广义Nekrasov矩阵新的判定条件

New Decision Conditions for Generalized Ne-krasov Matrix

摘要: 广义Nekrasov矩阵是一类应用广泛的特殊矩阵。本文主要研究Nekrasov矩阵的判定问题，通过构造新的放缩因子，从而给出了一组新的广义Nekrasov矩阵的判定条件，推广和改进了已有的结果，并用数值算例说明改进后方法的优越性。

Abstract: Generalized Nekrasov matrices are a kind of special matrices which are widely used. This paper mainly studies the Nekrasov matrices decision problem. In this paper, a new set of iterative decision conditions for generalized Nekrasov matrices are given by constructing new reduction factors. The existing results are extended and improved, and the superiority of the improved method is illus-trated by numerical examples.

1. 引言

近年来，广义Nekrasov矩阵是矩阵理论，经济数学，数值分析等众领域有广泛应用的一类重要特殊矩阵，因此，判定一个矩阵是否为广义Nekrasov矩阵有着重要的意义。许多学者对广义Nekrasov矩阵判定条件的研究已经取得诸多研究成果(见文 [1] - [10] )。其中，文 [2] 分别从构造新的递进判别系数和二次划分指标集这两种不同的思路出发，提出了判定广义Nekrasov矩阵的新方法；文 [3] 通过构造特殊的正对角矩阵，采用对非占优行区间的划分判定的方法，得到了广义Nekrasov矩阵的一类判别法；文 [5] 引入了参数p，为判定条件构造了小于1的迭代递进系数，从而拓宽了判定范围；文 [4] 改进了文 [3] 的主要结论，利用不同的划分方式对非占优行下指标集进行划分，构造特殊的正对角矩阵，结合不等式的放缩技巧，给出了新的判定条件和迭代算法，使得对矩阵进行判定所需的迭代次数较少。本文从矩阵的元素出发，通过构造新的放缩因子，进而给出新的广义Nekrasov矩阵判定条件，改进了文 [4] 的主要结果，并用数值方法说明了该判定方法的优越性和实用性。

用表示n阶复方阵集， $\langle n\rangle =\left\{1,2,\cdots ,n\right\}$ 为自然数集，设 $A=\left(a_{ij}\right)\in C^{n\times n}$ ，记

$R_1\left(A\right)={\displaystyle \underset{j>1}{\sum }\left|a_{1j}\right|},\left(j\in \langle n\rangle \right),$

$R_i\left(A\right)={\displaystyle \underset{j<i}{\sum }\left|a_{ij}\right|}\frac{R_j\left(A\right)}{\left|a_{jj}\right|}+{\displaystyle \underset{j>i}{\sum }\left|a_{ij}\right|},\left(2\le i\le n,j\in \langle n\rangle \right).$

$l_1\left(A\right)=0,l_i\left(A\right)={\displaystyle \underset{j<i}{\sum }\left|a_{ij}\right|\frac{R_j\left(A\right)}{\left|a_{jj}\right|}},\left(2\le i\le n,j\in \langle n\rangle \right).$

将 $\langle n\rangle =\left\{1,2,\cdots ,n\right\}$ 进行划分，并按照矩阵的行下标区域所满足的条件进行划分

$N_1=\left\{i\in \langle n\rangle :0<\left|a_{ii}\right|\le R_i\left(A\right)\right\},$

$N_2=\left\{i\in \langle n\rangle :\left|a_{ii}\right|>R_i\left(A\right)\right\}.$

定义1 [3] 设矩阵 $A=\left(a_{ij}\right)\in C^{n\times n}$ ，若 $\forall i\in \langle n\rangle$ ，都有 $\left|a_{ii}\right|>R_i\left(A\right)$ ，则称A为弱Nekrasov矩阵，记作 $A\in N^0$ ；若不等式严格成立，则称A为Nekrasov矩阵，记作 $A\in N$ ；若存在正对角矩阵D，使 $AD\in N$ ,则称A为广义Nekrasov矩阵，记作 $A\in N^{*}$ 。

