ABSTRACT

$\mathcal{H}$ -tensors have wide applications in science and engineering, but it is difficult to determine whether a given tensor is an $\mathcal{H}$ -tensor or not in practice. In this paper, we give some practical conditions for $\mathcal{H}$ -tensors by constructing different positive diagonal matrices and applying some techniques of inequalities. As an application, some sufficient conditions of the positive definiteness for an even-order real symmetric tensor are given. Advantages of results obtained are illustrated by numerical examples.

Keywords:$\mathcal{H}$ -Tensors, Real Symmetric Tensors, Irreducible, Nonzero Elements Chain, Positive Definiteness

$\mathcal{H}$ -张量的新判定及其应用

柏冬健^*，徐玉梅，吴念

贵州民族大学数据科学与信息工程学院，贵州 贵阳

收稿日期：2020年5月1日；录用日期：2020年5月14日；发布日期：2020年5月21日

摘 要

$\mathcal{H}$ -张量在科学和工程实际中具有重要应用，但在实际中要判定 $\mathcal{H}$ -张量是比较困难的。通过构造不同的正对角阵，结合不等式的放缩技巧，给出了一些比较实用的新判别条件。作为应用，给出了判定偶数阶实对称张量正定性的条件，相应数值算例说明了新结果的有效性。

关键词 :$\mathcal{H}$ -张量，实对称张量，不可约，非零元素链，正定性

Copyright © 2020 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

张量是矩阵的高阶推广，广泛出现在图像处理、自动控制、医疗成像、超图论、高阶统计、弹性材料研究和数据分析等学科和工程中。近年来，很多专家和学者都对其进行了广泛探讨 [1] - [17]。本文在文 [13] 的基础上，继续讨论 $\mathcal{H}$ -张量的判定问题，得到了一些新的判定条件。同时，利用新得到的 $\mathcal{H}$ -张量的判定条件，给出了偶数阶实对称张量，即偶次齐次多项式正定性的判定方法。最后，给出了一些数值算例来说明新结果的有效性。

2. 预备知识

记 $ℂ\left(ℝ\right)$ 为复(实)数域， $\left[n\right]:=\left\{1,2,\cdots ,n\right\}$。一个复(实)m阶n维张量 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$ 由 $n^m$ 个复(实)元素构成 [1] [2] [3] [4] [5]，其中

$a_{i_1}{{}_{i_2\cdots }}_{i_m}\in ℂ\left(ℝ\right),i_j\in \left[n\right],j\in \left[m\right].$

显然，2阶张量即为矩阵。此外，张量被称为对称的 [6] [7]，若

$a_{i_1}{{}_{i_2\cdots }}_{i_m}=a_{\pi \left(i_1{i}_2\cdots i_m\right)},\forall \pi \in {\Pi }_m,$

其中 ${\Pi }_m$ 为m个指标的置换群。若 $a_{i_1}{{}_{i_2\cdots }}_{i_m}\ge 0$，那么称张量 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$ 为非负张量。

定义2.1 [8]：张量 被称作单位张量，其中

${\delta }_{i_1}{{}_{i_2\cdots }}_{i_m}=\{\begin{array}{l}1,i_1=i_2=\cdots =i_m\\ 0,其它\end{array}$

定义2.2 [6]：给定一个m阶n维张量 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$，若存在一个复数 $\lambda$ 和一个非零复向量 $x={\left(x_1,x_2,\cdots ,x_n\right)}^{\text{T}}\in {ℂ}^n$，满足

$\mathcal{A}{x}^{m-1}=\lambda x^{\left[m-1\right]},$

那么称为张量 $\mathcal{A}$ 的特征值，x为张量 $\mathcal{A}$ 的关于特征值 $\lambda$ 的特征向量，其中 $\mathcal{A}{x}^{m-1}$ 和 $x^{\left[m-1\right]}$ 的第i个分量分别为

${\left(\mathcal{A}{x}^{m-1}\right)}_i={\displaystyle \underset{i_2,\cdots ,i_m\in \left[n\right]}{\sum }{a}_i{{}_{i_2\cdots }}_{i_m}{x}_{i_2}\cdots x_{i_n},}{\left(x^{\left[m-1\right]}\right)}_i=x_i^{m-1}.$

