MB-矩阵子直和仍为MB-矩阵的条件

Conditions That Subdirect Sums of MB-Matrices Is Still MB-Matrices

摘要: 通过将MB-矩阵分裂成一个非奇异M-矩阵和一个秩1非负矩阵之和，获得MB-矩阵的子直和仍为MB-矩阵的一些充要条件和充分条件，最后用数值例子对所给结论进行了说明和解释。

Abstract: By splitting an MB-matrix A into a sum of a nonsingular M-matrix and a nonnegative rank 1 matrix, some sufficient and necessary conditions and some sufficient conditions are given such that the subdirect sum of two MB-matrices is still an MB-matrix. Some examples are also given to illustrate the results.

1. 引言

1999年Fallat和Johnson首先在文献 [1] 中提出矩阵的子直和的概念，由于其在诸如马可夫链的递增许瓦兹迭代及分裂和重叠的递增许瓦兹迭代等研究中的重要性，引起了学者的关注和研究，并取得了一些重要研究成果，如文献 [2] [3] [4] 分别对非奇异M-矩阵及其逆的子直和、H-矩阵和双对角占优矩阵的子直和等进行了研究。本文在文献 [2] 和 [5] 的基础上对MB-矩阵的子直和进行研究，试图得到MB-矩阵的子直和仍为MB-矩阵的一些新的条件。

2. 预备知识

本节先给出一些基本概念、定理与符号，以备后用。

设 $A=\left(a_{ij}\right)\in R^{m\times n}$ ，如果对于所有的 $i=1,\cdots ,m;j=1,\cdots ,n$ 都有 $a_{ij}>0\left(a_{ij}\ge 0\right)$ ，则称 $A$ 为正(非负)矩阵，记为 $A>O\left(A\ge O\right)$ 。

定义2.1. [6] 设 $A=\left(a_{ij}\right)\in R^{n\times n}$ ，如果对于所有的 $1\le i,j\le n$ ， $i\ne j$ 都有 $a_{ij}\le 0$ ，则 $A$ 称为Z-矩阵。如果 $A$ 是Z-矩阵且 $A^{-1}\ge O$ ，则称 $A$ 为M-矩阵。

定义2.2. [7] 设 $A=\left(a_{ij}\right)\in R^{n\times n}$ ，将 $A$ 分裂为 $A=A^z+A^r$ ，其中

$A^z=\left[\begin{array}{cccc}{a}_{11}-{\beta }_1^A& a_{12}-{\beta }_1^A& \cdots & a_{1n}-{\beta }_1^A\\ a_{21}-{\beta }_2^A& a_{22}-{\beta }_2^A& \cdots & a_{2n}-{\beta }_2^A\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}-{\beta }_n^A& a_{n2}-{\beta }_n^A& \cdots & a_{nn}-{\beta }_n^A\end{array}\right]$ ， $A^r=\left[\begin{array}{cccc}{\beta }_1^A& {\beta }_1^A& \cdots & {\beta }_1^A\\ {\beta }_2^A& {\beta }_2^A& \cdots & {\beta }_2^A\\ ⋮& ⋮& \ddots & ⋮\\ {\beta }_n^A& {\beta }_n^A& \cdots & {\beta }_n^A\end{array}\right]$ (1)

${\beta }_i^A=\max \left\{0,a_{ij}|\forall j\ne i\right\}$ 。显然 $A^z$ 是Z-矩阵， $A^r$ 是秩1非负矩阵。若 $A^z$ 为M-矩阵，则称 $A$ 为MB-矩阵。

定义2.3. [2] 设 $A\in R^{n_1\times n_1},B\in R^{n_2\times n_2}$ ， $k$ 是整数且 $1\le k\le \min \left\{n_1,n_2\right\}$ ， $n=n_1+n_2-k$ ， $A,B$ 分块如下：

$A=\left[\begin{array}{cc}{A}_{11}& A_{12}\\ A_{21}& A_{22}\end{array}\right]$ ， $B=\left[\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\end{array}\right]$ (2)

其中 $A_{22}$ ， $B_{11}$ 都是 $k$ 阶方阵。定义矩阵

$M=\left[\begin{array}{ccc}{A}_{11}& A_{12}& O\\ A_{21}& A_{22}+B_{11}& B_{12}\\ O& B_{21}& B_{22}\end{array}\right]\in R^{n\times n}$ (3)

称其为 $A$ 和 $B$ 的 $n$ $\left(n=n_1+n_2-k\right)$ 阶k-子直和，记为 $M=A{\oplus }_{k}B$ 。

将 $A=\left(a_{ij}\right)\in R^{n_1\times n_1},B=\left(b_{ij}\right)\in R^{n_2\times n_2}$ 和 $M=\left(m_{ij}\right)\in R^{n\times n}$ 按定义2.2中的(1)式分别分裂为：