引理1 [3] 设矩阵 $A=\left(a_{ij}\right)\in C^{n\times n}$ ,若存在 ${\tilde{N}}_2\subseteq N_2$ ， ${\tilde{N}}_1=\langle n\rangle -{\tilde{N}}_2$ ，使 $\left(\left|a_{ii}\right|-{\alpha }_i\right)\left(\left|a_{jj}\right|-{\beta }_j\right)>{\beta }_i{\alpha }_j$ ，有,其中 ${\alpha }_i=l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in {\tilde{N}}_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}$ ， ${\beta }_i={\displaystyle \underset{\begin{array}{l}t\in {\tilde{N}}_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}$ ，则A为广义Nekrasov矩阵。

引理2 [4] 设矩阵为Nekrasov矩阵，则 $\left|a_{ii}\right|\ne 0,\forall i\in \langle n\rangle$ 。

引理3 [5] 设矩阵为Nekrasov矩阵，则 $N_2\ne \varnothing$ 。

引理4 [6] 设矩阵，对矩阵 $B=AD$ ,若 $D=diag\left\{d_1,d_2,\cdots ,d_n\right\}$ 为正对角矩阵，且对 $\forall i\in \langle n\rangle$ ，当 $d_i\le 1$ 时，有 $l_i\left(B\right)\le l_i\left(A\right)$ 。

若 $N_1$ 为空集，则A为广义Nekrasov矩阵。若 $N_2$ 为空集，则A不是广义Nekrasov矩阵。所以本文假设 $N_1$ ， $N_2$ 不为空集，且对角元 $\left|a_{ii}\right|\ne 0$ 。

2020年，在文献 [4] 的定理2.1给出了如下结果：

定理1 [4] 设 $A=\left(a_{ij}\right)\in C^{n\times n}$ ，若

$\begin{array}{l}\left(\left|a_{ii}\right|-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in {\hat{N}}_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in {\hat{N}}_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|\frac{R_t^{\left(1\right)}\left(A\right)}{\left|a_{tt}\right|}}\right)\left(R_j\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{jt}\right|\frac{R_t\left(A\right)}{\left|a_{tt}\right|}}\right)\\ >\left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|\frac{R_t\left(A\right)}{\left|a_{tt}\right|}}\right)\left(l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in {\hat{N}}_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in {\hat{N}}_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|\frac{R_t^{\left(1\right)}\left(A\right)}{\left|a_{tt}\right|}}\right)\forall i\in {\hat{N}}_1^{\left(1\right)},j\in N_2,\end{array}$

$\begin{array}{l}\left(R_i^{\left(1\right)}\left(A\right)-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in {\hat{N}}_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in {\hat{N}}_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|\frac{R_t^{\left(1\right)}\left(A\right)}{\left|a_{tt}\right|}}\right)\left(R_j\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{jt}\right|\frac{R_t\left(A\right)}{\left|a_{tt}\right|}}\right)\\ >\left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|\frac{R_t\left(A\right)}{\left|a_{tt}\right|}}\right)\left(l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in {\hat{N}}_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in {\hat{N}}_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|\frac{R_t^{\left(1\right)}\left(A\right)}{\left|a_{tt}\right|}}\right)\forall i\in {\hat{N}}_2^{\left(1\right)},j\in N_2,\end{array}$

则A为广义Nekrasov矩阵。

其中

${\hat{N}}_1^{\left(1\right)}=\left\{i\in \langle n\rangle |0<\left|a_{ii}\right|\le R_i^{\left(1\right)}\left(A\right)\right\},{\hat{N}}_2^{\left(1\right)}=\left\{i\in \langle n\rangle |R_i\left(A\right)\ge \left|a_{ii}\right|>R_i^{\left(1\right)}\left(A\right)\right\},$

$R_i^{\left(1\right)}\left(A\right)=l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}i\in N_1\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}i\in N_2\\ t>i\end{array}}{\sum }\left|a_{it}\right|}\frac{R_t\left(A\right)}{\left|a_{tt}\right|}.$

