H-张量的新判定准则及其应用

New Criterions for H-Tensors and Its Applications

摘要: H-张量在科学与工程实践等领域中有着重要的应用，但在实际中要判定H-张量是比较困难的。本文通过构造不同的正对角阵，结合不等式的放缩技巧，给出了H-张量比较实用的新判别条件。作为应用，给出了判定偶次齐次多项式正定性的新方法，并给出相应的数值算例，表明了新结论的有效性。

Abstract: H-tensors have important applications in science and engineering, but it is difficult to determine whether a given tensor is an H-tensor or not in practice. In this paper, by constructing different pos-itive diagonal matrices and combining the technique of inequality reduction, new practical condi-tions for H-tensors are given. As applications, new methods for determining the positive definite-ness of even homogeneous polynomials are presented, and the validity of new results is verified by some numerical examples.

1. 引言

张量是高阶广义矩阵，广泛应用于信号和图像处理、高阶统计学、自动控制、医学成像、超图理论、弹性材料科学和工程研究与数据分析等领域中。近年来，许多专家学者对一般张量 [1] - [6] 或特殊结构张量的理论、性质及应用进行了广泛探讨 [7] - [18]。本文继续讨论H-张量的判定问题，得到了只与张量元素有关的新判别不等式，拓广了文献 [11] [14] [15] [16] 的结果。同时，获得了偶数阶实对称张量，即偶数阶齐次多项式正定性的新判定条件。最后，利用数值算例说明了新条件的有效性。

2. 预备知识

记 $C\left(R\right)$ 为复(实)数集， $N=\left[n\right]=\left\{1,2,\cdots ,n\right\}$。一个m阶n维张量 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 由 $n^m$ 个元素构成，其中 $a_{i_1}{{}_{i_2\cdots }}_{i_m}\in C$， $i_j\in N$， $j\in \left[m\right]$ [3] [4]。若 $a_{i_1{i}_2\cdots i_m}=a_{\pi \left(i_1{i}_2\cdots i_m\right)}$， $\forall \pi \in {\Pi }_m$，则称 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 为对称张量 [5]，其中 ${\Pi }_m$ 为m个指标的置换群。称张量 $I=\left({\delta }_{i_1{i}_2\cdots i_m}\right)$ 为单位张量 [5]，若

${\delta }_{i_1{i}_2\cdots i_m}\text{=}\{\begin{array}{l}1,i_1=i_2=\cdots =i_m,\\ 0,其它.\end{array}$

若

$f\left(x\right)={\displaystyle \underset{i_1,\cdots ,i_m\in \left[n\right]}{\sum }{a}_{i_1{i}_2\cdots i_m}{x}_{i_1}\cdots x_{i_m}}>0$， $x={\left(x_1,x_2,\cdots ,x_n\right)}^T\in R^n$， $x\ne 0$，

则称m阶n次齐次多项式 $f\left(x\right)$ 是正定的 [3]。 $f\left(x\right)$ 也可以表示为m阶n维对称张量 $A$ 与 $x^m$ 的乘积，如下

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

若 $f\left(x\right)$ 是正定的，则对称张量 $A$ 也是正定的 [3]。

定义1 [8] 设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维张量，若存在正向量 $x={\left(x_1,x_2,\cdots ,x_n\right)}^T\in R^n$，满足

$\left|a_{ii\cdots i}\right|x_i^{m-1}>{\displaystyle \underset{\begin{array}{c}{i}_2,i_3,\cdots ,i_m\in \left[n\right]\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|x_{i_2}\cdots x_{i_m}}$， $\forall i\in N$，

则称 $A$ 为H-张量。

定义2 [5] 设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维张量，若存在一个非空子集 $I\subset N$，满足

$a_{i_1{i}_2\cdots i_m}=0$， $\forall i_1\in I$， $\forall i_2,\cdots ,i_m\notin I$，

