B-矩阵线性互补问题解的误差界的新估计式

A New Estimator of Error Bounds for B-Matrix Linear Complementary Problems

摘要: 利用严格对角占优M-矩阵逆矩阵的无穷大范数范围，综合运用不等式放缩技巧，得到了B-矩阵线性互补问题解的误差界的一个新估计式。理论证明新估计式改进了现有文献的有关结果，数值例子说明了新估计式的可行性和有效性。

Abstract: By using the infinite norm range of the strictly diagonally dominant M-matrix inverse matrix, a new estimator of the error bounds of the solutions of B-matrix linear complementarity problems is ob-tained by using the inequality reduction technique. It is proved that the new estimation improves the results of the existing literature. Numerical examples show the feasibility and effectiveness of the new estimation.

1. 引言

线性互补问题在许多领域具有广泛的应用，例如二次规划问题、市场均衡问题、最优控制问题等 [1] [2] [3]。线性互补问题的数学模型为求 $x\in R^n$，满足

$x\ge 0,Mx+q\ge 0,{\left(Mx+q\right)}^{\text{T}}x=0,$

简记为 $LCP\left(M,q\right)$。其中 $M=\left(m_{ij}\right)\in R^{n\times n}$ 为给定的实矩阵， $q\in R^n$ 为给定的实向量。

线性互补问题的解很大程度上取决于矩阵M的性质。若矩阵M为P-矩阵，则线性互补问题有唯一解。2006年，文 [4] 给出结果：设M是P-矩阵，则有

${\Vert x-x^{\ast }\Vert }_{\infty }\le \underset{d\in {\left[0,1\right]}^n}{\max }{\Vert {\left(I-D+DM\right)}^{-1}\Vert }_{\infty }{\Vert r\left(x\right)\Vert }_{\infty },$

其中 $x^{\ast }$ 是 $LCP\left(M,q\right)$ 的解， $r\left(x\right)=\min \left\{x,Mx+q\right\},D=diag\left(d_1,d_2,\cdots ,d_n\right),\left(0\le d_i\le 1\right)$。近年来，国内外学者在此基础上给出了很多特殊矩阵类线性互补问题解的误差界估计式 [5] - [15]。本文将继续讨论P-矩阵的子类B-矩阵线性互补问题解的误差界，给出B-矩阵线性互补问题解的误差界的一个新估计式，并通过理论分析和数值实例说明新估计式的有效性和可行性。

2. 预备知识

令 $N^{+}$ 表示全体正整数的集合， $A=\left(a_{ij}\right)\in R^{n\times n}$，若

$\left|a_{ii}\right|>{\displaystyle \underset{j=1,j\ne i}{\overset{n}{\sum }}\left|a_{ij}\right|},\forall i\in N=\left\{1,2,\cdots ,n\right\}$

则称A为严格对角占优矩阵；若 $a_{ij}\le 0,i\ne j,i,j\in N$，则称A为Z-矩阵 [11]；若A为Z-矩阵且 $A^{-1}\ge 0$，则称A为M-矩阵 [11]。

定义1 [3] 设 $A=\left(a_{ij}\right)\in R^{n\times n},x\in R^n$，若A满足 $\max x_i{\left(Ax\right)}_i\ge 0,\forall x\ne 0$ 则称A为P-矩阵。

定义2 [12] 设 $A=\left(a_{ij}\right)\in R^{n\times n}$，若对任意的 $i,j\in N$

${\displaystyle \underset{k\in N}{\sum }{a}_{ik}}>0,\frac{1}{n}\left({\displaystyle \underset{k\in N}{\sum }{a}_{ik}}\right)>a_{ij},j\ne i.$

则称A为B-矩阵。

2009年，文 [6] 给出：设 $M=\left(m_{ij}\right)\in R^{n\times n}$ 是B-矩阵， $M=B^{+}+C$，这里

$B^{+}=\left(b_{ij}\right)=\left(\begin{array}{ccc}{m}_{11}-r_1^{+}& \cdots & m_{1n}-r_1^{+}\\ ⋮& \ddots & ⋮\\ m_{n1}-r_n^{+}& \cdots & m_{nn}-r_n^{+}\end{array}\right),C=\left(\begin{array}{ccc}{r}_1^{+}& \cdots & r_1^{+}\\ ⋮& \ddots & ⋮\\ r_n^{+}& \cdots & r_n^{+}\end{array}\right),$ (1)