记m阶n次齐次多项式 $f\left(x\right)$ 为

$f\left(x\right)={\displaystyle \underset{i_1,\cdots ,i_m\in \left[n\right]}{\sum }{a}_i{{}_{{}_1{i}_2\cdots }}_{i_m}{x}_{i_1}\cdots x_{i_m},}$ (1)

其中 $x={\left(x_1,x_2,\cdots ,x_n\right)}^{\text{T}}\in {ℝ}^n$。当m为偶数时， $f\left(x\right)$ 是正定的，若

式(1)中的齐次多项式 $f\left(x\right)$ 可以表示为m阶n维对称张量 $\mathcal{A}$ 与 $x^m$ 的乘积 [9]，如下

$f\left(x\right)=\mathcal{A}{x}^m={\displaystyle \underset{i_1,\cdots ,i_m\in \left[n\right]}{\sum }{a}_i{{}_{{}_1{i}_2\cdots }}_{i_m}{x}_{i_1}\cdots x_{i_m},}$ (2)

当 $f\left(x\right)$ 是正定时，对称张量 $\mathcal{A}$ 也是正定的。

定义2.3 [10]：设 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$ 为m阶n维张量，如果对任意的 $i\in \left[n\right]$，

(3)

则称 $\mathcal{A}$ 是对角占优张量。若对于任意的 $i\in \left[n\right]$，

$\left|a_{ii\cdots i}\right|>{\displaystyle \underset{\begin{array}{l}{i}_2,\cdots ,i_m\in \left[n\right]\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|},$ (4)

则称 $\mathcal{A}$ 是严格对角占优张量。

定义2.4 [11]：m阶n维张量 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$ 与矩阵 $X=diag\left(x_1,x_2,\cdots ,x_n\right)$ 的乘积可表示为：

$\mathcal{B}=\left(b_{i_1\cdots i_m}\right)=\mathcal{A}{X}^{m-1},b_{i_1{i}_2\cdots i_m}=a_{i_1{i}_2\cdots i_m}{x}_{i_2}{x}_{i_3}\cdots x_{i_m},i_j\in \left[n\right],j\in \left[m\right].$

假设 $\Lambda$ 表示 $\left[n\right]$ 的任意非空子集，令

${\Lambda }^{m-1}:=\left\{i_2{i}_3\cdots i_m:i_j\in \Lambda ,j=2,3,\cdots ,m\right\},$

$\left[n\right]\backslash {\Lambda }^{m-1}:=\left\{i_2{i}_3\cdots i_m:i_2{i}_3\cdots i_m\in {\left[n\right]}^{m-1}且i_2{i}_3\cdots i_m\notin {\Lambda }^{m-1}\right\}.$

给定一个m阶n维张量 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$，令

$R_i\left(\mathcal{A}\right)={\displaystyle \underset{\begin{array}{l}{i}_2,\cdots ,i_m\in \left[n\right]\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}={\displaystyle \underset{i_2,\cdots ,i_m\in \left[n\right]}{\sum }\left|a_{i{i}_2\cdots i_m}\right|-\left|a_{ii\cdots i}\right|},$

${\Lambda }_1:=\left\{i\in \left[n\right]:0<\left|a_{ii\cdots i}\right|=R_i\left(\mathcal{A}\right)\right\},{\Lambda }_2=\left\{i\in \left[n\right]:0<\left|a_{ii\cdots i}\right|<R_i\left(\mathcal{A}\right)\right\},$

${\Lambda }_3=\left\{i\in [n]:\left|a_{ii\cdots i}\right|>R_i\left(\mathcal{A}\right)\right\},{\Lambda }_0^{m-1}={\Lambda }^{m-1}\backslash \left({\Lambda }_2^{m-1}\cup {\Lambda }_3^{m-1}\right).$

引理2.1 [12]：若 $\mathcal{A}$ 为严格对角占优张量，则 $\mathcal{A}$ 为 $\mathcal{H}$ -张量。

引理2.2 [13]：设 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$ 为m阶n维张量。若 $\mathcal{A}$ 是不可约的，

$\left|a_{ii\cdots i}\right|\ge R_i\left(\mathcal{A}\right),\forall i\in \left[n\right],$