我们对该定理的条件进行改进，得到判定范围更广的新条件。

2. 主要结果

$r=\underset{i\in N_2}{\max }\frac{l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+\frac{R_t\left(A\right)}{\left|a_{tt}\right|}{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}}{\left|a_{ii}\right|},$

$Q_i\left(A\right)=l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+r{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|},\left(i\in N_2\right),$

$P_i\left(A\right)=l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+\frac{Q_t\left(A\right)}{\left|a_{tt}\right|}{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|},{\delta }_{1,i}=\frac{P_i\left(A\right)}{\left|a_{ii}\right|},\left(i\in N_2\right),$

$N_1^{\left(1\right)}=\left\{i\in N_1|0<\left|a_{ii}\right|\le l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right\},N_2^{\left(1\right)}=N_1\backslash N_1^{\left(1\right)},$

${\delta }_{2,i}=\frac{\left|a_{ii}\right|+l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}}{2\left|a_{ii}\right|}.$

定理2 设矩阵 $A=\left(a_{ij}\right)\in C^{n\times n}$

$\begin{array}{l}\left(\left|a_{ii}\right|-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}\right)\left(\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}\right)\\ >\left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right)\left(l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}\right),\forall i\in N_1^{\left(1\right)},j\in N_2,\end{array}$ (1)

$\begin{array}{l}\left(\left|a_{ii}\right|{\delta }_{2,i}-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}\right)\left(\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}\right)\\ >\left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right)\left(l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}\right)\text{,}\forall i\in N_2^{\left(1\right)},j\in N_2,\end{array}$ (2)

则A为广义Nekrasov矩阵。

证明：若存在某个 $j_0$ 使得 $R_{j_0}\left(A\right)=l_{j_0}\left(A\right)+{\displaystyle \underset{t>i_0}{\sum }\left|a_{j_{0}t}\right|}=0$ ，即 $\left|a_{j_{0}t}\right|=0,t\in \langle n\rangle ,t\ne j_0$ ，则 $\forall i\in N_1^{\left(1\right)},j_0\in N_2$ ，(1)式两边相等且等于0，与已知条件矛盾，故对 $\forall j\in N_2$ ， $R_j\left(A\right)=l_j\left(A\right)+{\displaystyle \underset{t>j}{\sum }\left|a_{jt}\right|}>0$ ，即 $\left|a_{jt}\right|\ne 0,t\in \langle n\rangle ,t\ne j$ 。

由r的定义可知， $0<r<1$ ，且对于 $\forall i\in N_2$ ，有 $0<r\le \frac{R_i\left(A\right)}{\left|a_{ii}\right|}<1$ ，则

$r\left|a_{ii}\right|\ge l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+\frac{R_t\left(A\right)}{\left|a_{tt}\right|}{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}\ge l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+r{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}=Q_i\left(A\right),$

故 $0<\frac{Q_i\left(A\right)}{\left|a_{ii}\right|}\le r<1$ ，再由 $P_i\left(A\right)$ 的定义

$0<\frac{P_i\left(A\right)}{\left|a_{ii}\right|}\le \frac{Q_i\left(A\right)}{\left|a_{ii}\right|}\le r<1,$ (3)

则 $\forall i\in N_2$ ， $0<{\delta }_{1,i}=\frac{P_i\left(A\right)}{\left|a_{ii}\right|}<1$ 。

对于 $\forall i\in N_2^{\left(1\right)}$ ，有 $\left|a_{ii}\right|>l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}$ ，

$\begin{array}{c}{\delta }_{2,i}-1=\frac{\left|a_{ii}\right|+l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}}{2\left|a_{ii}\right|}-1=\frac{1}{2}+\frac{l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}}{2\left|a_{ii}\right|}-1\\ =\frac{l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}}{2\left|a_{ii}\right|}-\frac{1}{2}=\frac{1}{2}\left(\frac{l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}}{\left|a_{ii}\right|}-1\right)<0,\end{array}$

