非奇异 $H$ -矩阵是一类在计算数学、数学物理、控制论等领域具有广泛应用的特殊矩阵, 其数值判定一直是矩阵计算研究的重要课题.自文[1]给出非奇异 $H$ -矩阵的若干判定条件以来, 众多文献研究了这类问题的改进和推广(见文[2-10]).最近, 文[3]给出了一组非奇异 $H$ -矩阵的新判据, 其中的定理1改进了文[2]中的定理1.本文继续这方面的研究, 建立了一类判别非奇异 $H$ -矩阵的新条件, 改进了文[2]中的定理1, 并通过数值实例说明了本文所给非奇异 $H$ -矩阵的新判据的有效性.另外, 通过实例可见, 本文所给新判据还局部优于文[3]中的定理1.

本文用 $\mathbb{C}^{n\times n}$表示所有 $n\times n$阶复矩阵集合.设 $A=(a_{ij})\in\mathbb{C}^{n\times n}$, 记

$ \begin{equation*} \label{eq:1} \left\{ \begin{aligned} &N=\{1, 2, \cdots, n\}, \ R_i=R_i(A)=\sum\limits_{j\ne i}{|a_{ij}|, ~ i\in N}, \\ & \ N_{1}=\biggl\{{i\in N:0<|a_{ii}|}\leq\frac{R_{i}}{2}\biggr\}, ~ N_{2}=\biggl\{i\in N:\frac{R_{i}}{2}<|a_{ii}|<R_{i}\biggl\}, \\ & \ N_{3}=\{i\in N:0<|a_{ii}|=R_{i}\}, \ N_{4}=\{i\in N:|a_{ii}|>R_{i}\}. \end{aligned} \right. \end{equation*} $

定义1.1^[1]  设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}$, 如果 $|a_{ii}|>R_{i}(A)~(i\in N)$, 则称 $A$为严格对角占优矩阵, 记为 $A\in D$; 若存在正对角阵 $X$, 使得 $AX\in D$, 则称 $A$为广义严格对角占优矩阵(也称 $A$为非奇异 $H$ -矩阵), 记为 $A\in \overline{D}$.

若 $N_{1}\cup N_{2}\cup N_{3}=\emptyset$, 则 $A\in D$; 若 $A\in \overline{D}$, 则 $A$的主对角线元素非零, 且 $A$至少有一个严格对角占优行, 即 $N_{4}\neq\emptyset$.故本文总假设 $N_{1}\cup N_{2}\cup N_{3}$和 $N_{4}$均非空, 且矩阵 $A$的主对角线元素非零.

定义1.2^[4]  设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}$不可约, 如果 $|a_{ii}|\geq R_{i}(A)~(i\in N)$, 且其中至少有一个严格不等式成立, 则称 $A$为不可约对角占优矩阵.

定义1.3^[5]  设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}$, 如果 $|a_{ii}|\geq R_{i}(A)~(i\in N)$, 且其中至少有一个严格不等式成立, 又对每个等式成立的下标 $i$, 都存在非零元素链 $a_{ij_{1}}a_{j_{1}j_{2}}\cdots a_{j_{k-1}j_{k}}\neq0, $使得 $|a_{j_{k}j_{k}}|>R_{j_{k}}(A)$, 则称 $A$为具非零元素链对角占优矩阵.

引理1.1^[4]  设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}$为不可约对角占优矩阵, 则 $A\in \overline{D}.$

引理1.2^[5]  设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}$为具非零元素链对角占优矩阵, 则 $A\in \overline{D}.$

根据文[6]中的引理1, 本文总假定所讨论的矩阵中每一行的非对角元的模和为正.

本文首先引进文[3]中的记号.设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}$, 记

$ \begin{eqnarray*}&& r=\max\limits_{i\in N_{4}}\left(\frac{\sum\limits_{t\in N_{1}}{|a_{it}|}+\sum\limits_{t\in N_{2}}{|a_{it}|}+\sum\limits_{i\in N_{3}}{|a_{it}|}}{|a_{ii}|-\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}}\right), \\ && P_{i}=\sum\limits_{t\in N_{1}}{|a_{it}|}+\sum\limits_{t\in N_{2}}{|a_{it}|}+\sum\limits_{t\in N_{3}}{|a_{it}|}+r\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}~(\forall i\in N_{4}), \\ && x_{i}=\frac{|a_{ii}|}{R_i}~(i\in N_{1}), \ y_{i}=\frac{R_{i}-|a_{ii}|}{R_{i}}~(i\in N_{2}), \ z_{i}=\frac{P_{i}}{|a_{ii}|}~(i \in N_{4}), \ w_{i}=\frac{R_{i}}{|a_{ii}|}~(i \in N_{4}), \\ && h=\max\limits_{i\in N_{4}}\left(\frac{\sum\limits_{t\in N_{1}}{|a_{it}|}+\sum\limits_{t\in N_{2}}{|a_{it}|}+\sum\limits_{t\in N_{3}}{|a_{it}|}}{P_{i}-\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}z_{t}}\right).\end{eqnarray*} $

