广义严格α-链对角占优矩阵的细分迭代新判据

A New Criterion for Subdivision Iteration of Generalized Strictly α-Chain Diagonally Dominant Matrices

摘要: 广义严格对角占优矩阵在经济价值模型矩阵和反网络分析的系数矩阵以及最优化的线性互补等诸多领域中有着广泛的实际应用。本文依据α-链对角占优矩阵与广义严格对角占优矩阵的关系，通过不等式放缩技巧以及对矩阵行、列指标集进行细分，引入新的迭代因子，给出了一组判定广义严格α-链对角占优矩阵的细分迭代新判据。该判定条件推广和改进了已有的结果，并用数值算例说明了改进后的有效性。

Abstract: Generalized strictly diagonally dominant matrices are widely used in many fields such as economic value model matrix, inverse network analysis coefficient matrix and optimization of linear comple-mentarity. According to the relationship between α-chain diagonally dominant matrix and general-ized strictly diagonally dominant matrix, this paper presents a set of new criteria for subdivision it-eration of generalized strictly α-chain diagonally dominant matrix by means of inequality scaling and subdivision of matrix index set of row and column, and introduces new iteration factors. The results are generalized and improved by this criterion, and the effectiveness of the improved re-sults is illustrated by numerical examples.

1. 引言

广义严格对角占优矩阵(也称非奇异H-矩阵)是一类活跃在计算数学和矩阵理论中的特殊矩阵，不仅在学术上还是在实际应用中都占据着重要地位。如何高效判定一个矩阵是否为广义严格对角占优矩阵，一直是国内外许多数学工作者关注的热点话题之一。近年来国内外许多数学工作者相继提出了一些判定方法卓有成效 [1] - [14]，文献 [1] 根据广义严格α-链对角占优矩阵的性质，以及引入迭代因子，给出判定广义严格α-链对角占优矩阵的迭代判定条件。文献 [4] 通过对矩阵的行、列指标集作划分，根据矩阵自身元素、行和及列和，构造相应的正对角矩阵，得到一组广义严格对角占优矩阵的新的判定准则。在文献 [5] [6] 中，均利用矩阵指标集N的自由的k-级划分给出广义严格对角占优矩阵的判定条件。本文针对文献 [1] 的主要结果，根据α-链对角占优矩阵与广义严格对角占优矩阵的关系，利用矩阵指标集m-级细分的思想，结合不等式放缩技巧，对迭代因子进一步压缩，得到一组更小的正对角因子，给出了判定广义严格α-链对角占优矩阵的细分迭代新判据，推广和改进了已有的结果，并用数值算例验证说明了这一点。

本文采用如下记号和定义：

用 $C^{n\times n}\left(R^{n\times n}\right)$ 表示n阶复(实)矩阵的集合，设 $A=\left(a_{ij}\right)\in C^{n\times n}$， $N=\left(1,2,\cdots ,n\right)$， $\alpha \in \left(0,1\right]$，记

$R_i=R_i\left(A\right)={\displaystyle \underset{j\ne i}{\sum }\left|a_{ij}\right|},C_i=C_i\left(A\right)={\displaystyle \underset{j\ne i}{\sum }\left|a_{ji}\right|}\text{,}i,j\in N,Z^{+}=\left\{1,2,\cdots \right\},Z=\left\{0,1,2,\cdots \right\}.$

$\begin{array}{l}{N}_1=\left\{i\in N:0<\left|a_{ii}\right|<R_i^{\alpha }{C}_i^{1-\alpha }\right\},N_2=\left\{i\in N:\left|a_{ii}\right|=R_i^{\alpha }{C}_i^{1-\alpha }\right\},N_3=\left\{i\in N:\left|a_{ii}\right|>R_i^{\alpha }{C}_i^{1-\alpha }\right\},\\ {N^{\prime }}_1=\left\{i\in N:0<\left|a_{ii}\right|\le R_i^{\alpha }{C}_i^{1-\alpha }\right\},N=N_1\cup N_2\cup N_3.\end{array}$

将 $N_1$ 进一步划分为

$N_1=N_1^{\left(1\right)}\cup N_1^{\left(2\right)}\cup \cdots \cup N_1^{\left(m\right)},$

其中m是任意正整数，且

$N_1^{\left(1\right)}=\left\{i\in N:0<\left|a_{ii}\right|<\frac{1}{m}{R}_i^{\alpha }{C}_i^{1-\alpha }\right\},$

$N_1^{\left(k\right)}=\left\{i\in N:\frac{k-1}{m}{R}_i^{\alpha }{C}_i^{1-\alpha }\le \left|a_{ii}\right|<\frac{k}{m}{R}_i^{\alpha }{C}_i^{1-\alpha }\right\},k=2,3,\cdots ,m.$

