矩阵代数（四）- 分块矩阵

小结

1. 分块矩阵
2. 分块矩阵运算
3. 分块矩阵的逆

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分块矩阵

矩阵$\mathbf{A}=\left[\begin{array}{cccccc}3& 0& -1& 5& 9& -2\\ -5& 2& 4& 0& -3& 1\\ -8& -6& 3& 1& 7& -4\end{array}\right]$，也可写成$2\times 3$分块矩阵$\mathbf{A}=\left[\begin{array}{ccc}{\mathbf{A}}_1{}_1& {\mathbf{A}}_1{}_2& {\mathbf{A}}_1{}_3\\ {\mathbf{A}}_2{}_1& {\mathbf{A}}_2{}_2& {\mathbf{A}}_2{}_3\end{array}\right]$的形状，它的元素是分块（子矩阵）
$\begin{array}{rlrlrl}& {\mathbf{A}}_1{}_1=\left[\begin{array}{ccc}3& 0& -1\\ -5& 2& 4\end{array}\right]& & {\mathbf{A}}_1{}_2=\left[\begin{array}{cc}5& 9\\ 0& -3\end{array}\right]& & {\mathbf{A}}_1{}_3=\left[\begin{array}{c}-2\\ 1\end{array}\right]\\ & {\mathbf{A}}_1{}_2=\left[\begin{array}{cc}5& 9\\ 0& -3\end{array}\right]& & {\mathbf{A}}_2{}_2=\left[\begin{array}{c}1\\ 7\end{array}\right]& & {\mathbf{A}}_2{}_3=\left[\begin{array}{c}-4\end{array}\right]\end{array}$

加法与标量乘法

若矩阵$\mathbf{A}$与$\mathbf{B}$有相同维数且以相同方式分块，则自然有矩阵的和$\mathbf{A}+\mathbf{B}$也以同样方式分块。这时$\mathbf{A}+\mathbf{B}$的每一个分块恰好是$\mathbf{A}$和$\mathbf{B}$对应分块的（矩阵）和。分块矩阵乘以一个标量也可以逐块计算。

分块矩阵的乘法

设$\mathbf{A}=\left[\begin{array}{ccccc}2& -3& 1& 0& -4\\ 1& 5& -2& 3& 1\\ 0& -4& -2& 7& -1\end{array}\right]=\left[\begin{array}{cc}{\mathbf{A}}_1{}_1& {\mathbf{A}}_1{}_2\\ {\mathbf{A}}_2{}_1& {\mathbf{A}}_2{}_2\end{array}\right]$，$\mathbf{B}=\left[\begin{array}{cc}6& 4\\ -2& 1\\ -3& 7\\ -1& 3\\ 5& 2\end{array}\right]=\left[\begin{array}{cc}{\mathbf{B}}_1& {\mathbf{B}}_2\end{array}\right]$
$\mathbf{A}$的5列被分成3列一组和2列一组。$\mathbf{B}$的5行按同样方法分块—被分成3行一组和2行一组。我们称$\mathbf{A}$和$\mathbf{B}$的分块是与分块乘法相一致的。$\mathbf{A}\mathbf{B}$的乘积可以被写成$\mathbf{A}\mathbf{B}=\left[\begin{array}{cc}{\mathbf{A}}_1{}_1& {\mathbf{A}}_1{}_2\\ {\mathbf{A}}_2{}_1& {\mathbf{A}}_2{}_2\end{array}\right]\left[\begin{array}{cc}{\mathbf{B}}_1& {\mathbf{B}}_2\end{array}\right]=\left[\begin{array}{c}{\mathbf{A}}_1{}_1{\mathbf{B}}_1+{\mathbf{A}}_1{}_2{\mathbf{B}}_2\\ {\mathbf{A}}_2{}_1{\mathbf{B}}_1+{\mathbf{A}}_2{}_2{\mathbf{B}}_2\end{array}\right]$
${\mathbf{A}}_1{}_1{\mathbf{B}}_1=\left[\begin{array}{ccc}2& -3& 1\\ 1& 5& -2\end{array}\right]\left[\begin{array}{cc}6& 4\\ -2& 1\\ -3& 7\end{array}\right]=\left[\begin{array}{cc}15& 12\\ 2& -5\end{array}\right]$
${\mathbf{A}}_1{}_2{\mathbf{B}}_2=\left[\begin{array}{cc}0& -4\\ 3& -1\end{array}\right]\left[\begin{array}{cc}-1& 3\\ 5& 2\end{array}\right]=\left[\begin{array}{cc}-20& -8\\ -8& 7\end{array}\right]$
${\mathbf{A}}_1{}_1{\mathbf{B}}_1+{\mathbf{A}}_1{}_2{\mathbf{B}}_2=\left[\begin{array}{cc}15& 12\\ 2& -5\end{array}\right]+\left[\begin{array}{cc}-20& -8\\ -8& 7\end{array}\right]=\left[\begin{array}{cc}-5& 4\\ -6& 2\end{array}\right]$

