行列式


3　積


$(7)\qquad |AB|=|A||B|$

$簡単に証明できそうですが意外に手強いです。$
$まず、\ n=2\ のときとき成りたつことを確認しましょう。$

\[ |A|= \left| \begin{array}{rr} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{array} \right| ,\qquad |B|= \left| \begin{array}{rr} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{array} \right| \quad とおくと \] \begin{eqnarray*} |AB| &=& \left| \begin{array}{rr} 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{array} \right|\\ \\ &=& \left| \begin{array}{rr} a_{11}b_{11} \hspace{3em} & a_{11}b_{12} \hspace{3em} \ \ \\ a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22}\ \ \\ \end{array} \right| + \left| \begin{array}{rr} a_{12}b_{21} \hspace{3em} & a_{12}b_{22} \hspace{3em} \ \ \\ a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22}\ \ \\ \end{array} \right|\\ \\ &=& a_{11}\left| \begin{array}{rr} b_{11} \hspace{4em} & b_{12} \hspace{4em} \ \ \\ a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22}\ \ \\ \end{array} \right| + a_{12}\left| \begin{array}{rr} b_{21} \hspace{4em} & b_{22} \hspace{4em} \ \ \\ a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22}\ \ \\ \end{array} \right|\\ \\ &=& a_{11}\Big\{\left| \begin{array}{rr} b_{11} \hspace{1em} & b_{12}\hspace{1em} \ \ \\ a_{21}b_{11} & a_{21}b_{12} \ \ \\ \end{array} \right| + \left| \begin{array}{rr} b_{11} \hspace{1em} & b_{12} \hspace{1em} \ \ \\ a_{22}b_{21} & a_{22}b_{22} \ \ \\ \end{array} \right|\Big\} + a_{12}\Big\{\left| \begin{array}{rr} b_{21} \hspace{1em} & b_{22}\hspace{1em} \ \ \\ a_{21}b_{11} & a_{21}b_{12} \ \ \\ \end{array} \right| + \left| \begin{array}{rr} b_{21} \hspace{1em} & b_{22} \hspace{1em} \ \ \\ a_{22}b_{21} & a_{22}b_{22} \ \ \\ \end{array} \right|\Big\}\\ \\ &=& a_{11}\Big\{a_{21}\left| \begin{array}{rr} b_{11} & b_{12} \ \ \\ b_{11} & b_{12} \ \ \\ \end{array} \right| + a_{22}\left| \begin{array}{rr} b_{11} & b_{12} \ \ \\ b_{21} & b_{22} \ \ \\ \end{array} \right|\Big\} + a_{12}\Big\{a_{21}\left| \begin{array}{rr} b_{21} & b_{22} \ \ \\ b_{11} & b_{12} \ \ \\ \end{array} \right| + a_{22}\left| \begin{array}{rr} b_{21} & b_{22} \ \ \\ b_{21} & b_{22} \ \ \\ \end{array} \right|\Big\}\\ \\ & & 1項目と4項目は第1行と第2行が一致しているので行列式の値は0\\ \\ &=& a_{11}a_{22}\left| \begin{array}{rr} b_{11} & b_{12} \ \ \\ b_{21} & b_{22} \ \ \\ \end{array} \right| + a_{12}a_{21}\left| \begin{array}{rr} b_{21} & b_{22} \ \ \\ b_{11} & b_{12} \ \ \\ \end{array} \right|\\ \\ &=& a_{11}a_{22}\left| \begin{array}{rr} b_{11} & b_{12} \ \ \\ b_{21} & b_{22} \ \ \\ \end{array} \right| - a_{12}a_{21}\left| \begin{array}{rr} b_{11} & b_{12} \ \ \\ b_{21} & b_{22} \ \ \\ \end{array} \right|\\ \\ &=& (a_{11}a_{22}-a_{12}a_{21})\left| \begin{array}{rr} b_{11} & b_{12} \ \ \\ b_{21} & b_{22} \ \ \\ \end{array} \right|\\ \\ &=&|A||B| \end{eqnarray*}

