ABSTRACT

In this paper, the generalized inverse of a class of block matrix is discussed. By using the minus inverse of block matrix, the solution formula of a special matrix equation is given. Based on the properties of the minus inverse and the block matrix operation, the calculation process of the generalized inverse of a class of 2 × 2 block matrices is presented. And the method of solving the generalized inverse of a kind of block matrix by elementary transformation method is discussed. Then the method is applied to solving the equations. Finally, a numerical example is discussed, and the generalized inverse of 2 × 2 block matrix is extended to a special case of 4 × 4 block matrix.

Keywords:Partitioned Matrix, Generalized Inverse Matrix, Computing Method, Equations, Elementary Transformation

一类分块矩阵方程的广义逆求解方法

赖金凤，刘唐伟，唐阿敏，陈硕

东华理工大学，江西 南昌

收稿日期：2020年7月15日；录用日期：2020年7月28日；发布日期：2020年8月4日

摘 要

本文讨论了一类分块矩阵的广义逆，利用分块矩阵的减号逆，给出了一类特殊矩阵方程的求解公式。基于减号逆的性质，结合分块矩阵的运算，给出了一类2 × 2分块矩阵的广义逆的计算过程，探讨了利用初等变换法求解一类分块矩阵的广义逆的方法，并应用于方程组的求解。最后给出了数值计算例子，并将2 × 2分块矩阵的广义逆的计算方法推广应用到了一类特殊的4 × 4分块矩阵的情形。

关键词 :分块矩阵，广义逆矩阵，计算方法，方程组，初等变换

Copyright © 2020 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

在微分方程数值求解及优化问题求解中，经常会出现分块矩阵方程，广义逆矩阵在分块矩阵方程的求解中具有重要作用，分块矩阵的广义逆的求解具有重要意义 [1]。在已有研究中，很多学者对分块矩阵的一些广义逆给出了不同表达式 [2] [3]。但是，某些特殊块矩阵的广义逆仍然很难计算 [4]。本文探讨了2 × 2分块矩阵的广义逆的计算，推导了有效的计算公式，并应用于分块矩阵方程的求解，该方法能较大地降低矩阵方程求解的计算量。

2. 一类2 × 2分块矩阵的广义逆

2.1. 逆矩阵的基本概念

块矩阵 [5] 的逆矩阵有多种，例如减号逆，加号逆 [6]，最小二乘广义逆 [7]，自反广义逆和最小范数广义逆 [8]。在本文中，我们考虑四分块矩阵的减号逆 [9]。

对于矩阵 $A\in F^{m\times n}$，我们考虑以下公式：

$AXA=A,\forall A\in F^{m\times n}$ (1)

对于一个 $m\times n$ 的矩阵A，如果存在一个矩阵X满足(1)，那么矩阵X称为A矩阵的减号逆 $A^{-1}$，对任意的 $A\in F^{m\times n}$，$A^{-1}$ 存在。

2.2. 2 × 2分块矩阵减号逆

在求解如下离散方程组(2)时 [10]，会涉及到分块矩阵的求逆问题 [11]

(2)

我们考虑以下 $\text{2}\times \text{2}$ 块矩阵的广义逆，设系数矩阵T为

$T=\left[\begin{array}{cc}A& C^{\text{T}}\\ C& B\end{array}\right]$ (3)

设矩阵T减号逆为

$\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]$

方程(2)中U，P未知，系数矩阵T和右端项F已知，有

$\{\begin{array}{l}AU-C^{\text{T}}P=0,\\ CU-BP=F.\end{array}$ (4)

根据减号逆的性质有

$\left[\begin{array}{cc}A& C^{\text{T}}\\ C& B\end{array}\right]\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]\left[\begin{array}{cc}A& C^{\text{T}}\\ C& B\end{array}\right]=\left[\begin{array}{cc}A& C^{\text{T}}\\ C& B\end{array}\right]$,

从而

$\{\begin{array}{l}A{X}_{1}A+A{X}_{2}C+C^{\text{T}}{X}_{3}A+C^{\text{T}}{X}_{4}C=A,\\ A{X}_1{C}^{\text{T}}+A{X}_{2}B+C^{\text{T}}{X}_3{C}^{\text{T}}+C^{\text{T}}{X}_{4}B=C^{\text{T}},\\ C{X}_{1}A+C{X}_{2}C+B{X}_{3}A+B{X}_{4}C=C,\\ C{X}_1{C}^{\text{T}}+C{X}_{2}B+B{X}_3{C}^{\text{T}}+B{X}_{4}B=B.\end{array}$ (5)

