2.5 矩阵分块法

矩阵的分块

对于行数和列数较高的矩阵 $\mathit{A}$，为简化运算，用若干条纵线和横线将矩阵 $\mathit{A}$ 划分成多个小矩阵，每个小矩阵称为矩阵 $\mathit{A}$ 的子块，以子块为元素的形式上的矩阵称为分块矩阵。

例如：

$$\begin{array}{c}\mathit{A}=\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\end{array}\right)=\left(\begin{array}{cc}{\mathit{A}}_{11}& {\mathit{A}}_{12}\\ {\mathit{A}}_{21}& {\mathit{A}}_{22}\end{array}\right),\\ {\mathit{A}}_{11}=\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right),{\mathit{A}}_{12}=\left(\begin{array}{cc}{a}_{13}& a_{14}\\ a_{23}& a_{24}\end{array}\right),\\ {\mathit{A}}_{21}=\left(\begin{array}{cc}{a}_{31}& a_{32}\end{array}\right),{\mathit{A}}_{22}=\left(\begin{array}{cc}{a}_{33}& a_{34}\end{array}\right),\end{array}$$

这里的 ${\mathit{A}}_{11},{\mathit{A}}_{12},{\mathit{A}}_{21},{\mathit{A}}_{22}$ 称为子块。

分块矩阵的运算

前提条件：内外均合法。

分块矩阵的数乘

$$\lambda \mathit{A}=\left(\begin{array}{cccc}\lambda a_{11}& \lambda a_{12}& \lambda a_{13}& \lambda a_{14}\\ \lambda a_{21}& \lambda a_{22}& \lambda a_{23}& \lambda a_{24}\\ \lambda a_{31}& \lambda a_{32}& \lambda a_{33}& \lambda a_{34}\end{array}\right)=\left(\begin{array}{cc}\lambda {\mathit{A}}_{11}& \lambda {\mathit{A}}_{12}\\ \lambda {\mathit{A}}_{21}& \lambda {\mathit{A}}_{22}\end{array}\right)$$

没什么好说的。

分块矩阵的乘法

$$\begin{array}{rl}\mathit{A}& ={\left(\begin{array}{cc}{\mathit{A}}_{11}& {\mathit{A}}_{12}\\ {\mathit{A}}_{21}& {\mathit{A}}_{22}\end{array}\right)}_{m\times n}\mathit{B}={\left(\begin{array}{cc}{\mathit{B}}_{11}& {\mathit{B}}_{12}\\ {\mathit{B}}_{21}& {\mathit{B}}_{22}\end{array}\right)}_{n\times t}\\ \mathit{A}\mathit{B}& =\left(\begin{array}{cc}{\mathit{A}}_{11}& {\mathit{A}}_{12}\\ {\mathit{A}}_{21}& {\mathit{A}}_{22}\end{array}\right)\left(\begin{array}{cc}{\mathit{B}}_{11}& {\mathit{B}}_{12}\\ {\mathit{B}}_{21}& {\mathit{B}}_{22}\end{array}\right)\\ & =\left(\begin{array}{cc}{\mathit{A}}_{11}{\mathit{B}}_{11}+{\mathit{A}}_{12}{\mathit{B}}_{21}& {\mathit{A}}_{11}{\mathit{B}}_{12}+{\mathit{A}}_{12}{\mathit{B}}_{22}\\ {\mathit{A}}_{21}{\mathit{B}}_{11}+{\mathit{A}}_{22}{\mathit{B}}_{21}& {\mathit{A}}_{21}{\mathit{B}}_{12}+{\mathit{A}}_{22}{\mathit{B}}_{22}\end{array}\right)\end{array}$$

要求其中每项矩阵乘积都合法。

TIP

题目基本上会分好块，或者给出的矩阵很好分块。

例 1

设 $\mathit{A}=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -1& 2& 1& 0\\ 1& 1& 0& 1\end{array}\right),\mathit{B}=\left(\begin{array}{cccc}1& 0& 1& 0\\ -1& 2& 0& 1\\ 1& 0& 4& 1\\ -1& -1& 2& 0\end{array}\right)$，求 $\mathit{A}\mathit{B}$。

解 对两个矩阵进行分块。不妨设

$$\begin{array}{c}\mathit{A}=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -1& 2& 1& 0\\ 1& 1& 0& 1\end{array}\right)=\left(\begin{array}{cc}\mathit{E}& \mathit{O}\\ {\mathit{A}}^{\prime }& \mathit{E}\end{array}\right)\\ \mathit{B}=\left(\begin{array}{cccc}1& 0& 1& 0\\ -1& 2& 0& 1\\ 1& 0& 4& 1\\ -1& -1& 2& 0\end{array}\right)=\left(\begin{array}{cc}{\mathit{B}}_{11}& \mathit{E}\\ {\mathit{B}}_{21}& {\mathit{B}}_{22}\end{array}\right)\end{array}$$