$A=A^z+A^r$ ， $B=B^z+B^r$ ， $M=A{\oplus }_{k}B=M^z+M^r$

将 $A^z,A^r,B^z,B^r$ 按(2)式分块为：

$A^z=\left[\begin{array}{cc}{A}_{11}^z& A_{12}^z\\ A_{21}^z& A_{22}^z\end{array}\right]$ ， $A^r=\left[\begin{array}{cc}{A}_{11}^r& A_{12}^r\\ A_{21}^r& A_{22}^r\end{array}\right]$ ， $B^z=\left[\begin{array}{cc}{B}_{11}^z& B_{12}^z\\ B_{21}^z& B_{22}^z\end{array}\right]$ ， $B^r=\left[\begin{array}{cc}{B}_{11}^r& B_{12}^r\\ B_{21}^r& B_{22}^r\end{array}\right]$

定义矩阵

$\bar{M}=\left[\begin{array}{ccc}{A}_{11}^z& A_{12}^z& -A_{13}^r\\ A_{21}^z-B_{13}^r& A_{22}^z+B_{11}^z& B_{12}^z-A_{23}^r\\ -B_{23}^r& B_{21}^z& B_{22}^z\end{array}\right]\in R^{n\times n}$ (4)

其中

$A_{11}^z\in R^{\left(n_1-k\right)\times \left(n_1-k\right)}$ ， $A_{12}^z\in R^{\left(n_1-k\right)\times k}$ ， $A_{21}^z\in R^{k\times \left(n_1-k\right)}$ ， $A_{22}^z\in R^{k\times k}$ ，

$B_{11}^z\in R^{k\times k}$ ， $B_{12}^z\in R^{k\times \left(n-n_1\right)}$ ， $B_{21}^z\in R^{\left(n-n_1\right)\times k}$ ， $B_{22}^z\in R^{\left(n-n_1\right)\times \left(n-n_1\right)}$ ，

$A_{11}^r\in R^{\left(n_1-k\right)\times \left(n_1-k\right)}$ ， $A_{12}^r\in R^{\left(n_1-k\right)\times k}$ ， $A_{13}^r\in R^{\left(n_1-k\right)\left(n-n_1\right)}$ 的第i行为 $\left({\beta }_i^A,{\beta }_i^A,\cdots ,{\beta }_i^A\right)$ ，

$A_{21}^r\in R^{k\times \left(n_1-k\right)}$ ， $A_{22}^r\in R^{k\times k}$ ， $A_{23}^r\in R^{k\times \left(n-n_1\right)}$ 的第i行为 $\left({\beta }_i^A,{\beta }_i^A,\cdots ,{\beta }_i^A\right)$ ，

$B_{11}^r\in R^{k\times k}$ ， $B_{12}^r\in R^{k\times \left(n-n_1\right)}$ ， $B_{13}^r\in R^{k\times \left(n_1-k\right)}$ 的第i行为 $\left({\beta }_i^B,{\beta }_i^B,\cdots ,{\beta }_i^B\right)$ ，

$B_{21}^r\in R^{\left(n-n_1\right)\times k}$ ， $B_{22}^r\in R^{\left(n-n_1\right)\times \left(n-n_1\right)}$ ， $B_{23}^r\in R^{\left(n-n_1\right)\times \left(n_1-k\right)}$ 的第i行为 $\left({\beta }_i^B,{\beta }_i^B,\cdots ,{\beta }_i^B\right)$ 。

容易验证这里的 $\bar{M}$ 就是 [5] 中的 $\bar{M}$ ，于是由文献 [5] 知 $M^z\ge \bar{M}$ ，且都为Z-矩阵。

当 $A^z,B^z$ 为非奇异矩阵时，将 ${\left(A^z\right)}^{-1},{\left(B^z\right)}^{-1}$ 按(2)分块为：

${\left(A^z\right)}^{-1}=\left[\begin{array}{cc}\widehat{A_{11}^z}& \widehat{A_{12}^z}\\ \widehat{A_{21}^z}& \widehat{A_{22}^z}\end{array}\right]$ ， ${\left(B^z\right)}^{-1}=\left[\begin{array}{cc}\widehat{B_{11}^z}& \widehat{B_{12}^z}\\ \widehat{B_{21}^z}& \widehat{B_{22}^z}\end{array}\right]$ (*)

其中 $\widehat{A_{11}^z}\in R^{\left(n_1-k\right)\times \left(n_1-k\right)}$ ， $\widehat{A_{12}^z}\in R^{\left(n_1-k\right)\times k}$ ， $\widehat{A_{21}^z}\in R^{k\times \left(n_1-k\right)}$ ， $\widehat{A_{22}^z}\in R^{k\times k}$ ， $\widehat{B_{11}^z}\in R^{k\times k}$ ， $\widehat{B_{12}^z}\in R^{k\times \left(n-n_1\right)}$ ， $\widehat{B_{21}^z}\in R^{\left(n-n_1\right)\times k}$ ， $\widehat{B_{22}^z}\in R^{\left(n-n_1\right)\times \left(n-n_1\right)}$ 。