在(5)中的第1、3个等式左右同乘以U，第2、4个等式左右同乘以P有

$\{\begin{array}{l}A{X}_{1}AU+A{X}_{2}CU+C^{\text{T}}{X}_{3}AU+C^{\text{T}}{X}_{4}CU=AU,\\ A{X}_1{C}^{\text{T}}P+A{X}_{2}BP+C^{\text{T}}{X}_3{C}^{\text{T}}P+C^{\text{T}}{X}_{4}BP=C^{\text{T}}P,\\ C{X}_{1}AU+C{X}_{2}CU+B{X}_{3}AU+B{X}_{4}CU=CU,\\ C{X}_1{C}^{\text{T}}P+C{X}_{2}BP+B{X}_3{C}^{\text{T}}P+B{X}_{4}BP=BP.\end{array}$ (6)

因为 $AU-C^{\text{T}}P=0$，$CU-BP=F$，可得

$\{\begin{array}{l}A{X}_1{C}^{\text{T}}P+A{X}_{2}CU+C^{\text{T}}{X}_3{C}^{\text{T}}P+C^{\text{T}}{X}_{4}CU=AU,\\ A{X}_1{C}^{\text{T}}P+A{X}_{2}BP+C^{\text{T}}{X}_3{C}^{T}P+C^{\text{T}}{X}_{4}BP=C^{\text{T}}P,\\ C{X}_1{C}^{\text{T}}P+C{X}_{2}CU+B{X}_3{C}^{\text{T}}P+B{X}_{4}CU=CU,\\ C{X}_1{C}^{\text{T}}P+C{X}_{2}BP+B{X}_3{C}^{\text{T}}P+B{X}_{4}BP=BP.\end{array}$ (7)

在(7)中第一个等式减去第二个等式，第三个等式减去第四个等式得

$\{\begin{array}{l}A{X}_{2}F+C^{\text{T}}{X}_{4}F=0,\\ C{X}_{2}F+B{X}_{4}F=0.\end{array}$ (8)

化成矩阵形式得

$\left(\begin{array}{cccc}A& C^{\text{T}}& 0& 0\\ 0& 0& C& B\end{array}\right)\left[\begin{array}{cccc}{X}_2& 0& 0& 0\\ 0& X_4& 0& 0\\ 0& 0& X_2& 0\\ 0& 0& 0& X_4\end{array}\right]\left[\begin{array}{cc}F& 0\\ F& 0\\ 0& F\\ 0& F\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& F\end{array}\right]$ (9)

下面讨论 ${\left(\begin{array}{cccc}A& C^{\text{T}}& 0& 0\\ 0& 0& C& B\end{array}\right)}^{-}$ 的求解：

设 $A\in C^{p\times m}$，$C\in C^{q\times m}$，形如 $\left(A⋮C^{\text{T}}\right)$ 的矩阵叫做列分块矩阵，以下讨论列分块矩阵的减号逆。

对矩阵 $\left(A⋮C^{\text{T}}\right)$ 作初等变换 [7] 有

$\left(\begin{array}{c}{A}^{-}\\ E_m\end{array}\right)\left(A⋮C^{\text{T}}\right)\left(\begin{array}{cc}{E}_p& -A^{-}{C}^{\text{T}}\\ 0& E_q\end{array}\right)=\left(\begin{array}{cc}{A}^{-}A& A^{-}\left(E_n-A^{-}A\right)C^{\text{T}}\\ A& \left(E_n-A^{-}A\right)C^{\text{T}}\end{array}\right)$, (10)

令

$D_1=\left(E_n-A^{-}A\right)C^{\text{T}}$，$H=\left(\begin{array}{cc}{A}^{-}A& A^{-}\left(E_n-A^{-}A\right)C^{\text{T}}\\ A& \left(E_n-A^{-}A\right)C^{\text{T}}\end{array}\right)$，