$r_i^{+}=\max \left\{0,m_{ij}|j\ne i\right\}$ 则

$\underset{d\in {\left[0,1\right]}^n}{\max }\Vert {\left(I-D+DM\right)}^{-1}\Vert \le \frac{n-1}{\min \left\{\beta ,1\right\}}$ (2)

其中 $\beta =\underset{i\in N}{\min }\left\{{\beta }_i\right\}$，且 ${\beta }_i=b_{ii}-{\displaystyle \underset{j\ne i}{\sum }\left|b_{ij}\right|}$。

2016年，文 [9] 给出了优于(2)式的新估计式：设 $M=\left(m_{ij}\right)\in R^{n\times n}$ 是B-矩阵， $M=B^{+}+C$，其中 $B^{+}=\left(b_{ij}\right)$ 形如(1)式，则有

$\underset{d\in {\left[0,1\right]}^n}{\max }\Vert {\left(I-D+DM\right)}^{-1}\Vert \le {\displaystyle \underset{i=1}{\overset{n}{\sum }}\frac{n-1}{\min \left\{{\bar{\beta }}_i,1\right\}}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\left(1+\frac{1}{{\bar{\beta }}_j}{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}\right)}$ (3)

其中

${\bar{\beta }}_i=b_{ii}-{\displaystyle \underset{j=i+1}{\overset{n}{\sum }}\left|b_{ij}\right|l_i\left(B^{+}\right)},$ (4)

$l_k=\underset{k\le i\le n}{\max }\left\{\frac{1}{\left|b_{ii}\right|}{\displaystyle \underset{j=k,\ne i}{\overset{n}{\sum }}\left|b_{ij}\right|}\right\},{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\left(1+\frac{1}{{\bar{\beta }}_j}{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}\right)}=1,i=1$

2016年，文 [10] 又给出了新的估计式：设 $M=\left(m_{ij}\right)\in R^{n\times n}$ 是B-矩阵，且 $M=B^{+}+C$，其中 $B^{+}=\left(b_{ij}\right)$ 形如(1)式，则有：

$\underset{d\in {\left[0,1\right]}^n}{\max }\Vert {\left(I-D+DM\right)}^{-1}\Vert \le {\displaystyle \underset{i=1}{\overset{n}{\sum }}\frac{n-1}{\min \left\{{\tilde{\beta }}_i,1\right\}}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\left(\frac{b_{jj}}{{\tilde{\beta }}_j}\right)}$ (5)

其中 ${\tilde{\beta }}_i=b_{ii}-{\displaystyle \underset{j=i+1}{\overset{i-1}{\sum }}\left|b_{ij}\right|}>0$，且 ${\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{b_{jj}}{{\tilde{\beta }}_j}}=1\left(i=1\right)$。 (6)

3. 主要结果

引理1 [9] 设 $\gamma >0$ 和 $\eta \ge 0$，则对任意的 $x\in \left[0,1\right]$，

$\frac{1}{1-x+\gamma x}\le \frac{1}{\min \left\{\gamma ,1\right\}},\frac{\eta x}{1-x+\gamma x}\le \frac{\eta }{\gamma }$

引理2 [10] 设 $A=\left(a_{ij}\right)\in R^{n\times n}$ 且 $a_{ii}>{\displaystyle \underset{j=i+1}{\overset{n}{\sum }}\left|a_{ij}\right|},\forall i\in N$，则对任意的 $x_i\in \left[0,1\right],i\in N$

$\frac{1-x_i+a_{ii}{x}_i}{1-x_i+a_{ii}{x}_i-{\displaystyle \underset{j=i+1}{\overset{n}{\sum }}\left|a_{ij}\right|x_i}}\le \frac{a_{ii}}{a_{ii}-{\displaystyle \underset{j=i+1}{\overset{n}{\sum }}\left|a_{ij}\right|}}.$

引理3 [10] $H:={\left(h_1,h_2,\cdots ,h_n\right)}^{\text{T}}e,e=\left(1,1,\cdots ,1\right),h_1,h_2,\cdots ,h_n\ge 0,{\Vert {\left(I-H\right)}^{-1}\Vert }_{\infty }\le n-1$