文[2]和文[3]给出了如下重要结果:

定理1.1^[2]  设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}$, 若

$ \begin{eqnarray}&& |a_{ii}|x_{i}>\sum\limits_{t\in N_{1}, t\neq i}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}+\sum\limits_{t\in N_{4}}{|a_{it}|}w_{t}, \ i\in N_{1}, \end{eqnarray} $

(1.1)

$ \begin{eqnarray} && |a_{ii}|y_{i}>\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}, t\neq i}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}+\sum\limits_{t\in N_{4}}{|a_{it}|}w_{t}, \ i\in N_{2}, \end{eqnarray} $

(1.2)

$ \begin{eqnarray} && |a_{ii}|>\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}, t\neq i}{|a_{it}|}+\sum\limits_{t\in N_{4}}{|a_{it}|}w_{t}, \ i\in N_{3}, \end{eqnarray} $

(1.3)

则 $A\in \overline{D}.$

定理1.2^[3]  设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}$, 若

$ \begin{eqnarray}&& |a_{ii}|x_{i}>\sum\limits_{t\in N_{1}, t\neq i}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}+h\sum\limits_{t\in N_{4}}{|a_{it}|}z_{t}, \ i\in N_{1}, \end{eqnarray} $

(1.4)

$ \begin{eqnarray} && |a_{ii}|y_{i}>\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}, t\neq i}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}+h\sum\limits_{t\in N_{4}}{|a_{it}|}z_{t}, \ i\in N_{2}, \end{eqnarray} $

(1.5)

且 $|a_{ii}|\neq \sum\limits_{t\in N_{3}, t\neq i}{|a_{it}|}\ (i\in N_{3}), $则 $A\in \overline{D}.$

下面将给出非奇异 $H$ -矩阵的一组新判定, 并通过数值实例来说明其有效性.

为了便于叙述, 本文继续引入如下几个新记号.设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}, $记

$ \begin{eqnarray*}&& r'=\max\limits_{i\in N_{4}}\left(\frac{\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}}{|a_{ii}|-\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}}\right), \\ && P'_{i}=\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}+r'\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}~(i\in N_{4}), \\ && z'_{i}=\frac{P'_{i}}{|a_{ii}|}~(i\in N_{4}), \ h'=\max\limits_{i\in N_{4}}\left(\frac{\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}}{P'_{i}-\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}z'_{t}}\right), \\ && F_{i}=\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}, t\neq i}{|a_{it}|}+h'\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}z'_{t}\ (i\in N_{3}\cup N_{4}).\end{eqnarray*} $

定理2.1  设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}$, 若

$ \begin{align} \label{H-w-2.1} &|a_{ii}|x_{i}>\sum\limits_{t\in N_{1}, t\neq i}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}+\sum\limits_{t\in N_{4}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}, \ i\in N_{1}, \end{align} $

(2.1)

$ \begin{align} &|a_{ii}|y_{i}>\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}, t\neq i}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}+\sum\limits_{t\in N_{4}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}, \ i\in N_{2}, \label{H-w-2.2} \end{align} $

(2.2)

且 $|a_{ii}|\neq \sum\limits_{t\in N_{3}, t\neq i}{|a_{it}|}~(i\in N_{3}), $则 $A\in \overline{D}.$

证  显然, $0<x_{i}\leq1/2~(i\in N_{1}), \ 0<y_{i}<1~(i\in N_{2}), \ 0\leq r'<1~(i\in N_{4})$.从而

$ \begin{align}\label{H-w-2.3} r'|a_{ii}|\geq\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}+r'\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}=P'_{i}, \ i\in N_{4}, \end{align} $

(2.3)

即 $0\leq z'_{i}=P'_{i}/|a_{ii}|\leq r'<1~~(i\in N_{4})$, 进而由式(2.3) 可得

$ \frac{\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}}{P'_{i} -\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}z'_{t}}=\frac{P'_{i}-r'\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}}|}{P'_{i} -\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}z'_{t}}\leq1, \ i\in N_{4}. $

故 $0\leq h'\leq1$, 且

$ \begin{align}\label{H-w-2.4} h'P'_{i}\geq\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}+h'\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}z'_{t}, \ i\in N_{4}. \end{align} $