则有

$$\begin{array}{rl}\mathit{A}\mathit{B}& =\left(\begin{array}{cc}\mathit{E}& \mathit{O}\\ {\mathit{A}}^{\prime }& \mathit{E}\end{array}\right)\left(\begin{array}{cc}{\mathit{B}}_{11}& \mathit{E}\\ {\mathit{B}}_{21}& {\mathit{B}}_{22}\end{array}\right)\\ & =\left(\begin{array}{cc}\mathit{E}{\mathit{B}}_{11}+\mathit{O}{\mathit{B}}_{21}& {\mathit{E}}^2+\mathit{O}{\mathit{B}}_{2}2\\ {\mathit{A}}^{\prime }{\mathit{B}}_{11}+\mathit{E}{\mathit{B}}_{21}& {\mathit{A}}^{\prime }\mathit{E}+\mathit{E}{\mathit{B}}_{22}\end{array}\right)\\ & =\left(\begin{array}{cc}{\mathit{B}}_{11}& \mathit{E}\\ {\mathit{A}}^{\prime }{\mathit{B}}_{11}+{\mathit{B}}_{21}& {\mathit{A}}^{\prime }+{\mathit{B}}_{22}\end{array}\right)\\ {\mathit{A}}^{\prime }{\mathit{B}}_{11}+{\mathit{B}}_{21}& =\left(\begin{array}{cc}-1& 2\\ 1& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ -1& 2\end{array}\right)+\left(\begin{array}{cc}1& 0\\ -1& -1\end{array}\right)\\ & =\left(\begin{array}{cc}-1-2& 0+4\\ 1-1& 0+2\end{array}\right)+\left(\begin{array}{cc}1& 0\\ -1& -1\end{array}\right)\\ & =\left(\begin{array}{cc}-2& 4\\ -1& 1\end{array}\right)\\ {\mathit{A}}^{\prime }+{\mathit{B}}_{22}& =\left(\begin{array}{cc}-1& 2\\ 1& 1\end{array}\right)+\left(\begin{array}{cc}4& 1\\ 2& 0\end{array}\right)\\ & =\left(\begin{array}{cc}3& 3\\ 3& 1\end{array}\right)\\ \Rightarrow \mathit{A}\mathit{B}& =\left(\begin{array}{cccc}1& 0& 1& 0\\ -1& 2& 0& 1\\ -2& 4& 3& 3\\ -1& 1& 3& 1\end{array}\right)\end{array}$$

分块矩阵的逆矩阵

1. 主对角型：${\left(\begin{array}{c}\mathit{A}\\ & \mathit{B}\\ & & \mathit{C}\end{array}\right)}^{-1}=\left(\begin{array}{c}{\mathit{A}}^{-1}\\ & {\mathit{B}}^{-1}\\ & & {\mathit{C}}^{-1}\end{array}\right)$
2. 副对角型：${\left(\begin{array}{ccc}& & \mathit{A}\\ & \mathit{B}\\ \mathit{C}\end{array}\right)}^{-1}=\left(\begin{array}{ccc}& & {\mathit{C}}^{-1}\\ & {\mathit{B}}^{-1}\\ {\mathit{A}}^{-1}\end{array}\right)$

简记为主不变，副对调。

例 2

设 $\mathit{A}=\left(\begin{array}{ccc}5& 0& 0\\ 0& 3& 1\\ 0& 2& 1\end{array}\right)$，求 $|\mathit{A}|,{\mathit{A}}^{-1},{\mathit{A}}^{\ast }$。

解 对矩阵分块，有 $\mathit{A}=\left(\begin{array}{ccc}5& 0& 0\\ 0& 3& 1\\ 0& 2& 1\end{array}\right)$。设 $\mathit{B}=\left(\begin{array}{cc}3& 1\\ 2& 1\end{array}\right)$。

则 $|\mathit{A}|=5\times (3-2)=5$。（该知识点在 1.6 矩阵分块行列式）

$$\begin{array}{c}{\mathit{B}}^{-1}=\frac{1}{|\mathit{B}|}{\mathit{B}}^{\ast }=\frac{1}{3-2}\left(\begin{array}{cc}1& -1\\ -2& 3\end{array}\right)=\left(\begin{array}{cc}1& -1\\ -2& 3\end{array}\right)\\ \Rightarrow {\mathit{A}}^{-1}=\left(\begin{array}{ccc}\frac{1}{5}& 0& 0\\ 0& 1& -1\\ 0& -2& 3\end{array}\right)\\ \Rightarrow \mathit{A}\ast =|\mathit{A}|{\mathit{A}}^{-1}=5\left(\begin{array}{ccc}1& 0& 0\\ 0& 5& -5\\ 0& -10& 15\end{array}\right)\end{array}$$