有 $A{A}^{-}{D}_1=D_1=A{A}^{-}\left(E_n-A{A}^{-}\right)C^{\text{T}}=\left(A{A}^{-}-A{A}^{-}A{A}^{-}\right)C^{\text{T}}=\left(A{A}^{-}-A{A}^{-}\right)C^{\text{T}}=0$，

令 $X=\left(\begin{array}{cc}E& 0\\ -D_1{}^{-}A& D_1{}^{-}\end{array}\right)$，有

$\begin{array}{c}HXH=\left(\begin{array}{cc}{A}^{-}A& A^{-}\left(E_n-A{A}^{-}\right)C^{\text{T}}\\ A& \left(E_n-A{A}^{-}\right)C^{\text{T}}\end{array}\right)\left(\begin{array}{cc}E& 0\\ -D_1^{-}A& D_1^{-}\end{array}\right)\left(\begin{array}{cc}{A}^{-}A& A^{-}\left(E_n-A{A}^{-}\right)C^{\text{T}}\\ A& \left(E_n-A{A}^{-}\right)C^{\text{T}}\end{array}\right)\\ =\left(\begin{array}{cc}{A}^{-}A& A^{-}{D}_1\\ A& D_1\end{array}\right)\left(\begin{array}{cc}E& 0\\ -D_1^{-}A& D_1^{-}\end{array}\right)\left(\begin{array}{cc}{A}^{-}A& A^{-}{D}_1\\ A& D_1\end{array}\right)\\ =\left(\begin{array}{cc}\left(A^{-}A-A^{-}{D}_1{D}_1^{-}A\right)A^{-}A+A^{-}{D}_1{D}_1^{-}A& \left(A^{-}A-A^{-}{D}_1{D}_1^{-}A\right)A^{-}{D}_1+A^{-}{D}_1{D}_1^{-}{D}_1\\ \left(A-D_1{D}_1^{-}A\right)A^{-}A+D_1{D}_1^{-}A& \left(A-D_1{D}_1^{-}A\right)A^{-}{D}_1+D_1{D}_1^{-}{D}_1\end{array}\right)\\ =\left(\begin{array}{cc}{A}^{-}A& A^{-}{D}_1\\ A& D_1\end{array}\right)=H\end{array}$

又 $\left(\begin{array}{c}{A}^{-}\\ E_m\end{array}\right)$ 为列满秩阵， $\left(\begin{array}{cc}{E}_p& -A^{-}{C}^{\text{T}}\\ 0& E_q\end{array}\right)$ 为可逆阵。

又有

$\left(A⋮C^{\text{T}}\right)\left(\begin{array}{cc}{E}_p& -A^{-}{C}^{\text{T}}\\ 0& E_q\end{array}\right)X\left(\begin{array}{c}{A}^{-}\\ E_m\end{array}\right)\left(A⋮C^{\text{T}}\right)=\left(A⋮C^{\text{T}}\right)$, (11)

所以有

${\left(A⋮C^{\text{T}}\right)}^{-}=\left(\begin{array}{cc}{E}_p& -A^{-}{C}^{\text{T}}\\ 0& E_q\end{array}\right)X\left(\begin{array}{c}{A}^{-}\\ E_m\end{array}\right)=\left(\begin{array}{c}{A}^{-}-A^{-}{C}^{\text{T}}{D}_1^{-}\left(E_m-A{A}^{-}\right)\\ D_1^{-}\left(E_m-A{A}^{-}\right)\end{array}\right)$.(12)

同理有

${\left(C⋮B\right)}^{-}=\left(\begin{array}{c}{C}^{-}-C^{-}B{D}_2^{-}\left(E_m-C{C}^{-}\right)\\ D_2^{-}\left(E_m-C{C}^{-}\right)\end{array}\right)$, (13)

其中 $D_1=\left(E_n-A{A}^{-}\right)C^{\text{T}}$，$D_2=\left(E_n-C{C}^{-}\right)B$。

然后有

${\left[\begin{array}{cccc}A& C^{\text{T}}& 0& 0\\ 0& 0& C& B\end{array}\right]}^{-}=\left[\begin{array}{cc}{\left(\begin{array}{cc}A& C^{\text{T}}\end{array}\right)}^{-}& 0\\ 0& {\left(\begin{array}{cc}C& B\end{array}\right)}^{-}\end{array}\right]$

