From 837e260667243f197c465b1ac25d3191680500d7 Mon Sep 17 00:00:00 2001
From: raj_mathe <raj.dahya@math.uni-leipzig.de>
Date: Tue, 26 Jan 2021 23:29:02 +0100
Subject: [PATCH] =?UTF-8?q?master=20>=20master:=20=C3=9CB10?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
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---
 docs/loesungen.pdf |  Bin 604072 -> 638072 bytes
 docs/loesungen.tex | 1558 +++++++++++++++++++++++++++++++++-----------
 2 files changed, 1164 insertions(+), 394 deletions(-)

diff --git a/docs/loesungen.pdf b/docs/loesungen.pdf
index 9eb41132deca9dbb8f2e75abb941faa4df0af092..572112a420f7fa120379914ba4106739b814af39 100644
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diff --git a/docs/loesungen.tex b/docs/loesungen.tex
index 408362e..dbd2c02 100644
--- a/docs/loesungen.tex
+++ b/docs/loesungen.tex
@@ -63,6 +63,8 @@
 %%        |
 %%        ---- body/uebung/ueb9.tex;
 %%        |
+%%        ---- body/uebung/ueb10.tex;
+%%        |
 %%        ---- body/ska/ska4.tex;
 %%        |
 %%        ---- body/ska/ska5.tex;
@@ -1308,6 +1310,7 @@
 %% MATHE:
 %% ****************************************************************
 
+\def\cal#1{\mathcal{#1}}
 \def\reell{\mathbb{R}}
 \def\kmplx{\mathbb{C}}
 \def\Torus{\mathbb{T}}
@@ -1468,11 +1471,11 @@ durch
 
     \begin{mathe}[mc]{rclqrcl}
         A_{\alpha} &:= &\begin{matrix}{cccc}
-        1 &7 &2 &-1\\
-        1 &8 &6 &-3\\
-        2 &14 &\alpha &-2\\
+          1 &7 &2 &-1\\
+          1 &8 &6 &-3\\
+          2 &14 &\alpha &-2\\
         \end{matrix}
-        &\mathbf{b}_{\beta} &:= &\begin{vector}4\\0\\\beta\\\end{vector}
+        &\mathbf{b}_{\beta} &:= &\begin{vector}  4\\  0\\  \beta\\\end{vector}
     \end{mathe}
 
 gegeben sind.
@@ -1483,9 +1486,9 @@ Um die Lösungsmenge zu bestimmen führen wir das Gaußverfahren aus:
 
     \begin{mathe}[mc]{c}
         \begin{matrix}{cccc|c}
-        1 &7 &2 &-1 &4\\
-        1 &8 &6 &-3 &0\\
-        2 &14 &\alpha &-2 &\beta\\
+          1 &7 &2 &-1 &4\\
+          1 &8 &6 &-3 &0\\
+          2 &14 &\alpha &-2 &\beta\\
         \end{matrix}\\
     \end{mathe}
 
@@ -1501,9 +1504,9 @@ Um die Lösungsmenge zu bestimmen führen wir das Gaußverfahren aus:
 
     \begin{mathe}[mc]{c}
         \begin{matrix}{cccc|c}
-        \boxed{1} &7 &2 &-1 &4\\
-        0 &\boxed{1} &4 &-2 &-4\\
-        0 &0 &\boxed{\alpha - 4} &0 &\beta - 8\\
+          \boxed{1} &7 &2 &-1 &4\\
+          0 &\boxed{1} &4 &-2 &-4\\
+          0 &0 &\boxed{\alpha - 4} &0 &\beta - 8\\
         \end{matrix}\\
     \end{mathe}
 
@@ -1521,9 +1524,9 @@ Dies führt zu einem Fallunterschied:
 
         \begin{mathe}[mc]{c}
             \begin{matrix}{cccc|c}
-            \boxed{1} &7 &2 &-1 &4\\
-            0 &\boxed{1} &4 &-2 &-4\\
-            0 &0 &0 &0 &\beta - 8\\
+              \boxed{1} &7 &2 &-1 &4\\
+              0 &\boxed{1} &4 &-2 &-4\\
+              0 &0 &0 &0 &\beta - 8\\
             \end{matrix}\\
         \end{mathe}
 
@@ -1543,9 +1546,9 @@ Dies führt zu einem Fallunterschied:
 
                 \begin{mathe}[mc]{c}
                     \begin{matrix}{cccc|c}
-                    \boxed{1} &7 &2 &-1 &4\\
-                    0 &\boxed{1} &4 &-2 &-4\\
-                    0 &0 &0 &0 &0\\
+                      \boxed{1} &7 &2 &-1 &4\\
+                      0 &\boxed{1} &4 &-2 &-4\\
+                      0 &0 &0 &0 &0\\
                     \end{matrix}\\
                 \end{mathe}
 
@@ -1571,15 +1574,15 @@ Dies führt zu einem Fallunterschied:
                     Zusammengefasst erhalten wir die allgemeine Form der Lösung:
 
                     \begin{mathe}[mc]{rcl}
-                        \mathbf{x} &= &\begin{svector}x_{1}\\x_{2}\\x_{3}\\x_{4}\\\end{svector}\\
-                            &= &\begin{svector}32 + 26x_{3} + -13x_{4}\\-4 - 4x_{3} + 2x_{4}\\x_{3}\\x_{4}\\\end{svector}\\
-                            &= &\begin{svector}32 + 26x_{3} + -13x_{4}\\-4 - 4x_{3} + 2x_{4}\\0 + 1x_{3} + 0x_{4}\\0 + 0x_{3} + 1x_{4}\\\end{svector}\\
-                            &= &\begin{svector}32\\-4\\0\\0\\\end{svector}
-                            + \begin{svector}26x_{3}\\-4x_{3}\\1x_{3}\\0x_{3}\\\end{svector}
-                            + \begin{svector}-13x_{4}\\2x_{4}\\1x_{4}\\1x_{4}\\\end{svector}\\
-                            &= &\begin{svector}32\\-4\\0\\0\\\end{svector}
-                            + x_{3}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector}
-                            + x_{4}\cdot\begin{svector}-13\\2\\1\\1\\\end{svector}\\
+                        \mathbf{x} &= &\begin{svector}  x_{1}\\  x_{2}\\  x_{3}\\  x_{4}\\\end{svector}\\
+                            &= &\begin{svector}  32 + 26x_{3} + -13x_{4}\\  -4 - 4x_{3} + 2x_{4}\\  x_{3}\\  x_{4}\\\end{svector}\\
+                            &= &\begin{svector}  32 + 26x_{3} + -13x_{4}\\  -4 - 4x_{3} + 2x_{4}\\  0 + 1x_{3} + 0x_{4}\\  0 + 0x_{3} + 1x_{4}\\\end{svector}\\
+                            &= &\begin{svector}  32\\  -4\\  0\\  0\\\end{svector}
+                            + \begin{svector}  26x_{3}\\  -4x_{3}\\  1x_{3}\\  0x_{3}\\\end{svector}
+                            + \begin{svector}  -13x_{4}\\  2x_{4}\\  1x_{4}\\  1x_{4}\\\end{svector}\\
+                            &= &\begin{svector}  32\\  -4\\  0\\  0\\\end{svector}
+                            + x_{3}\cdot\begin{svector}  26\\  -4\\  1\\  0\\\end{svector}
+                            + x_{4}\cdot\begin{svector}  -13\\  2\\  1\\  1\\\end{svector}\\
                     \end{mathe}
 
                     mit $x_{3}$, $x_{4}$ frei wählbar.
@@ -1588,14 +1591,14 @@ Dies führt zu einem Fallunterschied:
                 Also erhalten wird in diesem Falle
                     $\boxed{
                         L_{\alpha,\beta}=\left\{
-                            \begin{svector}32\\-4\\0\\0\\\end{svector}
-                            + t_{1}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector}
-                            + t_{2}\cdot\begin{svector}-13\\2\\1\\1\\\end{svector}
+                            \begin{svector}  32\\  -4\\  0\\  0\\\end{svector}
+                            + t_{1}\cdot\begin{svector}  26\\  -4\\  1\\  0\\\end{svector}
+                            + t_{2}\cdot\begin{svector}  -13\\  2\\  1\\  1\\\end{svector}
                         \mid t_{1}, t_{2}\in\reell
                         \right\}
                     }$,
                 oder etwas kompakter formuliert,
-                    ${L_{\alpha,\beta}=\begin{svector}32\\-4\\0\\0\\\end{svector} + \vectorspacespan\left\{\begin{svector}26\\-4\\1\\0\\\end{svector}, \begin{svector}-13\\2\\1\\1\\\end{svector}\right\}}$.
+                    ${L_{\alpha,\beta}=\begin{svector}  32\\  -4\\  0\\  0\\\end{svector} + \vectorspacespan\left\{\begin{svector}  26\\  -4\\  1\\  0\\\end{svector}, \begin{svector}  -13\\  2\\  1\\  1\\\end{svector}\right\}}$.
         \end{enumerate}
 
     %% FALL 2
@@ -1614,12 +1617,12 @@ Dies führt zu einem Fallunterschied:
             das heißt
 
             \begin{mathe}[mc]{rcl}
-                \mathbf{x} &= &\begin{svector}32\\-4\\0\\0\\\end{svector}
-                    + x_{3}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector}
-                    + x_{4}\cdot\begin{svector}-13\\2\\1\\1\\\end{svector}\\
-                    &= &\begin{svector}32\\-4\\0\\0\\\end{svector}
-                    + \frac{\beta-8}{\alpha-4}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector}
-                    + x_{4}\cdot\begin{svector}-13\\2\\1\\1\\\end{svector},\\
+                \mathbf{x} &= &\begin{svector}  32\\  -4\\  0\\  0\\\end{svector}
+                    + x_{3}\cdot\begin{svector}  26\\  -4\\  1\\  0\\\end{svector}
+                    + x_{4}\cdot\begin{svector}  -13\\  2\\  1\\  1\\\end{svector}\\
+                    &= &\begin{svector}  32\\  -4\\  0\\  0\\\end{svector}
+                    + \frac{\beta-8}{\alpha-4}\cdot\begin{svector}  26\\  -4\\  1\\  0\\\end{svector}
+                    + x_{4}\cdot\begin{svector}  -13\\  2\\  1\\  1\\\end{svector},\\
             \end{mathe}
 
             wobei $x_{4}$ frei wählbar ist.
@@ -1628,14 +1631,14 @@ Dies führt zu einem Fallunterschied:
         Also erhalten wird in diesem Falle
             $\boxed{
                 L_{\alpha,\beta}=\left\{
-                    \begin{svector}32\\-4\\0\\0\\\end{svector}
-                    + \frac{\beta-8}{\alpha-4}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector}
-                    + t\cdot\begin{svector}-13\\2\\1\\1\\\end{svector}
+                    \begin{svector}  32\\  -4\\  0\\  0\\\end{svector}
+                    + \frac{\beta-8}{\alpha-4}\cdot\begin{svector}  26\\  -4\\  1\\  0\\\end{svector}
+                    + t\cdot\begin{svector}  -13\\  2\\  1\\  1\\\end{svector}
                 \mid t\in\reell
                 \right\}
             }$,
         oder etwas kompakter formuliert,
-            ${L_{\alpha,\beta}=\begin{svector}32\\-4\\0\\0\\\end{svector} + \frac{\beta-8}{\alpha-4}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector} + \vectorspacespan\left\{\begin{svector}-13\\2\\1\\1\\\end{svector}\right\}}$.
+            ${L_{\alpha,\beta}=\begin{svector}  32\\  -4\\  0\\  0\\\end{svector} + \frac{\beta-8}{\alpha-4}\cdot\begin{svector}  26\\  -4\\  1\\  0\\\end{svector} + \vectorspacespan\left\{\begin{svector}  -13\\  2\\  1\\  1\\\end{svector}\right\}}$.
 \end{enumerate}
 