这里的 $N_1^{\left(k\right)}$ 可能为空集。

定义1 [2] 设 $A=\left(a_{ij}\right)\in C^{n\times n}$，如果 $\left|a_{ii}\right|>R_i\left(i\in N\right)$，则称A为严格对角占优矩阵，记作 $A\in D$。若存在正对角矩阵X，使得 $AX\in D$，则称A为广义严格对角占优矩阵，记作 $A\in D^{*}$。

定义2 [2] 设 $A=\left(a_{ij}\right)\in C^{n\times n}$， $\alpha \in \left(0,1\right]$，如果 $\left|a_{ii}\right|>R_i^{\alpha }{C}_i^{1-\alpha }\left(i\in N\right)$，则称A为α-链对角占优矩阵，记作 $A\in D^{\alpha }$。若存在正对角矩阵X，使得 $AX\in D^{\alpha }$，则称A为广义严格α-链对角占优矩阵，记作 $A\in {\tilde{D}}^{\alpha }$。

定义3 [2] 设 $A=\left(a_{ij}\right)\in C^{n\times n}$， $\alpha \in \left(0,1\right]$，如果 $\left|a_{ii}\right|\ge R_i^{\alpha }{C}_i^{1-\alpha }\left(i\in N\right)$，且上式中至少有一个严格不等式成立。若A不可约，则称A为不可约α-链对角占优矩阵。若对每一个以等式成立的下标i，存在非零元素链 $a_{i{j}_1}{a}_{j_1{j}_2}\cdots a_{i_{k-1}{j}_k}$，使得 $k\in \left\{i\in N:\left|a_{ii}\right|>R_i^{\alpha }{C}_i^{1-\alpha }\right\}\ne \varphi$，则称A为具有非零元素链的α-链对角占优矩阵。

引理1 [2] 设 $A=\left(a_{ij}\right)\in C^{n\times n}$，若 $A\in {\tilde{D}}^{\alpha }$，则称A广义严格对角占优矩阵。

引理2 [3] 设 $A=\left(a_{ij}\right)\in C^{n\times n}$，若A为不可约α-链对角占优矩阵，则 $A\in D^{*}$。

引理3 [3] 设 $A=\left(a_{ij}\right)\in C^{n\times n}$，若A为具有非零元素链的α-链对角占优矩阵，则 $A\in D^{*}$。

若 $N_3\left(A\right)=\varphi$ 或存在 $i_0\in N$ 使得对角元 $a_{i_0{i}_0}=0$，则 $A\notin {\tilde{D}}^{\alpha }$。当某个 $R_i=0$ 或 $C_i=0$ 时，判别A是否为非奇异H-矩阵可转化为对A的主子矩阵的判别。因此，本文总假设 $\left|a_{ii}\right|\ne 0$， $R_i\ne 0$， $C_i\ne 0\left(i\in N\right)$ 且

规定 ${\displaystyle \underset{t\in \varphi }{\sum }\cdot }=0$。

文献 [1] 给出如下主要结果：

定理1 设 $A=\left(a_{ij}\right)\in C^{n\times n}$，若存在 $\alpha \in (0,1]$，使得

${\omega }_1={\displaystyle \underset{i\in {N^{\prime }}_1}{\sum }\frac{\left({\displaystyle \underset{t\in {N^{\prime }}_1,t\ne i}{\sum }\left|a_{it}\right|{\tilde{x}}_t}+g_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|\left({\delta }_{l+1,t}+\epsilon \right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left({\tilde{x}}_i\right)}^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\tilde{x}}_i^{\frac{1}{\alpha }}+\left({\displaystyle \underset{t\in {N^{\prime }}_1,t\ne i}{\sum }\left|a_{it}\right|{\tilde{x}}_t}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left({\tilde{x}}_i\right)}^{\frac{1-\alpha }{\alpha }}}}<\text{1,}$

其中

${\tilde{x}}_i=\frac{\left|a_{ii}\right|}{R_i^{\alpha }{C}_i^{1-\alpha }}\left(\forall i\in {N^{\prime }}_1\right),{\tilde{r}}_0=1,{\delta }_{l+1,i}=\frac{\left({\displaystyle \underset{t\in {N^{\prime }}_1}{\sum }\left|a_{it}\right|{\tilde{x}}_t+{\tilde{r}}_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|}}\right)C_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}}\left(\forall i\in N_3\right),$

${\tilde{r}}_{l+1}=\underset{i\in N_3}{\max }{\delta }_{l+1,i},g_l=\underset{i\in N_3}{\max }\frac{\left(\underset{i\in {N^{\prime }}_1}{\max }\left|a_{it}\right|{\tilde{x}}_t\right)C_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\delta }_{l+1,i}-\left({\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|{\delta }_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}}\left(l\in Z\right),$

这里 $\epsilon$ 是充分小的正数，则 $A\in D^{*}$。

2. 主要结果

进一步引进下面记号：

对于 $i\in N_1$，记

${\beta }_i={\displaystyle \underset{t\in N_1,t\ne i}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|},{\gamma }_i={\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|},$

