{vector} 2x_{1}\\ -x_{2}\\\end{vector}
|
||||
&= &\begin{vector} 2\\ 0\\\end{vector}x_{1}
|
||||
+ \begin{vector} 0\\ -1\\\end{vector}x_{2}
|
||||
&= &\underbrace{
|
||||
\begin{matrix}{rr}
|
||||
2 &0\\
|
||||
@ -6700,9 +6700,9 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
|
||||
|
||||
\begin{mathe}[mc]{rcccccl}
|
||||
\phi(x)
|
||||
&= &\begin{vector} x_{1}+x_{2}\\ 5x_{2}-x_{1}\\ 2x_{1}-4x_{2}\\\end{vector}
|
||||
&= &\begin{vector} 1\\ -1\\ 2\\\end{vector}x_{1}
|
||||
+ \begin{vector} 1\\ 5\\ -4\\\end{vector}x_{2}
|
||||
&= &\begin{vector} x_{1}+x_{2}\\ 5x_{2}-x_{1}\\ 2x_{1}-4x_{2}\\\end{vector}
|
||||
&= &\begin{vector} 1\\ -1\\ 2\\\end{vector}x_{1}
|
||||
+ \begin{vector} 1\\ 5\\ -4\\\end{vector}x_{2}
|
||||
&= &\underbrace{
|
||||
\begin{matrix}{rr}
|
||||
1 &1\\
|
||||
@ -6810,10 +6810,10 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
|
||||
|
||||
\begin{mathe}[mc]{rcccccl}
|
||||
\phi(x)
|
||||
&= &\begin{vector} 2x_{2}\\ 3x_{1}-4x_{3}\\\end{vector}
|
||||
&= &\begin{vector} 0\\ 3\\\end{vector}x_{1}
|
||||
+ \begin{vector} 2\\ 0\\\end{vector}x_{2}
|
||||
+ \begin{vector} 0\\ -4\\\end{vector}x_{3}
|
||||
&= &\begin{vector} 2x_{2}\\ 3x_{1}-4x_{3}\\\end{vector}
|
||||
&= &\begin{vector} 0\\ 3\\\end{vector}x_{1}
|
||||
+ \begin{vector} 2\\ 0\\\end{vector}x_{2}
|
||||
+ \begin{vector} 0\\ -4\\\end{vector}x_{3}
|
||||
&= &\underbrace{
|
||||
\begin{matrix}{rrr}
|
||||
0 &2 &0\\
|
||||
@ -7062,20 +7062,20 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
|
||||
|
||||
\begin{mathe}[mc]{rcl}
|
||||
M^{\cal{B}}_{\cal{B}}(\phi)
|
||||
&= &\begin{matrix}{ccccc}
|
||||
&= &\begin{matrix}{rrrrr}
|
||||
\choose{0}{0}a^{0} &\choose{1}{0}a^{1} &\choose{2}{0}a^{2} &\choose{3}{0}a^{3} &\choose{4}{0}a^{4}\\
|
||||
0 &\choose{1}{1}a^{0} &\choose{2}{1}a^{1} &\choose{3}{1}a^{2} &\choose{4}{1}a^{3}\\
|
||||
0 &0 &\choose{2}{2}a^{0} &\choose{3}{2}a^{1} &\choose{4}{2}a^{2}\\
|
||||
0 &0 &0 &\choose{3}{3}a^{0} &\choose{4}{3}a^{1}\\
|
||||
0 &0 &0 &0 &\choose{4}{4}a^{0}\\
|
||||
\end{matrix}\\
|
||||
&= &\begin{matrix}{ccccc}
|
||||
&= &\begin{matrix}{rrrrr}
|
||||
1 &a &a^{2} &a^{3} &a^{4}\\
|
||||
0 &1 &2a &3a^{2} &4a^{3}\\
|
||||
0 &0 &1 &3a &6a^{2}\\
|
||||
0 &0 &0 &1 &4a\\
|
||||
0 &0 &0 &0 &1\\
|
||||
\end{matrix}
|
||||
\end{matrix}.