(2.4)

(a.1) 因对任意的 $i\in N_{3}, $有 $|a_{ii}|\neq \sum\limits_{t\in N_{3}, t\neq i}{|a_{it}|}, $故存在充分小的正数 $\varepsilon_{1}, $使得

$ \begin{align}\label{H-w-2.5} |a_{ii}|>\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}, t\neq i}{|a_{it}|}+\sum\limits_{t\in N_{4}}{|a_{it}|}(h'z'_{t}+\varepsilon_{1}), ~ i\in N_{3}. \end{align} $

(2.5)

(a.2) 对任意的 $i\in N_{4}, $由式(2.4) 可得

$ \begin{align}\label{H-w-2.6} &|a_{ii}|(h'z'_{i}+\varepsilon_{1})=h'P'_{i}+\varepsilon_{1}|a_{ii}| >h'P'_{i}+\varepsilon_{1}\sum\limits_{t\in N_{4}, t\neq i}{|a_{it}|}\notag\\ \ge&\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}+\sum\limits_{t\in N_{4}, t\neq i}|a_{it}|(h'z'_{t}+\varepsilon_{1}). \end{align} $

(2.6)

取 $D_{1}=$diag $(d_{1}, d_{2}, \cdots, d_{n}), $ $B=(b_{ij})_{n\times n}=AD_{1}, $其中

$ d_{i}=x_{i}~ (i\in N_{1});\ d_{i}=y_{i}~ (i\in N_{2});\ d_{i}=1~ (i\in N_{3});\ d_{i}=h'z'_{i}+\varepsilon_{1}~ (i\in N_{4}). $

由式(2.5) 和式(2.6) 可知, 一定存在充分小的正数 $\varepsilon_{2}, $使得

$ \begin{align}\label{H-w-2.7} 0<R_{i}(B)+\varepsilon_{2}<|b_{ii}|, \ i\in N_{3}\cup N_{4}. \end{align} $

(2.7)

由式(2.1) 和式(2.2) 可知, 对上述充分小的正数 $\varepsilon_{1}$和 $\varepsilon_{2}, $有

$ \begin{align}\label{H-w-2.8} &|a_{ii}|x_{i}-\sum\limits_{t\in N_{1}, t\neq i}{|a_{it}|}x_{t}-\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}-\sum\limits_{t\in N_{3}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}-\sum\limits_{t\in N_{4}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}\notag\\ >&\sum\limits_{t\in N_{3}\bigcup N_{4}}{\frac{|a_{it}|}{|a_{tt}|}}\biggl(\varepsilon_{1}\sum\limits_{k\in N_{4}, k\neq t}{|a_{tk}|}+\varepsilon_{2}\biggr), \ i\in N_{1}; \end{align} $

(2.8)

$ \begin{align} &|a_{ii}|y_{i}-\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}-\sum\limits_{t\in N_{2}, t\neq i}{|a_{it}|}y_{t}-\sum\limits_{t\in N_{3}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}-\sum\limits_{t\in N_{4}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}\notag\\ >&\sum\limits_{t\in N_{3}\bigcup N_{4}}{\frac{|a_{it}|}{|a_{tt}|}}\biggl(\varepsilon_{1}\sum\limits_{k\in N_{4}, k\neq t}{|a_{tk}|}+\varepsilon_{2}\biggr), \ i\in N_{2}.\label{H-w-2.9} \end{align} $

(2.9)

进一步, 令 $D_{2}=$diag $(d_{1}', d_{2}', \cdots, d_{n}'), \ C=(c_{ij})_{n\times n}=BD_{2}$, 其中

$ d_{i}'=\frac{R_{i}(B)+\varepsilon_{2}}{|b_{ii}|}~ (i\in N_{3}\cup N_{4});\ d_{i}'=1~ (i\in N_{1}\cup N_{2}). $

(b.1) 对任意的 $i\in N_{3}\cup N_{4}, \ $由式(2.7) 可得

$ \begin{align*} R_{i}(C)&=\sum\limits_{t\in N_{3}\cup N_{4}, t\neq i}{|b_{it}|}\frac{R_{t}(B)+\varepsilon_{2}}{|b_{tt}|}+\sum\limits_{t\in N_{1}}{|b_{it}|}+\sum\limits_{t\in N_{2}}{|b_{it}|}\\ &< \sum\limits_{t\in N_{3}\bigcup N_{4}, t\neq i}{|b_{it}|}+\sum\limits_{t\in N_{1}}{|b_{it}|}+\sum\limits_{t\in N_{2}}{|b_{it}|}\\ &=R_{i}(B)={|b_{ii}|}d_{i}'-\varepsilon_{2}<{|b_{ii}|}d_{i}'={|c_{ii}|}. \end{align*} $