即

${\left[\begin{array}{cccc}A& C^{\text{T}}& 0& 0\\ 0& 0& C& B\end{array}\right]}^{-}=\left[\begin{array}{cc}\left(\begin{array}{c}{A}^{-}-A^{-}{C}^{\text{T}}{D}_1^{-}\left(E_m-A{A}^{-}\right)\\ D_1^{-}\left(E_m-A{A}^{-}\right)\end{array}\right)& 0\\ 0& \left(\begin{array}{c}{C}^{-}-C^{-}B{D}_2^{-}\left(E_m-C{C}^{-}\right)\\ D_2^{-}\left(E_m-C{C}^{-}\right)\end{array}\right)\end{array}\right]$

同理得

$\begin{array}{l}{\left[\begin{array}{cc}F& 0\\ F& 0\\ 0& F\\ 0& F\end{array}\right]}^{-}=\left[\begin{array}{cc}{\left[\begin{array}{c}F\\ F\end{array}\right]}^{-}& 0\\ 0& {\left[\begin{array}{c}F\\ F\end{array}\right]}^{-}\end{array}\right]\\ =\left[\begin{array}{cc}\left[\begin{array}{cc}{F}^{-}& 0\end{array}\right]& 0\\ 0& \left[\begin{array}{cc}{F}^{-}& 0\end{array}\right]\end{array}\right]\end{array}$. (14)

因此有

$\begin{array}{l}\left[\begin{array}{cccc}{X}_2& 0& 0& 0\\ 0& X_4& 0& 0\\ 0& 0& X_2& 0\\ 0& 0& 0& X_4\end{array}\right]={\left(\begin{array}{cccc}A& C^{\text{T}}& 0& 0\\ 0& 0& C& B\end{array}\right)}^{-}\left[\begin{array}{cc}0& 0\\ 0& F\end{array}\right]{\left[\begin{array}{cc}F& 0\\ F& 0\\ 0& F\\ 0& F\end{array}\right]}^{-}\\ =\left[\begin{array}{cc}\left(\begin{array}{c}{A}^{-}-A^{-}{C}^{\text{T}}{D}_1^{-}\left(E_m-A{A}^{-}\right)\\ D_1^{-}\left(E_m-A{A}^{-}\right)\end{array}\right)& 0\\ 0& \left(\begin{array}{c}{C}^{-}-C^{-}B{D}_2^{-}\left(E_m-C{C}^{-}\right)\\ D_2^{-}\left(E_m-C{C}^{-}\right)\end{array}\right)\end{array}\right]\left[\begin{array}{cc}0& 0\\ 0& F\end{array}\right]\left[\begin{array}{cc}\left[\begin{array}{cc}{F}^{-}& 0\end{array}\right]& 0\\ 0& \left[\begin{array}{cc}{F}^{-}& 0\end{array}\right]\end{array}\right]\end{array}$

设 $X_2=\left(C^{-}-C^{-}B{D}_2^{-}\left(E_m-C{C}^{-}\right)\right)F{F}^{-}$，$X_4=\left(D_2^{-}\left(E_m-C{C}^{-}\right)\right)F{F}^{-}$。

然后有

$\begin{array}{l}\left[\begin{array}{cccccccc}A& C^{\text{T}}& 0& 0& 0& 0& 0& 0\\ 0& 0& A& C^{\text{T}}& 0& 0& 0& 0\\ 0& 0& 0& 0& C& B& 0& 0\\ 0& 0& 0& 0& 0& 0& C& B\end{array}\right]\left[\begin{array}{cccccccc}{X}_1& 0& 0& 0& 0& 0& 0& 0\\ 0& X_3& 0& 0& 0& 0& 0& 0\\ 0& 0& X_1& 0& 0& 0& 0& 0\\ 0& 0& 0& X_3& 0& 0& 0& 0\\ 0& 0& 0& 0& X_1& 0& 0& 0\\ 0& 0& 0& 0& 0& X_3& 0& 0\\ 0& 0& 0& 0& 0& 0& X_1& 0\\ 0& 0& 0& 0& 0& 0& 0& X_3\end{array}\right]\left[\begin{array}{cccc}A& 0& 0& 0\\ A& 0& 0& 0\\ 0& C^{\text{T}}& 0& 0\\ 0& C^{\text{T}}& 0& 0\\ 0& 0& A& 0\\ 0& 0& A& 0\\ 0& 0& 0& C^{\text{T}}\\ 0& 0& 0& C^{\text{T}}\end{array}\right]\\ =\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& \left(C^{-}-C^{-}B{D}_2^{-}\left(E_m-C{C}^{-}\right)\right)F{F}^{-}& 0\\ 0& 0& 0& D_2^{-}\left(E_m-C{C}^{-}\right)F{F}^{-}\end{array}\right]\\ =\left[\begin{array}{cccc}A-A{X}_{2}C-C^{\text{T}}{X}_{4}C& 0& 0& 0\\ 0& C^{\text{T}}-A{X}_{2}B-C^{\text{T}}{X}_{4}B& 0& 0\\ 0& 0& C-C{X}_{2}C-B{X}_{4}C& 0\\ 0& 0& 0& B-C{X}_{2}B-B{X}_{4}B\end{array}\right]\end{array}$.(15)