定理2.1. [2] 设 $A=\left[a_{ij}\right]\in R^{n\times n}$ ，则如下3款成立：

1) 当 $A$ 为非奇异M-矩阵时，其主对角元为正。

2) 当 $A$ 为非奇异M-矩阵， $B=\left[b_{ij}\right]$ 为Z-矩阵且 $B\ge A$ 时， $B$ 为非奇异M-矩阵。

3) $A$ 为非奇异M-矩阵的充要条件为 $A$ 的每一个主子矩阵为非奇异M-矩阵。

引理2.1. [8] 设 $A=\left[\begin{array}{cc}D& E\\ F& G\end{array}\right]$ 非奇异，其中 $D\in R^{\left(n-k\right)\times \left(n-k\right)}$ ， $G\in R^{k\times k}$ ， $E\in R^{\left(n-k\right)\times k}$ ， $F\in R^{k\times \left(m-k\right)}$ 。则

1) 若 $D$ 非奇异且 $D^{-1}\ge 0$ ， $-E\ge 0$ ， $-F\ge 0$ ，则 $A^{-1}\ge 0$ 当且仅当 ${\left(A/D\right)}^{-1}\ge 0$ ；

2) 若 $G^{-1}\ge 0$ ， $-E\ge 0$ ， $-F\ge 0$ ，则 $A^{-1}\ge 0$ 当且仅当 ${\left(A/G\right)}^{-1}\ge 0$ 。

这里 $A/D$ 表示矩阵 $A$ 内 $D$ 的Schur补。

3. MB-矩阵的k-子直和

先给出 $\bar{M}$ 为非奇异的Z-矩阵的充要条件。

定理3.1. 设 $A^z,B^z$ 为M-矩阵，

$\det \hat{H}=\det \left\{\left(\widehat{B_{11}^z}+\widehat{A_{22}^z}\right)+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{11}^z}-A_{13}^r\widehat{B_{21}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{21}^z}\right\}\ne 0$

且 $B_{23}^r=O$ ，则 $\bar{M}$ 为非奇异的Z-矩阵。

证明：由(*)式得

${\left(A^z\right)}^{-1}\left[\begin{array}{cc}{I}_{n_1-k}& O\\ B_{13}^r& I_k\end{array}\right]=\left[\begin{array}{cc}{A}_{11}^z+\widehat{A_{12}^z}{B}_{13}^r& \widehat{A_{12}^z}\\ \widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r& \widehat{A_{22}^z}\end{array}\right]$

再由 $\det {\left(A^z\right)}^{-1}\det \left[\begin{array}{cc}{I}_{n_1-k}& O\\ B_{13}^r& I_k\end{array}\right]\ne 0$ 得

$\det \left[\begin{array}{cc}{A}_{11}^z+\widehat{A_{12}^z}{B}_{13}^r& \widehat{A_{12}^z}\\ \widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r& \widehat{A_{22}^z}\end{array}\right]\ne 0$

于是由 $A^z{\left(A^z\right)}^{-1}=I_{n_1}$ ，得

$A_{11}^z\widehat{A_{11}^z}+A_{12}^z\widehat{A_{12}^z}=I_{n_1-k}=I_{n-n_2}$ ， $A_{11}^z\widehat{A_{12}^z}+A_{12}^z\widehat{A_{22}^z}=0$ ，

$A_{21}^z\widehat{A_{11}^z}+A_{22}^z\widehat{A_{21}^z}=0$ ， $A_{21}^z\widehat{A_{12}^z}+A_{22}^z\widehat{A_{22}^z}=I_k$

由 ${\left(A^z\right)}^{-1}{A}^z=I_{n_1}$ ，得

$\widehat{A_{11}^z}{A}_{11}^z+\widehat{A_{12}^z}{A}_{12}^z=I_{n-n_2}$ ， $\widehat{A_{11}^z}{A}_{12}^z+\widehat{A_{12}^z}{A}_{22}^z=0$

$\widehat{A_{21}^z}{A}_{11}^z+\widehat{A_{22}^z}{A}_{21}^z=0$ ， $\widehat{A_{21}^z}{A}_{12}^z+\widehat{A_{22}^z}{A}_{22}^z=I_k$

由 $B^z{\left(B^z\right)}^{-1}=I_{n_2}$ ，得

$B_{11}^z\widehat{B_{11}^z}+B_{12}^z\widehat{B_{21}^z}=I_k$ ， $B_{11}^z\widehat{B_{12}^z}+B_{12}^z\widehat{B_{22}^z}=0$ ，

$B_{21}^z\widehat{B_{11}^z}+B_{22}^z\widehat{B_{21}^z}=0$ ， $B_{21}^z\widehat{B_{12}^z}+B_{22}^z\widehat{B_{22}^z}=I_{n-n_1}$

由 ${\left(B^z\right)}^{-1}{B}^z=I_{n_2}$ ，得

$\widehat{B_{11}^z}{B}_{11}^z+\widehat{B_{12}^z}{B}_{21}^z=I_k$ ， $\widehat{B_{11}^z}{B}_{12}^z+\widehat{B_{12}^z}{B}_{22}^z=0$ ，

$\widehat{B_{21}^z}{B}_{11}^z+\widehat{B_{22}^z}{B}_{21}^z=0$ ， $\widehat{B_{21}^z}{B}_{12}^z+\widehat{B_{22}^z}{B}_{22}^z=I_{n-n_1}$

