# ${}^{t}\left(AB\right)=$${}^{t}B$${}^{t}A$の証明

$l×m$ 行列$A=\left({a}_{ij}\right)$$m×n$行列$B=\left({b}_{jk}\right)$行列の積を行列 $F=\left({f}_{ik}\right)$ とする．

すなわち

$AB=F$

${f}_{ik}=\sum _{h=1}^{m}{a}_{ih}{b}_{hk}$ ･･････(1)

となる．また行列$A=\left({a}_{ij}\right)$転置行列を行列 $C=\left({c}_{ji}\right)$ ，行列$B=\left({b}_{jk}\right)$ の転置行列を$D=\left({d}_{kj}\right)$ ，行列 $F=\left({f}_{ik}\right)$ の転置行列を $G=\left({g}_{ki}\right)$ とする．

すなわち

${c}_{ji}={a}_{ji}$  ･･････(2)

${d}_{kj}={b}_{jk}$  ･･････(3)

${g}_{ki}={f}_{ik}$  ･･････(4)

とする．

${}^{t}\left(AB\right)=$${}^{t}F$

$=G$ ･･････(5)

また

${}^{t}B$${}^{t}A=DC$

$=\left(\sum _{h=1}^{m}{d}_{kh}{c}_{hi}\right)$

(2)，(3)より

$=\left(\sum _{k=1}^{m}{b}_{hk}{a}_{ih}\right)$

$=\left(\sum _{k=1}^{m}{a}_{ih}{b}_{hk}\right)$

$\sum _{k=1}^{m}{a}_{ih}{b}_{hk}={f}_{ik}={g}_{ki}$ より

$=\left({g}_{ki}\right)$

$=G$ ･･････(6)

が導かれる．

(5)，(6)より

${}^{t}\left(AB\right)=$${}^{t}B$${}^{t}A$

が成り立つ．

## ■具体例

$A=\left(\begin{array}{cc}{a}_{11}& {a}_{12}\\ {a}_{21}& {a}_{22}\end{array}\right)$$B=\left(\begin{array}{cc}{b}_{11}& {b}_{12}\\ {b}_{21}& {b}_{22}\end{array}\right)$とする．

${}^{t}\left(AB\right)=$${}^{t}\left(\begin{array}{cc}{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)$

$=\left(\begin{array}{cc}{a}_{11}{b}_{11}+{a}_{12}{b}_{21}& {a}_{21}{b}_{11}+{a}_{22}{b}_{21}\\ {a}_{11}{b}_{12}+{a}_{12}{b}_{22}& {a}_{21}{b}_{12}+{a}_{22}{b}_{22}\end{array}\right)$  ･･････(7)

${}^{t}B$${}^{t}A=$${}^{t}\left(\begin{array}{cc}{b}_{11}& {b}_{12}\\ {b}_{21}& {b}_{22}\end{array}\right){}^{t}\left(\begin{array}{cc}{a}_{11}& {a}_{12}\\ {a}_{21}& {a}_{22}\end{array}\right)$

$=\left(\begin{array}{cc}{b}_{11}& {b}_{21}\\ {b}_{12}& {b}_{22}\end{array}\right)\left(\begin{array}{cc}{a}_{11}& {a}_{21}\\ {a}_{12}& {a}_{22}\end{array}\right)$

$=\left(\begin{array}{cc}{b}_{11}{a}_{11}+{b}_{21}{a}_{12}& {b}_{11}{a}_{21}+{b}_{21}{a}_{12}\\ {b}_{12}{a}_{11}+{b}_{22}{a}_{12}& {b}_{12}{a}_{21}+{b}_{22}{a}_{22}\end{array}\right)$

$=\left(\begin{array}{cc}{a}_{11}{b}_{11}+{a}_{12}{b}_{12}& {a}_{21}{b}_{11}+{a}_{22}{b}_{21}\\ {a}_{11}{b}_{12}+{a}_{12}{b}_{22}& {a}_{21}{b}_{12}+{a}_{22}{b}_{22}\end{array}\right)$  ･･････(8)

(7)，(8)より

${}^{t}\left(AB\right)=$${}^{t}B$${}^{t}A$

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