矩阵代数

特别性质：

1.

若 \(A_{p\times q}~,~B_{q\times p}\) , 则

\[\left|\boldsymbol{I}_{p}+\boldsymbol{A B}\right|=\left|\boldsymbol{I}_{q}+\boldsymbol{B} \boldsymbol{A}\right| \]

证明:

\[\begin{array}{c} \because\left[\begin{array}{cc} \boldsymbol{I}_{p} & \boldsymbol{A} \\ \mathbf{0} & \boldsymbol{I}_{q} \end{array}\right]\left[\begin{array}{cc} \boldsymbol{I}_{p} & -\boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{I}_{q} \end{array}\right]=\left[\begin{array}{cc} \boldsymbol{I}_{p}+\boldsymbol{A B} & \mathbf{0} \\ \boldsymbol{B} & \boldsymbol{I}_{q} \end{array}\right] \\ {\left[\begin{array}{cc} \boldsymbol{I}_{p} & \mathbf{0} \\ -\boldsymbol{B} & \boldsymbol{I}_{q} \end{array}\right]\left[\begin{array}{cc} \boldsymbol{I}_{p} & -\boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{I}_{q} \end{array}\right]=\left[\begin{array}{lc} \boldsymbol{I}_{p} & -\boldsymbol{A} \\ \mathbf{0} & \boldsymbol{I}_{q}+\boldsymbol{A B} \end{array}\right]} \end{array}\]

所以上述两等式两边各取行列式，有

\[\left|\boldsymbol{I}_{p}+\boldsymbol{A B}\right|=\left|\boldsymbol{I}_{q}+\boldsymbol{B A}\right| \]

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2.

\(\left|A^{-1}\right|=\left|A\right|^{-1}\)

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3.

若 \(\boldsymbol{A}\) 和 \(\boldsymbol{B}\) 均为 \(p\) 阶非退化方阵，则

\[\left[\begin{array}{cc} \boldsymbol{A} & \mathbf{0} \\ \mathbf{0} & \boldsymbol{B} \end{array}\right]^{-1}=\left[\begin{array}{cc} \boldsymbol{A}^{-1} & \mathbf{0} \\ \mathbf{0} & \boldsymbol{B}^{-1} \end{array}\right]\]

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4.

\(\boldsymbol{A}\) 和 \(\boldsymbol{A}^{\prime}\) 有相同的特征值。

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5.

若 \(A_{p \times q}\) ， \(B_{q \times p}\) ，则 \(A B\) 和 \(B A\) 有相同的非零特征值。

证明:

\[\begin{array}{l} \because\left[\begin{array}{cc} \boldsymbol{I}_{p} & -\boldsymbol{A} \\ \mathbf{0} & \lambda \boldsymbol{I}_{q} \end{array}\right]\left[\begin{array}{cc} \lambda \boldsymbol{I}_{p} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{I}_{q} \end{array}\right]=\left[\begin{array}{cc} \lambda \boldsymbol{I}_{p}-\boldsymbol{A B} & \mathbf{0} \\ \lambda \boldsymbol{B} & \lambda \boldsymbol{I}_{q} \end{array}\right],\\ \left[\begin{array}{cc} \boldsymbol{I}_{p} & \mathbf{0} \\ -\boldsymbol{B} & \lambda \boldsymbol{I}_{q} \end{array}\right]\left[\begin{array}{cc} \lambda \boldsymbol{I}_{p} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{I}_{q} \end{array}\right]=\left[\begin{array}{cc} \lambda \boldsymbol{I}_{p} & \boldsymbol{A} \\ \mathbf{0} & \lambda \boldsymbol{I}_{q}-\boldsymbol{B} \boldsymbol{A} \end{array}\right] \\ \therefore\left|\begin{array}{cc} \lambda \boldsymbol{I}_{p}-\boldsymbol{A B} & \mathbf{0} \\ \lambda \boldsymbol{B} & \lambda \boldsymbol{I}_{q} \end{array}\right|=\left|\begin{array}{cc} \lambda \boldsymbol{I}_{p} & \boldsymbol{A} \\ \mathbf{0} & \lambda \boldsymbol{I}_{q}-\boldsymbol{B} \boldsymbol{A} \end{array}\right| \text { ，即 } \lambda^{q}\left|\lambda \boldsymbol{I}_{p}-\boldsymbol{A B}\right|=\lambda^{p}\left|\lambda \boldsymbol{I}_{q}-\boldsymbol{B} \boldsymbol{A}\right| \end{array}\]