故有

$\left[\begin{array}{ccc}\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r& \widehat{A_{12}^z}& 0\\ \widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r& \widehat{A_{22}^z}& 0\\ 0& 0& I_{n-n_1}\end{array}\right]\bar{M}\left[\begin{array}{ccc}{I}_{n-n_2}& 0& 0\\ 0& \widehat{B_{11}^z}& \widehat{B_{12}^z}\\ 0& \widehat{B_{21}^z}& \widehat{B_{22}^z}\end{array}\right]$

$=\left[\begin{array}{ccc}\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r& \widehat{A_{12}^z}& 0\\ \widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r& \widehat{A_{22}^z}& 0\\ 0& 0& I_{n-n_1}\end{array}\right]\left[\begin{array}{ccc}{A}_{11}^z& A_{12}^z& -A_{13}^r\\ A_{21}^z-B_{13}^r& A_{22}^z+B_{11}^z& B_{12}^z-A_{23}^r\\ -B_{23}^r& B_{21}^z& B_{22}^z\end{array}\right]\left[\begin{array}{ccc}{I}_{n-n_2}& 0& 0\\ 0& \widehat{B_{11}^z}& \widehat{B_{12}^z}\\ 0& \widehat{B_{21}^z}& \widehat{B_{22}^z}\end{array}\right]$ $=[\begin{array}{cc}\left(\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r\right)A_{11}^z+\widehat{A_{12}^z}\left(A_{21}^z-B_{13}^r\right)& \left(\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r\right)A_{12}^z+\widehat{A_{12}^z}\left(A_{22}^z+B_{11}^z\right)\\ \left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)A_{11}^z+\widehat{A_{22}^z}\left(A_{21}^z-B_{13}^r\right)& \left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)A_{12}^z+\widehat{A_{22}^z}\left(A_{22}^z+B_{11}^z\right)\\ -B_{23}^r& B_{21}^z\end{array}$ $\begin{array}{c}-\left(\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r\right)A_{13}^r+\widehat{A_{12}^z}\left(B_{12}^z-A_{23}^r\right)\\ -\left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)A_{13}^r+\widehat{A_{22}^z}\left(B_{12}^z-A_{23}^r\right)\\ B_{22}^z\end{array}]\left[\begin{array}{ccc}{I}_{n-n_2}& 0& 0\\ 0& \widehat{B_{11}^z}& \widehat{B_{12}^z}\\ 0& \widehat{B_{21}^z}& \widehat{B_{22}^z}\end{array}\right]$

$=[\begin{array}{c}\left(\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r\right)A_{11}^z+\widehat{A_{12}^z}\left(A_{21}^z-B_{13}^r\right)\\ \left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)A_{11}^z+\widehat{A_{22}^z}\left(A_{21}^z-B_{13}^r\right)\\ -B_{23}^r\end{array}$ $\begin{array}{l}\begin{array}{c}\left(\left(\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r\right)A_{12}^z+\widehat{A_{12}^z}\left(A_{22}^z+B_{11}^z\right)\right)\widehat{B_{11}^z}+\left(-\left(\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r\right)A_{13}^r+\widehat{A_{12}^z}\left(B_{12}^z-A_{23}^r\right)\right)\widehat{B_{21}^z}\\ \left(\left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)A_{12}^z+\widehat{A_{22}^z}\left(A_{22}^z+B_{11}^z\right)\right)\widehat{B_{11}^z}+\left(-\left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)A_{13}^r+\widehat{A_{22}^z}\left(B_{12}^z-A_{23}^r\right)\right)\widehat{B_{21}^z}\\ B_{21}^z\widehat{B_{11}^z}+B_{22}^z\widehat{B_{21}^z}\end{array}\\ \end{array}$ $\begin{array}{c}\left(\left(\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r\right)A_{12}^z+\widehat{A_{12}^z}\left(A_{22}^z+B_{11}^z\right)\right)\widehat{B_{12}^z}+\left(-\left(\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r\right)A_{13}^r+\widehat{A_{12}^z}\left(B_{12}^z-A_{23}^r\right)\right)\widehat{B_{22}^z}\\ \left(\left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)A_{12}^z+\widehat{A_{22}^z}\left(A_{22}^z+B_{11}^z\right)\right)\widehat{B_{12}^z}+\left(-\left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)A_{13}^r+\widehat{A_{22}^z}\left(B_{12}^z-A_{23}^r\right)\right)\widehat{B_{22}^z}\\ B_{21}^z\widehat{B_{12}^z}+B_{22}^z\widehat{B_{22}^z}\end{array}]$

