分块矩阵求逆公式

令 $A$和 $B$是互逆矩阵, 将 $A$和 $B$写成分块矩阵的形式, 有
$\left[ \begin{matrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix} \right] \left[ \begin{matrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{matrix} \right] =\left[ \begin{matrix} I_{11} & \\ & I_{22} \end{matrix} \right] \tag{1}$
式中, $A$和 $B$相应子块具有适当的维数, 对角块均为可逆方阵, 将上式进行展开, 有
$\begin{aligned} A_{11}B_{11} + A_{12}B_{21} &= I_{11}\\ A_{11}B_{12} + A_{12}B_{22} &= 0 \\ A_{21}B_{11} + A_{22}B_{21} &= 0 \\ A_{21}B_{12} + A_{22}B_{22} &= I_{22} \tag{2} \end{aligned}$

由式(2.3)有
$B_{21} = -A_{22}^{-1} A_{21} B_{11}$
代入式(2.1)有
$\begin{aligned} &A_{11}B_{11} - A_{12}A_{22}^{-1}A_{21}B_{11} = I_{11} \\ \Rightarrow & \left( A_{11} - A_{12}A_{22}^{-1}A_{21}\right) B_{11} = I_{11} \\ \Rightarrow & \begin{cases} B_{11} &= \left( A_{11} - A_{12}A_{22}^{-1}A_{21} \right)^{-1} \\ B_{21} &= -A_{22}^{-1} A_{21} \left( A_{11} - A_{12}A_{22}^{-1}A_{21} \right)^{-1} \end{cases} \end{aligned}$

同理, 由(2.2)有
$B_{12} = -A_{11}^{-1}A_{12}B_{22}$
代入式(2.4)
$\begin{aligned} &A_{22}B_{22} - A_{21}A_{11}^{-1}A_{12}B_{22} = I_{22} \\ \Rightarrow & \left( A_{22} - A_{21}A_{11}^{-1}A_{12} \right) B_{22} = I_{22} \\ \Rightarrow & \begin{cases} B_{22} &= \left( A_{22} - A_{21}A_{11}^{-1}A_{12} \right)^{-1} \\ B_{12} &= -A_{11}^{-1}A_{12} \left( A_{22} - A_{21}A_{11}^{-1}A_{12} \right)^{-1} \end{cases} \end{aligned}$

即
$\begin{cases} B_{11} &= \left( A_{11} - A_{12}A_{22}^{-1}A_{21} \right)^{-1} \\ B_{21} &= -A_{22}^{-1} A_{21} \left( A_{11} - A_{12}A_{22}^{-1}A_{21} \right)^{-1} \\ B_{22} &= \left( A_{22} - A_{21}A_{11}^{-1}A_{12} \right)^{-1} \\ B_{12} &= -A_{11}^{-1}A_{12} \left( A_{22} - A_{21}A_{11}^{-1}A_{12} \right)^{-1} \end{cases}$