引理4 [16] 设 $A=\left(a_{ij}\right)\in R^{n\times n}$ 是严格对角占优M-矩阵，则有

${\Vert A^{-1}\Vert }_{\infty }\le \max \left\{{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{1}{a_{ii}-{\displaystyle \underset{i<k\le n}{\sum }\left|a_{ik}\right|m_{ki}}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{u_j}{1-u_j{p}_j}}\right],{\displaystyle \underset{i=1}{\overset{n}{\sum }}\frac{p_i}{a_{ii-}{\displaystyle \underset{i<k\le n}{\sum }\left|a_{ik}\right|m_{ki}}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{1}{1-u_j{p}_j}}}}\right\}$

为叙述方便起见，本文引入以下符号：

$u_i\left(A\right)=\frac{1}{\left|a_{ii}\right|}{\displaystyle \underset{j=i+1}{\overset{n}{\sum }}\left|a_{ij}\right|},b_k\left(A\right)=\max \left\{\frac{{\displaystyle \underset{j\ne i+k,k\le j\le n}{\sum }\left|a_{i+k,j}\right|}}{\left|a_{i+k,i+k}\right|},i=1,2,\cdots ,n-k\right\},k=1,2,\cdots ,n,$

$p_k\left(A\right)=\max \left\{\frac{\left|a_{i+k,k}\right|+{\displaystyle \underset{h=k+1,h\ne i+k}{\overset{n}{\sum }}\left|a_{i+k,h}\right|b_k\left(A\right)}}{\left|a_{i+k,i+k}\right|},i=1,2,\cdots ,n-k\right\},k=1,2,\cdots ,n,$

$d_k\left(A\right)=\max \left\{\frac{{\displaystyle \underset{j\ne i+k-1,k\le j\le n}{\overset{n}{\sum }}\left|a_{i+k-1,j}\right|}}{\left|a_{i+k-1,i+k-1}\right|},i=1,2,\cdots ,n-k+1\right\},k=1,2,\cdots ,n,$

$r_{li}\left(A\right)=\frac{\left|a_{li}\right|}{\left|a_{ll}\right|-{\displaystyle \underset{k\ne l,i}{\sum }\left|a_{lk}\right|}},l\ne i,r_i\left(A\right)=\underset{l\ne i}{\max }\left\{r_{li}\left(A\right)\right\},i\in N,$

$m_{ji}\left(A\right)=\frac{\left|a_{ji}\right|+{\displaystyle \underset{k\ne j,i}{\sum }\left|a_{jk}\right|r_i\left(A\right)}}{\left|a_{jj}\right|},j\ne i,i\in N.$

3.1. 定理1

设 $M=\left(m_{ij}\right)\in R^{n\times n}$ 是B-矩阵，且 $M=B^{+}+C,B=\left(b_{ij}\right)$ 形如(1)式，则有：

$\underset{d\in {\left[0,1\right]}^n}{\max }{\Vert \left(I-D+DM\right){}^{-1}\Vert }_{\infty }\le \max \left\{{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{n-1}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}}{{\alpha }_j\left(B^{+}\right)}}\right],{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{\left(n-1\right)p_i\left(B^{+}\right)}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{\left|b_{jj}\right|}{{\alpha }_j\left(B^{+}\right)}}\right]}}\right\}.$ (7)

其中，

${\hat{\beta }}_i=\left|b_{ii}\right|-{\displaystyle \underset{k\ne i}{\sum }\left|b_{ik}\right|m_{ki}\left(B^{+}\right)},{\alpha }_i\left(B^{+}\right)=\left|b_{ii}\right|-{\displaystyle \underset{k=i+1}{\overset{n}{\sum }}\left|b_{ik}\right|p_i\left(B\right)}.$ (8)

证明：令 $M_D=I-D+DM$。则有

$M_D=I-D+DM=I-D+D\left(B^{+}+C\right)=B_D^{+}+C_D.$

其中 $B_D^{+}=I-D+D{B}^{+},C_D=DC$，由文献 [6] 中定理2.2的证明可知 $B_D^{+}$ 是主对角元素为正的严格对角占优M-矩阵，于是由引理3可得：

${\Vert M_D^{-1}\Vert }_{\infty }\le {\Vert {\left(I+{\left(B_D^{+}\right)}^{-1}{C}_D\right)}^{-1}\Vert }_{\infty }{\Vert {\left(B_D^{+}\right)}^{-1}\Vert }_{\infty }\le \left(n-1\right){\Vert {\left(B_D^{+}\right)}^{-1}\Vert }_{\infty }.$ (9)

