问题：设$A$为$m\times n$阶矩阵，$B$为$n\times m$阶，$m>n$,求证
\[ |\lambda I_{m}-AB|=\lambda^{m-n}|\lambda I_{n}-BA| \]
证明
当$\lambda=0$的时候，结论显然成立。下面证明$\lambda\neq 0$的情况。
注意到矩阵
\[ \left(
\begin{array}{cc}
\lambda E_{m} & A \\
B & E_{n} \\
\end{array}
\right)\]
有2种打洞方式。就是
\[ \left(
\begin{array}{cc}
E_{m} & -A \\
O & E_{n} \\
\end{array}
\right)\left(
\begin{array}{cc}
\lambda E_{m} & A \\
B & E_{n} \\
\end{array}
\right)
\left(\begin{array}{cc}
E_{m} & O \\
-B & E_{n} \\
\end{array}
\right)=
\left(
\begin{array}{cc}
\lambda E_{m}-AB & O \\
O & E_{n} \\
\end{array}
\right)\]
和
\[ \left(
\begin{array}{cc}
E_{m} & O \\
-\frac{1}{\lambda} & E_{n} \\
\end{array}
\right)
\left(
\begin{array}{cc}
\lambda E_{m} & A \\
B & E_{n} \\
\end{array}
\right)
\left(
\begin{array}{cc}
E_{m} & -\frac{1}{\lambda}A \\
O & E_{n} \\
\end{array}
\right)=
\left(
\begin{array}{cc}
\lambda E_{m} & O \\
O & E_{n}-\frac{1}{\lambda}BA\\
\end{array}
\right)\]
对上面两个式子取行列式，马上就可以得到
\[ |\lambda I_{m}-AB|=\lambda^{m-n}|\lambda I_{n}-BA| \]
$\square$  

我代数弱，大家不要笑啊。