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本帖最后由 hbghlyj 于 2022-6-12 16:55 编辑
math.chalmers.se/~rootzen/highdimensional/blockmatrixinverse.pdf
Consider a pair $A, B$ of $n \times n$ matrices, partitioned as
$$
A=\left(\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right), B=\left(\begin{array}{ll}
B_{11} & B_{12} \\
B_{21} & B_{22}
\end{array}\right),
$$
where $A_{11}$ and $B_{11}$ are $k \times k$ matrices. Suppose that $A$ is nonsingular and $B=A^{-1}$. In this note it will be shown how to derive the $B_{i j}$ 's in terms of the $A_{i j}$ 's, given that
\begin{equation}\label1
\det\left(A_{11}\right) \neq 0 \text { and } \det\left(A_{22}\right) \neq 0 .
\end{equation}
The latter conditions are sufficient for the nonsingularity of $A$. However, in general they are not necessary conditions. For example, consider the case
$$
A=\left(\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right)=\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right) .
$$
On the other hand, if $A$ is positive definite then the conditions \eqref{1} are necessary as well.
If $B=A^{-1}$ then
\begin{equation}\label2
\begin{aligned} A B &=\left(\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right)\left(\begin{array}{cc}B_{11} & B_{12} \\ B_{21} & B_{22}\end{array}\right) \\ &=\left(\begin{array}{ll}A_{11} B_{11}+A_{12} B_{21} & A_{11} B_{12}+A_{12} B_{22} \\ A_{21} B_{11}+A_{22} B_{21} & A_{21} B_{12}+A_{22} B_{22}\end{array}\right) \\ &=\left(\begin{array}{ll}I_{k} & O_{k, n-k} \\ O_{n-k, k} & I_{n-k}\end{array}\right) \end{aligned}\end{equation}
where as usual $I$ denotes the unit matrix and $O$ a zero matrix, with sizes indicated by the subscripts involved.
To solve \eqref{2}, we need to solve four matrix equations:
\begin{align}A_{11} B_{11}+A_{12} B_{21}&=I_{k}\label3
\\ A_{11} B_{12}+A_{12} B_{22}&=O_{k, n-k}\label4
\\ A_{21} B_{11}+A_{22} B_{21}&=O_{n-k, k}\label5
\\ A_{21} B_{12}+A_{22} B_{22}&=I_{n-k}\label6
\end{align}
It follows from \eqref{4} and \eqref{5} that
\begin{align}B_{12}&=-A_{11}^{-1} A_{12} B_{22}\label7 \\ B_{21}&=-A_{22}^{-1} A_{21} B_{11}\label8\end{align}
so that \eqref{3} and \eqref{6} become
\begin{array}{l}\left(A_{11}-A_{12} A_{22}^{-1} A_{21}\right) B_{11}=I_{k} \\ \left(A_{22}-A_{21} A_{11}^{-1} A_{12}\right) B_{22}=I_{n-k}\end{array}
Hence
\begin{aligned} B_{11} &=\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} \end{aligned}
Substituting these solutions in \eqref{7} and \eqref{8} it follows that
\begin{array}{l}B_{12}=-A_{11}^{-1} A_{12}\left(A_{22}-A_{21} A_{11}^{-1} A_{12}\right)^{-1} \\ B_{21}=-A_{22}^{-1} A_{21}\left(A_{11}-A_{12} A_{22}^{-1} A_{21}\right)^{-1}\end{array}
Thus,
$$A^{-1}=\left(\begin{array}{ll}\left(A_{11}-A_{12} A_{22}^{-1} A_{21}\right)^{-1} & -A_{11}^{-1} A_{12}\left(A_{22}-A_{21} A_{11}^{-1} A_{12}\right)^{-1} \\ -A_{22}^{-1} A_{21}\left(A_{11}-A_{12} A_{22}^{-1} A_{21}\right)^{-1} & \left(A_{22}-A_{21} A_{11}^{-1} A_{12}\right)^{-1}\end{array}\right)$$
Moreover, since $AA^{−1} = I_n$ implies $A^{−1}A = I_n$, we also have
$$A^{-1}=\left(\begin{array}{ll}\left(A_{11}-A_{12} A_{22}^{-1} A_{21}\right)^{-1} & -\left(A_{11}-A_{12} A_{22}^{-1} A_{21}\right)^{-1} A_{12} A_{22}^{-1} \\ -\left(A_{22}-A_{21} A_{11}^{-1} A_{12}\right)^{-1} A_{21} A_{11}^{-1} & \left(A_{22}-A_{21} A_{11}^{-1} A_{12}\right)^{-1}\end{array}\right)$$ |
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