master > master: Bemerkgung in Quiz9 überarbeitet
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@ -2715,8 +2715,7 @@ und daraus die Parameter abzulesen.
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1&-2&4&0\\
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0&11&-15&1\\
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0&0&-7&1\\
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\end{smatrix}
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\\
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\end{smatrix}\\
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\end{mathe}
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Wende die Zeilentransformation
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@ -2728,8 +2727,7 @@ und daraus die Parameter abzulesen.
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1&-2&4&0\\
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0&11&-8&0\\
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0&0&-7&1\\
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\end{smatrix}
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\\
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\end{smatrix}\\
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\end{mathe}
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Aus der Zeilenstufenform erschließt sich, dass $t_{4}$ frei ist.
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@ -7523,22 +7521,22 @@ Für jeden Fall berechnen wir $\ggT(a,b)$ mittels des Euklidischen Algorithmus
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$a$ &$b$ &Restberechnung (symbolisch) &Restberechnung (Werte)\\
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\hline
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\endhead
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$1529$ &$170$ &$a = b\cdot q_{1} + r_{1}$ &$1529 = 170\cdot 8 + 169$\\
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&&$b = r_{1}\cdot q_{2} + r_{2}$ &$170 = 169\cdot 1 + \boxed{\mathbf{1}}$\\
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&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$169 = 1\cdot 169 + 0$\\
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\hline
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$13758$ &$21$ &$a = b\cdot q_{1} + r_{1}$ &$13758 = 21\cdot 655 + \boxed{\mathbf{3}}$\\
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&&$b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 3\cdot 7 + 0$\\
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\hline
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$210$ &$45$ &$a = b\cdot q_{1} + r_{1}$ &$210 = 45\cdot 4 + 30$\\
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&&$b = r_{1}\cdot q_{2} + r_{2}$ &$45 = 30\cdot 1 + \boxed{\mathbf{15}}$\\
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&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$30 = 15\cdot 2 + 0$\\
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\hline
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$1209$ &$102$ &$a = b\cdot q_{1} + r_{1}$ &$1209 = 102\cdot 11 + 87$\\
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&&$b = r_{1}\cdot q_{2} + r_{2}$ &$102 = 87\cdot 1 + 15$\\
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&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$87 = 15\cdot 5 + 12$\\
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&&$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$15 = 12\cdot 1 + \boxed{\mathbf{3}}$\\
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&&$r_{3} = r_{4}\cdot q_{5} + r_{5}$ &$12 = 3\cdot 4 + 0$\\
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$1529$ &$170$ &$a = b\cdot q_{1} + r_{1}$ &$1529 = 170\cdot 8 + 169$\\
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&&$b = r_{1}\cdot q_{2} + r_{2}$ &$170 = 169\cdot 1 + \boxed{\mathbf{1}}$\\
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&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$169 = 1\cdot 169 + 0$\\
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\hline
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$13758$ &$21$ &$a = b\cdot q_{1} + r_{1}$ &$13758 = 21\cdot 655 + \boxed{\mathbf{3}}$\\
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&&$b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 3\cdot 7 + 0$\\
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\hline
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$210$ &$45$ &$a = b\cdot q_{1} + r_{1}$ &$210 = 45\cdot 4 + 30$\\
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&&$b = r_{1}\cdot q_{2} + r_{2}$ &$45 = 30\cdot 1 + \boxed{\mathbf{15}}$\\
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&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$30 = 15\cdot 2 + 0$\\
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\hline
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$1209$ &$102$ &$a = b\cdot q_{1} + r_{1}$ &$1209 = 102\cdot 11 + 87$\\
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&&$b = r_{1}\cdot q_{2} + r_{2}$ &$102 = 87\cdot 1 + 15$\\
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&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$87 = 15\cdot 5 + 12$\\
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&&$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$15 = 12\cdot 1 + \boxed{\mathbf{3}}$\\
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&&$r_{3} = r_{4}\cdot q_{5} + r_{5}$ &$12 = 3\cdot 4 + 0$\\
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\hline
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\hline
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\end{longtable}
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@ -7559,18 +7557,18 @@ Wir verwenden die Berechnungen aus der Tabelle in SKA \ref{ska:5:ex:6}.