$\begin{array}{l}=[\begin{array}{c}\widehat{A_{11}^z}{A}_{11}^z+\widehat{A_{12}^z}{B}_{13}^r{A}_{11}^z+\widehat{A_{12}^z}{A}_{21}^z-\widehat{A_{12}^z}{B}_{13}^r\\ \widehat{A_{21}^z}{A}_{11}^z+\widehat{A_{22}^z}{B}_{13}^r{A}_{11}^z+A_{22}^z{A}_{21}^z-\widehat{A_{22}^z}\widehat{B_{13}^r}\\ -B_{23}^z\end{array}\\ \end{array}$ $\begin{array}{c}\widehat{A_{11}^z}{A}_{12}^z\widehat{B_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r{A}_{12}^z\widehat{B_{11}^z}+\widehat{A_{12}^z}{A}_{22}^z\widehat{B_{11}^z}+\widehat{A_{12}^z}{B}_{11}^z\widehat{B_{11}^z}-\widehat{A_{11}^z}{A}_{13}^r\widehat{B_{21}^z}-\widehat{A_{12}^z}{B}_{13}^r{A}_{13}^r\widehat{B_{21}^z}+\widehat{A_{12}^z}{B}_{12}^z\widehat{B_{21}^z}-\widehat{A_{12}^z}{A}_{23}^r\widehat{B_{21}^z}\\ \widehat{A_{21}^z}{A}_{12}^z\widehat{B_{11}^z}+\widehat{A_{22}^z}{B}_{13}^r{A}_{12}^z\widehat{B_{11}^z}+\widehat{A_{22}^z}{A}_{22}^z\widehat{B_{11}^z}+\widehat{A_{22}^z}{B}_{11}^z\widehat{B_{11}^z}-\widehat{A_{21}^z}{A}_{13}^r\widehat{B_{21}^z}-\widehat{A_{22}^z}{B}_{13}^r{A}_{13}^r\widehat{B_{21}^z}+\widehat{A_{22}^z}{B}_{12}^z\widehat{B_{21}^z}-\widehat{A_{22}^z}{A}_{23}^r\widehat{B_{21}^z}\\ B_{21}^z\widehat{B_{11}^z}+B_{22}^z\widehat{B_{21}^z}\end{array}$ $\begin{array}{c}\widehat{A_{11}^z}{A}_{12}^z\widehat{B_{12}^z}+\widehat{A_{12}^z}{B}_{13}^r{A}_{12}^z\widehat{B_{12}^z}+\widehat{A_{12}^z}{A}_{22}^z\widehat{B_{12}^z}+\widehat{A_{12}^z}{B}_{11}^z\widehat{B_{12}^z}-\widehat{A_{11}^z}{A}_{13}^r\widehat{B_{22}^z}-\widehat{A_{12}^z}{B}_{13}^r{A}_{13}^r\widehat{B_{22}^z}+\widehat{A_{12}^z}{B}_{12}^z\widehat{B_{22}^z}-\widehat{A_{12}^z}{A}_{23}^r\widehat{B_{22}^z}\\ \widehat{A_{21}^z}{A}_{12}^z\widehat{B_{12}^z}+\widehat{A_{22}^z}{B}_{13}^r{A}_{12}^z\widehat{B_{12}^z}+\widehat{A_{22}^z}{A}_{22}^z\widehat{B_{12}^z}+\widehat{A_{22}^z}{B}_{11}^z\widehat{B_{12}^z}-\widehat{A_{21}^z}{A}_{13}^r\widehat{B_{22}^z}-\widehat{A_{22}^z}{B}_{13}^r{A}_{13}^r\widehat{B_{22}^z}+\widehat{A_{22}^z}{B}_{12}^z\widehat{B_{22}^z}-\widehat{A_{22}^z}{A}_{23}^r\widehat{B_{22}^z}\\ B_{21}^z\widehat{B_{12}^z}+B_{22}^z\widehat{B_{22}^z}\end{array}]$

$=[\begin{array}{cc}\widehat{A_{11}^z}{A}_{11}^z+\widehat{A_{12}^z}{A}_{21}^z& \left(\widehat{A_{11}^z}{A}_{12}^z+\widehat{A_{12}^z}{A}_{22}^z\right)\widehat{B_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{11}^z}-A_{13}^r\widehat{B_{21}^z}\right)+\widehat{A_{12}^z}\left(B_{11}^z\widehat{B_{11}^z}+B_{12}^z\widehat{B_{21}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{21}^z}\\ \widehat{A_{21}^z}{A}_{11}^z+\widehat{A_{22}^z}{A}_{21}^z& \left(\widehat{A_{21}^z}{A}_{12}^z+\widehat{A_{22}^z}{A}_{22}^z\right)\widehat{B_{11}^z}+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{11}^z}-A_{13}^r\widehat{B_{21}^z}\right)+\widehat{A_{22}^z}\left(B_{11}^z\widehat{B_{11}^z}+B_{12}^z\widehat{B_{21}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{21}^z}\\ -B_{23}^z& 0\end{array}$ $\begin{array}{c}\left(\widehat{A_{11}^z}{A}_{12}^z+\widehat{A_{12}^z}{A}_{22}^z\right)\widehat{B_{12}^z}+\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)+\widehat{A_{12}^z}\left(B_{11}^z\widehat{B_{12}^z}+B_{12}^z\widehat{B_{22}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\\ \left(\widehat{A_{21}^z}{A}_{12}^z+\widehat{A_{22}^z}{A}_{22}^z\right)\widehat{B_{12}^z}+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)+\widehat{A_{22}^z}\left(B_{11}^z\widehat{B_{12}^z}+B_{12}^z\widehat{B_{22}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\\ I_{n-n_1}\end{array}]$ $=\left[\begin{array}{ccc}{I}_{n-n_2}& \widehat{A_{12}^z}+\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{11}^z}-A_{13}^r\widehat{B_{21}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{21}^z}& \widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\\ 0& \hat{H}& \widehat{B_{12}^z}+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\\ -B_{23}^r& 0& I_{n-n_1}\end{array}\right]$