用同样的方法有

$\begin{array}{c}\left[\begin{array}{cc}{X}_1& 0\\ 0& X_3\end{array}\right]=\left[\begin{array}{cc}A& C^{\text{T}}\end{array}\right]\left(A-A{X}_{2}C-C^{\text{T}}{X}_{4}C\right){\left[\begin{array}{c}A\\ A\end{array}\right]}^{-}\\ =\left(\begin{array}{c}{A}^{-}-A^{-}{C}^{\text{T}}{D}_1^{-}\left(E_m-A{A}^{-}\right)\\ D_1^{-}\left(E_m-A{A}^{-}\right)\end{array}\right)\left(A-A{X}_{2}C-C^{\text{T}}{X}_{4}C\right)\left[\begin{array}{cc}{A}^{-}& 0\end{array}\right]\\ =\left[\begin{array}{cc}\left(A^{-}-A^{-}{C}^{\text{T}}{D}_1^{-}\left(E_m-A{A}^{-}\right)\right)\left(A-A{X}_{2}C-C^{\text{T}}{X}_{4}C\right)A^{-}& 0\\ \left(D_1^{-}\left(E_m-A{A}^{-}\right)\right)\left(A-A{X}_{2}C-C^{\text{T}}{X}_{4}C\right)A^{-}& 0\end{array}\right]\end{array}$ (16)

然后设

$\begin{array}{c}{X}_1=\left(A^{-}-A^{-}{C}^{\text{T}}{D}_1^{-}\left(E_m-A{A}^{-}\right)\right)\left(A-A{X}_{2}C-C^{\text{T}}{X}_{4}C\right)A^{-}\\ =M\left(A-A\left(C^{-}-C^{-}B{D}_2^{-}\left(E_m-C{C}^{-}\right)\right)F{F}^{-}C-C^{\text{T}}\left(D_2^{-}\left(E_m-C{C}^{-}\right)\right)F{F}^{-}C\right)A^{-}\end{array}$ (17)

$X_3=0$,

其中 $M=\left(A^{-}-A^{-}{C}^{\text{T}}{D}_1^{-}\left(E_m-A{A}^{-}\right)\right)$。

因此，有以下公式

${\left[\begin{array}{cc}A& C^{\text{T}}\\ C& B\end{array}\right]}^{-}=\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]=\left[\begin{array}{cc}{X}_1& \left(C^{-}-C^{-}B{D}_2^{-}\left(E_m-C{C}^{-}\right)\right)F{F}^{-}\\ 0& \left(D_2^{-}\left(E_m-C{C}^{-}\right)\right)F{F}^{-}\end{array}\right]$ (18)

其中

$X_1=\left(A^{-}-A^{-}{C}^{\text{T}}{D}_1^{-}\left(E_m-A{A}^{-}\right)\right)\left(A-A\left(C^{-}-C^{-}B{D}_2^{-}\left(E_m-C{C}^{-}\right)\right)F{F}^{-}C-C^{\text{T}}\left(D_2^{-}\left(E_m-C{C}^{-}\right)\right)F{F}^{-}C\right)A^{-}$