设$\mathbf{A}=\left[\begin{array}{ccc}-3& 1& 2\\ 1& -4& 5\end{array}\right]$和$\mathbf{B}=\left[\begin{array}{cc}a& b\\ c& d\\ e& f\end{array}\right]$。验证$\mathbf{A}\mathbf{B}=co{l}_1(\mathbf{A})ro{w}_1(\mathbf{B})+co{l}_2(\mathbf{A})ro{w}_2(\mathbf{B})+co{l}_3(\mathbf{A})ro{w}_3(\mathbf{B})$。
$\begin{array}{rl}& co{l}_1(\mathbf{A})ro{w}_1(\mathbf{B})=\left[\begin{array}{c}-3\\ 1\end{array}\right]\left[\begin{array}{cc}a& b\end{array}\right]=\left[\begin{array}{cc}-3a& -3b\\ a& b\end{array}\right]\\ & co{l}_2(\mathbf{A})ro{w}_2(\mathbf{B})=\left[\begin{array}{c}1\\ -4\end{array}\right]\left[\begin{array}{cc}c& d\end{array}\right]=\left[\begin{array}{cc}c& d\\ -4c& -4d\end{array}\right]\\ & co{l}_3(\mathbf{A})ro{w}_3(\mathbf{B})=\left[\begin{array}{c}2\\ 5\end{array}\right]\left[\begin{array}{cc}e& f\end{array}\right]=\left[\begin{array}{cc}2e& 2f\\ 5e& 5f\end{array}\right]\end{array}$
于是$ \sum_{k=1}^3{col_k(\boldsymbol{A})row_k(\boldsymbol{B})}=\begin{bmatrix}-3a + c + 2e & -3b + d + 2f \ a - 4c + 5e & b - 4d + 5f\end{bmatrix}$
这个矩阵恰好就是$\mathbf{A}\mathbf{B}$。

定理10（$\mathbf{A}\mathbf{B}$的列行展开）
若$\mathbf{A}$是$m\times n$矩阵，$\mathbf{B}$是$n\times p$矩阵，则$\begin{array}{rl}\mathbf{A}\mathbf{B}& =\left[\begin{array}{ccc}co{l}_1(\mathbf{A})& \cdots & co{l}_n(\mathbf{A})\end{array}\right]\left[\begin{array}{c}ro{w}_1(\mathbf{B})\\ ⋮\\ ro{w}_n(\mathbf{B})\end{array}\right]\\ & =co{l}_1(\mathbf{A})ro{w}_1(\mathbf{B})+\cdots +co{l}_n(\mathbf{A})ro{w}_n(\mathbf{B})\end{array}$

分块矩阵的逆

形如$\mathbf{A}=\left[\begin{array}{cc}{\mathbf{A}}_1{}_1& {\mathbf{A}}_1{}_2\\ 0& {\mathbf{A}}_2{}_2\end{array}\right]$的矩阵称为分块上三角矩阵。设${\mathbf{A}}_1{}_1$是$p\times p$矩阵，${\mathbf{A}}_2{}_2$是$q\times q$矩阵。求${\mathbf{A}}^{-1}$的表达式。
解：用$\mathbf{B}$表示${\mathbf{A}}^{-1}$且把它分块，使得$\left[\begin{array}{cc}{\mathbf{A}}_1{}_1& {\mathbf{A}}_1{}_2\\ 0& {\mathbf{A}}_2{}_2\end{array}\right]\left[\begin{array}{cc}{\mathbf{B}}_1{}_1& {\mathbf{B}}_1{}_2\\ {\mathbf{B}}_2{}_1& {\mathbf{B}}_2{}_2\end{array}\right]=\left[\begin{array}{cc}{\mathbf{I}}_{\mathbf{p}}& 0\\ 0& {\mathbf{I}}_{\mathbf{q}}\end{array}\right]$。
则有：$\begin{array}{rl}{\mathbf{A}}_1{}_1{\mathbf{B}}_1{}_1+{\mathbf{A}}_1{}_2{\mathbf{B}}_2{}_1& ={\mathbf{I}}_{\mathbf{p}}\\ {\mathbf{A}}_1{}_1{\mathbf{B}}_1{}_2+{\mathbf{A}}_1{}_2{\mathbf{B}}_2{}_2& =0\\ {\mathbf{A}}_2{}_2{\mathbf{B}}_2{}_1& =0\\ {\mathbf{A}}_2{}_2{\mathbf{B}}_2{}_2& ={\mathbf{I}}_{\mathbf{q}}\end{array}$
因$\mathbf{A}$可逆，且${\mathbf{A}}_2{}_2$所在行其余列全为$0$，可知${\mathbf{A}}_2{}_2$必有q个主元位置，故${\mathbf{B}}_2{}_2={{\mathbf{A}}_2{}_2}^{-1}$。
${\mathbf{B}}_2{}_1={{\mathbf{A}}_2{}_2}^{-1}0=0$。
于是${\mathbf{A}}_1{}_1{\mathbf{B}}_1{}_1+{\mathbf{A}}_1{}_2{\mathbf{B}}_2{}_1={\mathbf{A}}_1{}_1{\mathbf{B}}_1{}_1={\mathbf{I}}_{\mathbf{p}}$。同理${\mathbf{B}}_1{}_1={{\mathbf{A}}_1{}_1}^{-1}$.
最后${\mathbf{A}}_1{}_1{\mathbf{B}}_1{}_2=-{\mathbf{A}}_1{}_2{\mathbf{B}}_2{}_2=-{\mathbf{A}}_1{}_2{{\mathbf{A}}_2{}_2}^{-1}$。
${\mathbf{B}}_1{}_2=-{{\mathbf{A}}_1{}_1}^{-1}{\mathbf{A}}_1{}_2{{\mathbf{A}}_2{}_2}^{-1}$。
最终得：${\mathbf{A}}^{-1}=\mathbf{B}=\left[\begin{array}{cc}{{\mathbf{A}}_1{}_1}^{-1}& -{{\mathbf{A}}_1{}_1}^{-1}{\mathbf{A}}_1{}_2{{\mathbf{A}}_2{}_2}^{-1}\\ 0& {{\mathbf{A}}_2{}_2}^{-1}\end{array}\right]$

分块对角矩阵是一个分块矩阵，除了主对角上各分块外，其余全是零分块。这样的一个矩阵是可逆的当且仅当主对角线上各分块都是可逆的。