$\bf命题：$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
\[\left(
{\begin{array}{*{20}{c}}
{\lambda
{E_m}}&A\\
B&{{E_n}}
\end{array}} \right) \to \left(
{\begin{array}{*{20}{c}}
{\lambda {E_m} -
AB}&0\\
B&{{E_n}}
\end{array}} \right) \to \left(
{\begin{array}{*{20}{c}}
{\lambda {E_m} -
AB}&0\\
0&{{E_n}}
\end{array}} \right)\]
\[\left(
{\begin{array}{*{20}{c}}
{\lambda
{E_m}}&A\\
B&{{E_n}}
\end{array}} \right) \to \left(
{\begin{array}{*{20}{c}}
{\lambda {E_m}}&A\\
0&{\frac{1}{\lambda
}\left( {\lambda {E_n} - BA} \right)}
\end{array}} \right) \to \left(
{\begin{array}{*{20}{c}}
{\lambda {E_m}}&0\\
0&{\frac{1}{\lambda
}\left( {\lambda {E_n} - BA} \right)}
\end{array}} \right)\]