且至少有一个i使得严格不等式成立，则 $\mathcal{A}$ 为 $\mathcal{H}$ -张量。

引理2.3 [13]：设 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$ 为m阶n维张量。如果存在一个正对角矩阵X，使得 是 $\mathcal{H}$ -张量，则 $\mathcal{A}$ 为 $\mathcal{H}$ -张量。

引理2.4 [14]：设 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$ 为m阶n维张量。若

(i)， $\forall i\in \left[n\right]$，

(ii) ${\Lambda }_3=\left\{i\in [n]:\left|a_{ii\cdots i}\right|>R_i\left(\mathcal{A}\right)\right\}\ne \varnothing$，

(iii)，从i到j存在一个非零元素链使得 $j\in {\Lambda }_3$，

则 $\mathcal{A}$ 为 $\mathcal{H}$ -张量。

3. 主要结果

为了叙述方便，引入以下符号：对 $\forall i\in {\Lambda }_2$，记

${\alpha }_i={\displaystyle \underset{i_2{i}_3\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|},{\beta }_i={\displaystyle \underset{i_2{i}_3\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|},$

$x_i=\{\begin{array}{l}\frac{\left|a_{ii\cdots i}\right|-{\alpha }_i}{{\beta }_i},\left|a_{ii\cdots i}\right|>{\alpha }_i,\\ \frac{\left|a_{ii\cdots i}\right|-{\beta }_i}{{\alpha }_i},\left|a_{ii\cdots i}\right|>{\beta }_i,\\ \frac{\left|a_{ii\cdots i}\right|}{R_i\left(\mathcal{A}\right)},\left|a_{ii\cdots i}\right|\le \min \left\{{\alpha }_i,{\beta }_i\right\}.\end{array}$ $\delta =\underset{i\in {\Lambda }_2}{\max }(\; x\; i\; )$

再记

$r=\underset{i\in {\Lambda }_3}{\max }\left(\frac{\delta \left({\displaystyle \underset{i_2{i}_3\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2{i}_3\cdots i_m\in {\Lambda }_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)}{\left|a_{ii\cdots i}\right|-{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}}\right),$

$P_{i,r}\left(\mathcal{A}\right)=\delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+r{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\left(i\in {\Lambda }_3\right),$

$h=\underset{i\in {\Lambda }_3}{\max }\left(\frac{\delta \left({\displaystyle \underset{i_2,i_3,\cdots ,i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2{i}_3\cdots i_m\in {\Lambda }_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)}{P_{i,r}\left(\mathcal{A}\right)-{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{P_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}}\right).$

易知 0 ≤ r < 1 ，且对任意的 $i\in {\Lambda }_3$，

$r\left|a_{ii\cdots i}\right|\ge \delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+r{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}=P_{i,r}\left(\mathcal{A}\right),$

从而有 $0\le \frac{P_{i,r}\left(A\right)}{\left|a_{ii\cdots i}\right|}\le r<1,\forall i\in {\Lambda }_3$。注意到

$\frac{\delta \left({\displaystyle \underset{i_2{i}_3\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2{i}_3\cdots i_m\in {\Lambda }_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)}{P_{i,r}\left(\mathcal{A}\right)-{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{P_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}}=\frac{P_{i,r}\left(\mathcal{A}\right)-r{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}}{P_{i,r}\left(\mathcal{A}\right)-{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{P_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}}\le 1.$

故 $0\le h\le 1$，进而由h的定义可知，对于任意 $i\in {\Lambda }_3$，有

$h{P}_{i,r}\left(\mathcal{A}\right)\ge \delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}.$ (5)

定理3.1：设 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$ 为m阶n维张量。若对任意的 $i\in {\Lambda }_2$， $\mathcal{A}$ 满足

$\left|a_{ii\cdots i}\right|x_i>\delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|},$ (6)

且对 $\forall i\in {\Lambda }_1$，存在 $i_2{i}_3\cdots i_m\in {\Lambda }_3^{m-1}$，使得，则 $\mathcal{A}$ 是 $\mathcal{H}$ -张量。