(b.2) 对任意的 $i\in N_{1}$, 由式(2.8) 可得

$ \begin{align*} R_{i}(C)&=\sum\limits_{t\in N_{3}\bigcup N_{4}}{|b_{it}|}\frac{R_{t}(B)+\varepsilon_{2}}{|b_{tt}|}+\sum\limits_{t\in N_{1}, t\neq i}{|b_{it}|}+\sum\limits_{t\in N_{2}}{|b_{it}|}\\ &=\sum\limits_{t\in N_{3}}{|a_{it}|}\frac{1}{|a_{tt}|}\biggl[\sum\limits_{k\in N_{1}}{|a_{tk}|}x_{k}+\sum\limits_{k\in N_{2}}{|a_{tk}|}y_{k}+\sum\limits_{k\in N_{3}, k\neq t}{|a_{tk}|}\\ &\ \ \ \ +\sum\limits_{k\in N_{4}}{|a_{tk}|}(h'z'_{k}+\varepsilon_{1})+\varepsilon_{2}\biggr]+\sum\limits_{t\in N_{4}}{|a_{it}|}\frac{1}{|a_{tt}|}\biggl[\sum\limits_{k\in N_{1}}{|a_{tk}|}x_{k}+\sum\limits_{k\in N_{2}}{|a_{tk}|}y_{k}\\ &\ \ \ \ +\sum\limits_{k\in N_{3}}{|a_{tk}|}+\sum\limits_{k\in N_{4}, k\neq t}{|a_{tk}|}(h'z'_{k}+\varepsilon_{1})+\varepsilon_{2}\biggr]+\sum\limits_{t\in N_{1}, t\neq i}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}\\ &=\sum\limits_{t\in N_{3}}{|a_{it}|}\frac{1}{|a_{tt}|}\biggl[F_{t}+\varepsilon_{1}\sum\limits_{k\in N_{4}}{|a_{tk}|} +\varepsilon_{2}\biggr]\\ &\ \ \ \ +\sum\limits_{t\in N_{4}}{|a_{it}|}\frac{1}{|a_{tt}|}\biggl[F_{t}+\varepsilon_{1}\sum\limits_{k\in N_{4}, k\neq t}{|a_{tk}|} +\varepsilon_{2}\biggr]+\sum\limits_{t\in N_{1}, t\neq i}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}\\ &=\sum\limits_{t\in N_{1}, t\neq i}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}+\sum\limits_{t\in N_{4}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}\\ &\ \ \ \ +\varepsilon_{1}\sum\limits_{t\in N_{3}\bigcup N_{4}}\frac{|a_{it}|}{|a_{tt}|}\biggl(\sum\limits_{k\in N_{4}, k\neq t}{|a_{tk}|}\biggr)+\varepsilon_{2}\sum\limits_{t\in N_{3}\bigcup N_{4}}\frac{|a_{it}|}{|a_{tt}|}\\ &<|a_{ii}|x_{i}=|b_{ii}|=|c_{ii}|. \end{align*} $

(b.3) 对任意的 $i\in N_{2}$, 如同(b.2) 的证法, 由式(2.9) 亦可证得 $R_{i}(C)<|c_{ii}|, \ i\in N_{2}$.综上所述, $C=AD_{1}D_{2}\in D, \ $故 $A\in \overline{D}.$证毕.

注  由于

$ 0\leq\frac{F_{i}}{|a_{ii}|}\leq\frac{R_{i}}{|a_{ii}|}=1~ (i\in N_{3}), \ 0\leq \frac{F_{i}}{|a_{ii}|}\leq\frac{R_{i}}{|a_{ii}|}=w_{i}<1~ (i\in N_{4}), $

故定理2.1改进了定理1.1.另外, 定理2.1还局部优于定理1.2 (见例3.1).

由引理1.1-引理1.2及定理2.1的证法, 易得到在不可约和非零元素链下的相应结果.

定理2.2  设 $A=(a_{ij}) \in \mathbb{C}^{n\times n}$, 且

$ |a_{ii}|x_{i}\geq\sum\limits_{t\in N_{1}, t\neq i}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}+\sum\limits_{t\in N_{4}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}, \ i\in N_{1}, $

(2.10)

$ |a_{ii}|y_{i}\geq\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}, t\neq i}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}+\sum\limits_{t\in N_{4}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}, \ i\in N_{2}. $

(2.11)

若 $A$还满足下列条件之一, 则 $A\in \overline{D}$,

(ⅰ) $A$不可约, 且 $N_{3}\neq\emptyset$或式(2.10) 和式(2.11) 的诸式中至少有一个严格不等式成立.