由引理4可得：

${\Vert {\left(B_D^{+}\right)}^{-1}\Vert }_{\infty }\le \max \left\{{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{1}{1-d_i+b_{ii}{d}_i-{\displaystyle \underset{i<k\le n}{\sum }\left|b_{ik}\right|d_i{m}_{ki}}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{u_j}{1-u_j{p}_j}}\right],{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{p_i}{1-d_i+b_{ii}{d}_i-{\displaystyle \underset{i<k\le n}{\sum }\left|b_{ik}\right|d_i{m}_{ki}}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{1}{1-u_j{p}_j}}\right]}}\right\}.$ (10)

由引理1可得：

$b_k\left(B_D^{+}\right)=\max \left\{\frac{{\displaystyle \underset{j\ne i+k,k\le j\le n}{\sum }\left|b_{i+k,j}\right|d_{i+k}}}{1-d_{i+k}+b_{i+k,i+k}{d}_{i+k}}\right\}\le \max \left\{\frac{{\displaystyle \underset{j\ne i+k,k\le j\le n}{\sum }\left|b_{i+k,j}\right|}}{\left|b_{i+k,i+k}\right|}\right\}=b_k\left(B^{+}\right),$

$\begin{array}{c}{p}_k\left(B_D^{+}\right)=\max \left\{\frac{\left|b_{i+k,k}\right|d_{i+k}+{\displaystyle \underset{h=k+1,h\ne i+k}{\overset{n}{\sum }}\left|b_{i+k,h}\right|d_{i+k}{b}_k\left(B_D^{+}\right)}}{1-d_{i+k}+b_{i+k,i+k}{d}_{i+k}}\right\}\\ \le \max \left\{\frac{\left|b_{i+k,k}\right|+{\displaystyle \underset{h=k+1,h\ne i+k}{\overset{n}{\sum }}\left|b_{i+k,h}\right|b_k\left(B_D^{+}\right)}}{\left|b_{i+k,i+k}\right|}\right\}=p_k\left(B^{+}\right),\end{array}$

$u_i\left(B_D^{+}\right)=\frac{1}{1-d_i+b_{ii}{d}_i}{\displaystyle \underset{j=i+1}{\overset{n}{\sum }}\left|b_{ij}\right|d_i}\le \frac{1}{\left|b_{ii}\right|}{\displaystyle \underset{j=i+1}{\overset{n}{\sum }}\left|b_{ij}\right|}=u_i\left(B^{+}\right),$

$r_{li}\left(B_D^{+}\right)=\frac{\left|b_{li}\right|d_l}{\left(1-d_l+b_{ll}{d}_l\right)-{\displaystyle \underset{k\ne l,i}{\sum }\left|b_{lk}\right|d_l}}\le \frac{\left|b_{li}\right|}{\left|b_{ll}\right|-{\displaystyle \underset{k\ne l,i}{\sum }\left|b_{lk}\right|}}=r_{li}\left(B^{+}\right),$

$r_i\left(B_D^{+}\right)=\underset{l\ne i}{\max }\left\{r_{li}\left(B_D^{+}\right)\right\}\le \underset{l\ne i}{\max }\left\{r_{li}\left(B^{+}\right)\right\}=r_i\left(B^{+}\right),$

$m_{ji}\left(B_D^{+}\right)=\frac{\left|b_{ji}\right|d_j+{\displaystyle \underset{k\ne j,i}{\sum }\left|b_{jk}\right|d_j{r}_i\left(B_D^{+}\right)}}{1-d_j+b_{jj}{d}_j}\le \frac{\left|b_{ji}\right|+{\displaystyle \underset{k\ne j,i}{\sum }\left|b_{jk}\right|r_i\left(B_D^{+}\right)}}{\left|b_{jj}\right|}=m_{ji}\left(B^{+}\right),$

进而，由引理2可得：

$\frac{1}{\left(1-d_i+b_{ii}{d}_i\right)-{\displaystyle \underset{i<k\le n}{\sum }\left|b_{ik}\right|d_i{m}_{ki}\left(B_D^{+}\right)}}\le \frac{1}{\min \left\{\left|b_{ii}\right|-{\displaystyle \underset{i<k\le n}{\sum }\left|b_{ik}\right|m_{ki}\left(B^{+}\right)},1\right\}}=\frac{1}{\min \left\{{\hat{\beta }}_i,1\right\}},$