证明：由(6)式知，对对任意的 $i\in {\Lambda }_2$,

令

$T_i\equiv \frac{\left|a_{ii\cdots i}\right|x_i-\delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)-{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}}{{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}}.$ (7)

当 ${\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}=0$ 时，记 $T_i=+\infty$。由(7)式知 $T_i>0\left(\forall i\in {\Lambda }_2\right)$，且

$0\le \frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}<1\left(\forall j\in {\Lambda }_3\right),$

从而必有充分小的正数 $\epsilon$，使 $0<\epsilon <\underset{i\in {\Lambda }_2}{\min }{T}_i\le +\infty$，且 $\underset{j\in {\Lambda }_3}{\max }\left\{\frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}+\epsilon \right\}<1$。

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

$d_i=\{\begin{array}{l}{\left(\delta \right)}^{\frac{1}{m-1}},i\in {\Lambda }_1,\\ {\left(x_i\right)}^{\frac{1}{m-1}},i\in {\Lambda }_2,\\ \left(\frac{h{P}_{i,r}\left(A\right)}{\left|a_{ii\cdots i}\right|}+\epsilon \right){}^{\frac{1}{m-1}},i\in {\Lambda }_3.\end{array}$

(a) 对 $\forall i\in {\Lambda }_1$，存在 $i_2{i}_3\cdots i_m\in {\Lambda }_3^{m-1}$，使得 $a_{i{i}_2\cdots i_m}\ne 0$，且对任意 $j\in {\Lambda }_3$，总可以取到充分小的正数，使得 $0<\frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}+\epsilon \le r<\delta <1$，则

$\begin{array}{c}{R}_i\left(\mathcal{B}\right)\le \delta \left({\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_0^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|\left(\epsilon +\underset{j\in \left\{i_2{i}_3\cdots i_m\right\}}{\max }\frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\right)}\\ <\delta \left({\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_0^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2,\cdots ,i_m\in {\Lambda }_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2,\cdots ,i_m\in {\Lambda }_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)=\delta R_i\left(\mathcal{A}\right)=\left|a_{ii\cdots i}\right|\delta =\left|b_{ii\cdots i}\right|.\end{array}$

(b) 对 $\forall i\in {\Lambda }_2$，由 (7)式知

$\begin{array}{c}{R}_i\left(\mathcal{B}\right)\le \delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|\left(\epsilon +\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\right)}\\ =\epsilon {\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+\delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+h{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{P_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}\\ <T_i{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+\mu \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+h{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{P_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}\\ =\left|a_{ii\cdots i}\right|x_i=\left|b_{ii\cdots i}\right|.\end{array}$

(c) 对 $\forall i\in {\Lambda }_3$，由(5)式知

$\begin{array}{c}{R}_i\left(\mathcal{B}\right)\le \delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|\left(\epsilon +\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\right)}\\ =\epsilon {\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+\delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+h{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{P_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}\\ \le \epsilon {\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+h{P}_{i,r}\left(\mathcal{A}\right)<\epsilon \left|a_{ii\cdots i}\right|+h{P}_{i,r}\left(\mathcal{A}\right)=\left|b_{ii\cdots i}\right|.\end{array}$

综上所述，，即 $\mathcal{B}$ 是严格对角占优的。由引理2.1知 $\mathcal{B}$ 是 $\mathcal{H}$ -张量，进而由引理2.3知 $\mathcal{A}$ 是 $\mathcal{H}$ -张量。

定理3.2：设 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$ 为m阶n维张量， $\mathcal{A}$ 不可约，若对任意的 $i\in {\Lambda }_2$，

$\left|a_{ii\cdots i}\right|x_i\ge \delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{h{P}_{j,r}\left(A\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|},$ (8)

且(8)中至少有一个严格不等式成立，则 $\mathcal{A}$ 是 $\mathcal{H}$ -张量。

证明：类似定理3.1的证明方法。由于 $\mathcal{A}$ 是不可约的，则

$\delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)\ne 0,\forall i\in {\Lambda }_3.$