$D_1=\left(E_n-A{A}^{-}\right)C^{\text{T}}$,$D_2=\left(E_n-C{C}^{-}\right)B$,

从而可得

$\left[\begin{array}{c}U\\ -P\end{array}\right]={\left[\begin{array}{cc}A& C^{\text{T}}\\ C& B\end{array}\right]}^{-}\left[\begin{array}{c}0\\ F\end{array}\right]$。

2.3. 初等变换法 [12]

对于任意一个分块矩阵A，它的减号逆总存在，不一定唯一，并且

$A^{-}=\{\begin{array}{l}{A}^{\text{T}}{\left(A{A}^{\text{T}}\right)}^{-1}，当R\left(A\right)=m;\\ {\left(A{A}^{\text{T}}\right)}^{-1}{A}^{\text{T}},当R\left(A\right)=n;\\ C^{\text{T}}\left(C{C}^{\text{T}}\right)\left(D{D}^{-1}\right)D^{\text{T}},当A=DC为A满秩分解式时.\end{array}$

是A的一个减号逆。设A是 $m\times n$ 矩阵， $R\left(A\right)=r$，则存在可逆的m阶方阵P和n阶方阵Q，使

$A=P\left[\begin{array}{cc}{E}_r& 0\\ 0& 0\end{array}\right]Q$，即 $P^{-1}A{Q}^{-1}=\left[\begin{array}{cc}{E}_r& 0\\ 0& 0\end{array}\right]=D$，令 $P^{-1}=P_{\text{l}}\cdots P_1$,$Q^{-1}=Q_{\text{1}}\cdots Q_S$,$P_i$，$Q_j\left(i=1,\cdots ,l;j=1,\cdots ,s\right)$ 都是初等矩阵。则有

$P^{-1}A{Q}^{-1}=P_{\text{l}}\cdots P_{1}T{Q}_{\text{1}}\cdots Q_s=\left[\begin{array}{cc}{E}_r& 0\\ 0& 0\end{array}\right]$,

$P^{-1}=P_l\cdots P_1=P_l\cdots P\cdot E_m$,$Q^{-1}=Q_1\cdots Q_s=E_n\cdot Q_1\cdots Q_s$.

这一过程可对下面分块矩阵施行初等变换完成

$\left[\begin{array}{cc}{T}_{m\times n}& E_m\\ E_n& 0\end{array}\right]\stackrel{初等行变换}{\to }\left[\begin{array}{cc}{P}^{-1}{T}_{m\times n}& P^{-1}\\ E_n& 0\end{array}\right]\underset{初等列变换}{\to }\left[\begin{array}{cc}{P}^{-1}{T}_{m\times n}{Q}^{-1}& P^{-1}\\ Q^{-1}& 0\end{array}\right]=\left[\begin{array}{cc}\left[\begin{array}{cc}{E}_r& 0\\ 0& 0\end{array}\right]& P^{-1}\\ Q^{-1}& 0\end{array}\right]$

因此 $A^{-}=Q^{-1}{D}^{\text{T}}{P}^{-1}$，$G=Q^{-1}\left[\begin{array}{cc}{E}_r& C\\ D& I\end{array}\right]P^{-1}$ 是A的全部广义逆，其中C，D，I分别为任意的 $r\times \left(m-r\right)$，$\left(n-r\right)\times r$，$\left(n-r\right)\times \left(m-r\right)$ 矩阵。

3. 数值例子

当 $A=\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]$，$C=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$，$B=\left[\begin{array}{cc}3& 1\\ 1& 5\end{array}\right]$，$F=\left[\begin{array}{c}2\\ 3\end{array}\right]$ 时，求方程组 $\left[\begin{array}{cc}A& C^{\text{T}}\\ C& B\end{array}\right]\left[\begin{array}{c}U\\ -P\end{array}\right]=\left[\begin{array}{c}0\\ F\end{array}\right]$。

解：系数矩阵

$T=\left[\begin{array}{cc}A& C^{\text{T}}\\ C& B\end{array}\right]=\left[\begin{array}{cccc}1& 3& 1& 2\\ 2& 6& 2& 4\\ 1& 2& 3& 1\\ 2& 4& 1& 5\end{array}\right]$,$b=\left[\begin{array}{c}0\\ 0\\ 2\\ 3\end{array}\right]$,