$\frac{p_i\left(B_D^{+}\right)}{\left(1-d_i+b_{ii}{d}_i\right)-{\displaystyle \underset{i<k\le n}{\sum }\left|b_{ik}\right|d_i{m}_{ki}\left(B_D^{+}\right)}}\le \frac{p_i\left(B^{+}\right)}{\min \left\{{\hat{\beta }}_i,1\right\}},$

$\begin{array}{c}\frac{1}{1-u_j\left(B_D^{+}\right)p_j\left(B_D^{+}\right)}=\frac{1-d_j+b_{jj}{d}_j}{1-d_j+b_{jj}{d}_j-{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|d_j{p}_j\left(B_D^{+}\right)}}\\ \le \frac{\left|b_{jj}\right|}{\left|b_{jj}\right|-{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|p_j\left(B^{+}\right)}}=\frac{\left|b_{jj}\right|}{{\alpha }_j\left(B^{+}\right)},\end{array}$

$\begin{array}{c}\frac{u_j\left(B_D^{+}\right)}{1-u_j\left(B_D^{+}\right)p_j\left(B_D^{+}\right)}=\frac{{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|d_j}}{\left(1-d_j+b_{jj}{d}_j\right)-{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|d_j{p}_j\left(B_D^{+}\right)}}\\ \le \frac{{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}}{\left|b_{jj}\right|-{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|p_j\left(B^{+}\right)}}=\frac{{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}}{{\alpha }_j\left(B^{+}\right)},\end{array}$

由此可得：

${\Vert {\left(B_D^{+}\right)}^{-1}\Vert }_{\infty }\le \max \left\{{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{1}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}}{{\alpha }_j\left(B^{+}\right)}}\right],{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{p_i\left(B^{+}\right)}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{\left|b_{jj}\right|}{{\alpha }_j\left(B^{+}\right)}}\right]}}\right\},$ (11)

因此，由(10)、(11)式可得(7)式成立。

接下来，对(3)式、(5)式及(7)式进行比较。

3.2. 定理2

设 $M=\left(m_{ij}\right)\in R^{n\times n}$ 是B-矩阵，且 $M=B^{+}+C$，其中 $B^{+}=\left(b_{ij}\right)$ 形如(1)式，则有：

$\begin{array}{l}\max \left\{{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{n-1}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}}{{\alpha }_j\left(B^{+}\right)}}\right],{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{\left(n-1\right)p_i\left(B^{+}\right)}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{\left|b_{jj}\right|}{{\alpha }_j\left(B^{+}\right)}}\right]}}\right\}\\ \le {\displaystyle \underset{i=1}{\overset{n}{\sum }}\frac{n-1}{\min \left\{{\bar{\beta }}_i,1\right\}}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\left[1+\frac{1}{{\bar{\beta }}_j}{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}\right]}\le {\displaystyle \underset{i=1}{\overset{n}{\sum }}\frac{n-1}{\min \left\{{\tilde{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{\left|b_{jj}\right|}{{\tilde{\beta }}_j}.}}\end{array}$ (12)

其中 ${\hat{\beta }}_i$， ${\alpha }_i\left(B^{+}\right)$， ${\bar{\beta }}_i$ 及 ${\tilde{\beta }}_i$ 分别如(8)式，(4)式及(6)式所示。

证明：因为 $B^{+}$ 为具有正主对角元的严格对角占优矩阵，因此对任意的 $j\in N$，有

$0\le r_j\left(B^{+}\right)<1,0\le l_j\left(B^{+}\right)<1,0\le p_j\left(B^{+}\right)<1.$

且对任意的 $j=1,2,\cdots ,n-1$，

$\begin{array}{c}{m}_{ki}\left(B^{+}\right)-l_{ki}\left(B^{+}\right)=\frac{\left|b_{ki}\right|+{\displaystyle \underset{j=i+1,\ne k}{\overset{n}{\sum }}\left|b_{kj}\right|r_i\left(B^{+}\right)}}{\left|b_{kk}\right|}-\frac{\left|b_{ki}\right|+{\displaystyle \underset{j=i+1,\ne k}{\overset{n}{\sum }}\left|b_{kj}\right|}}{\left|b_{kk}\right|}\\ =\frac{{\displaystyle \underset{j=i+1,\ne k}{\overset{n}{\sum }}\left|b_{kj}\right|r_i\left(B^{+}\right)}-{\displaystyle \underset{j=i+1,\ne k}{\overset{n}{\sum }}\left|b_{kj}\right|}}{\left|b_{kk}\right|}\le 0,\end{array}$

