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

${\displaystyle l\times m}$ 行列${\displaystyle A=\left(a_{ij}\right)}$ と${\displaystyle m\times n}$行列${\displaystyle B=\left(b_{jk}\right)}$ の行列の積を行列 ${\displaystyle F=\left(f_{ik}\right)}$ とする．

すなわち

${\displaystyle AB=F}$

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

となる．また行列${\displaystyle A=\left(a_{ij}\right)}$ の転置行列を行列 ${\displaystyle C=\left(c_{ji}\right)}$ ，行列${\displaystyle B=\left(b_{jk}\right)}$ の転置行列を${\displaystyle D=\left(d_{kj}\right)}$ ，行列 ${\displaystyle F=\left(f_{ik}\right)}$ の転置行列を ${\displaystyle G=\left(g_{ki}\right)}$ とする．

すなわち

${\displaystyle c_{ji}=a_{ji}}$  ･･････(2)

${\displaystyle d_{kj}=b_{jk}}$  ･･････(3)

${\displaystyle g_{ki}=f_{ik}}$  ･･････(4)

とする．

以上のように定めると

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

${\displaystyle =G}$ ･･････(5) 

また

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

行列の積の行列の積の定義より

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

(2)，(3)より

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

積は交換方法則が成り立つので ${\displaystyle b_{hk}{a}_{ih}=a_{ih}{b}_{hk}}$ となる．よって

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

${\displaystyle {\displaystyle \sum _{k=1}^m{a}_{ih}{b}_{hk}=f_{ik}=g_{ki}}}$ より

${\displaystyle =\left(g_{ki}\right)}$

${\displaystyle =G}$ ･･････(6) 

が導かれる．

(5)，(6)より

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

が成り立つ．

■具体例

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

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

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

${\displaystyle {}^{t}B}$${\displaystyle {}^{t}A=}$${\displaystyle {}^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)}$

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

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

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

(7)，(8)より

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

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最終更新日： 2024年11月22日