$\begin{array}{c}{\delta }_{2,i}-\frac{l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}}{\left|a_{ii}\right|}=\frac{\left|a_{ii}\right|+l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}-2\left(l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right)}{2\left|a_{ii}\right|}\\ =\frac{\left|a_{ii}\right|-\left(l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right)}{2\left|a_{ii}\right|}>0,\end{array}$

所以对于 $\forall i\in N_2^{\left(1\right)}$ ，有 $0<\frac{l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}}{\left|a_{ii}\right|}<{\delta }_{2,i}<1$ 。

再由式(1)和(2)，有

$\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}>0,\forall j\in N_2,$ (4)

$\left|a_{ii}\right|-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}>0,\forall i\in N_1^{\left(1\right)},$ (5)

$\left|a_{ii}\right|{\delta }_{2,i}-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}>0,\forall i\in N_2^{\left(1\right)}.$ (6)

构造正对角矩阵 $X=diag\left(x_1,x_2,x_3\right)$ ，其中

$x_i=\{\begin{array}{ll}1\hfill & i\in N_1^{\left(1\right)},\hfill \\ {\delta }_{2,i}\hfill & i\in N_2^{\left(1\right)},\hfill \\ {\delta }_{1,i}\hfill & i\in N_2,\hfill \end{array}$

易知，对任意的 $i\in \langle n\rangle$ ，都有 $x_i\le 1$ ，所以X为正对角矩阵，记 $B=AX=\left(b_{ij}\right)$ ，则 $b_{ij}=a_{ij}{x}_j,\left(i,j\in \langle n\rangle \right)$ 。由引理4知，对任意的 $i\in \langle n\rangle$ ，都有 $l_i\left(B\right)\le l_i\left(A\right)$ ， $\forall i\in \langle n\rangle$ 。

设 ${\tilde{N}}_2=\left\{j\in \langle n\rangle :\left|b_{jj}\right|>R_j\left(B\right)\right\}$ ，若 ${\tilde{N}}_2=\varnothing$ ，则

$\left|a_{ii}\right|\le l_i\left(B\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)},\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)},\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}},\forall i\in N_1^{\left(1\right)},$ (7)

$\left|a_{ii}\right|{\delta }_{2,i}\le l_i\left(B\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)},\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)},\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}},\forall i\in N_2^{\left(1\right)},$ (8)

$\left|a_{ii}\right|{\delta }_{1,i}\le l_i\left(B\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)},\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)},\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}},\forall i\in N_2.$ (9)

由 $l_i\left(B\right)\le l_i\left(A\right)$ ， $\forall i\in \langle n\rangle$ 及不等式(7)，(8)，(9)得

$\left|a_{ii}\right|-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}\le {\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}},\forall i\in N_1^{\left(1\right)},$ (10)

$\left|a_{ii}\right|{\delta }_{2,i}-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)},\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)},\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}\le {\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}},\forall i\in N_2^{\left(1\right)},$ (11)

$\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}\le l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)},\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)},\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}},\forall j\in N_2,$ (12)

由式(4)，(5)和式(10)，(12)可得

$\begin{array}{l}\left(\left|a_{ii}\right|-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}\right)\left(\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}\right)\\ \le \left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right)\left(l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}\right),\forall i\in N_1^{\left(1\right)},j\in N_2,\end{array}$ (13)

由式(4)，(6)和(11)，(12)可得

$\begin{array}{l}\left(\left|a_{ii}\right|{\delta }_{1,i}-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}\right)\left(\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}\right)\\ \le \left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right)\left(l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}\right)\text{,}\forall i\in N_2^{\left(1\right)},j\in N_2,\end{array}$ (14)