则有 ${\hat{\beta }}_i\ge {\bar{\beta }}_i,\frac{1}{\min \left\{{\hat{\beta }}_i,1\right\}}\le \frac{1}{\min \left\{{\bar{\beta }}_i,1\right\}}.$ (13)

$\begin{array}{c}{p}_k\left(B^{+}\right)-l_k\left(B^{+}\right)=\frac{\left|b_{i+k,k}\right|+{\displaystyle \underset{h=k+1,\ne i+k}{\overset{n}{\sum }}\left|b_{i+k,h}\right|b_k}}{\left|b_{i+k,i+k}\right|}-\frac{\left|b_{i+k,k}\right|+{\displaystyle \underset{h=k+1,\ne i+k}{\overset{n}{\sum }}\left|b_{i+k,h}\right|}}{\left|b_{i+k,i+k}\right|}\\ =\frac{{\displaystyle \underset{h=k+1,\ne i+k}{\overset{n}{\sum }}\left|b_{i+k,h}\right|b_k}-{\displaystyle \underset{h=k+1,\ne i+k}{\overset{n}{\sum }}\left|b_{i+k,h}\right|}}{\left|b_{i+k,i+k}\right|}\le 0,\end{array}$

则有 ${\alpha }_i\left(B^{+}\right)\ge {\bar{\beta }}_i.$ (14)

由(13)式及(14)式可得

${\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{n-1}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}}{{\alpha }_j\left(B^{+}\right)}}\right]}\le {\displaystyle \underset{i=1}{\overset{n}{\sum }}\frac{n-1}{\min \left\{{\bar{\beta }}_i,1\right\}}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\left[1+\frac{1}{{\bar{\beta }}_i}{\displaystyle \underset{k=j+1}{\overset{n}{\prod }}\left|b_{jk}\right|}\right]},$ (15)

$\begin{array}{c}\frac{\left|b_{jj}\right|}{{\alpha }_j\left(B^{+}\right)}\le \frac{\left|b_{jj}\right|}{{\bar{\beta }}_j}=\frac{\left|b_{jj}\right|-{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}+{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}}{{\alpha }_j\left(B^{+}\right)}\\ =1+\frac{1}{{\bar{\beta }}_j}{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|l_j\left(B^{+}\right)}\le 1+\frac{1}{{\bar{\beta }}_j}{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|},\end{array}$ (16)

由(15)式及(16)式可得

${\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{\left(n-1\right)}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{\left|b_{jj}\right|}{{\alpha }_j\left(B^{+}\right)}}\right]\le {\displaystyle \underset{i=1}{\overset{n}{\sum }}\frac{n-1}{\min \left\{{\bar{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\left[1+\frac{1}{{\bar{\beta }}_i}{\displaystyle \underset{k=j+1}{\overset{n}{\prod }}\left|b_{jk}\right|}\right]}},}$ (17)

综上，由(15)式及(17)式可得

$\begin{array}{l}\max \left\{{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{n-1}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}}{{\alpha }_j\left(B^{+}\right)}}\right],{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{\left(n-1\right)p_i\left(B^{+}\right)}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{\left|b_{jj}\right|}{{\alpha }_j\left(B^{+}\right)}}\right]}}\right\}\\ \le {\displaystyle \underset{i=1}{\overset{n}{\sum }}\frac{n-1}{\min \left\{{\bar{\beta }}_i,1\right\}}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\left[1+\frac{1}{{\bar{\beta }}_j}{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}\right]}.\end{array}$ (18)

${\bar{\beta }}_i\ge {\tilde{\beta }}_i,\frac{1}{\min \left\{{\bar{\beta }}_i,1\right\}}\le \frac{1}{\min \left\{{\tilde{\beta }}_i,1\right\}},$

因此，

$1+\frac{1}{{\bar{\beta }}_j}{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}\le 1+\frac{1}{{\tilde{\beta }}_j}{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}=\frac{{\tilde{\beta }}_j+{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}}{{\tilde{\beta }}_j}=\frac{\left|b_{jj}\right|}{{\tilde{\beta }}_j},$ (19)