 Wir fassen die Lösung für alle Fälle zusammen:
@@ -1650,9 +1653,9 @@ Wir fassen die Lösung für alle Fälle zusammen:
 
 für alle $\alpha,\beta\in\reell$,
 wobei
-    $\mathbf{u} = \begin{svector}32\\-4\\0\\0\\\end{svector}$,
-    $\mathbf{v} = \begin{svector}26\\-4\\1\\0\\\end{svector}$,
-    $\mathbf{w} = \begin{svector}-13\\2\\1\\1\\\end{svector}$.
+    $\mathbf{u} = \begin{svector}  32\\  -4\\  0\\  0\\\end{svector}$,
+    $\mathbf{v} = \begin{svector}  26\\  -4\\  1\\  0\\\end{svector}$,
+    $\mathbf{w} = \begin{svector}  -13\\  2\\  1\\  1\\\end{svector}$.
 
 %% AUFGABE 1-2
 \let\altsectionname\sectionname
@@ -1969,12 +1972,12 @@ Für diese Aufgabe wird das Konzept der \emph{linearen Unabhängigkeit} aus Kapi
         {\scriptsize
         \begin{mathe}[mc]{c}
             \begin{matrix}{ccc|c}
-            -5 &0 &0 &-7\\
-            4 &-6 &-10 &6\\
-            -2 &-6 &-6 &9\\
-            -7 &4 &-1 &-5\\
-            4 &-5 &2 &-9\\
-            -5 &8 &-7 &-5\\
+              -5 &0 &0 &-7\\
+              4 &-6 &-10 &6\\
+              -2 &-6 &-6 &9\\
+              -7 &4 &-1 &-5\\
+              4 &-5 &2 &-9\\
+              -5 &8 &-7 &-5\\
             \end{matrix}
         \end{mathe}}
 
@@ -1983,9 +1986,9 @@ Für diese Aufgabe wird das Konzept der \emph{linearen Unabhängigkeit} aus Kapi
         {\scriptsize
         \begin{mathe}[bc]{c}
             \begin{matrix}{ccc|c}
-            4 &-6 &-10 &6\\
-            4 &-5 &2 &-9\\
-            -5 &8 &-7 &-5\\
+              4 &-6 &-10 &6\\
+              4 &-5 &2 &-9\\
+              -5 &8 &-7 &-5\\
             \end{matrix}.
         \end{mathe}}
 
@@ -2152,13 +2155,13 @@ Mit diesem Mittel können wir nun die Hauptaussage in der Aufgabe formulieren:
         {\scriptsize
         \begin{mathe}[mc]{rcl}
             \begin{matrix}{cccccccc|c}
-            \underbrace{0\,0\,\ldots\,0}_{\ell_{1}} &\gamma_{1} &\cdots\cdots &\ast &\cdots\cdots &\cdots\cdots &\ast &\cdots\cdots &b^{(N)}_{1}\\
-            0\,0\,\ldots\,0 &0 &\underbrace{0\,0\,\ldots\,0}_{\ell_{2}} &\gamma_{2} &\cdots\cdots &\cdots\cdots &\ast &\cdots\cdots &b^{(N)}_{2}\\
-            \vdots & & & & & & &\vdots\\
-            0\,0\,\ldots\,0 &0 &0\,0\,\ldots\,0 &0 &\cdots\cdots &\underbrace{0\,0\,\ldots\,0}_{\ell_{r}} &\gamma_{r} &\cdots\cdots &b^{(N)}_{r}\\
-            0\,0\,\ldots\,0 &0 &0\,0\,\ldots\,0 &0 &\cdots\cdots &0\,0\,\ldots\,0 &0 &\cdots\cdots &b^{(N)}_{r+1}\\
-            \vdots & & & & & & &\vdots\\
-            0\,0\,\ldots\,0 &0 &0\,0\,\ldots\,0 &0 &\cdots\cdots &0\,0\,\ldots\,0 &0 &\cdots\cdots &b^{(N)}_{m}\\
+              \underbrace{0\,0\,\ldots\,0}_{\ell_{1}} &\gamma_{1} &\cdots\cdots &\ast &\cdots\cdots &\cdots\cdots &\ast &\cdots\cdots &b^{(N)}_{1}\\
+              0\,0\,\ldots\,0 &0 &\underbrace{0\,0\,\ldots\,0}_{\ell_{2}} &\gamma_{2} &\cdots\cdots &\cdots\cdots &\ast &\cdots\cdots &b^{(N)}_{2}\\
+              \vdots & & & & & & &\vdots\\
+              0\,0\,\ldots\,0 &0 &0\,0\,\ldots\,0 &0 &\cdots\cdots &\underbrace{0\,0\,\ldots\,0}_{\ell_{r}} &\gamma_{r} &\cdots\cdots &b^{(N)}_{r}\\
+              0\,0\,\ldots\,0 &0 &0\,0\,\ldots\,0 &0 &\cdots\cdots &0\,0\,\ldots\,0 &0 &\cdots\cdots &b^{(N)}_{r+1}\\
+              \vdots & & & & & & &\vdots\\
+              0\,0\,\ldots\,0 &0 &0\,0\,\ldots\,0 &0 &\cdots\cdots &0\,0\,\ldots\,0 &0 &\cdots\cdots &b^{(N)}_{m}\\
             \end{matrix}
         \end{mathe}}
 
@@ -2388,10 +2391,10 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
         Betrachte die folgenden Vektoren in $\reell^{3}$:
 
             \begin{mathe}[mc]{rclqrclqrclqrcl}
-                \mathbf{v}          &= &\begin{svector}0\\0\\0\\\end{svector},
-                &\mathbf{v}^{\prime} &= &\begin{svector}1\\0\\0\\\end{svector},
-                &\mathbf{w}          &= &\begin{svector}0\\1\\0\\\end{svector},
-                &\mathbf{w}^{\prime} &= &\begin{svector}0\\1\\1\\\end{svector}.\\
+                \mathbf{v}          &= &\begin{svector}  0\\  0\\  0\\\end{svector},
+                &\mathbf{v}^{\prime} &= &\begin{svector}  1\\  0\\  0\\\end{svector},
+                &\mathbf{w}          &= &\begin{svector}  0\\  1\\  0\\\end{svector},
+                &\mathbf{w}^{\prime} &= &\begin{svector}  0\\  1\\  1\\\end{svector}.\\
             \end{mathe}
 
         Bis auf 2-Dimensionalität erfüllen diese die Voraussetzungen in \Cref{satz:main:ueb:2:ex:2a}.
@@ -2471,15 +2474,15 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
                 gelten muss, da $\gamma_{1}\neq\gamma_{2}$.
                 Eingesetzt in die erste Gleichung oben liefert
                     $2x+y=\gamma\cdot 0=0$.
-                Darum muss $\begin{svector}x\\y\\\end{svector}$
+                Darum muss $\begin{svector}  x\\  y\\\end{svector}$
                 das LGS $(A|\mathbf{b})$ lösen, wobei
 
                     \begin{mathe}[mc]{rclqrcl}
                         A &= &\begin{smatrix}
-                        1&-3\\
-                        2&1\\
+                          1 &-3\\
+                          2 &1\\
                         \end{smatrix},
-                        &\mathbf{b} &= &\begin{svector}7\\0\\\end{svector}
+                        &\mathbf{b} &= &\begin{svector}  7\\  0\\\end{svector}
                     \end{mathe}
 
                 \begin{algorithm}[\rtab][\rtab]
@@ -2487,8 +2490,8 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
 
                     \begin{mathe}[mc]{c}
                         \begin{matrix}{cc|c}
-                        1 &-3 &7\\
-                        2 &1 &0\\
+                          1 &-3 &7\\
+                          2 &1 &0\\
                         \end{matrix}\\
                     \end{mathe}
 
@@ -2498,8 +2501,8 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
 
                     \begin{mathe}[mc]{c}
                         \begin{matrix}{cc|c}
-                        1 &-3 &7\\
-                        0 &7 &-14\\
+                          1 &-3 &7\\
+                          0 &7 &-14\\
                         \end{matrix}\\
                     \end{mathe}
 
@@ -2632,10 +2635,10 @@ Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche ve
 Wir arbeiten im Vektorraum $\reell^{3}$ und betrachten die Vektoren
 
     \begin{mathe}[mc]{rclqrclqrclqrcl}
-        \mathbf{v}_{1} &= &\begin{svector}1\\3\\1\\\end{svector}
-        &\mathbf{v}_{2} &= &\begin{svector}-2\\5\\-2\\\end{svector}
-        &\mathbf{w}_{1} &= &\begin{svector}4\\-3\\-3\\\end{svector}
-        &\mathbf{w}_{2} &= &\begin{svector}0\\1\\1\\\end{svector}\\
+        \mathbf{v}_{1} &= &\begin{svector}  1\\  3\\  1\\\end{svector}
+        &\mathbf{v}_{2} &= &\begin{svector}  -2\\  5\\  -2\\\end{svector}
+        &\mathbf{w}_{1} &= &\begin{svector}  4\\  -3\\  -3\\\end{svector}
+        &\mathbf{w}_{2} &= &\begin{svector}  0\\  1\\  1\\\end{svector}\\
     \end{mathe}
 
