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Zur {\bfseries Eindeutigkeit}: Seien $r_{i}\in(0,\infty)$, $\alpha_{i}\in[0,2\pi)$ mit - $r_{i}\cdot\begin{svector} \cos(\alpha_{i})\\ \sin(\alpha_{i})\\\end{svector}=z$ + $r_{i}\cdot\begin{svector} \cos(\alpha_{i})\\ \sin(\alpha_{i})\\\end{svector}=z$ für $i\in\{1,2\}$. \textbf{Zu zeigen:} $r_{1}=r_{2}$ und $\alpha_{1}=\alpha_{2}$. Es gilt @@ -4294,10 +4294,10 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen \item \begin{claim*} Seien $z_{1},z_{2}\in\kmplx\ohne\{0\}$ mit Darstellungen - $z_{i}=r_{i}\cdot\begin{svector} \cos(\alpha_{i})\\ \sin(\alpha_{i})\\\end{svector}$ + $z_{i}=r_{i}\cdot\begin{svector} \cos(\alpha_{i})\\ \sin(\alpha_{i})\\\end{svector}$ für $i\in\{1,2\}$. Dann gilt die Rechenregel - $z_{1}z_{2}=r_{1}r_{2}\cdot\begin{svector} \cos(\alpha_{1}+\alpha_{2})\\ \sin(\alpha_{1}+\alpha_{2})\\\end{svector}$. + $z_{1}z_{2}=r_{1}r_{2}\cdot\begin{svector} \cos(\alpha_{1}+\alpha_{2})\\ \sin(\alpha_{1}+\alpha_{2})\\\end{svector}$. \end{claim*} \begin{proof} @@ -4305,10 +4305,10 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen \begin{longmathe}[mc]{RCL} z_{1}z_{2} - &= &\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}\\ - &= &\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}\\ - &= &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}\\ - &= &r_{1}r_{2}\begin{svector} \cos(\alpha_{1}+\alpha_{2})\\ \sin(\alpha_{1}+\alpha_{2})\\\end{svector}.\\ + &= &\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}\\ + &= &\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}\\ + &= &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}\\ + &= &r_{1}r_{2}\begin{svector} \cos(\alpha_{1}+\alpha_{2})\\ \sin(\alpha_{1}+\alpha_{2})\\\end{svector}.\\ \end{longmathe} Die letzte Vereinfachung folgt aus der trigonometrischen Additionsregel. @@ -4317,9 +4317,9 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen \item \begin{claim*}[de Moivre] 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