由(18)式及(19)式可得

$\begin{array}{l}\max \left\{{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{n-1}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}}{{\alpha }_j\left(B^{+}\right)}}\right],{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left[\frac{\left(n-1\right)p_i\left(B^{+}\right)}{\min \left\{{\hat{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{\left|b_{jj}\right|}{{\alpha }_j\left(B^{+}\right)}}\right]}}\right\}\\ \le {\displaystyle \underset{i=1}{\overset{n}{\sum }}\frac{n-1}{\min \left\{{\bar{\beta }}_i,1\right\}}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\left[1+\frac{1}{{\bar{\beta }}_j}{\displaystyle \underset{k=j+1}{\overset{n}{\sum }}\left|b_{jk}\right|}\right]}\le {\displaystyle \underset{i=1}{\overset{n}{\sum }}\frac{n-1}{\min \left\{{\tilde{\beta }}_i,1\right\}}{\displaystyle \underset{j=1}{\overset{i-1}{\prod }}\frac{\left|b_{jj}\right|}{{\tilde{\beta }}_j}.}}\end{array}$

4. 数值算例

例1. 考虑B-矩阵

$M=\left(\begin{array}{cccc}1.3& -0.1& 0.3& 0\\ -0.3& 1.4& -0.1& 0.4\\ 0.4& 0.2& 1.4& 0.3\\ 0.5& 0.5& 0.3& 1.5\end{array}\right).$

$M=B^{+}+C$，其中

$B^{+}=\left(\begin{array}{cccc}1& -0.4& 0& -0.3\\ -0.1& 1& -0.5& 0\\ 0& -0.2& 1& -0.1\\ 0& 0& -0.2& 1\end{array}\right).$

由(3)式可得

$\underset{d\in {\left[0,1\right]}^n}{\max }{\Vert \left(I-D+DM\right){}^{-1}\Vert }_{\infty }\le 40.5506,$

由(5)式可得

$\underset{d\in {\left[0,1\right]}^n}{\max }{\Vert \left(I-D+DM\right){}^{-1}\Vert }_{\infty }\le 74.4444,$

由(7)式可得

$\underset{d\in {\left[0,1\right]}^n}{\max }{\Vert \left(I-D+DM\right){}^{-1}\Vert }_{\infty }\le \mathrm{9.7961.}$

可见，(7)式优于(3)式和(5)式。

例2. 考虑B-矩阵 [9]

$M_k=\left(\begin{array}{cccc}1.5& 0.5& 0.4& 0.5\\ -0.1& 1.7& 0.7& 0.6\\ 0.8& -0.1\frac{k}{k+1}& 1.8& 0.7\\ 0& 0.7& 0.8& 1.8\end{array}\right).$

$M_k=B_k^{+}+C_k$，其中

$B_k^{+}=\left(\begin{array}{cccc}1& 0& -0.1& 0\\ -0.8& 1& 0& -0.1\\ 0& -0.1\frac{k}{k+1}-0.8& 1& -0.1\\ -0.8& -0.1& 0& 1\end{array}\right).$

由(2)式可得

$\underset{d\in {\left[0,1\right]}^n}{\max }{\Vert \left(I-D+DM\right){}^{-1}\Vert }_{\infty }\le \frac{3}{\min \left\{\beta ,1\right\}}=30\left(k+1\right),$

当 $k\to +\infty$ 时， $30\left(k+1\right)\to +\infty$，因此该数值结果会趋于正无穷。

由(3)式可得

$\underset{d\in {\left[0,1\right]}^n}{\max }{\Vert \left(I-D+DM\right){}^{-1}\Vert }_{\infty }\le 14.8064,$

由(5)式可得

$\underset{d\in {\left[0,1\right]}^n}{\max }{\Vert \left(I-D+DM\right){}^{-1}\Vert }_{\infty }\le 15.2675,$

由(7)式可得

$\underset{d\in {\left[0,1\right]}^n}{\max }{\Vert \left(I-D+DM\right){}^{-1}\Vert }_{\infty }\le \mathrm{3.6762.}$

由此可知，(7)式优于(2)式、(3)式和(5)式。

由数值算例的结果可知，定理1中的误差界新估计式是可行的、有效的，改进了文献 [9] [10] 中的结果。

5. 结论

理论证明本文所得B-矩阵线性互补问题解的误差界新估计式优于文献 [9] [10] 中的结果，数值算例亦说明了本文所得新估计式的有效性和可行性。

NOTES

^*通讯作者。