$x_i=\{\begin{array}{ll}\frac{\left|a_{ii}\right|-{\beta }_i^{\alpha }{C}_i^{1-\alpha }}{{\gamma }_i^{\alpha }{C}_i^{1-\alpha }},\hfill & \left|a_{ii}\right|>{\beta }_i^{\alpha }{C}_i^{1-\alpha },\hfill \\ \frac{\left|a_{ii}\right|-{\gamma }_i^{\alpha }{C}_i^{1-\alpha }}{{\beta }_i^{\alpha }{C}_i^{1-\alpha }},\hfill & \left|a_{ii}\right|>{\gamma }_i^{\alpha }{C}_i^{1-\alpha },\hfill \\ \frac{\left|a_{ii}\right|}{R_i^{\alpha }{C}_i^{1-\alpha }},\hfill & \left|a_{ii}\right|\le \min \left\{{\beta }_i^{\alpha }{C}_i^{1-\alpha },{\gamma }_i^{\alpha }{C}_i^{1-\alpha }\right\},\hfill \end{array}$

令 $x_{1i}^{\left(k\right)}=\frac{k}{m}\min \left(x_i\right)$， $i\in N_1^{\left(k\right)}$， $k=1,2,\cdots ,m$

$x_{3i}=\frac{R_i{C}_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}},i\in N_3,$

$x_{2i}=\underset{i\in N_2}{\max }\left(\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|x_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}-\left({\displaystyle \underset{t\in N_2,t\ne i}{\sum }\left|a_{it}\right|}\right)C_i^{\frac{1-\alpha }{\alpha }}}\right),r_0=1,$

$r_{l+1}=\underset{i\in N_3}{\max }\left(\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+r_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|x_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{x}_{3i}}\right),i\in N_3,l\in Z,$

$M_{l+1,i}=\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+r_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|x_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{x}_{3i}}\text{,}i\in N_3,l\in Z,$

$f_{l+1,i}=\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|M_{l+1,i}{x}_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}},i\in N_3,l\in Z,$

$h_l=\underset{i\in N_3}{\max }\frac{\left(\underset{t\in N_1^{\left(k\right)}}{\max }\left|a_{it}\right|x_{1t}^{\left(k\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}\right)C_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{f}_{l+1,i}-\left({\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}}\text{,}i\in N_3,l\in Z.$

2.1. 定理2

设 $A=\left(a_{ij}\right)\in C^{n\times n}$， $\alpha \in \left(0,1\right]$，若存在 $l\in Z$，使得

$\begin{array}{l}\tau ={\displaystyle \underset{i\in N_1{}^{\left(k\right)}}{\sum }\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|\left(f_{l+1,t}+\epsilon \right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1}{\alpha }}+\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|}{x}_{1t}^{\left(k\right)}\right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}}<1,\\ i\in N_1{}^{\left(k\right)},k=1,2,\cdots ,m.\end{array}$ (1)

且对于 $\forall i\in N_2$，存在 $t\in N_1\cup N_3$，使得 $\left|a_{it}\right|\ne 0$，这里 $\epsilon$ 是充分小的正数，则 $A\in D^{*}$。

证明

对于 $\forall i\in N_1^{\left(k\right)}$ $\left(k=1,2,\cdots ,m\right)$，有 $0<x_{1i}^{\left(k\right)}<1$。对于 $\forall i\in N_3$，有 $0<x_{3i}<1$。则根据 $x_{2i}$ 表达式可知 $0<x_{2i}<1$。

对于 $\forall i\in N_3$，由于 $r_0=1$， $x_{3i}=\frac{R_i{C}_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}}$ 有

${\left|a_{ii}\right|}^{\frac{1}{\alpha }}{x}_{3i}=R_i{C}_i^{\frac{1-\alpha }{\alpha }},$

$\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+r_0{\displaystyle \underset{t\in N_3,t\ne j}{\sum }\left|a_{it}\right|x_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}<R_i{C}_i^{\frac{1-\alpha }{\alpha }},$

可得

$\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+r_0{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|x_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{x}_{3i}}<1,$

由 $M_{1,i},r_1$ 的定义，可知 $0<M_{1,i}\le r_1<r_0=1$，根据 $r_2$ 的定义，有

$\begin{array}{l}\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+r_1{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|x_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\\ <\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+r_0{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|x_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\end{array}$

根据 $M_{2,i}$， $r_2$ 的定义，有

$M_{2,i}\le r_2\le r_1<r_0=1$

假设当 $l=s$ 时，有 $M_{s+1,i}\le r_{s+1}\le r_s<1$ 成立，则可知

$\begin{array}{l}\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+r_{s+1}{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|x_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\\ <\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+r_s{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|x_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\end{array}$