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$a$ &$b$ &Rest (symbolisch) &Rest (Werte)\\
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\hline
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\endhead
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$1529$ &$170$ &$r_{1} = a - 8\cdot b$ &$169 = 1\cdot a + -8\cdot b$\\
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&&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{1 = \mathbf{-1}\cdot a + \mathbf{9}\cdot b}$\\
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\hline
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$13758$ &$21$ &$r_{1} = a - 655\cdot b$ &$\boxed{3 = \mathbf{1}\cdot a + \mathbf{-655}\cdot b}$\\
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\hline
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$210$ &$45$ &$r_{1} = a - 4\cdot b$ &$30 = 1\cdot a + -4\cdot b$\\
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&&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{15 = \mathbf{-1}\cdot a + \mathbf{5}\cdot b}$\\
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\hline
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$1209$ &$102$ &$r_{1} = a - 11\cdot b$ &$87 = 1\cdot a + -11\cdot b$\\
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&&$r_{2} = b - 1\cdot r_{1}$ &$15 = -1\cdot a + 12\cdot b$\\
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&&$r_{3} = r_{1} - 5\cdot r_{2}$ &$12 = 6\cdot a + -71\cdot b$\\
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&&$r_{4} = r_{2} - 1\cdot r_{3}$ &$\boxed{3 = \mathbf{-7}\cdot a + \mathbf{83}\cdot b}$\\
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$1529$ &$170$ &$r_{1} = a - 8\cdot b$ &$169 = 1\cdot a + -8\cdot b$\\
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&&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{1 = \mathbf{-1}\cdot a + \mathbf{9}\cdot b}$\\
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\hline
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$13758$ &$21$ &$r_{1} = a - 655\cdot b$ &$\boxed{3 = \mathbf{1}\cdot a + \mathbf{-655}\cdot b}$\\
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\hline
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$210$ &$45$ &$r_{1} = a - 4\cdot b$ &$30 = 1\cdot a + -4\cdot b$\\
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&&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{15 = \mathbf{-1}\cdot a + \mathbf{5}\cdot b}$\\
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\hline
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$1209$ &$102$ &$r_{1} = a - 11\cdot b$ &$87 = 1\cdot a + -11\cdot b$\\
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&&$r_{2} = b - 1\cdot r_{1}$ &$15 = -1\cdot a + 12\cdot b$\\
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&&$r_{3} = r_{1} - 5\cdot r_{2}$ &$12 = 6\cdot a + -71\cdot b$\\
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&&$r_{4} = r_{2} - 1\cdot r_{3}$ &$\boxed{3 = \mathbf{-7}\cdot a + \mathbf{83}\cdot b}$\\
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\hline
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\hline
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\end{longtable}
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@ -8237,10 +8235,10 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
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Restberechnung (symbolisch) &Restberechnung (Werte)\\
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\hline
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\endhead
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$a = b\cdot q_{1} + r_{1}$ &$103 = 21\cdot 4 + 19$\\
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$b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 19\cdot 1 + 2$\\
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$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$19 = 2\cdot 9 + \boxed{\mathbf{1}}$\\
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$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$2 = 1\cdot 2 + 0$\\
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$a = b\cdot q_{1} + r_{1}$ &$103 = 21\cdot 4 + 19$\\
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$b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 19\cdot 1 + 2$\\
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$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$19 = 2\cdot 9 + \boxed{\mathbf{1}}$\\
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$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$2 = 1\cdot 2 + 0$\\
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\hline
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\hline
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\end{longtable}
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@ -8254,9 +8252,9 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
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Rest (symbolisch) &Rest (Werte)\\
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\hline
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\endhead
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$r_{1} = a - 4\cdot b$ &$19 = 1\cdot a + -4\cdot b$\\
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$r_{2} = b - 1\cdot r_{1}$ &$2 = -1\cdot a + 5\cdot b$\\
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$r_{3} = r_{1} - 9\cdot r_{2}$ &$\boxed{1 = \mathbf{10}\cdot a + \mathbf{-49}\cdot b}$\\
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$r_{1} = a - 4\cdot b$ &$19 = 1\cdot a + -4\cdot b$\\
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$r_{2} = b - 1\cdot r_{1}$ &$2 = -1\cdot a + 5\cdot b$\\
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$r_{3} = r_{1} - 9\cdot r_{2}$ &$\boxed{1 = \mathbf{10}\cdot a + \mathbf{-49}\cdot b}$\\
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\hline
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\hline
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\end{longtable}
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@ -9054,7 +9052,7 @@ für alle linearen Unterräume, $U\subseteq V$.
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\textbf{zu zeigen}, dass ein $x\in U$ existiert mit
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\begin{mathe}[mc]{rcl}
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\eqtag[eq:0:quiz:9]{$\ast$}
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\eqtag[eq:0:quiz:9]
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(\psi\circ\phi)(x) &= &z.\\
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\end{mathe}
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@ -9102,15 +9100,18 @@ für alle linearen Unterräume, $U\subseteq V$.
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\end{einzug}
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\begin{rem*}
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Wir können in der Tat zeigen, die umgekehrte Richtung auch gilt:
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Wir können in der Tat zeigen, dass die umgekehrte Richtung auch gilt:
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Angenommen, $\psi\circ\phi$ sei surjektiv.
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Dann gilt
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$W\supseteq\psi(V)\supseteq\psi(\phi(U))=(\psi\circ\phi)(U)=W$,
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und somit $\psi(V)=W$,
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sodass \eqcref{it:1:quiz:9} gilt.
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Und für alle $y\in V$, wegen Surjektivität von $\psi\circ\phi$,
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existiert ein $x\in U$, so dass $\psi(y)=(\psi\circ\phi)(x)$.
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Daraus folgt
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Für \eqcref{it:2:quiz:9} brauchen wir nur die $\supseteq$-Inklusion zu zeigen,
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da die $\subseteq$-Inklusion offensichtlich wahr ist.
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Sei also $y\in V$ beliebig.
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Wegen Surjektivität von $\psi\circ\phi$ existiert nun ein $x\in U$,
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so dass $\psi(y)=(\psi\circ\phi)(x)$.
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Beobachte man, dass
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\begin{mathe}[mc]{rcccl}
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\psi(y-\phi(x))
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@ -9127,8 +9128,10 @@ für alle linearen Unterräume, $U\subseteq V$.
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&\in &\ker(\psi)+\range(\phi).\\
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\end{mathe}
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Also gilt die $\supseteq$-Inklusion in \eqcref{it:2:quiz:9}.
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Und offensichtlich gilt die $\subseteq$-Inklusion in \eqcref{it:1:quiz:9}.
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Damit haben wir bewiesen, dass $V\subseteq \ker(\psi)+\range(\phi)$
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(d.\,h. die $\supseteq$-Inklusion in \eqcref{it:2:quiz:9}).\\
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Darum gilt:
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$\psi\circ\phi$ surjektiv $\Rightarrow$ \eqcref{it:1:quiz:9}+\eqcref{it:2:quiz:9} gelten.
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\end{rem*}
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%% ********************************************************************************
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