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

$d_i=\{\begin{array}{l}{\left(\delta \right)}^{\frac{1}{m-1}},i\in {\Lambda }_1,\\ {\left(x_i\right)}^{\frac{1}{m-1}},i\in {\Lambda }_2,\\ \left(\frac{h{P}_{i,r}\left(A\right)}{\left|a_{ii\cdots i}\right|}+\epsilon \right){}^{\frac{1}{m-1}},i\in {\Lambda }_3.\end{array}$

令，则 $\mathcal{B}$ 不可约。类似于定理3.1的证明，可得 $\left|b_{ii\cdots i}\right|\ge R_i\left(\mathcal{B}\right)\left(\forall i\in \left[n\right]\right)$，且对 $\forall i\in {\Lambda }_2$，(8)中至少有一个严格不等式成立，即存在一个 $i_0\in {\Lambda }_2$，使得 $\left|b_{i_0{i}_0\cdots i_0}\right|>R_{i_0}\left(\mathcal{B}\right)$。由引理2.2知 $\mathcal{B}$ 是 $\mathcal{H}$ -张量，进而由引理2.3知 $\mathcal{A}$ 也是 $\mathcal{H}$ -张量。

定理3.3：设 $\mathcal{A}=\left(a_{i_1}{{}_{i_2\cdots }}_{i_m}\right)$ 为m阶n维张量。若对任意的 $i\in {\Lambda }_2$，

$\left|a_{ii\cdots i}\right|x_i\ge \delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|},$

$\begin{array}{l}K\left(\mathcal{A}\right)=[i\in {\Lambda }_2:\left|a_{ii\cdots i}\right|x_i>\delta \left({\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{l}{i}_2\cdots i_m\in {\Lambda }_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\right)\\ \begin{array}{c}\\ \\ {}_{}\end{array}+{\displaystyle \underset{i_2\cdots i_m\in {\Lambda }_3^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{h{P}_{j,r}\left(\mathcal{A}\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}]\ne \varnothing ,\end{array}$

且对 $\forall i\in \left[n\right]\backslash K$，存在从i到j的非零元素链使得 ，则 $\mathcal{A}$ 是 $\mathcal{H}$ -张量。

证明：构造正对角矩阵 $D=diag\left(d_1,d_2,\cdots ,d_n\right)$，记，其中

$d_i=\{\begin{array}{l}{\left(\delta \right)}^{\frac{1}{m-1}},i\in {\Lambda }_1,\\ {\left(x_i\right)}^{\frac{1}{m-1}},i\in {\Lambda }_2,\\ \left(\frac{h{P}_{i,r}\left(\mathcal{A}\right)}{\left|a_{ii\cdots i}\right|}\right){}^{\frac{1}{m-1}},i\in {\Lambda }_3.\end{array}$

类似于的定理3.1的证明，可得 $\left|b_{ii\cdots i}\right|\ge R_i\left(\mathcal{B}\right)\left(\forall i\in \left[n\right]\right)$，至少存在一个 $i\in {\Lambda }_2$，使得 $\left|b_{ii\cdots i}\right|>R_i\left(\mathcal{B}\right)$。另外，如果 $\left|b_{ii\cdots i}\right|=R_i\left(\mathcal{B}\right)$，那么 $i\in \left[n\right]\backslash K$。假设 $\mathcal{A}$ 中存在一个从i到j非零元素链使得 $k\in K$，那么中也存在从i到j的非零元素链使得k满足 $\left|b_{kk\cdots k}\right|>R_k\left(\mathcal{B}\right)$。因此， $\mathcal{B}$ 满足引理2.4的条件，所以是 $\mathcal{H}$ -张量，进而由引理2.3知 $\mathcal{A}$ 是 $\mathcal{H}$ -张量。

例3.1：给定，其中

$A\left(1,:,:\right)=\left(\begin{array}{ccc}12& 1& 0\\ 1& 6& 0\\ 1& 0& 12\end{array}\right),A\left(2,:,:\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 10& 2\\ 0& 2& 2\end{array}\right),A\left(3,:,:\right)=\left(\begin{array}{ccc}0& 0& 0\\ 1& 1& 0\\ 0& 0& 8\end{array}\right).$

由张量 $\mathcal{A}$ 的元素得到

$\left|a_{111}\right|=12,R_1=21,\left|a_{222}\right|=10,R_2=7,\left|a_{333}\right|=8,R_3=2,$