99

$\bf命题：$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

\[{\rm{ }}\left| {\lambda {E_m} - AB} \right| = {\lambda ^{m - n}}\left| {\lambda {E_n} - BA} \right|\]

方法二：等价标准形

由$r\left( A \right) = r$知，存在可逆阵$P,Q$，使得
\[PAQ = \left( {\begin{array}{*{20}{c}}
{{E_r}}&0\\
0&0
\end{array}} \right)\]令\[{Q^{ - 1}}B{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}
{{B_1}}&{{B_3}}\\
{{B_4}}&{{B_2}}
\end{array}} \right)\]
其中${B_1}$为$r \times r$矩阵，于是有
\[PAB{P^{ - 1}} = PAQ \cdot {Q^{ - 1}}B{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}
{{B_1}}&{{B_3}}\\
0&0
\end{array}} \right)\]
且\[{Q^{ - 1}}BAQ = {Q^{ - 1}}B{P^{ - 1}} \cdot PAQ = \left( {\begin{array}{*{20}{c}}
{{B_1}}&0\\
{{B_4}}&0
\end{array}} \right)\]
从而可知
\[\left| {\lambda {E_m} - AB} \right| = \left| {\begin{array}{*{20}{c}}
{\lambda {E_r} - {B_1}}&{ - {B_3}}\\
0&{\lambda {E_{m - r}}}
\end{array}} \right| = {\lambda ^{m - r}}\left| {\lambda {E_r} - {B_1}} \right|\]
且\[\left| {\lambda {E_n} - BA} \right| = \left| {\begin{array}{*{20}{c}}
{\lambda {E_r} - {B_1}}&0\\
{ - {B_4}}&{\lambda {E_{n - r}}}
\end{array}} \right| = {\lambda ^{n - r}}\left| {\lambda {E_r} - {B_1}} \right|\]

故结论成立

$\bf注1：$$\left| {\lambda {E_m} - AB} \right| = \left| {{P^{ - 1}}} \right| \cdot \left| {\lambda {E_m} - AB} \right| \cdot \left| P \right| = \left| {{P^{ - 1}}\lambda {E_m}P - {P^{ - 1}}ABP} \right| = \left| {\lambda {E_m} - {P^{ - 1}}ABP} \right|$