Wenn Sie das Buch noch nicht kennen, dann können Sie hier weitere Informationen finden.

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*}