 \textbf{Zu berechnen:}
@@ -2690,9 +2693,9 @@ wobei
                     \mathbf{w}_{2}
                 \right)
             &= &\begin{smatrix}
-            1&-2&4&0\\
-            3&5&-3&1\\
-            1&-2&-3&1\\
+              1 &-2 &4 &0\\
+              3 &5 &-3 &1\\
+              1 &-2 &-3 &1\\
             \end{smatrix}\\
         \end{mathe}
 
@@ -2709,9 +2712,9 @@ und daraus die Parameter abzulesen.
 
     \begin{mathe}[mc]{c}
         \begin{smatrix}
-        1&-2&4&0\\
-        0&11&-15&1\\
-        0&0&-7&1\\
+          1 &-2 &4 &0\\
+          0 &11 &-15 &1\\
+          0 &0 &-7 &1\\
         \end{smatrix}\\
     \end{mathe}
 
@@ -2721,9 +2724,9 @@ und daraus die Parameter abzulesen.
 
     \begin{mathe}[mc]{c}
         \begin{smatrix}
-        1&-2&4&0\\
-        0&11&-8&0\\
-        0&0&-7&1\\
+          1 &-2 &4 &0\\
+          0 &11 &-8 &0\\
+          0 &0 &-7 &1\\
         \end{smatrix}\\
     \end{mathe}
 
@@ -2743,7 +2746,7 @@ und daraus die Parameter abzulesen.
     Also ist die homogene Lösung gegeben durch
 
     \begin{mathe}[mc]{rcl}
-        \mathbf{t} &= &\beta\begin{svector}28\\-8\\-11\\-77\\\end{svector},
+        \mathbf{t} &= &\beta\begin{svector}  28\\  -8\\  -11\\  -77\\\end{svector},
         \quad\text{mit $\beta\in\reell$ frei wählbar}.
     \end{mathe}
 \end{algorithm}
@@ -2763,7 +2766,7 @@ Wir können nun \eqcref{eq:0:ueb:3:ex:1} fortsetzen und erhalten
                     \mathbf{\xi}=t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
                     \,\text{und}\,
                     \exists{\beta\in\reell:~}
-                        \mathbf{t}=\beta\begin{svector}28\\-8\\-11\\-77\\\end{svector}\\
+                        \mathbf{t}=\beta\begin{svector}  28\\  -8\\  -11\\  -77\\\end{svector}\\
             &\Longleftrightarrow
                 &\exists{\beta\in\reell:~}
                 \mathbf{\xi}=\beta\cdot(
@@ -2779,10 +2782,10 @@ Es gilt
 
     \begin{mathe}[mc]{rcccccl}
         \mathbf{u}
-            &= &28\begin{svector}1\\3\\1\\\end{svector}
-                -8\begin{svector}-2\\5\\-2\\\end{svector}
-            &= &\begin{svector}44\\44\\44\\\end{svector}
-            &= &44\begin{svector}1\\1\\1\\\end{svector}.\\
+            &= &28\begin{svector}  1\\  3\\  1\\\end{svector}
+                -8\begin{svector}  -2\\  5\\  -2\\\end{svector}
+            &= &\begin{svector}  44\\  44\\  44\\\end{svector}
+            &= &44\begin{svector}  1\\  1\\  1\\\end{svector}.\\
     \end{mathe}
 
 Aus \eqcref{eq:1:ueb:3:ex:1} ergibt sich der zu berechnende Untervektorraum
@@ -2793,8 +2796,8 @@ als
         \cap\vectorspacespan\{\mathbf{w}_{1},\mathbf{w}_{2}\}
         &= &U
         &= &\vectorspacespan\{\mathbf{u}\}
-        &= &\vectorspacespan\{44\begin{svector}1\\1\\1\\\end{svector}\}
-        &= &\vectorspacespan\{\begin{svector}1\\1\\1\\\end{svector}\}.\\
+        &= &\vectorspacespan\{44\begin{svector}  1\\  1\\  1\\\end{svector}\}
+        &= &\vectorspacespan\{\begin{svector}  1\\  1\\  1\\\end{svector}\}.\\
     \end{mathe}
 
 %% AUFGABE 3-2
@@ -4161,7 +4164,7 @@ für alle $A,B\in R$.
 Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
 
     \begin{mathe}[mc]{rcl}
-        z\in\kmplx &\mapsto &\begin{svector}\ReTeil(z)\\\ImTeil(z)\\\end{svector}\in\reell^{2},\\
+        z\in\kmplx &\mapsto &\begin{svector}  \ReTeil(z)\\  \ImTeil(z)\\\end{svector}\in\reell^{2},\\
         \mathbf{x}\in\reell^{2} &\mapsto &x_{1}+\imageinh x_{2}\in\kmplx.\\
     \end{mathe}
 
@@ -4170,12 +4173,12 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
     \item
         \begin{claim*}
             Für alle $z\in\kmplx\ohne\{0\}$ existieren eindeutige Werte $r\in(0,\infty)$ und $\alpha\in[0,2\pi)$,
-            dann $z=r\cdot\begin{svector}\cos(\alpha)\\\sin(\alpha)\\\end{svector}$ (unter der o.\,s. Identifizierung).
+            dann $z=r\cdot\begin{svector}  \cos(\alpha)\\  \sin(\alpha)\\\end{svector}$ (unter der o.\,s. Identifizierung).
         \end{claim*}
 
         \begin{proof}
-            Unter der Identifizierung können wir $z=\begin{svector}x\\y\\\end{svector}$ schreiben, wobei $x,y\in\reell$.
-            Da $z\neq 0=\begin{svector}0\\0\\\end{svector}$, muss entweder $x\neq 0$ oder $y\neq 0$ gelten.
+            Unter der Identifizierung können wir $z=\begin{svector}  x\\  y\\\end{svector}$ schreiben, wobei $x,y\in\reell$.
+            Da $z\neq 0=\begin{svector}  0\\  0\\\end{svector}$, muss entweder $x\neq 0$ oder $y\neq 0$ gelten.
 
             Zur {\bfseries Existenz}:
             Sei $r:=\sqrt{x^{2}+y^{2}}$. Dann $r>0$ weil $(x,y)\neq (0,0)$.\\
@@ -4223,15 +4226,15 @@ Wir identifizieren $\kmplx$ mit $\reell^{2}$ mittel der Abbildungen
                 \end{kompaktenum}
 
             Darum gilt in allen Fällen
-                $r\cdot\begin{svector}\cos(\alpha)\\\sin(\alpha)\\\end{svector}
-                =\begin{svector}r\cos(\alpha)\\r\sin(\alpha)\\\end{svector}
-                =\begin{svector}x\\y\\\end{svector}
+                $r\cdot\begin{svector}  \cos(\alpha)\\  \sin(\alpha)\\\end{svector}
+                =\begin{svector}  r\cos(\alpha)\\  r\sin(\alpha)\\\end{svector}
+                =\begin{svector}  x\\  y\\\end{svector}
                 =z$.
 
             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
@@ -4291,10 +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}
@@ -4302,10 +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.
@@ -4314,9 +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*}
 
@@ -4328,10 +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
@@ -4339,11 +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 $n$.
@@ -4359,21 +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*}
@@ -4710,9 +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*}
 
@@ -4721,9 +4724,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{rcl}
                     A &:= &\begin{matrix}{ccc}
-                    1 &3 &2\\
-                    2 &2 &1\\
-                    2 &1 &-1\\
+                      1 &3 &2\\
+                      2 &2 &1\\
+                      2 &1 &-1\\
                     \end{matrix}
                 \end{mathe}
 
@@ -4739,9 +4742,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{c}
                     \begin{matrix}{ccc}
-                    1 &2 &2\\
-                    0 &4 &5\\
-                    0 &3 &5\\
+                      1 &2 &2\\
+                      0 &4 &5\\
+                      0 &3 &5\\
                     \end{matrix}\\
                 \end{mathe}
 
@@ -4751,9 +4754,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{c}
                     \begin{matrix}{ccc}
-                    \boxed{1} &2 &2\\
-                    0 &\boxed{4} &5\\
-                    0 &0 &\boxed{5}\\
+                      \boxed{1} &2 &2\\
+                      0 &\boxed{4} &5\\
+                      0 &0 &\boxed{5}\\
                     \end{matrix}\\
                 \end{mathe}
             \end{algorithm}
@@ -4768,9 +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*}
 
@@ -4779,9 +4782,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{rcl}
                     A &:= &\begin{matrix}{ccc}
-                    1 &3 &2\\
-                    2 &2 &1\\
-                    2 &1 &4\\
+                      1 &3 &2\\
+                      2 &2 &1\\
+                      2 &1 &4\\
                     \end{matrix}
                 \end{mathe}
 
@@ -4794,9 +4797,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{c}
                     \begin{matrix}{ccl}
-                    \boxed{1} &2 &2\\
-                    0 &\boxed{4} &5(=0)\\
-                    0 &0 &5(=0)\\
+                      \boxed{1} &2 &2\\
+                      0 &\boxed{4} &5(=0)\\
+                      0 &0 &5(=0)\\
                     \end{matrix}\\
                 \end{mathe}
 
@@ -4811,9 +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*}
 
@@ -4822,9 +4825,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{rcl}
                     A &:= &\begin{matrix}{ccc}
-                    1 &1+\imageinh &\imageinh\\
-                    \imageinh &-\imageinh &1-\imageinh\\
-                    0 &1-2\imageinh &2-\imageinh\\
+                      1 &1+\imageinh &\imageinh\\
+                      \imageinh &-\imageinh &1-\imageinh\\
+                      0 &1-2\imageinh &2-\imageinh\\
                     \end{matrix}
                 \end{mathe}
 
@@ -4838,9 +4841,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{c}
                     \begin{matrix}{ccc}
-                    1 &1+\imageinh &\imageinh\\
-                    0 &2+\imageinh &1+2\imageinh\\
-                    0 &1-2\imageinh &2-\imageinh\\
+                      1 &1+\imageinh &\imageinh\\
+                      0 &2+\imageinh &1+2\imageinh\\
+                      0 &1-2\imageinh &2-\imageinh\\
                     \end{matrix}\\
                 \end{mathe}
 
@@ -4850,9 +4853,9 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{c}
                     \begin{matrix}{ccc}
-                    \boxed{1} &1+\imageinh &\imageinh\\
-                    0 &\boxed{2+\imageinh} &1+2\imageinh\\
-                    0 &0 &0\\
+                      \boxed{1} &1+\imageinh &\imageinh\\
+                      0 &\boxed{2+\imageinh} &1+2\imageinh\\
+                      0 &0 &0\\
                     \end{matrix}\\
                 \end{mathe}
             \end{algorithm}
@@ -4868,9 +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*}
 
