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$\bf命题：$设$A \in {M_{m \times n}}\left( F \right),B \in {M_{n \times m}}\left( F \right),m \ge n,\lambda \ne 0$，则

\[{\rm{ }}\left| {\lambda {E_m} - AB} \right| = {\lambda ^{m - n}}\left| {\lambda {E_n} - BA} \right|\]
方法一：初等变换法
\[\left( {\begin{array}{*{20}{c}}
{\lambda {E_m}}&A\\
B&{{E_n}}
\end{array}} \right) \to \left( {\begin{array}{*{20}{c}}
{\lambda {E_m} - AB}&0\\
B&{{E_n}}
\end{array}} \right) \to \left( {\begin{array}{*{20}{c}}
{\lambda {E_m} - AB}&0\\
0&{{E_n}}
\end{array}} \right)\]
\[\left( {\begin{array}{*{20}{c}}
{\lambda {E_m}}&A\\
B&{{E_n}}
\end{array}} \right) \to \left( {\begin{array}{*{20}{c}}
{\lambda {E_m}}&A\\
0&{\frac{1}{\lambda }\left( {\lambda {E_n} - BA} \right)}
\end{array}} \right) \to \left( {\begin{array}{*{20}{c}}
{\lambda {E_m}}&0\\
0&{\frac{1}{\lambda }\left( {\lambda {E_n} - BA} \right)}
\end{array}} \right)\]

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原文：http://www.cnblogs.com/ly758241/p/3706320.html