由此式知，当 $\det \hat{H}\ne 0$ 且 $B_{23}^r=O$ 时， $\bar{M}$ 为非奇异的。

现在讨论MB-矩阵的子直和为MB-矩阵的充分条件。

容易验证

${\left[\begin{array}{ccc}{I}_{n-n_2}& F& Y\\ 0& \hat{H}& Q\\ 0& 0& I_{n-n_1}\end{array}\right]}^{-1}\left[\begin{array}{ccc}{I}_{n-n_2}& C& D\\ O& {\hat{H}}^{-1}& E\\ O& O& I_{n-n_1}\end{array}\right]=\left[\begin{array}{ccc}{I}_{n-n_2}& 0& 0\\ 0& I_k& 0\\ 0& 0& I_{n-n_1}\end{array}\right]$

其中

$C=-\left[\widehat{A_{12}^z}+\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{11}^z}-A_{13}^r\widehat{B_{21}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{21}^z}\right]\times {\hat{H}}^{-1}$

$\begin{array}{c}D=-\left[\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\right]\\ +\left[\widehat{A_{12}^z}+\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{11}^z}-A_{13}^r\widehat{B_{21}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{21}^z}\right]\\ \times {\hat{H}}^{-1}\times \left[\widehat{B_{12}^z}+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\right]\end{array}$

$E=-{\hat{H}}^{-1}\times \left[\widehat{B_{12}^z}+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\right]$

$F=\widehat{A_{12}^z}+\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{11}^z}-A_{13}^r\widehat{B_{21}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{21}^z}$

$Y=\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{22}^z}$

$Q=\widehat{B_{12}^z}+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{22}^z}$

从而有

$\left[\begin{array}{ccc}\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r& \widehat{A_{12}^z}& 0\\ \widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r& \widehat{A_{22}^z}& 0\\ 0& 0& I_{n-n_1}\end{array}\right]\bar{M}\left[\begin{array}{ccc}{I}_{n-n_2}& 0& 0\\ 0& \widehat{B_{11}^z}& \widehat{B_{12}^z}\\ 0& \widehat{B_{21}^z}& \widehat{B_{22}^z}\end{array}\right]$

$=[\begin{array}{cc}{I}_{n-n_2}& \widehat{A_{12}^z}+\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{11}^z}-A_{13}^r\widehat{B_{21}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{21}^z}\\ 0& \hat{H}\\ -B_{23}^r& 0\end{array}$ $\begin{array}{c}\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\\ \widehat{B_{12}^z}+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\\ I_{n-n_1}\end{array}]$

对上式两边同时取逆得：

${\left[\begin{array}{ccc}{I}_{n-n_2}& 0& 0\\ 0& \widehat{B_{11}^z}& \widehat{B_{12}^z}\\ 0& \widehat{B_{21}^z}& \widehat{B_{22}^z}\end{array}\right]}^{-1}{\left(\bar{M}\right)}^{-1}{\left[\begin{array}{ccc}\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r& \widehat{A_{12}^z}& 0\\ \widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r& \widehat{A_{22}^z}& 0\\ 0& 0& I_{n-n_1}\end{array}\right]}^{-1}$

$\begin{array}{l}=\{[\begin{array}{cc}{I}_{n-n_2}& \widehat{A_{12}^z}+\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{11}^z}-A_{13}^r\widehat{B_{21}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{21}^z}\\ 0& \hat{H}\\ -B_{23}^r& 0\end{array}\\ \end{array}$

${\begin{array}{c}\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\\ \widehat{B_{12}^z}+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\\ I_{n-n_1}\end{array}]\}}^{-1}$

${\begin{array}{c}\widehat{A_{12}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{11}^z}{A}_{13}^r+\widehat{A_{12}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\\ \widehat{B_{12}^z}+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{12}^z}-A_{13}^r\widehat{B_{22}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{22}^z}\\ I_{n-n_1}\end{array}]\}}^{-1}$

进而可得：

${\left(\bar{M}\right)}^{-1}=\left[\begin{array}{ccc}{I}_{n-n_2}& O& O\\ O& \widehat{B_{11}^z}& \widehat{B_{12}^z}\\ O& \widehat{B_{21}^z}& \widehat{B_{22}^z}\end{array}\right]\left[\begin{array}{ccc}{I}_{n-n_2}& C& D\\ O& {\hat{H}}^{-1}& E\\ O& O& I_{n-n_1}\end{array}\right]\left[\begin{array}{ccc}\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r& \widehat{A_{12}^z}& O\\ \widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r& \widehat{A_{22}^z}& O\\ O& O& I_{n-n_1}\end{array}\right]$