@@ -4879,12 +4882,12 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{rcl}
                     A &:= &\begin{smatrix}
-                    1&1&0\\
-                    0&1&1\\
-                    0&0&1\\
-                    1&-1&-1\\
-                    0&1&2\\
-                    0&-2&-1\\
+                      1 &1 &0\\
+                      0 &1 &1\\
+                      0 &0 &1\\
+                      1 &-1 &-1\\
+                      0 &1 &2\\
+                      0 &-2 &-1\\
                     \end{smatrix}
                 \end{mathe}
 
@@ -4898,12 +4901,12 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{c}
                     \begin{smatrix}
-                    1&1&0\\
-                    0&1&1\\
-                    0&0&1\\
-                    0&2&1\\
-                    0&1&2\\
-                    0&-2&-1\\
+                      1 &1 &0\\
+                      0 &1 &1\\
+                      0 &0 &1\\
+                      0 &2 &1\\
+                      0 &1 &2\\
+                      0 &-2 &-1\\
                     \end{smatrix}\\
                 \end{mathe}
 
@@ -4913,12 +4916,12 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{c}
                     \begin{smatrix}
-                    1&1&0\\
-                    0&1&1\\
-                    0&-2&-1\\
-                    0&2&1\\
-                    0&1&2\\
-                    0&0&1\\
+                      1 &1 &0\\
+                      0 &1 &1\\
+                      0 &-2 &-1\\
+                      0 &2 &1\\
+                      0 &1 &2\\
+                      0 &0 &1\\
                     \end{smatrix}\\
                 \end{mathe}
 
@@ -4931,12 +4934,12 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{c}
                     \begin{smatrix}
-                    1&1&0\\
-                    0&1&1\\
-                    0&0&1\\
-                    0&0&1\\
-                    0&0&1\\
-                    0&0&1\\
+                      1 &1 &0\\
+                      0 &1 &1\\
+                      0 &0 &1\\
+                      0 &0 &1\\
+                      0 &0 &1\\
+                      0 &0 &1\\
                     \end{smatrix}\\
                 \end{mathe}
 
@@ -4949,12 +4952,12 @@ Dies ist zufälligerweise auch das Nullelement von $V$.
 
                 \begin{mathe}[mc]{c}
                     \begin{smatrix}
-                    1&1&0\\
-                    0&1&1\\
-                    0&0&1\\
-                    0&0&0\\
-                    0&0&0\\
-                    0&0&0\\
+                      1 &1 &0\\
+                      0 &1 &1\\
+                      0 &0 &1\\
+                      0 &0 &0\\
+                      0 &0 &0\\
+                      0 &0 &0\\
                     \end{smatrix}\\
                 \end{mathe}
             \end{algorithm}
@@ -5042,14 +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).
@@ -5076,9 +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
@@ -5204,7 +5207,7 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
                 &\Longleftrightarrow
                     &\underbrace{
                         \begin{matrix}{cccc}
-                        1 &-1 &-2 &2\\
+                          1 &-1 &-2 &2\\
                         \end{matrix}
                     }_{=:A_{1}}
                     \mathbf{x}=\zerovector\\
@@ -5214,7 +5217,7 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
                 &\Longleftrightarrow
                     &\underbrace{
                         \begin{matrix}{cccc}
-                        1 &-1 &-1 &-1\\
+                          1 &-1 &-1 &-1\\
                         \end{matrix}
                     }_{=:A_{2}}
                     \mathbf{x}=\zerovector\\
@@ -5227,8 +5230,8 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
                 &\Longleftrightarrow
                     &\underbrace{
                         \begin{matrix}{cccc}
-                        1 &-1 &-2 &2\\
-                        1 &-1 &-1 &-1\\
+                          1 &-1 &-2 &2\\
+                          1 &-1 &-1 &-1\\
                         \end{matrix}
                     }_{=:A_{3}}
                     \mathbf{x}=\zerovector\\
@@ -5247,7 +5250,7 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
             \begin{mathe}[mc]{rcl}
                 A_{1} &= &\begin{matrix}{cccc}
-                1 &-1 &-2 &2\\
+                  1 &-1 &-2 &2\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -5266,13 +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}
 
@@ -5283,9 +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}
@@ -5297,7 +5300,7 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
             \begin{mathe}[mc]{rcl}
                 A_{2} &= &\begin{matrix}{cccc}
-                1 &-1 &-2 &2\\
+                  1 &-1 &-2 &2\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -5314,13 +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}
 
@@ -5331,9 +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}
@@ -5346,12 +5349,12 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
             \begin{mathe}[mc]{rclcl}
                 A_{3}
                     &= &\begin{matrix}{cccc}
-                    1 &-1 &-2 &2\\
-                    1 &-1 &-1 &-1\\
+                      1 &-1 &-2 &2\\
+                      1 &-1 &-1 &-1\\
                     \end{matrix}
                     &\rightsquigarrow &\begin{matrix}{cccc}
-                    1 &-1 &-2 &2\\
-                    0 &0 &1 &-3\\
+                      1 &-1 &-2 &2\\
+                      0 &0 &1 &-3\\
                     \end{matrix}\\
             \end{mathe}
 
@@ -5371,10 +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.
@@ -5387,8 +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}
@@ -5403,21 +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}
 
@@ -5435,10 +5438,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                 \begin{mathe}[mc]{rcl}
                     \begin{matrix}{ccccc}
-                    1 &2 &-2 &1 &1\\
-                    1 &0 &0 &0 &0\\
-                    0 &1 &0 &1 &0\\
-                    0 &0 &1 &0 &1\\
+                      1 &2 &-2 &1 &1\\
+                      1 &0 &0 &0 &0\\
+                      0 &1 &0 &1 &0\\
+                      0 &0 &1 &0 &1\\
                     \end{matrix}\\
                 \end{mathe}
 
@@ -5448,10 +5451,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
             \begin{mathe}[mc]{rcl}
                 \begin{matrix}{ccccc}
-                1 &2 &-2 &1 &1\\
-                0 &2 &-2 &1 &1\\
-                0 &1 &0 &1 &0\\
-                0 &0 &1 &0 &1\\
+                  1 &2 &-2 &1 &1\\
+                  0 &2 &-2 &1 &1\\
+                  0 &1 &0 &1 &0\\
+                  0 &0 &1 &0 &1\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -5461,10 +5464,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
             \begin{mathe}[mc]{rcl}
                 \begin{matrix}{ccccc}
-                1 &2 &-2 &1 &1\\
-                0 &2 &-2 &1 &1\\
-                0 &0 &2 &1 &-1\\
-                0 &0 &1 &0 &1\\
+                  1 &2 &-2 &1 &1\\
+                  0 &2 &-2 &1 &1\\
+                  0 &0 &2 &1 &-1\\
+                  0 &0 &1 &0 &1\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -5474,10 +5477,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
             \begin{mathe}[mc]{rcl}
                 \begin{matrix}{ccccc}
-                1 &2 &-2 &1 &1\\
-                0 &2 &-2 &1 &1\\
-                0 &0 &2 &1 &-1\\
-                0 &0 &0 &1 &-3\\
+                  1 &2 &-2 &1 &1\\
+                  0 &2 &-2 &1 &1\\
+                  0 &0 &2 &1 &-1\\
+                  0 &0 &0 &1 &-3\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -5489,10 +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}
 
@@ -5517,10 +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}
 
@@ -5850,8 +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
@@ -5859,8 +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.
@@ -5884,11 +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
@@ -5918,10 +5921,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
                         v_{1}\:v_{2}\:v_{3}\:u_{1}\:u_{2}
                     \right)
                     &= &\begin{smatrix}
-                    1&-1&2&1&-1\\
-                    2&3&-1&2&-2\\
-                    -1&0&-1&-1&1\\
-                    -2&-2&1&1&2\\
+                      1 &-1 &2 &1 &-1\\
+                      2 &3 &-1 &2 &-2\\
+                      -1 &0 &-1 &-1 &1\\
+                      -2 &-2 &1 &1 &2\\
                     \end{smatrix}\\
                 \end{mathe}
 
@@ -5937,10 +5940,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                 \begin{mathe}[mc]{rcl}
                     \begin{matrix}{ccccc}
-                    1 &-1 &2 &1 &-1\\
-                    0 &5 &-5 &0 &0\\
-                    0 &-1 &1 &0 &0\\
-                    0 &-4 &5 &3 &0\\
+                      1 &-1 &2 &1 &-1\\
+                      0 &5 &-5 &0 &0\\
+                      0 &-1 &1 &0 &0\\
+                      0 &-4 &5 &3 &0\\
                     \end{matrix}.\\
                 \end{mathe}
 
@@ -5952,10 +5955,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                 \begin{mathe}[mc]{rcl}
                     \begin{matrix}{ccccc}
-                    1 &-1 &2 &1 &-1\\
-                    0 &5 &-5 &0 &0\\
-                    0 &0 &0 &0 &0\\
-                    0 &0 &1 &3 &0\\
+                      1 &-1 &2 &1 &-1\\
+                      0 &5 &-5 &0 &0\\
+                      0 &0 &0 &0 &0\\
+                      0 &0 &1 &3 &0\\
                     \end{matrix}.\\
                 \end{mathe}
 
@@ -5965,10 +5968,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                 \begin{mathe}[mc]{rcl}
                     \begin{matrix}{ccccc}
-                    \boxed{1} &-1 &2 &1 &-1\\
-                    0 &\boxed{5} &-5 &0 &0\\
-                    0 &0 &\boxed{1} &3 &0\\
-                    0 &0 &0 &0 &0\\
+                      \boxed{1} &-1 &2 &1 &-1\\
+                      0 &\boxed{5} &-5 &0 &0\\
+                      0 &0 &\boxed{1} &3 &0\\
+                      0 &0 &0 &0 &0\\
                     \end{matrix}.\\
                 \end{mathe}
             \end{algorithm}
@@ -6047,10 +6050,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
                         u_{1}\:u_{2}\:v_{1}\:v_{2}\:v_{3}
                     \right)
                     &= &\begin{smatrix}
-                    1&-1&1&-1&2\\
-                    2&-2&2&3&-1\\
-                    -1&1&-1&0&-1\\
-                    1&2&-2&-2&1\\
+                      1 &-1 &1 &-1 &2\\
+                      2 &-2 &2 &3 &-1\\
+                      -1 &1 &-1 &0 &-1\\
+                      1 &2 &-2 &-2 &1\\
                     \end{smatrix}.\\
                 \end{mathe}
 