式(13)，(14)分别与条件(1)，(2)矛盾，故 ${\tilde{N}}_2\ne \varnothing$ 。

对 $\forall i\in N_1^{\left(1\right)}\subseteq N_1,j\in N_2$ ，由式(1)得

$\begin{array}{l}\left(\left|b_{ii}\right|-{\alpha }_i\left(B\right)\right)\left(\left|b_{jj}\right|-{\beta }_j\left(B\right)\right)\\ =\left(\left|a_{ii}\right|-l_i\left(B\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}\right)\left(\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}\right)\\ \ge \left(\left|a_{ii}\right|-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}\right)\left(\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}\right)\\ >\left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right)\left(l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}\right)\\ \ge \left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right)\left(l_j\left(B\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}\right)\\ >{\beta }_i\left(B\right){\alpha }_j\left(B\right).\end{array}$

对 $\forall i\in N_2^{\left(1\right)}\subseteq N_1,j\in N_2$ ，由式(2)得

$\begin{array}{l}\left(\left|b_{ii}\right|-{\alpha }_i\left(B\right)\right)\left(\left|b_{jj}\right|-{\beta }_j\left(B\right)\right)\\ =\left(\left|a_{ii}\right|{\delta }_{2,i}-l_i\left(B\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}\right)\left(\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}\right)\\ \ge \left(\left|a_{ii}\right|{\delta }_{2,i}-l_i\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}\right)\left(\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}\right)\\ >\left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right)\left(l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}\right)\\ \ge \left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}\right)\left(l_j\left(B\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}\right)\\ >{\beta }_i\left(B\right){\alpha }_j\left(B\right).\end{array}$

综上所述，对 $\forall i\in N_1,j\in N_2$ ，有 $\left(\left|b_{ii}\right|-{\alpha }_i\left(B\right)\right)\left(\left|b_{jj}\right|-{\beta }_j\left(B\right)\right)>{\beta }_i\left(B\right){\alpha }_j\left(B\right)$ 。由引理1可得 $B\in N$ ，所以 $A\in N^{*}$ ，即A为广义Nekrasov矩阵。

定理2：设矩阵 $A=\left(a_{ij}\right)\in C^{n\times n}$ ，若 $\forall i\in N_1^{\left(1\right)}$ ，存在 $t\in N_2^{\left(1\right)}$ ，使得 $\left|a_{it}\right|\ne 0$ ，且满足

$\left|a_{ii}\right|>l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)},\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)},\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}},$ (15)

则A为广义Nekrasov矩阵。

证明：由定理2的证明可得 $\forall i\in N_2$ ， $0<{\delta }_{1,i}<1$ ， $\forall i\in N_2^{\left(1\right)}$ ，有 $0<\frac{l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}}{\left|a_{ii}\right|}<{\delta }_{2,i}<1$ 。

构造正对角矩阵 $D=diag\left(d_1,d_2,d_3\right)$ ，其中

$d_i=\{\begin{array}{ll}1\hfill & i\in N_1^{\left(1\right)},\hfill \\ {\delta }_{2,i}\hfill & i\in N_2^{\left(1\right)},\hfill \\ {\delta }_{1,i}\hfill & i\in N_2.\hfill \end{array}$

易知，对任意的 $i\in \langle n\rangle$ ，都有 $d_i\le 1$ ，所以D为正对角矩阵，记 $B=AD=\left(b_{ij}\right)$ ，则 $b_{ij}=a_{ij}{d}_j$ 。由引理4知，对任意的 $i\in \langle n\rangle$ ，都有 $l_i\left(B\right)\le l_i\left(A\right)$ ，下证 $B\in N$ 。

对 $\forall i\in N_1^{\left(1\right)}$ ，由式(15)得

$\left|b_{ii}\right|=\left|a_{ii}\right|>l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}=R_i\left(B\right).$

对 $\forall i\in N_2$ ，由式(3)得

$\left|b_{ii}\right|={\delta }_{1,i}\left|a_{ii}\right|=P_i\left(A\right)=l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+\frac{Q_i\left(A\right)}{\left|a_{ii}\right|}{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}>l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}=R_i\left(B\right).$