即

${\left(\bar{M}\right)}^{-1}=\left[\begin{array}{ccc}\widehat{A_{11}^z}+\widehat{A_{12}^z}{B}_{13}^r+C\left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)& \widehat{A_{12}^z}+C\widehat{A_{22}^z}& D\\ \widehat{B_{11}^z}\hat{H}\left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)& \widehat{B_{11}^z}{\hat{H}}^{-1}\widehat{A_{22}^z}& \widehat{B_{11}^z}E+\widehat{B_{12}^z}\\ \widehat{B_{21}^z}{\hat{H}}^{-1}\left(\widehat{A_{21}^z}+\widehat{A_{22}^z}{B}_{13}^r\right)& \widehat{B_{21}^z}{\hat{H}}^{-1}\widehat{A_{22}^z}& \widehat{B_{21}^z}E+\widehat{B_{22}^z}\end{array}\right]$ (5)

定理3.2. 设 $A^z,B^z$ 为M-矩阵，

$\det \hat{H}=\det \left\{\left(\widehat{B_{11}^z}+\widehat{A_{22}^z}\right)+\widehat{A_{22}^z}{B}_{13}^r\left(A_{12}^z\widehat{B_{11}^z}-A_{13}^r\widehat{B_{21}^z}\right)-\left(\widehat{A_{21}^z}{A}_{13}^r+\widehat{A_{22}^z}{A}_{23}^r\right)\widehat{B_{21}^z}\right\}\ne 0$

且 $B_{23}^r=O$ ，则当(5)式中的每一个分块为非负矩阵时， $M=A{\oplus }_{k}B$ 为MB-矩阵。

证明：当 $\bar{M}$ 满足上述条件时，由定义2.1知 $\bar{M}$ 为非奇异M-矩阵。再由 $M^z\ge \bar{M}$ 及定理2.1的性质(2)知 $M^z$ 为非奇异M-矩阵，于是由定义2.2知 $M=A{\oplus }_{k}B$ 为MB-矩阵。

下面对 $B_{23}^r\ne O$ 时的情况进行讨论。设 $A=\left[\begin{array}{cc}D& E\\ F& G\end{array}\right]\in R^{n\times n}$ ，其中 $D\in R^{k\times k}$ 且非奇异，矩阵 $A$ 内 $D$ 的Schur补记为 $A/D$ ，即 $A/D=G-F{D}^{-1}E$ 。同样如果 $G$ 为非奇异时， $A/G=D-E{G}^{-1}F$ 。

定理3.4. 设 $A,B$ 为MB-矩阵，若

${\left(\tilde{D}\right)}^{-1}\ge 0$ ， ${\left(\tilde{G}-\tilde{F}{\left(\tilde{D}\right)}^{-1}\tilde{E}\right)}^{-1}\ge 0$

则 $M=A{\oplus }_{k}B$ 为MB-矩阵，其中

$\tilde{D}=A_{11}^z-A_{13}^r{\left(B_{22}^z\right)}^{-1}{B}_{23}^r$ ， $\tilde{E}=A_{12}^z+A_{13}^r{\left(B_{22}^z\right)}^{-1}{B}_{21}^z\le 0$ ，

$\tilde{F}=A_{21}^z-B_{13}^r+B_{12}^z{\left(B_{22}^z\right)}^{-1}{B}_{23}^r-A_{23}^r{\left(B_{22}^z\right)}^{-1}{B}_{23}^r\le 0$ ， $\tilde{G}=A_{22}^z+B_{11}^z-B_{12}^z{\left(B_{22}^z\right)}^{-1}{B}_{21}^z+A_{23}^r{\left(B_{22}^z\right)}^{-1}{B}_{21}^z$

证明：将 $\bar{M}$ 分块为 $\bar{M}=\left[\begin{array}{cc}X& \tilde{Q}\\ \tilde{S}& T\end{array}\right]$ ，其中 $X=\left(\begin{array}{cc}{A}_{11}^z& A_{12}^z\\ A_{21}^z-B_{13}^r& A_{22}^z+B_{11}^z\end{array}\right)$ ， $\tilde{Q}={\left[\begin{array}{cc}-A_{13}^r& B_{12}^z-A_{23}^r\end{array}\right]}^{\text{T}}$ ， $\tilde{S}=\left[\begin{array}{cc}-B_{23}^r& B_{21}^z\end{array}\right]$ ， $T=B_{22}^z$ 。由 $A^z,\begin{array}{c}{B}^z\end{array}$ 为M-矩阵，定义2.2及定理2.1得 $\tilde{Q}\le 0$ ， $T^{-1}\ge 0$ ， $\tilde{S}\le 0$ 。由引理2.1得 ${\left(\bar{M}\right)}^{-1}\ge 0$ 当且仅当 ${\left(\bar{M}/T\right)}^{-1}\ge 0$ ，其中 $\bar{M}/T=X-\tilde{Q}{T}^{-1}\tilde{S}$ ，即

$\bar{M}/T=\left[\begin{array}{cc}{A}_{11}^z-A_{13}^r{\left(B_{22}^z\right)}^{-1}{B}_{23}^r& A_{12}^z+A_{13}^r{\left(B_{22}^z\right)}^{-1}{B}_{21}^z\\ A_{21}^z-B_{13}^r+B_{12}^z{\left(B_{22}^z\right)}^{-1}{B}_{23}^r-A_{23}^r{\left(B_{22}^z\right)}^{-1}{B}_{23}^r& A_{22}^z+B_{11}^z-B_{12}^z{\left(B_{22}^z\right)}^{-1}{B}_{21}^z+A_{23}^r{\left(B_{22}^z\right)}^{-1}{B}_{21}^z\end{array}\right]$