@@ -6067,10 +6070,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                 \begin{mathe}[mc]{rcl}
                     \begin{matrix}{ccccc}
-                    1 &-1 &1 &-1 &2\\
-                    0 &0 &0 &5 &-5\\
-                    0 &0 &0 &-1 &1\\
-                    0 &3 &-3 &-1 &-1\\
+                      1 &-1 &1 &-1 &2\\
+                      0 &0 &0 &5 &-5\\
+                      0 &0 &0 &-1 &1\\
+                      0 &3 &-3 &-1 &-1\\
                     \end{matrix}.\\
                 \end{mathe}
 
@@ -6080,10 +6083,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                 \begin{mathe}[mc]{rcl}
                     \begin{matrix}{ccccc}
-                    1 &-1 &1 &-1 &2\\
-                    0 &3 &-3 &-1 &-1\\
-                    0 &0 &0 &-1 &1\\
-                    0 &0 &0 &5 &-5\\
+                      1 &-1 &1 &-1 &2\\
+                      0 &3 &-3 &-1 &-1\\
+                      0 &0 &0 &-1 &1\\
+                      0 &0 &0 &5 &-5\\
                     \end{matrix}.\\
                 \end{mathe}
 
@@ -6093,10 +6096,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                 \begin{mathe}[mc]{rcl}
                     \begin{matrix}{ccccc}
-                    \boxed{1} &-1 &1 &-1 &2\\
-                    0 &\boxed{3} &-3 &-1 &-1\\
-                    0 &0 &0 &\boxed{-1} &1\\
-                    0 &0 &0 &0 &0\\
+                      \boxed{1} &-1 &1 &-1 &2\\
+                      0 &\boxed{3} &-3 &-1 &-1\\
+                      0 &0 &0 &\boxed{-1} &1\\
+                      0 &0 &0 &0 &0\\
                     \end{matrix}.\\
                 \end{mathe}
             \end{algorithm}
@@ -6133,9 +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 $\phi:\reell^{3}\to\reell^{4}$
@@ -6161,10 +6164,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                 \begin{mathe}[mc]{rcl}
                     A &:= &\begin{smatrix}
-                    1&2&-2\\
-                    -2&-5&6\\
-                    3&7&-9\\
-                    1&0&-3\\
+                      1 &2 &-2\\
+                      -2 &-5 &6\\
+                      3 &7 &-9\\
+                      1 &0 &-3\\
                     \end{smatrix}
                 \end{mathe}
 
@@ -6187,10 +6190,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                     \begin{mathe}[mc]{rcl}
                         \begin{matrix}{ccccc}
-                        1 &2 &-2\\
-                        0 &-1 &2\\
-                        0 &1 &-3\\
-                        0 &-2 &-1\\
+                          1 &2 &-2\\
+                          0 &-1 &2\\
+                          0 &1 &-3\\
+                          0 &-2 &-1\\
                         \end{matrix}.\\
                     \end{mathe}
 
@@ -6202,10 +6205,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                     \begin{mathe}[mc]{rcl}
                         \begin{matrix}{ccccc}
-                        1 &2 &-2\\
-                        0 &-1 &2\\
-                        0 &0 &-1\\
-                        0 &0 &-5\\
+                          1 &2 &-2\\
+                          0 &-1 &2\\
+                          0 &0 &-1\\
+                          0 &0 &-5\\
                         \end{matrix}.\\
                     \end{mathe}
 
@@ -6215,10 +6218,10 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
 
                     \begin{mathe}[mc]{rcl}
                         \begin{matrix}{ccccc}
-                        \boxed{1} &2 &-2\\
-                        0 &\boxed{-1} &2\\
-                        0 &0 &\boxed{-1}\\
-                        0 &0 &0\\
+                          \boxed{1} &2 &-2\\
+                          0 &\boxed{-1} &2\\
+                          0 &0 &\boxed{-1}\\
+                          0 &0 &0\\
                         \end{matrix}.\\
                     \end{mathe}
 
@@ -6349,6 +6352,773 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
                 ist dies zur Injektivität von $\psi\circ\phi$ äquivalent.
         \end{proof}
 