$これを \ \Sigma \ 記号をつかって表してみると$

\begin{eqnarray*} |AB| &=& \left| \begin{array}{rr} \sum _{k_1=1}^2 a_{1k_1}b_{k_11} & \sum _{k_1=1}^2 a_{1k_1}b_{k_12}\ \ \\ \sum _{k_2=1}^2 a_{2k_2}b_{k_21} & \sum _{k_2=1}^2 a_{2k_2}b_{k_22}\ \ \\ \end{array} \right|\\ \\ &=& \sum _{k_1=1}^2 a_{1k_1} \left| \begin{array}{rr} b_{k_11} \hspace{4em} & b_{k_12} \hspace{4em} \ \ \\ \sum _{k_2=1}^2 a_{2k_2}b_{k_21} & \sum _{k_2=1}^2 a_{2k_2}b_{k_22}\ \ \\ \end{array} \right|\\ \\ &=& \sum _{k_1=1}^2 \sum _{k_2=1}^2 a_{1k_1}a_{2k_2} \left| \begin{array}{rr} b_{k_11} & b_{k_12}\ \ \\ b_{k_21} & b_{k_22}\ \ \\ \end{array} \right|\\ \\ \end{eqnarray*} \[ \qquad k_1=k_2 \quad のときは \qquad \left| \begin{array}{rr} b_{k_11} & b_{k_12}\ \ \\ b_{k_21} & b_{k_22}\ \ \\ \end{array} \right| =0 \quad だから k_1,\ k_2 \ \ は異なる数 \quad よって \quad (k_1,\ k_2)=(1,\ 2),\ (2,\ 1) \] \[これは、\{1,\ 2\} \ の順列であるから \quad 置換 \quad \sigma= \left( \begin{array}{rrr} 1 & 2 \\ k_1 & k_2 \\ \end{array} \right) \quad とすると \quad \sum _{k_2=1}^2 \sum _{k_1=1}^2 a_{1k_1}a_{2k_2}=\sum _{\sigma \in S_2} a_{1k_1}a_{2k_2} \] $よって$
\begin{eqnarray*} |AB| &=&\sum _{\sigma \in S_2} a_{1k_1}a_{2k_2} \left| \begin{array}{rr} b_{k_11} & b_{k_12}\ \ \\ b_{k_21} & b_{k_22}\ \ \\ \end{array} \right|\\ \\ &=& a_{11}a_{22} \left| \begin{array}{rr} b_{11} & b_{12}\ \ \\ b_{21} & b_{22}\ \ \\ \end{array} \right| + a_{12}a_{21} \left| \begin{array}{rr} b_{21} & b_{22}\ \ \\ b_{11} & b_{12}\ \ \\ \end{array} \right|\\ \\ &=& a_{11}a_{22} \left| \begin{array}{rr} b_{11} & b_{12}\ \ \\ b_{21} & b_{22}\ \ \\ \end{array} \right| - a_{12}a_{21} \left| \begin{array}{rr} b_{11} & b_{12}\ \ \\ b_{21} & b_{22}\ \ \\ \end{array} \right|\\ \\ &=&\sum _{\sigma \in S_2} a_{1k_1}a_{2k_2} \cdot \varepsilon(\sigma ) \left| \begin{array}{rr} b_{11} & b_{12}\ \ \\ b_{21} & b_{22}\ \ \\ \end{array} \right|\\ \\ &=&\sum _{\sigma \in S_2}\varepsilon(\sigma ) a_{1k_1}a_{2k_2}\cdot \left| \begin{array}{rr} b_{11} & b_{12}\ \ \\ b_{21} & b_{22}\ \ \\ \end{array} \right|\\ \\ &=&|A||B| \end{eqnarray*}