根据 $M_{s+2,i},r_{s+2}$ 的定义有 $M_{s+2,i}\le r_{s+2}\le r_{s+1}<1$ 成立，故由数学归纳法可知

$M_{l+1,i}\le r_{l+1}\le r_l\le \cdots \le r_1<r_0=1,i\in N_3,l\in Z.$

对于 $\forall i\in N_3$，有 $0<x_{3i}<1$，则

$M_{l+1,i}{x}_{3i}\le r_{l+1}{x}_{3i}<1,i\in N_3,l\in Z,$

由 $x_{3i,i},f_{l+1,i}$ 的定义知

$f_{l+1,i}\le M_{l+1,i}{x}_{3i}\le r_{l+1}{x}_{3i}<1,i\in N_3,l\in Z.$ (2)

由 $h_l$ 和 $f_{l+1,i}$ 的表达式，以及上式可知

$\begin{array}{l}0<h_l\le \underset{i\in N_3}{\max }\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{f}_{l+1,i}-\left({\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}}\\ =\underset{i\in N_3}{\max }\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}}{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}}+{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|\left(M_{l+1,t}{x}_{3i}-f_{l+1,t}\right)}\right)C_i^{\frac{1-\alpha }{\alpha }}}\\ \le \text{1,}\end{array}$

由定理假设知 $0<\tau <1$，记

$\begin{array}{l}{d}_i=\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|}{x}_{1t}^{\left(k\right)}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|\left(f_{l+1,t}+\epsilon \right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1}{\alpha }}+\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}+\frac{1-\tau }{n_1},\\ i\in N_1^{\left(k\right)},k=1,2,\cdots ,m,\end{array}$ (3)

其中 $n_1$ 是 $N_1^{\left(k\right)}\left(k=1,2,\cdots ,m\right)$ 中所含元素的个数，则 ${\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }{d}_i}=1$，从而 $0<d_i\le 1\left(i\in N_1^{\left(k\right)},k=1,2,\cdots ,m\right)$。由(3)式易知

$\begin{array}{l}{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1}{\alpha }}{d}_i>\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|\left(f_{l+1,t}+\epsilon \right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}\\ -d_i\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}\\ \ge \left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|\left(f_{l+1,t}+\epsilon \right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }},\end{array}$ (4)

构造正对角矩阵 $X=diag\left(x_1,x_2,\cdots ,x_n\right)$，并记 $B=AX=\left(b_{ij}\right)$，其中

$x_i=\{\begin{array}{l}{d}_i{x}_{1i}^{\left(k\right)}\text{,}i\in N_1^{\left(k\right)},k=1,2,\cdots ,m,\\ x_{2i}\text{,}i\in N_2,\\ h_l\left(f_{l+1,i}+\epsilon \right),i\in N_3.\end{array}$

1) 对任意的 $i\in N_1^{\left(k\right)}\left(k=1,2,\cdots ,m\right)$，由(4)式可得

$\begin{array}{l}{\left|b_{ii}\right|}^{\frac{1}{\alpha }}-R_i\left(B\right)C_i^{\frac{1-\alpha }{\alpha }}\left(B\right)\\ ={\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}{d}_i\right)}^{\frac{1}{\alpha }}-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|\left(f_{l+1,t}+\epsilon \right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}{d}_i\right)}^{\frac{1-\alpha }{\alpha }}\\ =\left({\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1}{\alpha }}{d}_i-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|\left(f_{l+1,t}+\epsilon \right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}\right)d_i{}^{\frac{1-\alpha }{\alpha }}\\ >0.\end{array}$

2) 对任意的 $i\in N_2$，由 $x_{2i}$ 的定义可知

${\left|a_{ii}\right|}^{\frac{1}{\alpha }}{x}_{2i}>\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)+{\displaystyle \underset{t\in N_2,t\ne i}{\sum }\left|a_{it}\right|x_{2t}}+{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|x_{3t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\text{,}$

从而

$\begin{array}{l}{\left|b_{ii}\right|}^{\frac{1}{\alpha }}-R_i\left(B\right)C_i^{\frac{1-\alpha }{\alpha }}\left(B\right)\\ ={\left|a_{ii}\right|}^{\frac{1}{\alpha }}{x}_{2i}^{\frac{1}{\alpha }}-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2,t\ne i}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|\left(f_{l+1,t}+\epsilon \right)}}\right)C_i^{\frac{1-\alpha }{\alpha }}{x}_{2i}^{\frac{1-\alpha }{\alpha }}\\ =\left({\left|a_{ii}\right|}^{\frac{1}{\alpha }}{x}_{2i}-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2,t\ne i}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|\left(f_{l+1,t}+\epsilon \right)}}\right)C_i^{\frac{1-\alpha }{\alpha }}\right)x_{2i}^{\frac{1-\alpha }{\alpha }}\\ >0.\end{array}$

3) 对任意的 $i\in N_3$，由 $h_l$ 的表达式可得

$\left(\underset{t\in N_1^{\left(k\right)}}{\max }\left|a_{it}\right|x_{1t}^{\left(k\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}\le {\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l{f}_{l+1,i},$