+%% ********************************************************************************
+%% FILE: body/uebung/ueb10.tex
+%% ********************************************************************************
+
+\setcounternach{chapter}{10}
+\chapter[Woche 10]{Woche 10}
+    \label{ueb:10}
+
+\textbf{ACHTUNG.}
+Diese Lösungen dienen \emph{nicht} als Musterlösungen sondern eher als Referenz.
+Hier wird eingehender gearbeitet, als generell verlangt wird.
+Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche vergleichen kann.
+
+%% AUFGABE 10-1
+\let\altsectionname\sectionname
+\def\sectionname{Aufgabe}
+\section[Aufgabe 1]{}
+    \label{ueb:10:ex:1}
+\let\sectionname\altsectionname
+
+    Seien $V,W,X$ Vektorräume über einem Körper $K$
+    und seien ${\phi:V\to W}$ und ${\psi:W\to X}$ linear.
+    Des Weiteren sei angenommen, $V,W,X$ seien endlich dimensional.
+
+    \begin{enumerate}{\bfseries (a)}
+        %% AUFGABE 10-1(a)
+        \item\voritemise
+            \begin{schattierteboxdunn}
+            \begin{claim*}
+                $\rank(\psi\circ\phi)\leq\min\{\rank(\psi),\rank(\phi)\}$.
+            \end{claim*}
+            \end{schattierteboxdunn}
+
+            \begin{proof}
+                Sei $d=\rank(\psi\circ\phi)=\dim(\range(\psi\circ\phi))$.
+                \textbf{Zu zeigen:}
+                    ${d\leq\rank(\psi)\textoverset{Defn}{=}\dim(\range(\psi))}$
+                und
+                    ${d\leq\rank(\phi)\textoverset{Defn}{=}\dim(\range(\phi))}$.
+                Da $d=\dim(\range(\psi\circ\phi))$, existiert eine Basis
+
+                    \begin{mathe}[mc]{c}
+                        (x_{1},x_{2},\ldots,x_{d})\subseteq\range(\psi\circ\phi)\\
+                    \end{mathe}
+
+                für $\range(\psi\circ\phi)$.
+                Insbesondere sind $x_{1},x_{2},\ldots,x_{d}$ linear unabhängig.
+                Da jedes $x_{i}\in\range(\psi\circ\phi)$, existieren $v_{i}\in V$,
+                so dass $(\psi\circ\phi)(v_{i})=x_{i}$ für alle ${i\in\{1,2,\ldots,d\}}$.
+                Setze außerdem $w_{i}:=\phi(v_{i})$ für jedes ${i\in\{1,2,\ldots,d\}}$.
+                Darum $x_{i}=(\psi\circ\phi)(v_{i})=\psi(\phi(v_{i}))=\psi(w_{i})$ für alle ${i\in\{1,2,\ldots,d\}}$.
+
+                Da ${x_{i}=\psi(w_{i})\in\range(\psi)}$ für alle ${i\in\{1,2,\ldots,d\}}$,
+                folgt aus der linearen Unabhängigkeit von
+                    ${(x_{1},x_{2},\ldots,x_{d})}$,
+                dass \fbox{$d\leq\dim(\range(\psi))$}
+                (siehe \cite[Satz~5.3.4]{sinn2020}).
+
+                Es bleibt noch zu zeigen, dass $d\leq\rank(\phi)$ gilt.
+                Da $w_{i}\in\range(\phi)$ für alle ${i\in\{1,2,\ldots,d\}}$,
+                reicht es aus wie oben zu zeigen,
+                dass ${(w_{1},w_{2},\ldots,w_{d})}$ linear unabhängig sind.
+                Seien also $c_{1},c_{2},\ldots,c_{d}\in K$ beliebig
+                mit $\sum_{i=1}^{d}c_{i}w_{i}=\zerovector$.
+                \textbf{Zu zeigen:} $c_{1},c_{2},\ldots,c_{d}=0_{K}$.
+                Es gilt nun
+
+                    \begin{mathe}[mc]{rclql}
+                        \sum_{i=1}^{d}c_{i}x_{i}
+                            &= &\sum_{i=1}^{d}c_{i}\psi(w_{i})
+                                &\text{(siehe oben)}\\
+                            &= &\psi(\sum_{i=1}^{d}c_{i}w_{i})
+                                &\text{wegen Linearität}\\
+                            &= &\psi(\zerovector)
+                                &\text{per Voraussetzung}\\
+                            &= &\zerovector
+                                &\text{(siehe \cite[Lemma~6.1.2]{sinn2020})}.\\
+                    \end{mathe}
+
+                Wegen linearer Unabhängigkeit von ${(x_{1},x_{2},\ldots,x_{d})}$,
+                folgt hieraus, dass $c_{1},c_{2},\ldots,c_{d}=0_{K}$.
+                Damit haben wir die lineare Unabhängigkeit von ${(w_{1},w_{2},\ldots,w_{d})}$ bewiesen.
+                Da $w_{i}\in\range(\phi)$ für alle ${i\in\{1,2,\ldots,d\}}$,
+                erschließt sich aus der linearen Unabhängigkeit
+                    \fbox{$d\leq\dim(\range(\phi))$}
+                (siehe \cite[Satz~5.3.4]{sinn2020}).
+            \end{proof}
+        %% AUFGABE 10-1(b)
+        \item\voritemise
+            \begin{schattierteboxdunn}
+            \begin{claim*}
+                Falls $\psi$ injektiv ist,
+                so gilt $\rank(\psi\circ\phi)=\rank(\phi)$.
+            \end{claim*}
+            \end{schattierteboxdunn}
+
+            \begin{proof}
+                Laut Teil (a) wissen wir bereits, dass $\rank(\psi\circ\phi)\leq\rank(\phi)$.
+                Es bleibt \textbf{zu zeigen}, dass $\rank(\psi\circ\phi)\geq\rank(\phi)$.
+                Sei $d:=\rank(\phi)\textoverset{Defn}{=}\rank(\range(\phi))$.
+                Dann existiert eine Basis
+
+                    \begin{mathe}[mc]{c}
+                        (w_{1},w_{2},\ldots,w_{d})\subseteq\range(\phi)\\
+                    \end{mathe}
+
+                für $\range(\phi)$.
+                Da jedes $w_{i}\in\range(\phi)$, existieren $v_{i}\in V$,
+                so dass $\phi(v_{i})=w_{i}$ für alle ${i\in\{1,2,\ldots,d\}}$.
+                Setze außerdem
+                    $x_{i}:=\psi(w_{i})=\psi(\phi(v_{i}))=(\psi\circ\phi)(v_{i})\in\range(\psi\circ\phi)$
+                für jedes ${i\in\{1,2,\ldots,d\}}$.
+
+                Wir zeigen nun, dass
+                    $(x_{1},x_{2},\ldots,x_{d})$
+                linear unabhängig ist.
+                Seien also $c_{1},c_{2},\ldots,c_{d}\in K$ beliebig
+                mit $\sum_{i=1}^{d}c_{i}x_{i}=\zerovector$.
+                \textbf{Zu zeigen:} $c_{1},c_{2},\ldots,c_{d}=0_{K}$.
+                Es gilt nun
+
+                    \begin{mathe}[mc]{rclql}
+                        \psi(\sum_{i=1}^{d}c_{i}w_{i})
+                            &= &\sum_{i=1}^{d}c_{i}\psi(w_{i})
+                                &\text{wegen Linearität}\\
+                            &= &\sum_{i=1}^{d}c_{i}x_{i}
+                                &\text{per Konstruktion}\\
+                            &= &\zerovector
+                                &\text{per Voraussetzung}.\\
+                    \end{mathe}
+
+                Darum $\sum_{i=1}^{d}c_{i}w_{i}\in\ker(\psi)=\{\zerovector\}$
+                wegen \uline{Injektivität} von $\psi$ (siehe \cite[Lemma~6.1.4]{sinn2020}).
+                Also $\sum_{i=1}^{d}c_{i}w_{i}=\zerovector$,
+                woraus sich ergibt, dass $c_{1},c_{2},\ldots,c_{d}=0_{K}$,
+                weil ${(w_{1},w_{2},\ldots,w_{d})}$ linear unabhängig ist.
+                Darum haben wir die lineare Unabhängigkeit von $(x_{1},x_{2},\ldots,x_{d})$ bewiesen.
+
+                Da $(x_{1},x_{2},\ldots,x_{d})$ linear unabhängig ist
+                und $x_{i}\in\range(\psi\circ\phi)$ für alle ${i\in\{1,2,\ldots,d\}}$,
+                folgt \fbox{$%
+                    d\leq\dim(\range(\psi\circ\phi))
+                    \textoverset{Defn}{=}\rank(\psi\circ\phi)%
+                $}
+                (siehe \cite[Satz~5.3.4]{sinn2020}).
+            \end{proof}
+        %% AUFGABE 10-1(c)
+        \item\voritemise
+            \begin{schattierteboxdunn}
+            \begin{claim*}
+                Falls $\phi$ surjektiv ist,
+                so gilt $\rank(\psi\circ\phi)=\rank(\psi)$.
+            \end{claim*}
+            \end{schattierteboxdunn}
+
+            \begin{proof}
+                Es gilt
+
+                    \begin{mathe}[bc]{rcl}
+                        \rank(\psi\circ\phi)
+                        &\textoverset{Defn}{=}
+                            &\dim(\range(\psi\circ\phi))\\
+                        &= &\dim((\psi\circ\phi)(V))\\
+                        &= &\dim(\psi(\phi(V)))\\
+                        &= &\dim(\psi(\range(\phi)))\\
+                        &= &\dim(\psi(W)),
+                            \quad\text{da $\phi$ surjektiv ist}\\
+                        &= &\dim(\range(\psi))\\
+                        &\textoverset{Defn}{=}
+                            &\rank(\psi).\\
+                    \end{mathe}
+
+            \end{proof}
+    \end{enumerate}
+
+%% AUFGABE 10-2
+\clearpage
+\let\altsectionname\sectionname
+\def\sectionname{Aufgabe}
+\section[Aufgabe 2]{}
+    \label{ueb:10:ex:2}
+\let\sectionname\altsectionname
+
+    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}.
+    \end{mathe}
+
+    Zuerst beoachten wir dass
+        $\cal{A}:=(v_{1},\,v_{2})$
+    und
+        $\cal{B}:=(w_{1},\,w_{2},\,w_{3})$
+    Basen für $\reell^{2}$ bzw. $\reell^{3}$:
+
+    Setze
+        ${A:=(v_{1}\ v_{2})=\begin{smatrix}
+          2 &-1\\
+          1 &1\\
+        \end{smatrix}}$
+    und
+        ${B:=(w_{1}\ w_{2}\ w_{3})=\begin{smatrix}
+          1 &1 &0\\
+          1 &-1 &3\\
+          0 &2 &-1\\
+        \end{smatrix}}$.
+    Anwendung des Gaußverfahrens liefert
+
+    \begin{mathe}[mc]{rclcl}
+        A
+            &\xrightarrow{
+                Z_{2}\mapsfrom 2\cdot Z_{2} - Z_{1}
+            }
+                &
+                \begin{matrix}{rr}
+                  \boxed{2} &-1\\
+                  0 &\boxed{3}\\
+                \end{matrix}
+                \\
+        B
+            &\xrightarrow{
+                Z_{2}\mapsfrom Z_{2} - Z_{1}
+            }
+                &
+                \begin{matrix}{rrr}
+                  1 &1 &0\\
+                  0 &-2 &3\\
+                  0 &2 &-1\\
+                \end{matrix}
+            &\xrightarrow{
+                Z_{3}\mapsfrom Z_{3} + Z_{2}
+            }
+                &
+                \begin{matrix}{rrr}
+                  \boxed{1} &1 &0\\
+                  0 &\boxed{-2} &3\\
+                  0 &0 &\boxed{2}\\
+                \end{matrix}
+                \\
+    \end{mathe}
+
+    Darum gilt
+        ${\rank(A)=2=\dim(\reell^{2})}$
+        und
+        ${\rank(B)=3=\dim(\reell^{3})}$,
+    woraus sich ergibt,
+    dass
+        $\cal{A}$ eine Basis für $\reell^{2}$
+    und $\cal{B}$ eine Basis für $\reell^{3}$
+    sind (vgl. \cite[Korollar~5.4.4]{sinn2020}).
+
+    \begin{enumerate}{\bfseries (a)}
+        %% AUFGABE 10-2(a)
+        \item
+            Für $x\in\reell^{2}$ sei $\phi(x)\in\reell^{2}$ definiert durch
+
+                \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}
+                        &= &\underbrace{
+                            \begin{matrix}{rr}
+                              2 &0\\
+                              0 &-1\\
+                            \end{matrix}
+                            }_{=:C}
+                            x.\\
+                \end{mathe}
+
+            Also gilt $\phi=\phi_{C}$.
+            Folglich gilt $M^{\cal{A}}_{\cal{A}}(\phi)=A^{-1}CA$.
+            Um dies zu berechnen betrachten wir das augmentierte System
+            $\left(A \vert CA\right)$
+            und führen darauf das Gaußverfahren, bis die linke Hälfte wie die Identitätsmatrix aussieht.
+            Dann steht in der rechten Hälfte die Matrix $A^{-1}CA$.
+            Es gilt
+
+                \begin{mathe}[mc]{rcccl}
+                    CA &= &\begin{matrix}{rr}
+                      2 &0\\
+                      0 &-1\\
+                    \end{matrix} \begin{matrix}{rr}
+                      2 &-1\\
+                      1 &1\\
+                    \end{matrix}
+                        &= &
+                            \begin{matrix}{rr}
+                              4 &-2\\
+                              -1 &-1\\
+                            \end{matrix}
+                            .\\
+                \end{mathe}
+
+            Und das augmentierte System, $\left(A \vert CA\right)$, wird wie folgt reduziert:
+
+                \begin{mathe}[mc]{rcl}
+                    \begin{matrix}{rr|rr}
+                      2 &-1 &4 &-2\\
+                      1 &1 &-1 &-1\\
+                    \end{matrix}
+                        &\xrightarrow{
+                            Z_{2}\mapsfrom 2\cdot Z_{2} - Z_{1}
+                        }
+                            &
+                            \begin{matrix}{rr|rr}
+                              2 &-1 &4 &-2\\
+                              0 &3 &-6 &0\\
+                            \end{matrix}
+                        \\
+                            &\xrightarrow{
+                                Z_{1}\mapsfrom 3\cdot Z_{1} + Z_{2}
+                            }
+                                &
+                                \begin{matrix}{rr|rr}
+                                  6 &0 &6 &-6\\
+                                  0 &3 &-6 &0\\
+                                \end{matrix}
+                        \\
+                        &\xrightarrow{
+                            \substack{
+                                Z_{1}\mapsfrom Z_{1}:6\\
+                                Z_{2}\mapsfrom Z_{2}:3\\
+                            }
+                        }
+                            &
+                            \begin{matrix}{rr|rr}
+                              1 &0 &1 &-1\\
+                              0 &1 &-2 &0\\
+                            \end{matrix}
+                            .\\
+                \end{mathe}
+
+            Darum gilt $M^{\cal{A}}_{\cal{A}}(\phi)=A^{-1}CA=\boxed{\begin{matrix}{rr}
+              1 &-1\\
+              -2 &0\\
+            \end{matrix}}$.
+
+        %% AUFGABE 10-2(b)
+        \item
+            Für $x\in\reell^{2}$ sei $\phi(x)\in\reell^{3}$ definiert durch
+
+                \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}
+                        &= &\underbrace{
+                            \begin{matrix}{rr}
+                              1 &1\\
+                              -1 &5\\
+                              2 &-4\\
+                            \end{matrix}
+                            }_{=:C}
+                            x.\\
+                \end{mathe}
+
+            Also gilt $\phi=\phi_{C}$.
+            Folglich gilt $M^{\cal{A}}_{\cal{B}}(\phi)=B^{-1}CA$.
+            Um dies zu berechnen betrachten wir das augmentierte System
+            $\left(B \vert CA\right)$
+            und führen darauf das Gaußverfahren, bis die linke Hälfte wie die Identitätsmatrix aussieht.
+            Dann steht in der rechten Hälfte die Matrix $B^{-1}CA$.
+            Es gilt
+
+                \begin{mathe}[mc]{rcccl}
+                    CA &= &\begin{matrix}{rr}
+                      1 &1\\
+                      -1 &5\\
+                      2 &-4\\
+                    \end{matrix} \begin{matrix}{rr}
+                      2 &-1\\
+                      1 &1\\
+                    \end{matrix}
+                        &= &
+                            \begin{matrix}{rrr}
+                              3 &0\\
+                              3 &6\\
+                              0 &-6\\
+                            \end{matrix}
+                            .\\
+                \end{mathe}
+
+            Und das augmentierte System, $\left(B \vert CA\right)$, wird wie folgt reduziert:
+
