# 転置行列の行列式の値が元の行列の行列式の値と等しいことの証明

$A=\left(\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right)$ のとき，行列式の定義より

$|A|=|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|$$=\sum _{n}\mathrm{sgn}\left(\begin{array}{cccc}1& 2& \cdots & n\\ {i}_{1}& {i}_{2}& \cdots & {i}_{n}\end{array}\right){a}_{1{i}_{1}}{a}_{2{i}_{2}}\cdots {a}_{n{i}_{n}}$・・・・・・(1)

${}^{t}A=\left(\begin{array}{cccc}{b}_{11}& {b}_{12}& \cdots & {b}_{1n}\\ {b}_{21}& {b}_{22}& \cdots & {b}_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ {b}_{n1}& {b}_{n2}& \cdots & {b}_{nn}\end{array}\right)$  のとき同様に

$|{}^{t}A|=|\begin{array}{cccc}{b}_{11}& {b}_{12}& \cdots & {b}_{1n}\\ {b}_{21}& {b}_{22}& \cdots & {b}_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ {b}_{n1}& {b}_{n2}& \cdots & {b}_{nn}\end{array}|$$=\sum _{n}\mathrm{sgn}\left(\begin{array}{cccc}1& 2& \cdots & n\\ {i}_{1}& {i}_{2}& \cdots & {i}_{n}\end{array}\right){b}_{1{i}_{1}}{b}_{2{i}_{2}}\cdots {b}_{n{i}_{n}}$・・・・・・(2)

となる．

(2)の以下のように式変形をする．

${b}_{1{i}_{1}}={a}_{{i}_{1}1}$,${b}_{2{i}_{2}}={a}_{{i}_{2}2}$,$\cdots$,${b}_{n{i}_{n}}={b}_{{i}_{n}n}$より

$|{}^{t}A|=\sum _{n}sgn\left(\begin{array}{cccc}1& 2& \cdots & n\\ {i}_{1}& {i}_{2}& \cdots & {i}_{n}\end{array}\right){a}_{i1}{a}_{i2}\cdots {a}_{{i}_{n}n}$

$=\sum _{n}sgn\left(\begin{array}{cccc}{k}_{1}& {k}_{2}& \cdots & {k}_{n}\\ 1& 2& \cdots & n\end{array}\right){a}_{1{k}_{1}}{a}_{2{k}_{2}}\cdots {a}_{n{k}_{n}}$

$\left(\begin{array}{cccc}1& 2& \cdots & n\\ {i}_{1}& {i}_{2}& \cdots & {i}_{n}\end{array}\right)$$\left(\begin{array}{cccc}{k}_{1}& {k}_{2}& \cdots & {k}_{n}\\ 1& 2& \cdots & n\end{array}\right)$置換の相等$\left(\begin{array}{cccc}1& 2& \cdots & n\\ {i}_{1}& {i}_{2}& \cdots & {i}_{n}\end{array}\right)=\left(\begin{array}{cccc}{k}_{1}& {k}_{2}& \cdots & {k}_{n}\\ 1& 2& \cdots & n\end{array}\right)$ であり，対応する数字の対は変わらない．よって，

$sgn\left(\begin{array}{cccc}{k}_{1}& {k}_{2}& \cdots & {k}_{n}\\ 1& 2& \cdots & n\end{array}\right)=sgn\left(\begin{array}{cccc}1& 2& \cdots & n\\ {i}_{1}& {i}_{2}& \cdots & {i}_{n}\end{array}\right)$ より

$=\sum _{n}sgn\left(\begin{array}{cccc}1& 2& \cdots & n\\ {k}_{1}& {k}_{2}& \cdots & {k}_{n}\end{array}\right){a}_{1{k}_{1}}{a}_{2{k}_{2}}\cdots {a}_{n{k}_{n}}$

さらに，${k}_{1}\to {i}_{1},{k}_{2}\to {i}_{2},\cdots ,{k}_{n}\to {i}_{n}$  に置き換える．

$=\sum _{n}sgn\left(\begin{array}{cccc}1& 2& \cdots & n\\ {i}_{1}& {i}_{2}& \cdots & {i}_{n}\end{array}\right){a}_{1{i}_{1}}{a}_{2{i}_{2}}\cdots {a}_{n{i}_{n}}$

となり(2)は(1)と等しくなる．

したがって

$|{}^{t}A|=|A|$

となる．

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