利用初等变换理论做分块矩阵的初等变换

$\left[\begin{array}{cc}T& E_4\\ E_4& 0\end{array}\right]\stackrel{初等变换}{\to }\left[\begin{array}{cccccccc}1& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& -2& 1& 0& 0\\ 0& 1& 0& 0& 1& 0& -1/5& -2/5\\ 0& 0& 1& 0& 0& 0& 2/5& -1/5\\ 1& -3& -1& -16/5& 0& 0& 0& 0\\ 0& 1& 0& 1/5& 0& 0& 0& 0\\ 0& 0& 1& 3/5& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0\end{array}\right]\stackrel{记作}{\to }\left[\begin{array}{cc}D& P^{-1}\\ Q^{-1}& 0\end{array}\right]$

满足关系式 $P^{-1}T{Q}^{-1}=D$，因此可以求出T的一个减号逆为

$T^{-}=Q^{-1}{D}^{\text{T}}{P}^{-1}=\left[\begin{array}{cccc}-2& 0& 0.2& 1.4\\ 1& 0& -0.2& -0.4\\ 0& 0& 0.4& -0.2\\ 0& 0& 0& 0\end{array}\right]$,

再求原方程的通解

$\left[\begin{array}{c}U\\ -P\end{array}\right]=T^{-}b+\left(E-A^{-}A\right)\xi =\left[\begin{array}{c}4.6\\ -1.6\\ 0.2\\ 0\end{array}\right]+k_4\left[\begin{array}{c}-3.2\\ 0.2\\ 0.6\\ 1\end{array}\right]$,

其中 $k_4$ 为任意实数， $\xi ={\left(k_1,k_2,k_3,k_4\right)}^{\text{T}}$。

验证：由于广义逆是不唯一的，利用matlab直接计算，得到方程一个解

$\left[\begin{array}{c}U\\ -P\end{array}\right]={\left[\begin{array}{cc}A& C^{\text{T}}\\ C& B\end{array}\right]}^{-}\left[\begin{array}{c}0\\ F\end{array}\right]=\left[\begin{array}{c}0.4983\\ -1.3436\\ 0.9691\\ 1.2818\end{array}\right]$

当 $k_4=\text{1}\text{.128}$，$\left[\begin{array}{c}U\\ -P\end{array}\right]=T^{-}b+\left(E-A^{-}A\right)\xi =\left[\begin{array}{c}4.6\\ -1.6\\ 0.2\\ 0\end{array}\right]+\text{1}.\text{128}\left[\begin{array}{c}-3.2\\ 0.2\\ 0.6\\ 1\end{array}\right]=\left[\begin{array}{c}0.4983\\ -1.3436\\ 0.9691\\ 1.2818\end{array}\right]$，说明该方法有效。

4. 2 × 2分块矩阵求解的推广

已知矩阵方程 [12]

$\left[\begin{array}{cccc}{A}_1& 0& 0& 0\\ 0& A_2& C_1& 0\\ 0& C_1& B_1& 0\\ 0& 0& 0& B_2\end{array}\right]\left[\begin{array}{c}{U}_1\\ U_2\\ -P_1\\ -P_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ F_1\\ F_2\end{array}\right]$ (20)

其中系数矩阵 $T=\left[\begin{array}{cccc}{A}_1& 0& 0& 0\\ 0& A_2& C_1& 0\\ 0& C_1& B_1& 0\\ 0& 0& 0& B_2\end{array}\right]$ 已知，下面求解T的广义逆。

令 $A=\left[\begin{array}{cc}{A}_1& 0\\ 0& A_2\end{array}\right]$,$B=\left[\begin{array}{cc}{B}_1& 0\\ 0& B_2\end{array}\right]$,$C=\left[\begin{array}{cc}0& C_1\\ 0& 0\end{array}\right]$,$F=\left[\begin{array}{c}{F}_1\\ F_2\end{array}\right]$。

有 $A^{-}=\left[\begin{array}{cc}{A}_1^{-}& 0\\ 0& A_2^{-}\end{array}\right]$,$B^{-}=\left[\begin{array}{cc}{B}_1^{-}& 0\\ 0& B_2^{-}\end{array}\right]$,$C^{-}=\left[\begin{array}{cc}0& C_1^{-}\\ 0& 0\end{array}\right]$。

$D_1=\left(E-A{A}^{-}\right)C^{\text{T}}=\left[\begin{array}{cc}0& 0\\ C_1-A_2{A}_2^{-}{C}_1& 0\end{array}\right]$,$D_2=\left(E_n-C{C}^{-}\right)B=\left[\begin{array}{cc}{B}_1& 0\\ 0& B_2\end{array}\right]$,

$D_1^{-}=\left[\begin{array}{cc}0& 0\\ C_{\text{1}}^{-}-C_1^{-}{A}_2{A}_2^{-}& 0\end{array}\right]$,$D_2^{-}=\left[\begin{array}{cc}{B}_1^{-}& 0\\ 0& B_2^{-}\end{array}\right]$.