从而

$\begin{array}{l}\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\\ \le \left(\underset{t\in N_1^{\left(k\right)}}{\max }\left|a_{it}\right|x_{1t}^{\left(k\right)}{\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }{d}_t}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}\\ =\left(\underset{t\in N_1^{\left(k\right)}}{\max }\left|a_{it}\right|x_{1t}^{\left(k\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}\\ \le {\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l{f}_{l+1,i},\end{array}$

故对任意的 $\epsilon >0$，有

$h_l\epsilon >\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}-{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l{f}_{l+1,i}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}-{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|C_i^{\frac{1-\alpha }{\alpha }}}},$

于是

$\begin{array}{l}{\left|b_{ii}\right|}^{\frac{1}{\alpha }}-R_i\left(B\right)C_i^{\frac{1-\alpha }{\alpha }}\left(B\right)\\ ={\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l^{\frac{1}{\alpha }}{\left(f_{l+1,i}+\epsilon \right)}^{\frac{1}{\alpha }}-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|}\left(f_{l+1,t}+\epsilon \right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{h}_l^{\frac{1-\alpha }{\alpha }}{\left(f_{l+1,i}+\epsilon \right)}^{\frac{1-\alpha }{\alpha }}\\ =\left({\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l\left(f_{l+1,i}+\epsilon \right)-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|}\left(f_{l+1,t}+\epsilon \right)}\right)C_i^{\frac{1-\alpha }{\alpha }}\right)h_l^{\frac{1-\alpha }{\alpha }}{\left(f_{l+1,i}+\epsilon \right)}^{\frac{1-\alpha }{\alpha }}\\ =(\epsilon \left({\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l-h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|C_i^{\frac{1-\alpha }{\alpha }}}\right)+{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l{f}_{l+1,i}\\ -\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }})h_l^{\frac{1-\alpha }{\alpha }}{\left(f_{l+1,i}+\epsilon \right)}^{\frac{1-\alpha }{\alpha }}\\ >0.\end{array}$

综上所诉，我们有 $\left|b_{ii}\right|>R_i^{\alpha }\left(B\right)C_i^{1-\alpha }\left(B\right)\left(i\in N\right)$，所以则 $A\in D^{*}$，证毕。

注 本文定理2推广了文献 [1] 中定理1的条件。通过将N划分三个区间，并将其中非占优指标集进一步细分为 $N_1=N_1^{\left(1\right)}\cup N_1^{\left(2\right)}\cup \cdots \cup N_1^{\left(m\right)}$，并结合迭代得到广义严格对角占优矩阵的新判据。事实上，

$0<x_{1i}^{\left(k\right)}=\frac{k}{m}\min \left(x_i\right)\le {\tilde{x}}_i<1\left(\forall i\in N_1^{\left(k\right)},k=1,2,\cdots ,m\right)$， $0<x_{2i}<1\left(\forall i\in N_2\right)$，则本文定理2在迭代判定时相比文献 [1] 中的定理1的判定范围更广，后面的数值算例可以详细说明。

2.2. 定理3

设 $A=\left(a_{ij}\right)\in C^{n\times n}$ 且不可约， $\alpha \in \left(0,1\right]$，若存在 $l\in Z$，使得

$\tau ={\displaystyle \underset{i\in N_1^{\left(k\right)}}{\sum }\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1}{\alpha }}+\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}}\le 1,i\in N_1^{\left(k\right)},k=1,2,\cdots ,m.$ (5)

且上式不等式中至少有一个严格不等式成立，则 $A\in D^{*}$。

证明

由 $A=\left(a_{ij}\right)\in C^{n\times n}$ 且不可约知，对于 $\forall i\in N_1^{\left(k\right)}\left(k=1,2,\cdots ,m\right)$，有 $0<x_{1i}^{\left(k\right)}<1$。对于 $\forall i\in N_3$，有 $0<x_{3i}<1$。则根据 $x_{2i}$ 表达式可知 $0<x_{2i}<1$。对于 $\forall i\in N_3$，有 $0<h_l\le 1$， $0<h_l{f}_{l+1,i}<1$。

由定理假设知 $0<\tau \le 1$，记

$d_i=\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1}{\alpha }}+\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}+\frac{1-\tau }{n_1},i\in N_1^{\left(k\right)},k=1,2,\cdots ,m,$ (6)

其中 $n_1$ 是 $N_1^{\left(k\right)}\left(k=1,2,\cdots ,m\right)$ 中所含元素的个数，则 ${\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }{d}_i}=1$，从而 $0<d_i\le 1\left(i\in N_1^{\left(k\right)},k=1,2,\cdots ,m\right)$。由(6)式易知

$\begin{array}{c}{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1}{\alpha }}{d}_i>\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}\\ -d_i\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}\\ \ge \left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }},\end{array}$ (7)

