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b/docs/loesungen.tex @@ -2715,8 +2715,7 @@ und daraus die Parameter abzulesen. 1&-2&4&0\\ 0&11&-15&1\\ 0&0&-7&1\\ - \end{smatrix} - \\ + \end{smatrix}\\ \end{mathe} Wende die Zeilentransformation @@ -2728,8 +2727,7 @@ und daraus die Parameter abzulesen. 1&-2&4&0\\ 0&11&-8&0\\ 0&0&-7&1\\ - \end{smatrix} - \\ + \end{smatrix}\\ \end{mathe} Aus der Zeilenstufenform erschließt sich, dass $t_{4}$ frei ist. @@ -7523,22 +7521,22 @@ Für jeden Fall berechnen wir $\ggT(a,b)$ mittels des Euklidischen Algorithmus $a$ &$b$ &Restberechnung (symbolisch) &Restberechnung (Werte)\\ \hline \endhead - $1529$ &$170$ &$a = b\cdot q_{1} + r_{1}$ &$1529 = 170\cdot 8 + 169$\\ - &&$b = r_{1}\cdot q_{2} + r_{2}$ &$170 = 169\cdot 1 + \boxed{\mathbf{1}}$\\ - &&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$169 = 1\cdot 169 + 0$\\ - \hline - $13758$ &$21$ &$a = b\cdot q_{1} + r_{1}$ &$13758 = 21\cdot 655 + \boxed{\mathbf{3}}$\\ - &&$b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 3\cdot 7 + 0$\\ - \hline - $210$ &$45$ &$a = b\cdot q_{1} + r_{1}$ &$210 = 45\cdot 4 + 30$\\ - &&$b = r_{1}\cdot q_{2} + r_{2}$ &$45 = 30\cdot 1 + \boxed{\mathbf{15}}$\\ - &&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$30 = 15\cdot 2 + 0$\\ - \hline - $1209$ &$102$ &$a = b\cdot q_{1} + r_{1}$ &$1209 = 102\cdot 11 + 87$\\ - &&$b = r_{1}\cdot q_{2} + r_{2}$ &$102 = 87\cdot 1 + 15$\\ - &&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$87 = 15\cdot 5 + 12$\\ - &&$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$15 = 12\cdot 1 + \boxed{\mathbf{3}}$\\ - &&$r_{3} = r_{4}\cdot q_{5} + r_{5}$ &$12 = 3\cdot 4 + 0$\\ + $1529$ &$170$ &$a = b\cdot q_{1} + r_{1}$ &$1529 = 170\cdot 8 + 169$\\ + &&$b = r_{1}\cdot q_{2} + r_{2}$ &$170 = 169\cdot 1 + \boxed{\mathbf{1}}$\\ + &&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$169 = 1\cdot 169 + 0$\\ + \hline + $13758$ &$21$ &$a = b\cdot q_{1} + r_{1}$ &$13758 = 21\cdot 655 + \boxed{\mathbf{3}}$\\ + &&$b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 3\cdot 7 + 0$\\ + \hline + $210$ &$45$ &$a = b\cdot q_{1} + r_{1}$ &$210 = 45\cdot 4 + 30$\\ + &&$b = r_{1}\cdot q_{2} + r_{2}$ &$45 = 30\cdot 1 + \boxed{\mathbf{15}}$\\ + &&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$30 = 15\cdot 2 + 0$\\ + \hline + $1209$ &$102$ &$a = b\cdot q_{1} + r_{1}$ &$1209 = 102\cdot 11 + 87$\\ + &&$b = r_{1}\cdot q_{2} + r_{2}$ &$102 = 87\cdot 1 + 15$\\ + &&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$87 = 15\cdot 5 + 12$\\ + &&$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$15 = 12\cdot 1 + \boxed{\mathbf{3}}$\\ + &&$r_{3} = r_{4}\cdot q_{5} + r_{5}$ &$12 = 3\cdot 4 + 0$\\ \hline \hline \end{longtable} @@ -7559,18 +7557,18 @@ Wir verwenden die Berechnungen aus der Tabelle in SKA \ref{ska:5:ex:6}. $a$ &$b$ &Rest (symbolisch) &Rest (Werte)\\ \hline \endhead - $1529$ &$170$ &$r_{1} = a - 8\cdot b$ &$169 = 1\cdot a + -8\cdot b$\\ - &&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{1 = \mathbf{-1}\cdot a + \mathbf{9}\cdot b}$\\ - \hline - $13758$ &$21$ &$r_{1} = a - 655\cdot b$ &$\boxed{3 = \mathbf{1}\cdot a + \mathbf{-655}\cdot b}$\\ - \hline - $210$ &$45$ &$r_{1} = a - 4\cdot b$ &$30 = 1\cdot a + -4\cdot b$\\ - &&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{15 = \mathbf{-1}\cdot a + \mathbf{5}\cdot b}$\\ - \hline - $1209$ &$102$ &$r_{1} = a - 11\cdot b$ &$87 = 1\cdot a + -11\cdot b$\\ - &&$r_{2} = b - 1\cdot r_{1}$ &$15 = -1\cdot a + 12\cdot b$\\ - &&$r_{3} = r_{1} - 5\cdot r_{2}$ &$12 = 6\cdot a + -71\cdot b$\\ - &&$r_{4} = r_{2} - 1\cdot r_{3}$ &$\boxed{3 = \mathbf{-7}\cdot a + \mathbf{83}\cdot b}$\\ + $1529$ &$170$ &$r_{1} = a - 8\cdot b$ &$169 = 1\cdot a + -8\cdot b$\\ + &&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{1 = \mathbf{-1}\cdot a + \mathbf{9}\cdot b}$\\ + \hline + $13758$ &$21$ &$r_{1} = a - 655\cdot b$ &$\boxed{3 = \mathbf{1}\cdot a + \mathbf{-655}\cdot b}$\\ + \hline + $210$ &$45$ &$r_{1} = a - 4\cdot b$ &$30 = 1\cdot a + -4\cdot b$\\ + &&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{15 = \mathbf{-1}\cdot a + \mathbf{5}\cdot b}$\\ + \hline + $1209$ &$102$ &$r_{1} = a - 11\cdot b$ &$87 = 1\cdot a + -11\cdot b$\\ + &&$r_{2} = b - 1\cdot r_{1}$ &$15 = -1\cdot a + 12\cdot b$\\ + &&$r_{3} = r_{1} - 5\cdot r_{2}$ &$12 = 6\cdot a + -71\cdot b$\\ + &&$r_{4} = r_{2} - 1\cdot r_{3}$ &$\boxed{3 = \mathbf{-7}\cdot a + \mathbf{83}\cdot b}$\\ \hline \hline \end{longtable} @@ -8237,10 +8235,10 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$. Restberechnung (symbolisch) &Restberechnung (Werte)\\ \hline \endhead - $a = b\cdot q_{1} + r_{1}$ &$103 = 21\cdot 4 + 19$\\ - $b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 19\cdot 1 + 2$\\ - $r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$19 = 2\cdot 9 + \boxed{\mathbf{1}}$\\ - $r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$2 = 1\cdot 2 + 0$\\ + $a = b\cdot q_{1} + r_{1}$ &$103 = 21\cdot 4 + 19$\\ + $b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 19\cdot 1 + 2$\\ + $r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$19 = 2\cdot 9 + \boxed{\mathbf{1}}$\\ + $r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$2 = 1\cdot 2 + 0$\\ \hline \hline \end{longtable} @@ -8254,9 +8252,9 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$. Rest (symbolisch) &Rest (Werte)\\ \hline \endhead - $r_{1} = a - 4\cdot b$ &$19 = 1\cdot a + -4\cdot b$\\ - $r_{2} = b - 1\cdot r_{1}$ &$2 = -1\cdot a + 5\cdot b$\\ - $r_{3} = r_{1} - 9\cdot r_{2}$ &$\boxed{1 = \mathbf{10}\cdot a + \mathbf{-49}\cdot b}$\\ + $r_{1} = a - 4\cdot b$ &$19 = 1\cdot a + -4\cdot b$\\ + $r_{2} = b - 1\cdot r_{1}$ &$2 = -1\cdot a + 5\cdot b$\\ + $r_{3} = r_{1} - 9\cdot r_{2}$ &$\boxed{1 = \mathbf{10}\cdot a + \mathbf{-49}\cdot b}$\\ \hline \hline \end{longtable} @@ -9054,7 +9052,7 @@ für alle linearen Unterräume, $U\subseteq V$. \textbf{zu zeigen}, dass ein $x\in U$ existiert mit \begin{mathe}[mc]{rcl} - \eqtag[eq:0:quiz:9]{$\ast$} + \eqtag[eq:0:quiz:9] (\psi\circ\phi)(x) &= &z.\\ \end{mathe} @@ -9102,15 +9100,18 @@ für alle linearen Unterräume, $U\subseteq V$. \end{einzug} \begin{rem*} - Wir können in der Tat zeigen, die umgekehrte Richtung auch gilt: + Wir können in der Tat zeigen, dass die umgekehrte Richtung auch gilt: Angenommen, $\psi\circ\phi$ sei surjektiv. Dann gilt $W\supseteq\psi(V)\supseteq\psi(\phi(U))=(\psi\circ\phi)(U)=W$, und somit $\psi(V)=W$, sodass \eqcref{it:1:quiz:9} gilt. - Und für alle $y\in V$, wegen Surjektivität von $\psi\circ\phi$, - existiert ein $x\in U$, so dass $\psi(y)=(\psi\circ\phi)(x)$. - Daraus folgt + Für \eqcref{it:2:quiz:9} brauchen wir nur die $\supseteq$-Inklusion zu zeigen, + da die $\subseteq$-Inklusion offensichtlich wahr ist. + Sei also $y\in V$ beliebig. + Wegen Surjektivität von $\psi\circ\phi$ existiert nun ein $x\in U$, + so dass $\psi(y)=(\psi\circ\phi)(x)$. + Beobachte man, dass \begin{mathe}[mc]{rcccl} \psi(y-\phi(x)) @@ -9127,8 +9128,10 @@ für alle linearen Unterräume, $U\subseteq V$. &\in &\ker(\psi)+\range(\phi).\\ \end{mathe} - Also gilt die $\supseteq$-Inklusion in \eqcref{it:2:quiz:9}. - Und offensichtlich gilt die $\subseteq$-Inklusion in \eqcref{it:1:quiz:9}. + Damit haben wir bewiesen, dass $V\subseteq \ker(\psi)+\range(\phi)$ + (d.\,h. die $\supseteq$-Inklusion in \eqcref{it:2:quiz:9}).\\ + Darum gilt: + $\psi\circ\phi$ surjektiv $\Rightarrow$ \eqcref{it:1:quiz:9}+\eqcref{it:2:quiz:9} gelten. \end{rem*} %% ********************************************************************************