$F^{-}={\left[\begin{array}{c}{F}_1\\ F_2\end{array}\right]}^{-}={\left({\left[\begin{array}{cc}{F}_1^{\text{T}}& F_2^{\text{T}}\end{array}\right]}^{\text{T}}\right)}^{-}=\left[\begin{array}{cc}{F}_1^{-}-F_1^{-}{F}_2{P}^{-}\left(E-F_1{F}_1^{-}\right)& P^{-}\left(E-F_1{F}_1^{-}\right)\end{array}\right]$

其中 $P=\left(E-F_1{F}_1^{-}\right)F_2$，$P^{-}=F_2^{-}-F_2^{-}{F}_1{F}_1^{-}$。

令

$\left[\begin{array}{cccc}{A}_1& 0& 0& 0\\ 0& A_2& C_1& 0\\ 0& C_1& B_1& 0\\ 0& 0& 0& B_2\end{array}\right]=\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]$

代入公式(19)可得

$X_1=\left[\begin{array}{cc}{A}_1^{-}-A_1^{-}{C}_1{C}_1^{-}\left(E-A_2{A}_2^{-}\right)\left(E-A_1{A}_1^{-}\right)T_3{A}_1^{-}& 0\\ 0& A_2^{-}-A_2^{-}{C}_1{B}_1^{-}{F}_1{T}_1{C}_1{A}_2^{-}\end{array}\right]$

$X_2=\left[\begin{array}{cc}{C}_1^{-}\left(E-B_2{B}_2^{-}\right)F_2\left(F_1^{-}-F_1^{-}{F}_2{P}^{-}\left(E-F_1{F}_1^{-}\right)\right)& C_1^{-}\left(E-B_2{B}_2^{-}\right)F_2\left(P^{-}\left(E-F_1{F}_1^{-}\right)\right)\\ 0& 0\end{array}\right]$

$X_3=0$

$X_4=\left[\begin{array}{cc}{B}_1^{-}{F}_1\left(F_1^{-}-F_1^{-}{F}_2{P}^{-}\left(E-F_1{F}_1^{-}\right)\right)& B_1^{-}{F}_1\left(P^{-}\left(E-F_1{F}_1^{-}\right)\right)\\ B_2^{-}{F}_2\left(F_1^{-}-F_1^{-}{F}_2{P}^{-}\left(E-F_1{F}_1^{-}\right)\right)& B_2^{-}{F}_2\left(P^{-}\left(E-F_1{F}_1^{-}\right)\right)\end{array}\right]$

其中 $T_3=A_1-C_1^{-}+C_1^{-}{B}_2{B}_2^{-}{F}_2{T}_1{C}_1$，$T_1=F_1^{-}-F_1^{-}{F}_2{P}^{-}\left(E-F_1{F}_1^{-}\right)$。

5. 结论

通过分块矩阵的广义逆探讨矩阵方程的求解，也是代数方程求解的探索方向之一。在本文中，探讨了一类 $\text{2}\times \text{2}$ 分块矩阵的广义逆的具体计算过程，应用初等变换法求解一类分块矩阵的广义逆，给出了数值计算例子，应用于一些特殊方程的求解，并对 $\text{2}\times \text{2}$ 分块矩阵的广义逆的计算进行了推广与应用。

基金项目

江西省教育厅科技项目(项目号：GJJ160557)。

文章引用

赖金凤,刘唐伟,唐阿敏,陈 硕. 一类分块矩阵方程的广义逆求解方法
The Generalized Inverse Computing Method of a Class of Block Matrixequations[J]. 应用数学进展, 2020, 09(08): 1115-1123. https://doi.org/10.12677/AAM.2020.98131

参考文献