$これで \ n\ 次の証明の見通しがたちました。$

$(証明)$

\begin{eqnarray*} |AB| &=& \left| \begin{array}{rr} \sum _{k_1=1}^n a_{1k_1}b_{k_11} & \cdots & \sum _{k_1=1}^n a_{1k_1}b_{k_1n}\ \ \\ \vdots \hspace{3em} & & \vdots \hspace{3em}\ \ \\ \sum _{k_n=1}^n a_{nk_n}b_{k_n1} & \cdots & \sum _{k_n=1}^n a_{nk_n}b_{k_nn}\ \ \\ \end{array} \right|\\ \\ &=& \sum _{k_1=1}^n a_{1k_1} \left| \begin{array}{rr} b_{k_11} \hspace{4em} & \cdots & b_{k_1n} \hspace{4em} \ \ \\ \vdots \hspace{3em} & & \vdots \hspace{3em}\ \ \\ \sum _{k_n=1}^n a_{nk_n}b_{k_n1} & \cdots & \sum _{k_n=1}^n a_{nk_n}b_{k_nn}\ \ \\ \end{array} \right|\\ \\ &=& \sum _{k_1=1}^n \sum _{k_2=1}^n \cdots \sum _{k_n=1}^n a_{1k_1}a_{2k_2} \cdots a_{nk_n} \left| \begin{array}{rr} b_{k_11} & \cdots & b_{k_1n} \ \ \\ \vdots & & \vdots \ \ \\ b_{k_n1} & \cdots & b_{k_nn}\ \ \\ \end{array} \right|\\ \\ \end{eqnarray*} \[ \qquad k_1=k_2= \cdots =k_n \quad のときは \quad \left| \begin{array}{rr} b_{k_11} & \cdots & b_{k_1n} \ \ \\ \vdots & & \vdots \ \ \\ b_{k_n1} & \cdots & b_{k_nn}\ \ \\ \end{array} \right| =0 \quad だから\\ \] $\qquad k_1,k_2,\cdots k_n はすべて異なる数 \quad すなわち \quad \{1,\ 2,\ \cdots \ n \}\ の順列である。$

\[ 置換 \quad \sigma= \left( \begin{array}{rrr} 1 & 2 & \cdots & n\\ p_1 & p_2 & \cdots & p_n \\ \end{array} \right) \quad とすると\\ \]
\[ \sum _{k_1=1}^n \sum _{k_2=1}^n \cdots \sum _{k_n=1}^n a_{1k_1}a_{2k_2} \cdots a_{nk_n}= \sum _{\sigma \in S_n}a_{1k_1}a_{2k_2} \cdots a_{nk_n} \]
$よって$
\begin{eqnarray*} |AB| &=&\sum _{\sigma \in S_n}a_{1k_1}a_{2k_2} \cdots a_{nk_n} \left| \begin{array}{rr} b_{k_11} & \cdots & b_{k_1n} \ \ \\ \vdots & & \vdots \ \ \\ b_{k_n1} & \cdots & b_{k_nn}\ \ \\ \end{array} \right|\\ \\ &=&\sum _{\sigma \in S_n}a_{1k_1}a_{2k_2} \cdots a_{nk_n} \cdot \varepsilon(\sigma ) \left| \begin{array}{rr} b_{11} & & \cdots & b_{1n} \ \ \\ \vdots & & & \vdots \ \ \\ b_{n1} & & \cdots & b_{nn} \ \ \\ \end{array} \right|\\ \\ &=&\sum _{\sigma \in S_n}\varepsilon(\sigma ) a_{1k_1}a_{2k_2} \cdots a_{nk_n}\cdot \left| \begin{array}{rr} b_{11} & & \cdots & b_{1n} \ \ \\ \vdots & & & \vdots \ \ \\ b_{n1} & & \cdots & b_{nn} \ \ \\ \end{array} \right|\\ \\ &=&|A||B| \end{eqnarray*}

　

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