对任意的 $i\in N_2$，由 $x_{2i}$ 的定义及(2)式可知

${\left|a_{ii}\right|}^{\frac{1}{\alpha }}{x}_{2i}-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2,t\ne i}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\ge 0,$ (8)

构造正对角矩阵 $X=diag\left(x_1,x_2,\cdots ,x_n\right)$，并记 $B=AX=\left(b_{ij}\right)$，其中

$x_i=\{\begin{array}{l}{d}_i{x}_{1i}^{\left(k\right)}\text{,}i\in N_1^{\left(k\right)},k=1,2,\cdots ,m,\\ x_{2i}\text{,}i\in N_2,\\ h_l{f}_{l+1,i},i\in N_3.\end{array}$

1) 对任意的 $i\in N_1^{\left(k\right)}\left(k=1,2,\cdots ,m\right)$，由(7)式可得

$\begin{array}{l}{\left|b_{ii}\right|}^{\frac{1}{\alpha }}-R_i\left(B\right)C_i^{\frac{1-\alpha }{\alpha }}\left(B\right)\\ ={\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}{d}_i\right)}^{\frac{1}{\alpha }}-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}{d}_i\right)}^{\frac{1-\alpha }{\alpha }}\\ =\left({\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1}{\alpha }}{d}_i-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}\right)d_i{}^{\frac{1-\alpha }{\alpha }}\\ \ge 0.\end{array}$

2) 对任意的 $i\in N_2$，由(8)式可得

$\begin{array}{l}{\left|b_{ii}\right|}^{\frac{1}{\alpha }}-R_i\left(B\right)C_i^{\frac{1-\alpha }{\alpha }}\left(B\right)\\ ={\left|a_{ii}\right|}^{\frac{1}{\alpha }}{x}_{2i}^{\frac{1}{\alpha }}-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2,t\ne i}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}{x}_{2i}^{\frac{1-\alpha }{\alpha }}\\ =\left({\left|a_{ii}\right|}^{\frac{1}{\alpha }}{x}_{2i}-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2,t\ne i}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\right)x_{2i}^{\frac{1-\alpha }{\alpha }}\\ \ge 0.\end{array}$

3) 对任意的 $i\in N_3$，由 $h_l$ 的表达式可得

$\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(\underset{t\in N_1^{\left(k\right)}}{\max }\left|a_{it}\right|x_{1t}^{\left(k\right)}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}\le {\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l{f}_{l+1,i},$

从而

$\begin{array}{l}\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\\ \le \left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(\left(\underset{t\in N_1^{\left(k\right)}}{\max }\left|a_{it}\right|x_{1t}^{\left(k\right)}\right){\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }{d}_t}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\\ =\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left(\underset{t\in N_1^{\left(k\right)}}{\max }\left|a_{it}\right|x_{1t}^{\left(k\right)}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\\ \le {\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l{f}_{l+1,i},\end{array}$

于是

$\begin{array}{l}{\left|b_{ii}\right|}^{\frac{1}{\alpha }}-R_i\left(B\right)C_i^{\frac{1-\alpha }{\alpha }}\left(B\right)\\ ={\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l^{\frac{1}{\alpha }}{f}_{l+1,i}^{\frac{1}{\alpha }}-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}{h}_l^{\frac{1-\alpha }{\alpha }}{f}_{l+1,i}^{\frac{1-\alpha }{\alpha }}\\ =\left({\left|a_{ii}\right|}^{\frac{1}{\alpha }}{h}_l{f}_{l+1,i}-\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)}}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}{d}_t}\right)+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2i}}+h_l{\displaystyle \underset{t\in N_3,t\ne i}{\sum }\left|a_{it}\right|f_{l+1,t}}}\right)C_i^{\frac{1-\alpha }{\alpha }}\right)h_l^{\frac{1-\alpha }{\alpha }}{f}_{l+1,i}^{\frac{1-\alpha }{\alpha }}\\ \ge 0.\end{array}$

综上所诉，我们有 $\left|b_{ii}\right|\ge R_i^{\alpha }\left(B\right)C_i^{1-\alpha }\left(B\right)\left(i\in N\right)$，且由假设知至少有一个严格不等式成立。由矩阵A不可约知矩阵B不可约，则为不可约α-链对角占优矩阵。由引理2知 $A\in D^{*}$，证毕。

2.3. 定理4

设 $A=\left(a_{ij}\right)\in C^{n\times n}$， $\alpha \in \left(0,1\right]$，若存在 $l\in Z$，使得

$\tau ={\displaystyle \underset{i\in N_1^{\left(k\right)}}{\sum }\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{it}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{it}\right|f_{l+1,t}}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}{{\left|a_{ii}\right|}^{\frac{1}{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1}{\alpha }}+\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne i}{\sum }\left|a_{it}\right|x_{1t}^{\left(k\right)}}\right)}\right)C_i^{\frac{1-\alpha }{\alpha }}{\left(x_{1i}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}}\le 1,i\in N_1^{\left(k\right)},k=1,2,\cdots ,m.$ (9)

