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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

$\bf注1：$命题中的结论称为$\bf行列式的降阶公式$

$\bf注2：$$\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|$

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