Lösung für Aufgabe 5.2.8
Beweisen Sie, dass $(M_{2}(\R),\cdot)$ eine Halbgruppe ist.Seien $A,B,C\in M_2(\R)$. Dann gilt
\begin{eqnarray*}
(AB)C &=&
\begin{pmatrix}
a_{11} b_{11} + a_{12} b_{21}& a_{11} b_{12} + a_{12} b_{22}\\a_{21} b_{11} + a_{22} b_{21}& a_{21} b_{12} + a_{22} b_{22}
\end{pmatrix}
\begin{pmatrix}
c_{11}&c_{12}\\c_{21}&c_{22}
\end{pmatrix}\\
&=&
\begin{pmatrix}
(a_{11} b_{11} + a_{12} b_{21}) c_{11} + (a_{11} b_{12} + a_{12} b_{22}) c_{21}& (a_{11} b_{11} + a_{12} b_{21}) c_{12} + (a_{11} b_{12} + a_{12} b_{22}) c_{22}\\(a_{21} b_{11} + a_{22} b_{21}) c_{11} + (a_{21} b_{12} + a_{22} b_{22}) c_{21}& (a_{21} b_{11} + a_{22} b_{21}) c_{12} + (a_{21} b_{12} + a_{22} b_{22}) c_{22}
\end{pmatrix}\\
&=&
\begin{pmatrix}
a_{11} b_{11} c_{11} + a_{12} b_{21} c_{11} + a_{11} b_{12} c_{21} + a_{12} b_{22} c_{21}& a_{11} b_{11} c_{12} + a_{12} b_{21} c_{12} + a_{11} b_{12} c_{22} + a_{12} b_{22} c_{22}\\a_{21} b_{11} c_{11} + a_{22} b_{21} c_{11} + a_{21} b_{12} c_{21} + a_{22} b_{22} c_{21}& a_{21} b_{11} c_{12} + a_{22} b_{21} c_{12} + a_{21} b_{12} c_{22} + a_{22} b_{22} c_{22}
\end{pmatrix}\\
&=&
\begin{pmatrix}
a_{11} (b_{11} c_{11} + b_{12} c_{21}) + a_{12} (b_{21} c_{11} + b_{22} c_{21})& a_{11} (b_{11} c_{12} + b_{12} c_{22}) + a_{12} (b_{21} c_{12} + b_{22} c_{22})\\a_{21} (b_{11} c_{11} + b_{12} c_{21}) + a_{22} (b_{21} c_{11} + b_{22} c_{21})& a_{21} (b_{11} c_{12} + b_{12} c_{22}) + a_{22} (b_{21} c_{12} + b_{22} c_{22})
\end{pmatrix}\\
&=&
\begin{pmatrix}
a_{11}&a_{12}\\a_{21}&a_{22}
\end{pmatrix}
\begin{pmatrix}
b_{11} c_{11} + b_{12} c_{21}& b_{11} c_{12} + b_{12} c_{22}\\b_{21} c_{11} + b_{22} c_{21}& b_{21} c_{12} + b_{22} c_{22}
\end{pmatrix}\\
&=& A(BC)
\end{eqnarray*}