且对上式不等式成立的i，都存在非零元素链 $a_{i{j}_1}{a}_{j_1{j}_2}\cdots a_{j_{r}j}$ 满足

$\tau ={\displaystyle \underset{j\in N_1^{\left(k\right)}}{\sum }\frac{\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne j}{\sum }\left|a_{jt}\right|x_{1t}^{\left(k\right)}}\right)}+{\displaystyle \underset{t\in N_2}{\sum }\left|a_{jt}\right|x_{2t}}+h_l{\displaystyle \underset{t\in N_3}{\sum }\left|a_{jt}\right|f_{l+1,t}}\right)C_j^{\frac{1-\alpha }{\alpha }}{\left(x_{1j}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}{{\left|a_{jj}\right|}^{\frac{1}{\alpha }}{\left(x_{1j}^{\left(k\right)}\right)}^{\frac{1}{\alpha }}+\left({\displaystyle \underset{k=1}{\overset{m}{\sum }}\left({\displaystyle \underset{t\in N_1^{\left(k\right)},t\ne j}{\sum }\left|a_{jt}\right|x_{1t}^{\left(\left(k\right)\right)}}\right)}\right)C_j^{\frac{1-\alpha }{\alpha }}{\left(x_{1j}^{\left(k\right)}\right)}^{\frac{1-\alpha }{\alpha }}}}\le 1,j\in N_1^{\left(k\right)},k=1,2,\cdots ,m.$

则 $A\in D^{*}$。

3. 数值算例

例1 设矩阵

$A=\left(\begin{array}{cccccc}5& 2& 1& 1& 2& 3\\ 1& 6.5& 1& 2& 3& 2\\ 1& 1& 5.3& 2& 2& 4\\ 1& 1& 1& 8& 2& 3\\ 0& 0& 0& 2& 20& 2\\ 0& 1& 0& 1& 2& 23\end{array}\right),$

在判定矩阵A是否为广义严格对角占优矩阵时，利用文献 [1] 中的记号，取 $\alpha =0.5$， $\epsilon =0.0001$ 时，有 $N_1=\left\{1,2,3,4\right\}$， $N_2=\left\{5,6\right\}$，则当 $k=2$ 时，有 ${\omega }_1=1.3144>1$，故无法用文献 [1] 中的定理1来判定。利用文献 [4] 的记号有 $N_{11}=\varphi$，所以不能用文献 [4] 定理1判定。同理利用文 [5] [6] 中记号，均可验证不满足文 [5] [6] 的判定条件。

由于

${\displaystyle \underset{i=1}{\overset{6}{\sum }}\frac{\underset{j\ne i}{\max }\left|a_{ij}\right|}{\left|a_{ii}\right|+\underset{j\ne i}{\max }\left|a_{ij}\right|}}=1.5645>1,$

故矩阵A不满足文献 [7] 中定理1的条件，所以不能用文献 [7] 来判定。使用文献 [8] 中记号，有

${\sigma }_1={\displaystyle \underset{i=1}{\overset{6}{\sum }}\frac{R_i+C_i}{2\left|a_{ii}\right|+R_i+C_i}}=2.6688>1,$

${\sigma }_{\infty }={\displaystyle \underset{i=1}{\overset{6}{\sum }}\frac{\underset{j\ne i}{\max }\left(\left|a_{ij}\right|+\left|a_{ji}\right|\right)}{2\left|a_{ii}\right|+\underset{j\ne i}{\max }\left(\left|a_{ij}\right|+\left|a_{ji}\right|\right)}}=1.0623>1.$

可知不满足文献 [8] 中定理3和定理4的条件，所以不能用文献 [8] 定理3或4来判定。同理，使用文献 [9] 的记号，通过计算可知无论如何选取 $N_1$ 和 $N_2$，矩阵A都不满足文献 [9] 定理2的条件，故无法用文献 [9] 的定理2来判定矩阵A是否为广义严格对角占优矩阵。

取 $\alpha =0.5$， $m=1$，则 $N_1=N_1^{\left(1\right)}=\left\{1,2,3\right\}$， $N_2=\left\{4\right\}$， $N_3=\left\{5,6\right\}$，令 $\epsilon =0.0001$， $l=0$ 时，有

$\begin{array}{l}{x}_{11}^{\left(1\right)}=0.3464,x_{12}^{\left(1\right)}=0.3354,\\ x_{13}^{\left(1\right)}=0.3052,x_{24}=0.1906,\\ f_{1,5}=0.0114,f_{1,6}=0.0148,h_0=0.9903,\end{array}$

计算得 $\tau =0.8300<1$。故本文定理1可判定矩阵A为广义严格对角占优矩阵。事实上，取正对角矩阵 $X=diag\left\{0.1106,0.0629,0.0987,0.1906,0.0114,0.0148\right\}$，有 $AX\in D^{*}$，则矩阵A为广义严格对角占优矩阵。