+                \begin{mathe}[mc]{rcl}
+                    \begin{matrix}{rrr|rr}
+                      1 &1 &0 &3 &0\\
+                      1 &-1 &3 &3 &6\\
+                      0 &2 &-1 &0 &-6\\
+                    \end{matrix}
+                        &\xrightarrow{
+                            Z_{2}\mapsfrom Z_{1} - Z_{2}
+                        }
+                            &
+                            \begin{matrix}{rrr|rr}
+                              1 &1 &0 &3 &0\\
+                              0 &2 &-3 &0 &-6\\
+                              0 &2 &-1 &0 &-6\\
+                            \end{matrix}
+                            \\
+                        &\xrightarrow{
+                            \substack{
+                                Z_{1}\mapsfrom 2\cdot Z_{1}-Z_{2}\\
+                                Z_{3}\mapsfrom Z_{3}-Z_{2}\\
+                            }
+                        }
+                            &
+                            \begin{matrix}{rrr|rr}
+                              2 &0 &3 &6 &6\\
+                              0 &2 &-3 &0 &-6\\
+                              0 &0 &2 &0 &0\\
+                            \end{matrix}
+                            \\
+                        &\xrightarrow{
+                            \substack{
+                                Z_{1}\mapsfrom 2\cdot Z_{1} - 3\cdot Z_{3}\\
+                                Z_{2}\mapsfrom 2\cdot Z_{1} + 3\cdot Z_{3}\\
+                            }
+                        }
+                            &
+                            \begin{matrix}{rrr|rr}
+                              4 &0 &0 &12 &12\\
+                              0 &4 &0 &0 &-12\\
+                              0 &0 &2 &0 &0\\
+                            \end{matrix}
+                            \\
+                        &\xrightarrow{
+                            \substack{
+                                Z_{1}\mapsfrom Z_{1}:4\\
+                                Z_{2}\mapsfrom Z_{2}:4\\
+                                Z_{3}\mapsfrom Z_{3}:2\\
+                            }
+                        }
+                            &
+                            \begin{matrix}{rrr|rr}
+                              1 &0 &0 &3 &3\\
+                              0 &1 &0 &0 &-3\\
+                              0 &0 &1 &0 &0\\
+                            \end{matrix}
+                            .\\
+                \end{mathe}
+
+            Darum gilt $M^{\cal{A}}_{\cal{B}}(\phi)=B^{-1}CA=\boxed{\begin{matrix}{rr}
+              3 &3\\
+              0 &-3\\
+              0 &0\\
+            \end{matrix}}$.
+
+        \item
+            Für $x\in\reell^{3}$ sei $\phi(x)\in\reell^{2}$ definiert durch
+
+                \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}
+                        &= &\underbrace{
+                            \begin{matrix}{rrr}
+                              0 &2 &0\\
+                              3 &0 &-4\\
+                            \end{matrix}
+                            }_{=:C}
+                            x.\\
+                \end{mathe}
+
+            Also gilt $\phi=\phi_{C}$.
+            Folglich gilt $M^{\cal{B}}_{\cal{A}}(\phi)=A^{-1}CB$.
+            Um dies zu berechnen betrachten wir das augmentierte System
+            $\left(A \vert CB\right)$
+            und führen darauf das Gaußverfahren, bis die linke Hälfte wie die Identitätsmatrix aussieht.
+            Dann steht in der rechten Hälfte die Matrix $A^{-1}CB$.
+            Es gilt
+
+                \begin{mathe}[mc]{rcccl}
+                    CB &= &\begin{matrix}{rrr}
+                      0 &2 &0\\
+                      3 &0 &-4\\
+                    \end{matrix} \begin{matrix}{rrr}
+                      1 &1 &0\\
+                      1 &-1 &3\\
+                      0 &2 &-1\\
+                    \end{matrix}
+                        &= &
+                            \begin{matrix}{rrr}
+                              2 &-2 &6\\
+                              3 &-5 &4\\
+                            \end{matrix}
+                            .\\
+                \end{mathe}
+
+            Und das augmentierte System, $\left(A \vert CB\right)$, wird wie folgt reduziert:
+
+                \begin{mathe}[mc]{rcl}
+                    \begin{matrix}{rr|rrr}
+                      2 &-1 &2 &-2 &6\\
+                      1 &1 &3 &-5 &4\\
+                    \end{matrix}
+                        &\xrightarrow{
+                            Z_{2}\mapsfrom 2\cdot Z_{2} - Z_{1}
+                        }
+                            &
+                            \begin{matrix}{rr|rrr}
+                              2 &-1 &2 &-2 &6\\
+                              0 &3 &4 &-8 &2\\
+                            \end{matrix}
+                            \\
+                        &\xrightarrow{
+                            Z_{1}\mapsfrom 3\cdot Z_{1}+Z_{2}
+                        }
+                            &
+                            \begin{matrix}{rr|rrr}
+                              6 &0 &10 &-14 &20\\
+                              0 &3 &4 &-8 &2\\
+                            \end{matrix}
+                            \\
+                        &\xrightarrow{
+                            \substack{
+                                Z_{1}\mapsfrom Z_{1}:6\\
+                                Z_{2}\mapsfrom Z_{2}:3\\
+                            }
+                        }
+                            &
+                            \begin{matrix}{rr|rrr}
+                              1 &0 &5/3 &-7/3 &10/3\\
+                              0 &1 &4/3 &-8/3 &2/3\\
+                            \end{matrix}
+                            .\\
+                \end{mathe}
+
+            Darum gilt $M^{\cal{B}}_{\cal{A}}(\phi)=A^{-1}CB=\boxed{\begin{matrix}{rrr}
+              5/3 &-7/3 &10/3\\
+              4/3 &-8/3 &2/3\\
+            \end{matrix}}$.
+    \end{enumerate}
+
+%% AUFGABE 10-3
+\clearpage
+\let\altsectionname\sectionname
+\def\sectionname{Aufgabe}
+\section[Aufgabe 3]{}
+    \label{ueb:10:ex:3}
+\let\sectionname\altsectionname
+
+    Seien $d\in\ntrlpos$ und $a\in\reell$.
+    Betrachet sei die Abbildung
+
+        \begin{mathe}[mc]{rcccl}
+            \phi &: &\reell[x]_{\leq d} &\to &\reell[x]_{\leq d}\\
+                &: &f(x) &\mapsto &f(x+a)\\
+        \end{mathe}
+
+    Bevor wir die uns den Aufgaben widmen,
+    machen wir von der binomischen Formel mehrfach Gebrauch
+    und beobachten wir
+    für
+        $f\in\reell[x]_{\leq d}$
+        der Form $f=\sum_{k=0}^{d}\alpha_{k}x^{k}$,
+        wobei $\alpha_{0},\alpha_{1},\ldots,\alpha_{d}\in\reell$,
+    dass
+
+        \begin{mathe}[mc]{rcl}
+            \eqtag[eq:0:ueb:10:ex:3]
+            \phi(f)(x)
+                &= &f(x+a)\\
+                &= &\sum_{k=0}^{d}\alpha_{k}(x+a)^{k}\\
+                &= &\sum_{k=0}^{d}\alpha_{k}\sum_{i=0}^{k}\choose{k}{i}a^{k-i}x^{i}
+                    \quad\text{(Anwendung der bin. Formel)}\\
+                &= &\sum_{k=0}^{d}\sum_{i=0}^{k}\alpha_{k}\choose{k}{i}a^{k-i}x^{i}\\
+                &= &\sum_{i=0}^{d}
+                    \big(
+                    \underbrace{
+                        \sum_{k=i}^{d}
+                        \choose{k}{i}a^{k-i}\alpha_{k}
+                    }_{=:\alpha'_{i}}
+                    \big)
+                    x^{i}.\\
+        \end{mathe}
+
+    Insbesondere ist es zumindest klar,
+    dass $\phi(f)\in\reell[x]_{\leq d}$
+    für alle $f\in\reell[x]_{\leq d}$
+    gilt. D.\,h. $\phi$ ist wohldefiniert.
+
+    \begin{enumerate}{\bfseries (a)}
+        %% AUFGABE 10-3(a)
+        \item
+
+            \begin{claim*}
+                $\phi$ ist linear.
+            \end{claim*}
+
+            \begin{proof}[Ansatz I]
+                Wir beweisen die Aussage direkt.
+                Für diesen Beweis ist es wichtig, dass wir die Objekte des Vektorraumes,
+                    $\reell[x]_{\leq d}$
+                wirklich als Funktionen und nicht als abstrakte Objekt betrachten.
+
+                Es reicht aus \textbf{zu zeigen},
+                dass $\phi$ dem (LIN)-Axiom genügt (siehe \cite[\S{}6.1]{sinn2020}).
+                Seien also $c,c'\in\reell$ und $f,g\in\reell[x]_{\leq d}$ beliebig.
+                \textbf{Zu zeigen:} $\phi(cf+c'g)=c\phi(f)+c'\phi(g)$.
+                Für alle $x\in\reell$ berechnen wir
+
+                    \begin{mathe}[mc]{rcl}
+                        \eqtag[eq:1:ueb:10:ex:3]
+                        (\phi(cf+c'g))(x)
+                            &\textoverset{Defn}{=}
+                                &(cf+c'g)(x+a)\\
+                            &\textoverset{($\ast$)}{=}
+                                &(cf)(x+a)+(c'g)(x+a)\\
+                            &\textoverset{($\ast$)}{=}
+                                &c\cdot (f(x+a)) + c'\cdot (g(x+a))\\
+                            &\textoverset{Defn}{=}
+                                &c\cdot(\phi(f)(x)) + c'\cdot(\phi(g)(x))\\
+                            &\textoverset{($\ast$)}{=}
+                                &(c\cdot\phi(f))(x) + (c'\cdot\phi(g))(x)\\
+                            &\textoverset{($\ast$)}{=}
+                                &(c\cdot\phi(f)+c'\cdot\phi(g))(x)\\
+                    \end{mathe}
+
+                Hier gelten die mit ($\ast$) gekennzeichneten Gleichungen,
+                per Definition von (punktweise) Skalarmultiplikation und Vektoraddition
+                innerhalb des Vektorraumes, der aus $\reell$-wertigen Funktionen besteht.
+
+                Da nun \eqcref{eq:1:ueb:10:ex:3} für alle $x\in\reell$ gilt,
+                haben wir beweisen, dass $\phi(cf+c'g)=c\phi(f)+c'\phi(g)$
+                für alle $c,c'\in\reell$ und $f,g\in\reell[x]_{\leq d}$
+                Darum gilt das (LIN)-Axiom.
+                Somit ist $\phi$ linear.
+            \end{proof}
+
+            \begin{proof}[Ansatz II]
+                In diesem Ansatz machen wir von Darstellungsmatrizen Gebrauch.
+                Da nun ${\cal{B}=\{x^{0},x^{1},\ldots,x^{d}\}}$ eine Basis von ${V:=\reell[x]_{\leq d}}$ ist
+                und $\phi$ eine wohldefinierte Funktion von $V$ nach $V$ ist,
+                können wir o.\,E. $\phi$ als Abbildung von $\reell^{d+1}$ nach $\reell^{d+1}$ betrachten.
+                Bezeichen wir diese Abbildung als $\tilde{\phi}$.
+                Da $\reell^{d+1}$ zu $V$ isomorph ist,
+                reicht es aus, um die Linearität von $\phi$ zu beweisen,
+                \textbf{zu zeigen},
+                dass eine Matrix $C\in\reell^{(d+1)\times(d+1)}$ existiert,
+                so dass $\tilde{\phi}$ gleich $\phi_{C}$ ist,
+                was auf jeden Fall linear ist (siehe \cite[\S{}6.2]{sinn2020}).
+                Zu diesem Zwecke konstruieren wir die Matrix,
+                    ${C\in\reell^{(d+1)\times(d+1)}}$,
+                mit Einträgen
+
+                    \begin{mathe}[mc]{rcl}
+                        C_{ij} &=
+                            &\begin{cases}[m]{ccl}
+                                \choose{j}{i}a^{j-i} &: &j\geq i\\
+                                0                    &: &j<i\\
+                            \end{cases}\\
+                    \end{mathe}
+
+                für ${i,j\in\{0,1,\ldots,d\}}$.
+                Laut \eqcref{eq:0:ueb:10:ex:3} gilt für alle
+                    ${\mathbf{\alpha}\in\reell^{d+1}}$
+                unter Betrachtung des entsprechenden Objekts
+                    ${f=\sum_{k=0}^{d}\alpha_{k}x^{k}\in\reell[x]_{\leq d}}$
+
+                    \begin{mathe}[mc]{rcl}
+                        \tilde{\phi}(\mathbf{\alpha}) &= &\mathbf{\alpha}'\\
+                    \end{mathe}
+
+                wobei $\mathbf{\alpha}'\in\reell^{d+1}$ durch
+
+                    \begin{mathe}[mc]{rcl}
+                        \mathbf{\alpha}'_{i}
+                            &\eqcrefoverset{eq:0:ueb:10:ex:3}{=}
+                                &\sum_{j=i}^{d}\choose{j}{i}a^{j-i}\alpha_{j}\\
+                    \end{mathe}
+
+                für alle ${i\in\{0,1,\ldots,d+1\}}$ gegeben ist.
+                Beachte nun, dass
+
+                    \begin{mathe}[mc]{rcccccl}
+                        \sum_{j=i}^{d}\choose{j}{i}a^{j-i}\alpha_{j}
+                            &= &\sum_{j=i}^{d}C_{ij}\alpha_{j}
+                            &= &\sum_{j=0}^{d}C_{ij}\alpha_{j}
+                            &= &\text{das $i$-te Element von}\,C\cdot\mathbf{\alpha}\\
+                    \end{mathe}
+
+                für alle ${i\in\{0,1,\ldots,d+1\}}$, da $C_{ij}=0$ für $0\leq j<i$.
+                Darum gilt
+                    $\tilde{\phi}(\mathbf{\alpha})
+                    =C\cdot\mathbf{\alpha}
+                    =\phi_{C}(\mathbf{\alpha})$
+                für alle $\mathbf{\alpha}\in\reell^{d+1}$.
+                Also, $\tilde{\phi}=\phi_{C}$.
+                Darum ist $\phi$ wie oben erklärt linear.
+            \end{proof}
+
+        %% AUFGABE 10-3(b)
+        \item
+            Laut Teil (a) ist $\phi$ linear, und somit hat $\phi$
+            eine (eindeutige) Matrizendarstellung
+                $M^{\cal{B}}_{\cal{B}}(\phi)\in\reell^{(d+1)\times (d+1)}$.
+            Wir haben dies im 2. Ansatz im Allgemeinen explizit ausgerechnet
+            (siehe die Konstruktion von $C$ in diesem Beweis).
+            Für $d=4$ ist dies nun
+
+                \begin{mathe}[mc]{rcl}
+                    M^{\cal{B}}_{\cal{B}}(\phi)
+                        &= &\begin{matrix}{ccccc}
+                          \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}
+                          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{mathe}
+
+            Als \textbf{Alternative} können wir (davon ausgehend, dass $\phi$ linear ist)
+            einfach die Abbildung auf die Basiselemente anwenden,
+            diese wiederum in Bezug auf die Basis umschreiben
+            und daraus die Darstellungsmatrix ablesen:
+
+                \begin{mathe}[mc]{rclclcl}
+                        \phi(x^{0})
+                            &= &(x+a)^{0}
+                            &= &
+                    1\cdot x^{0}
+                            &\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}\\
+                        \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}\\
+                        \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}\\
+                        \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}\\
+                \end{mathe}
+
+            Wenn wir diese Spalten in eine Matrix aufführen,
+            ergibt sich genau die o.\,s. Matrizendarstellung von $\phi$.
+
+            \textbf{Bemerkung.} Als kleine Bestätigung, dass dies »nicht offensichtlich falsch« sei,
+            beachte, dass sich diese Matrix der Identitätsmatrix ($\cong$ Identitätsabbbildung)
+            nähert als ${a\longrightarrow 0}$ in $\reell$, was zu erwarten ist.
+    \end{enumerate}
+
 \setcounternach{part}{2}
 \part{Selbstkontrollenaufgaben}
 