对 $\forall i\in N_2^{\left(1\right)}$ ，有

$\left|b_{ii}\right|=\left|a_{ii}\right|{\delta }_{2,i}>\frac{l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}}{\left|a_{ii}\right|}\left|a_{ii}\right|>l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}=R_i\left(B\right).$

综上所述，对 $i\in \langle n\rangle$ ，都有 $R_i\left(B\right)<\left|b_{ii}\right|$ ，所以 $B\in N$ ,故 $A\in N^{*}$ ，即A为广义Nekrasov矩阵。

注1：对任意的 $i\in N_2$ ，都有 $0<{\delta }_{1,i}\le \frac{R_i\left(A\right)}{\left|a_{ii}\right|}<1$ ；同时对任意 $i\in N_2^{\left(1\right)}$ ，都有 $0<\frac{l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1,\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}}}{\left|a_{ii}\right|}<{\delta }_{2,i}<1$ ，故相较于定理1来说，本文将非占优行区间快速缩小，同时将 ${\delta }_{2,i}$ 进行放大，逆向对非占有行进一步细分，获得了更加简捷快速的判定条件。后面的数值算例说明其判定范围优于定理1。

注2：由定理2的证明过程可得 $\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}>0,\forall j\in N_2$ ，令

$\begin{array}{c}\mu =\underset{j\in N_2}{\max }\frac{l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}}}{\left|a_{jj}\right|{\delta }_{1,j}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{1,t}}}\\ =\underset{j\in N_2}{\max }\frac{l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}}}{l_j\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>j\end{array}}{\sum }\left|a_{jt}\right|{\delta }_{2,t}}}+\left(\frac{Q_t\left(A\right)}{\left|a_{tt}\right|}-{\delta }_{1,t}\right){\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{jt}\right|}}.\end{array}$

易知 $0<\mu \le 1$ ，由式(1)和式(2)得

$\left|a_{ii}\right|>l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}+\mu {\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}},\forall i\in N_1^{\left(1\right)},$

${\delta }_{2,t}\left|a_{ii}\right|>l_i\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{2,t}}+\mu {\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{it}\right|{\delta }_{1,t}},\forall i\in N_2^{\left(1\right)},$

则定理2改进了文献 [3] 的定理1。

3. 数值算例

例1

$A=\left[\begin{array}{ccccccc}5& 1& 1& 2& 0& 2& 1\\ 1& 20& 0.1& 0.1& 3& 0& 0\\ 3& 4& 20& 0& 1& 1& 0\\ 0& 0.1& 1& 8& 1& 0.1& 1\\ 0& 0& 1& 3& 8& 3& 2\\ 1.8& 0& 0& 0& 0& 20& 1\\ 0& 0& 0.2& 0& 0& 0& 30\end{array}\right].$

经过Matlab程序计算，可得，根据定理1， $N_1^{\left(1\right)}=\varnothing$ ， $N_2^{\left(1\right)}=\left\{1\right\}$ ， $R_1^{\left(1\right)}=1.0856$ ， $i=1$ ， $j=7$ 时，

$\begin{array}{l}\left(R_1^{\left(1\right)}\left(A\right)-l_1\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>1\end{array}}{\sum }\left|a_{1t}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>1\end{array}}{\sum }\left|a_{1t}\right|\frac{R_t^{\left(1\right)}\left(A\right)}{\left|a_{tt}\right|}}\right)\left(R_7\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>i\end{array}}{\sum }\left|a_{7t}\right|\frac{R_t\left(A\right)}{\left|a_{tt}\right|}}\right)=0.1049\\ =\left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>1\end{array}}{\sum }\left|a_{1t}\right|\frac{R_t\left(A\right)}{\left|a_{tt}\right|}}\right)\left(l_7\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>7\end{array}}{\sum }\left|a_{jt}\right|}+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>7\end{array}}{\sum }\left|a_{7t}\right|\frac{R_t^{\left(1\right)}\left(A\right)}{\left|a_{tt}\right|}}\right).\end{array}$