|
||||
\end{mathe}
|
||||
|
||||
Als \textbf{Alternative} können wir (davon ausgehend, dass $\phi$ linear ist)
|
||||
@ -7088,27 +7088,27 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
|
||||
&= &(x+a)^{0}
|
||||
&= &
|
||||
1\cdot x^{0}
|
||||
&\cong &\begin{svector} 1\\ 0\\ 0\\ 0\\ 0\\\end{svector}\\
|
||||
&\cong &\begin{svector} 1\\ 0\\ 0\\ 0\\ 0\\\end{svector}\\
|
||||
\phi(x^{1})
|
||||
&= &(x+a)^{1}
|
||||
&= &
|
||||
a\cdot x^{0} + 1\cdot x^{1}
|
||||
&\cong &\begin{svector} a\\ 1\\ 0\\ 0\\ 0\\\end{svector}\\
|
||||
&\cong &\begin{svector} a\\ 1\\ 0\\ 0\\ 0\\\end{svector}\\
|
||||
\phi(x^{2})
|
||||
&= &(x+a)^{2}
|
||||
&= &
|
||||
a^{2}\cdot x^{0} + 2a\cdot x^{1} + 1\cdot x^{2}
|
||||
&\cong &\begin{svector} a^{2}\\ 2a\\ 1\\ 0\\ 0\\\end{svector}\\
|
||||
&\cong &\begin{svector} a^{2}\\ 2a\\ 1\\ 0\\ 0\\\end{svector}\\
|
||||
\phi(x^{3})
|
||||
&= &(x+a)^{3}
|
||||
&= &
|
||||
a^{3}\cdot x^{0} + 3a^{2}\cdot x^{1} + 3a\cdot x^{2} + 1\cdot x^{3}
|
||||
&\cong &\begin{svector} a^{3}\\ 3a^{2}\\ 3a\\ 1\\ 0\\\end{svector}\\
|
||||
&\cong &\begin{svector} a^{3}\\ 3a^{2}\\ 3a\\ 1\\ 0\\\end{svector}\\
|
||||
\phi(x^{4})
|
||||
&= &(x+a)^{4}
|
||||
&= &
|
||||
a^{4}\cdot x^{0} + 4a^{3}\cdot x^{1} + 6a^{2}\cdot x^{2} + 4a\cdot x^{3} + 1\cdot x^{4}
|
||||
&\cong &\begin{svector} a^{4}\\ 4a^{3}\\ 6a^{2}\\ 4a\\ 1\\\end{svector}\\
|
||||
&\cong &\begin{svector} a^{4}\\ 4a^{3}\\ 6a^{2}\\ 4a\\ 1\\\end{svector}\\
|
||||
\end{mathe}
|
||||
|
||||
Wenn wir diese Spalten in eine Matrix aufführen,
|
||||
@ -9120,7 +9120,7 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
|
||||
&= &\boxed{11}.\\
|
||||
\end{mathe}
|
||||
|
||||
Also ist \fbox{$\mathbf{x}=\begin{svector} 10\\ 11\\\end{svector}$}
|
||||
Also ist \fbox{$\mathbf{x}=\begin{svector} 10\\ 11\\\end{svector}$}
|
||||
die Lösung des LGS in $\intgr/13\intgr$.
|
||||
\end{algorithm}
|
||||
|
||||
@ -9251,8 +9251,8 @@ Sei $L$ die Gerade $\{\mathbf{v}+t\mathbf{w}\mid t\in\reell\}\subseteq\reell^{3}
|
||||
wobei
|
||||
|
||||
\begin{mathe}[mc]{rclqrcl}
|
||||
\mathbf{v} &= &\begin{svector} -4\\ 2\\ 5\\\end{svector},
|
||||
&\mathbf{w} &= &\begin{svector} 2\\ -6\\ 12\\\end{svector}.\\
|
||||
\mathbf{v} &= &\begin{svector} -4\\ 2\\ 5\\\end{svector},
|
||||
&\mathbf{w} &= &\begin{svector} 2\\ -6\\ 12\\\end{svector}.\\
|
||||
\end{mathe}
|
||||
|
||||
\begin{enumerate}{\bfseries (1)}
|
||||
@ -9260,7 +9260,7 @@ wobei
|
||||
\item
|
||||
|
||||
\begin{claim*}
|
||||
Der Punkt, $\mathbf{x}=\begin{svector} -3\\ -1\\ 11\\\end{svector}$, liegt in der Geraden, $L$.
|
||||
Der Punkt, $\mathbf{x}=\begin{svector} -3\\ -1\\ 11\\\end{svector}$, liegt in der Geraden, $L$.
|
||||
\end{claim*}
|
||||
|
||||
\begin{proof}
|
||||
@ -9276,7 +9276,7 @@ wobei
|
||||
\mathbf{x}-\mathbf{v}=t\mathbf{w}\\
|
||||
&\Longleftrightarrow
|
||||
&\exists{t\in\reell:~}
|
||||
\begin{svector} 1\\ -3\\ 6\\\end{svector}=t\begin{svector} 2\\ -6\\ 12\\\end{svector}\\
|
||||
\begin{svector} 1\\ -3\\ 6\\\end{svector}=t\begin{svector} 2\\ -6\\ 12\\\end{svector}\\
|
||||
\end{mathe}
|
||||
|
||||
Nun ist die letzte Aussage wahr,
|
||||
@ -9291,7 +9291,7 @@ wobei
|
||||
Z.\,B. können wir
|
||||
|
||||
\begin{mathe}[mc]{rcl}
|
||||
\mathbf{w}_{\perp} &= &\begin{svector} 3\\ -1\\ 0\\\end{svector}\\
|
||||
\mathbf{w}_{\perp} &= &\begin{svector} 3\\ -1\\ 0\\\end{svector}\\
|
||||
\end{mathe}
|
||||
|
||||
wählen. Dann gilt $\brkt{\mathbf{w},\mathbf{w}_{\perp}}=0$,
|
||||
@ -9467,7 +9467,7 @@ Wir betrachten die Komposition ${g\circ f:X\to Z}$
|
||||
|
||||
\begin{claim*}
|
||||
Seien $n\in\ntrlpos$ und $p\in\mathbb{P}$ mit $n<p\leq 2n$.