@@ -7518,22 +8288,22 @@ 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}
@@ -7554,18 +8324,18 @@ 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}
@@ -8135,8 +8905,8 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
 
             \begin{mathe}[mc]{c}
                 \begin{matrix}{cc|c}
-                1+\imageinh &1 &0\\
-                -2 &2-\imageinh &1\\
+                  1+\imageinh &1 &0\\
+                  -2 &2-\imageinh &1\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -8146,8 +8916,8 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
 
             \begin{mathe}[mc]{c}
                 \begin{matrix}{cc|c}
-                2 &1-\imageinh &0\\
-                -2 &2-\imageinh &1\\
+                  2 &1-\imageinh &0\\
+                  -2 &2-\imageinh &1\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -8157,8 +8927,8 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
 
             \begin{mathe}[mc]{c}
                 \begin{matrix}{cc|c}
-                2 &1-\imageinh &0\\
-                0 &3-2\imageinh &1\\
+                  2 &1-\imageinh &0\\
+                  0 &3-2\imageinh &1\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -8232,10 +9002,10 @@ 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}
@@ -8249,9 +9019,9 @@ 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}
@@ -8273,8 +9043,8 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
 
             \begin{mathe}[mc]{c}
                 \begin{matrix}{cc|c}
-                2 &-3 &0\\
-                10 &7 &-5\\
+                  2 &-3 &0\\
+                  10 &7 &-5\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -8284,8 +9054,8 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
 
             \begin{mathe}[mc]{c}
                 \begin{matrix}{cc|c}
-                2 &-3 &0\\
-                0 &22 &-5\\
+                  2 &-3 &0\\
+                  0 &22 &-5\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -8295,8 +9065,8 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
 
             \begin{mathe}[bc]{c}
                 \begin{matrix}{cc|c}
-                14 &1 &-5\\
-                0 &22 &-5\\
+                  14 &1 &-5\\
+                  0 &22 &-5\\
                 \end{matrix}\\
             \end{mathe}
         \end{algorithm}
@@ -8308,8 +9078,8 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
 
             \begin{mathe}[mc]{c}
                 \begin{matrix}{cc|c}
-                \mathbf{3} &1 &6\\
-                0 &0 &\mathbf{6}\\
+                  \mathbf{3} &1 &6\\
+                  0 &0 &\mathbf{6}\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -8323,8 +9093,8 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
 
             \begin{mathe}[mc]{c}
                 \begin{matrix}{cc|c}
-                1 &1 &8\\
-                0 &9 &8\\
+                  1 &1 &8\\
+                  0 &9 &8\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -8334,8 +9104,8 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
 
             \begin{mathe}[mc]{c}
                 \begin{matrix}{cc|c}
-                \mathbf{9} &0 &-1\\
-                0 &\mathbf{9} &8\\
+                  \mathbf{9} &0 &-1\\
+                  0 &\mathbf{9} &8\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -8350,7 +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}
 
@@ -8446,8 +9216,8 @@ Einfacher ist also natürlich die Anwendung von dem Lemma von B\'ezout.
 
             \begin{mathe}[mc]{c}
                 \begin{matrix}{cc|c}
-                -1 &a &3\\
-                a &-4 &0\\
+                  -1 &a &3\\
+                  a &-4 &0\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -8457,8 +9227,8 @@ Einfacher ist also natürlich die Anwendung von dem Lemma von B\'ezout.
 
             \begin{mathe}[mc]{c}
                 \begin{matrix}{cc|c}
-                1 &a &3\\
-                0 &a^{2}-4 &3a\\
+                  1 &a &3\\
+                  0 &a^{2}-4 &3a\\
                 \end{matrix}\\
             \end{mathe}
 
@@ -8481,8 +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)}
@@ -8490,7 +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}
@@ -8506,7 +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,
@@ -8521,7 +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 $\brkt{\mathbf{w},\mathbf{w}_{\perp}}=0$,
@@ -8697,7 +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}
@@ -8709,23 +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}
 
 %% ********************************************************************************
@@ -8831,9 +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 $\mathbb{F}_{5}$.
@@ -8848,9 +9618,9 @@ Wir betrachten die Komposition ${g\circ f:X\to Z}$
 
                 \begin{mathe}[mc]{rcl}
                     A &= &\begin{matrix}{ccc}
-                    1 &0 &2\\
-                    2 &2 &1\\
-                    2 &1 &1\\
+                      1 &0 &2\\
+                      2 &2 &1\\
+                      2 &1 &1\\
                     \end{matrix}.\\
                 \end{mathe}
 
@@ -8865,9 +9635,9 @@ Wir betrachten die Komposition ${g\circ f:X\to Z}$
 
                 \begin{mathe}[mc]{rcl}
                     \begin{matrix}{ccc}
-                    1 &0 &2\\
-                    0 &2 &2\\
-                    0 &1 &2\\
+                      1 &0 &2\\
+                      0 &2 &2\\
+                      0 &1 &2\\
                     \end{matrix}.\\
                 \end{mathe}
 
@@ -8877,9 +9647,9 @@ Wir betrachten die Komposition ${g\circ f:X\to Z}$
 
                 \begin{mathe}[mc]{rcl}
                     \begin{matrix}{ccc}
-                    1 &0 &2\\
-                    0 &2 &2\\
-                    0 &0 &2\\
+                      1 &0 &2\\
+                      0 &2 &2\\
+                      0 &0 &2\\
                     \end{matrix}.\\
                 \end{mathe}
             \end{algorithm}