由上述结果可得，定理1无法判定矩阵A为广义Nekrasov矩阵。同样可以验证文献 [3] 和文献 [4] 都无法判断矩阵A为广义Nekrasov矩阵，而定理2可以直接判定A是广义Nekrasov矩阵。

可得 $N_1=\left\{1\right\}$ ， $N_2=\left\{2,3,4,5,6,7\right\}$ ，根据定理2， $N_1^{\left(1\right)}=\varnothing$ ， $N_2^{\left(1\right)}=\left\{1\right\}$ ，当 $i=1,j=7$ 时，

$\begin{array}{l}\left(\left|a_{11}\right|{\delta }_{2,t}-l_1\left(A\right)-{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>1\end{array}}{\sum }\left|a_{1t}\right|}-{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>1\end{array}}{\sum }\left|a_{1t}\right|{\delta }_{2,t}}\right)\left(\left|a_{66}\right|{\delta }_{1,6}-{\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>6\end{array}}{\sum }\left|a_{6t}\right|{\delta }_{1,t}}\right)=0.1973\\ >\left({\displaystyle \underset{\begin{array}{l}t\in N_2,\\ t>1\end{array}}{\sum }\left|a_{1t}\right|{\delta }_{1,t}}\right)\left(l_6\left(A\right)+{\displaystyle \underset{\begin{array}{l}t\in N_1^{\left(1\right)}\\ t>6\end{array}}{\sum }\left|a_{6t}\right|+{\displaystyle \underset{\begin{array}{l}t\in N_2^{\left(1\right)}\\ t>6\end{array}}{\sum }\left|a_{6t}\right|{\delta }_{2,t}}}\right)=0.0226,\end{array}$

其中正对角矩阵 $X=diag\left\{0.2173,0.3298,0.1616,0.1771,0.0001,0.0025\right\}$ 。

$B=\left[\begin{array}{ccccccc}2.6518& 1.3036& 0& 0& 0& 0& 0\\ 0.6630& 4.3455& 0.0330& 0.0162& 0.5314& 0& 0\\ 1.9889& 0.8691& 6.5970& 0& 0.1762& 0.00012& 0\\ 0& 0.0217& 0.3298& 1.2930& 0.1771& 0.0001& 0.0025\\ 0& 0& 0.3298& 0.4849& 1.4270& 0.0037& 0.005\\ 0& 0& 0& 0& 0& 0.0025& 0.0025\\ 0& 0& 0.0660& 0& 0& 0& 0.0744\end{array}\right].$

例2

$A=\left[\begin{array}{ccccccccc}40& 0& 1& 2& 0& 1& 0& 0& 3\\ 1& 30& 9& 2& 4& 0& 0& 1& 4\\ 3& 1& 30& 2& 0& 1& 0& 4& 9\\ 0& 2& 0& 30& 12& 7& 1& 0& 0\\ 0& 6& 1& 4& 39& 0& 1& 3& 1\\ 2& 0& 0& 4& 1& 10& 2& 0& 6\\ 1& 2& 1& 0& 1& 0& 9& 4& 3\\ 2& 0& 2& 1& 0.1& 0& 6& 30& 0\\ 1& 4& 3& 2& 0& 0& 0& 2& 29\end{array}\right].$

经过Matlab计算，可得 $N_1=\left\{1\right\}$ ， $N_2=\left\{2,3,4,5,6,7\right\}$ ，根据定理2，有 $N_1^{\left(1\right)}=\varnothing$ ， $N_2^{\left(1\right)}=\left\{6,7\right\}$ ，当 $i=6,j=8$ 时，不满足定理1的条件，但是利用本文定理2可以判断矩阵A为广义Nekrasov矩阵。

4. 结论

通过进行数值实验，可以得出定理2对广义Nekrasov矩阵的判定范围比定理1更加广泛，故本文给出的主要判定条件改进了现有的结果。

致谢

感谢庹清老师对本文章的悉心指导。

基金项目

国家自然科学基金项目(11461027)；湖南省研究生科研创新项目(CX20231071)。