(ⅱ) $N_{1}\cup N_{2}\cup N_{3}-W_{1}-W_{2}\neq\emptyset$, 且对任意的 $i\in W_{1}\cup W_{2}\cup N_{4}$, 存在非零元素链 $a_{is_{1}}a_{s_{1}s_{2}}\cdots a_{s_{k}k}\neq0$, 使得 $k\in [N_{1}\cup N_{2}\cup N_{3}-W_{1}-W_{2}]$, 其中

$ \begin{eqnarray*}&& \overline{N_{3}}=\biggl\{i:|a_{ii}|=\sum\limits_{t\in N_{3}, t\neq i}{|a_{it}|}, \ i\in N_{3}\biggr\}, \\ &&W_{1}=\biggl\{i:|a_{ii}|x_{i}=\sum\limits_{t\in N_{1}, t\neq i}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}\cup N_{4}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}, \ i\in N_{1}\biggr\}\cup\overline{N_{3}}, \\ &&W_{2}=\biggl\{i:|a_{ii}|y_{i}=\sum\limits_{t\in N_{1}}{|a_{it}|}x_{t}+\sum\limits_{t\in N_{2}, t\neq i}{|a_{it}|}y_{t}+\sum\limits_{t\in N_{3}\cup N_{4}}{|a_{it}|}\frac{F_{t}}{|a_{tt}|}, \ i\in N_{2}\biggr\}\cup\overline{N_{3}}.\end{eqnarray*} $

例3.1  设矩阵

$ \begin{equation*} A = \begin{bmatrix} 4 &1& 1 & 0 &{0.5}&{1.5}\\ {1.5}&4& 1 & 1 & 0 &{0.5}\\ {0.5}&0&{1.5}& 0 &{0.5}& 1 \\ 0 &0&{0.5}&{2.5}& 0 &{9.5}\\ 1 &1& 0 & 2 & 20 & 4 \\ 0 &0& 1 & 0 & 5 & 40 \\ \end{bmatrix}, \end{equation*} $

则 $N_{1}=\{4\}, \ N_{2}=\{3\}, \ N_{3}=\{1, 2\}, \ N_{4}=\{5, 6\}$.由于

$ |a_{44}|x_{4}=\frac{25}{40}<\frac{42}{40}=|a_{43}|y_{3}+|a_{41}|+|a_{42}|+|a_{45}|w_{5}+|a_{46}|w_{6}, $

故矩阵 $A$不满足定理1.1 (即文[2]中定理1) 的条件.又因为

$ |a_{44}|x_{4}=\frac{200}{320}<\frac{211}{320}=|a_{43}|y_{3}+|a_{41}|+|a_{42}|+|a_{45}|z_{5}+|a_{46}|z_{6}, $

故矩阵 $A$不满足定理1.2 (即文[3]中定理1) 的条件.而

$ \begin{eqnarray*}&& |a_{44}|x_{4} =\frac{1600}{2560}>\frac{947}{2560}\\ &=&|a_{43}|y_{3}+|a_{41}|\frac{F_{1}}{|a_{11}|}+|a_{42}| \frac{F_{2}}{|a_{22}|}+|a_{45}|\frac{F_{5}}{|a_{55}|} +|a_{46}|\frac{F_{6}}{|a_{66}|}, \\ && |a_{33}|y_{3} =\frac{76800}{204800}>\frac{22107}{204800}\\ &=&|a_{34}|x_{4}+|a_{31}|\frac{F_{1}}{|a_{11}|}+|a_{32}| \frac{F_{2}}{|a_{22}|}+|a_{35}|\frac{F_{5}}{|a_{55}|} +|a_{36}|\frac{F_{6}}{|a_{66}|}, \\ && |a_{11}|=4\neq\sum\limits_{t\in N_{3}, t\neq 1}{|a_{1t}|}=1, \ |a_{22}|=4\neq\sum\limits_{t\in N_{3}, t\neq 2}{|a_{2t}|}=1.5, \end{eqnarray*} $

故矩阵 $A$满足定理2.1的条件, 即 $A\in \overline{D}.\ $当取 $X=$diag $\Bigl(\frac{1}{2}, \frac{1}{2}, \frac{1}{4}, \frac{1}{4}, \frac{8}{91}, \frac{4}{91}\Bigr)$时, 有 $AX\in D.$