例2 设矩阵

$B=\left(\begin{array}{ccccccc}8.9& 3& 0& 2& 3& 6& 6\\ 1& 11.6& 2& 3& 1& 5& 3\\ 1& 1& 5& 2& 1& 0& 0\\ 1& 2& 1& 8.9& 1& 3& 2\\ 0& 1& 1& 0& 30& 5& 6\\ 1& 1& 0& 1& 2& 32& 3\\ 0& 1& 1& 0& 3& 2& 35\end{array}\right),$

在判定矩阵B是否为广义严格对角占优矩阵时，在文献 [1] 中，取 $\alpha =0.5$， $\epsilon =0.0001$ 时，有 $N_1=\left\{1,2,3,4\right\}$， $N_2=\left\{5,6,7\right\}$，则当 $k=2$ 时，有 ${\omega }_1=1.1848>1$，故无法用文献 [1] 中的定理1来判定。同理利用文 [4] [5] [6] 中记号，均可验证不满足文 [4] [5] [6] 的判定条件。

使用文献 [10] 中的记号有：

$\begin{array}{l}{N^{\prime }}_1=\left\{i\in N:0<\left|a_{ii}\right|\le R_i\right\},{N^{\prime }}_2=\left\{i\in N:\left|a_{ii}\right|>R_i\right\},\\ N_{+}=\left\{i\in {N^{\prime }}_2:\left|a_{ii}\right|\le C_i\right\},\end{array}$

则对该矩阵有 ${N^{\prime }}_1=\left\{1,2,3,4\right\}$， ${N^{\prime }}_2=\left\{5,6,7\right\}$， $N_{+}=\varphi$，不满足文献 [10] 中定理1的条件。又对任意 $\alpha \in \left(0,1\right)$，有

$\begin{array}{l}\frac{{\left[\left|a_{15}\right|+\left|a_{16}\right|+\left|a_{17}\right|+\left|a_{12}\right|\right]}^{\alpha }{C}_1^{1-\alpha }}{\left|a_{11}\right|-C_1^{1-\alpha }}+\frac{{\left[\left|a_{25}\right|+\left|a_{26}\right|+\left|a_{27}\right|+\left|a_{24}\right|\right]}^{\alpha }{C}_2^{1-\alpha }}{\left|a_{22}\right|-C_2^{1-\alpha }}\\ +\frac{{\left[\left|a_{35}\right|+\left|a_{36}\right|+\left|a_{37}\right|+\left|a_{34}\right|\right]}^{\alpha }{C}_3^{1-\alpha }}{\left|a_{33}\right|-C_3^{1-\alpha }}+\frac{{\left[\left|a_{45}\right|+\left|a_{46}\right|+\left|a_{47}\right|+\left|a_{42}\right|\right]}^{\alpha }{C}_4^{1-\alpha }}{\left|a_{44}\right|-C_4^{1-\alpha }}>1.\end{array}$

可见也不满足文献 [10] 中定理2的条件，所以不能用文献 [10] 来判定。

而在本文定理1的条件上，取 $\alpha =0.5$， $m=2$，则 $N_1=N_1^{\left(1\right)}\cup N_1^{\left(2\right)}$， $N_1^{\left(1\right)}=\varphi$， $N_1^{\left(2\right)}=\left\{1,2,4\right\}$， $N_2=\left\{3\right\}$， $N_3=\left\{5,6,7\right\}$，令 $\epsilon =0.0001$， $l=1$ 时，有

$\begin{array}{l}{x}_{11}^{\left(2\right)}=0.2580,x_{12}^{\left(2\right)}=0.3538,\\ x_{14}^{\left(2\right)}=0.3486,x_{23}=0.2936,\\ f_{2,5}=0.0103,f_{2,6}=0.0210,\\ f_{2,7}=0.0119,h_1=0.9869,\end{array}$

计算得 $\tau =0.9049<1$。故本文定理1可判定矩阵B为广义严格对角占优矩阵。事实上，取正对角矩阵 $X=diag\left\{0.0738,0.1105,0.2936,0.1070,0.0103,0.0208,0.0118\right\}$，有 $BX\in D^{*}$，则矩阵B为广义严格对角占优矩阵。

4. 结论

通过数值算例表明，在细分区间的基础下，本文的迭代判定条件比文献 [1] 定理1判定条件更好，且相比文献 [4] - [9] 的判定范围更广。因此，本文给出的广义严格α-链对角占优矩阵的细分迭代新判据，不仅拓宽了广义严格对角占优矩阵的判定范围，而且判定更有效率。

致谢

感谢庹清老师对本项目的悉心指导和帮助。

基金项目

吉首大学校级自筹科研基金项目(JGY2022076)和湖南省科研创新项目(21C0365)。

参考文献

NOTES

^*通讯作者。