|
||||
Dann gilt $p\divides\begin{svector} 2n\\ n\\\end{svector}$.
|
||||
Dann gilt $p\divides\begin{svector} 2n\\ n\\\end{svector}$.
|
||||
\end{claim*}
|
||||
|
||||
\begin{proof}
|
||||
@ -9479,23 +9479,23 @@ Wir betrachten die Komposition ${g\circ f:X\to Z}$
|
||||
\prod_{i=n+1}^{2n}i
|
||||
&= &\dfrac{\prod_{i=1}^{2n}i}{n!}
|
||||
&= &n!\dfrac{(2n)!}{n!(2n-n)!}
|
||||
&= &n!\begin{vector} 2n\\ n\\\end{vector}.\\
|
||||
&= &n!\begin{vector} 2n\\ n\\\end{vector}.\\
|
||||
\end{mathe}
|
||||
|
||||
Aus (i) und \eqcref{eq:1:quiz:5:ex:1} folgt also
|
||||
(ii)~$p\divides \begin{svector} 2n\\ n\\\end{svector}\cdot n!$.\\
|
||||
Beachte, dass $p$ eine Primzahl ist und ${n!,\begin{svector} 2n\\ n\\\end{svector}\in\intgr}$.\\
|
||||
(ii)~$p\divides \begin{svector} 2n\\ n\\\end{svector}\cdot n!$.\\
|
||||
Beachte, dass $p$ eine Primzahl ist und ${n!,\begin{svector} 2n\\ n\\\end{svector}\in\intgr}$.\\
|
||||
Aus (ii) und \cite[Satz 3.4.14]{sinn2020} folgt also
|
||||
$p\divides\begin{svector} 2n\\ n\\\end{svector}$ oder $p\divides n!$.\\
|
||||
$p\divides\begin{svector} 2n\\ n\\\end{svector}$ oder $p\divides n!$.\\
|
||||
|
||||
\fbox{Angenommen, $p\ndivides\begin{svector} 2n\\ n\\\end{svector}$.}\\
|
||||
\fbox{Angenommen, $p\ndivides\begin{svector} 2n\\ n\\\end{svector}$.}\\
|
||||
Dann muss laut des o.\,s. Arguments ${p\divides n!(=\prod_{i=1}^{n}i)}$ gelten.\\
|
||||
Eine weitere Anwendung von \cite[Satz 3.4.14]{sinn2020} liefert,
|
||||
dass ${p\divides i_{0}}$ für ein $i_{0}\in\{1,2,\ldots,n\}$.\\
|
||||
Aber dann gilt $1\leq p\leq i_{0}\leq n$.
|
||||
Das widerspricht der Voraussetzung, dass $n<p$.\\
|
||||
Darum stimmt die Annahme nicht.
|
||||
Das heißt, $p\divides\begin{svector} 2n\\ n\\\end{svector}$.
|
||||
Das heißt, $p\divides\begin{svector} 2n\\ n\\\end{svector}$.
|
||||
\end{proof}
|
||||
|
||||
%% ********************************************************************************
|
||||
@ -9601,9 +9601,9 @@ Wir betrachten die Komposition ${g\circ f:X\to Z}$
|
||||
Seien
|
||||
|
||||
\begin{mathe}[mc]{rclqrclqrcl}
|
||||
\mathbf{v}_{1} &= &\begin{svector} 1\\ 2\\ 2\\\end{svector}
|
||||
&\mathbf{v}_{2} &= &\begin{svector} 0\\ 2\\ 1\\\end{svector}
|
||||
&\mathbf{v}_{3} &= &\begin{svector} 2\\ 1\\ 1\\\end{svector}\\
|
||||
\mathbf{v}_{1} &= &\begin{svector} 1\\ 2\\ 2\\\end{svector}
|
||||
&\mathbf{v}_{2} &= &\begin{svector} 0\\ 2\\ 1\\\end{svector}
|
||||
&\mathbf{v}_{3} &= &\begin{svector} 2\\ 1\\ 1\\\end{svector}\\
|
||||
\end{mathe}
|
||||
|
||||
Vektoren im Vektorraum $\mathbb{F}_{5}^{3}$ über dem Körper $\mathbb{F}_{5}$.
|
||||
|
||||
Loading…
x
Reference in New Issue
Block a user