令 $\bar{M}/T=\left[\begin{array}{cc}\tilde{D}& \tilde{E}\\ \tilde{F}& \tilde{G}\end{array}\right]$ ，则当 ${\left(\tilde{D}\right)}^{-1}\ge 0$ 时，由引理2.1得 ${\left(\bar{M}/T\right)}^{-1}\ge 0$ 当且仅当

${\left(\left(\bar{M}/T\right)/\tilde{D}\right)}^{-1}={\left(\tilde{G}-\tilde{F}{\left(\tilde{D}\right)}^{-1}\tilde{E}\right)}^{-1}\ge 0$

从而当 ${\left(\tilde{G}-\tilde{F}{\left(\tilde{D}\right)}^{-1}\tilde{E}\right)}^{-1}\ge 0$ 时， ${\left(\bar{M}\right)}^{-1}\ge 0$ 。又因 $\bar{M}$ 为Z-矩阵，故 $\bar{M}$ 为M-矩阵。由 $M^z\ge \bar{M}$ 且 $M^z$ 为Z-矩阵得， $M^z$ 为M-矩阵，再由定义2.2得 $M=A{\oplus }_{k}B$ 为MB-矩阵。

下面我们将讨论 $A,B$ 均为MB-矩阵且 $A_{22}^z=B_{11}^z$ 的特殊情况，定理3.7给出了它们的子直和仍为MB-矩阵的充分条件。

定理3.7. 设 $A,B$ 为MB-矩阵且 $A_{22}^z=B_{11}^z$ ，若 $C$ 为非奇异M-矩阵，则 $M=A{\oplus }_{k}B$ 为MB-矩阵，其中

$C=\left[\begin{array}{ccc}{A}_{11}^z& A_{12}^z& -A_{13}^r\\ A_{21}^z-B_{13}^r& A_{22}^z& B_{12}^z-A_{23}^r\\ -B_{23}^r& B_{21}^z& B_{22}^z\end{array}\right]$

证明：构造一个Z-矩阵 $T\in R^{n\times n}$

$T=\left[\begin{array}{ccc}{A}_{11}^z& 2A_{12}^z& -A_{13}^r\\ A_{21}^z-B_{13}^r& 2A_{22}^z& B_{12}^z-A_{23}^r\\ -B_{23}^r& 2B_{21}^z& B_{22}^z\end{array}\right]$

则 $T=Cdiag\left(I,2I,I\right)$ 且 $T^{-1}=diag\left(I,\left(1/2\right)I,I\right)C^{-1}\ge O$ ， $T$ 是一个非奇异的M-矩阵。此时

$\bar{M}=\left[\begin{array}{ccc}{A}_{11}^z& A_{12}^z& -A_{13}^r\\ A_{21}^z-B_{13}^r& 2A_{22}^z& B_{12}^z-A_{23}^r\\ -B_{23}^r& B_{21}^z& B_{22}^z\end{array}\right]$

故 $\bar{M}\ge T$ ，于是 $\bar{M}$ 为非奇异的M-矩阵。由 $M^z\ge \bar{M}$ 得 $M^z$ 也为非奇异的M-矩阵。

故由定义2.2得 $M=A{\oplus }_{k}B$ 为MB-矩阵。

例3.3. 设矩阵 $A,B$ 及按定义2.2分裂为

$A=\left[\begin{array}{cccc}4& .& 1& 1\\ .& .& .& .\\ -1& .& 11& -3\\ -1& .& -13& 8\end{array}\right]=A^z+A^r=\left[\begin{array}{ccc}3& 0& 0\\ -1& 11& -3\\ -1& -13& 8\end{array}\right]+\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

$B=\left[\begin{array}{cccc}11& -2& .& -1\\ -13& 8& .& -3\\ .& .& .& .\\ 1& 1& .& 3\end{array}\right]\left[\begin{array}{ccc}11& -3& -1\\ -13& 8& -3\\ 1& 1& 3\end{array}\right]=B^z+B^r=\left[\begin{array}{ccc}11& -3& -1\\ -13& 8& -3\\ 0& 0& 2\end{array}\right]+\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 1& 1& 1\end{array}\right]$

按定理3.7取

$C=\left[\begin{array}{cccc}3& 0& 0& -1\\ -1& 11& -3& -1\\ -1& -13& 8& -3\\ -1& 0& 0& 2\end{array}\right]$

容易验证 $A,B$ 为MB-矩阵和 $C$ 为非奇异M-矩阵，则 $A,B$ 的子直和

$M=A{\oplus }_{2}B=\left[\begin{array}{cccc}4& 1& 1& 0\\ -1& 22& -6& -1\\ -1& -26& 16& -3\\ 0& 1& 1& 3\end{array}\right]$

由定理3.7得 $M=A{\oplus }_{2}B$ 为MB-矩阵。

致谢

由衷地感谢我的导师李耀堂教授在学习和科研上对我的指导和帮助。