则称 $A$ 是可约张量。否则，称 $A$ 是不可约张量。

定义3 [9] 设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维张量，若存在指标 $k_1,k_2,\cdots ,k_r$，满足

${\displaystyle \underset{\begin{array}{c}{i}_2,i_3,\cdots ,i_m\in \left[n\right]\\ {\delta }_{k_s{i}_2\cdots i_m}=0\\ k_{s+1}\in \left\{i_2,i_3,\cdots ,i_m\right\}\end{array}}{\sum }\left|a_{k_s{i}_2\cdots i_m}\right|}\ne 0$， $s=0,1,\cdots ,r$， $\forall i,j\in N\left(i\ne j\right)$，

其中 $k_0=i$， $k_{r+1}=j$，则称张量 $A$ 中有一条从指标i到指标j的非零元素链。

3. 主要结果

为讨论方便，给出如下记号：设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维张量，令

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

$N^{m-1}\backslash S^{m-1}=\left\{i_2{i}_3\cdots i_m:i_2{i}_3\cdots i_m\in N^{m-1}且i_2{i}_3\cdots i_m\notin S^{m-1}\right\}$，

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

$N_1=\left\{i\in N:0<\left|a_{ii\cdots i}\right|=r_i\left(A\right)\right\}$， $N_2=\left\{i\in N:0<\left|a_{ii\cdots i}\right|<r_i\left(A\right)\right\}$，

$N_3=\left\{i\in N:\left|a_{ii\cdots i}\right|>r_i\left(A\right)\right\}$， $N_0^{m-1}=N^{m-1}/\left(N_2^{m-1}\cup N_3^{m-1}\right)$，

${\beta }_i=\frac{r_i\left(A\right)-\left|a_{ii\cdots i}\right|}{r_i\left(A\right)}$， $M=\underset{\begin{array}{c}i\in N_2\\ j\in N_3\end{array}}{\max }\left\{{\beta }_i,\frac{r_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\right\}$，

$K=\underset{i\in N_3}{\max }\frac{M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2\cdots i_m\in N_2^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}}{r_i\left(A\right)-{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}}$，

$R_i\left(A\right)=M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{i_2\cdots i_m\in N_2^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}+K{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\left(i\in N_3\right)$.

引理1 [6] 若 $A$ 是严格对角占优的张量，则 $A$ 是H-张量。

引理2 [10] 设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维张量。若存在正对角阵 $X$，满足 $A{X}^{m-1}$ 是H-张量，则 $A$ 是H-张量。

引理3 [6] 设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维张量且不可约。若

$\left|a_{ii\cdots i}\right|\ge r_i\left(A\right)$， $\forall i\in N$，

且上式中至少有一个严格不等式成立，则 $A$ 是H-张量。

引理4 [9] 设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维张量。若

1) $\left|a_{ii\cdots i}\right|\ge r_i\left(A\right)$， $\forall i\in N$ ；

2) $J\left(A\right)=\left\{i\in N:\left|a_{ii\cdots i}\right|>r_i\left(A\right)\right\}\ne \varnothing$ ；

3) $\forall i\notin J\left(A\right)$，从指标i到指标j有一条非零元素链，满足 $j\in J\left(A\right)$ ；

则 $A$ 是H-张量。

定理1 设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维张量。若 $A$ 满足

$\begin{array}{c}\left|a_{ii\cdots i}\right|{\beta }_i>M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\right\}\left|a_{i{i}_2\cdots i_m}\right|}\end{array}$， $\forall i\in N_2$， (1)

且 $\left|a_{ii\cdots i}\right|\ne {\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_0^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}$ ( $\forall i\in N_1$ )，则 $A$ 是H-张量。

证明：由K的定义知

$K\left|a_{ii\cdots i}\right|\ge M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}}\left|a_{i{i}_2\cdots i_m}\right|+{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}=R_i\left(A\right)$.

因此

$K\ge \frac{R_i\left(A\right)}{\left|a_{ii\cdots i}\right|}$， $\forall i\in N_3$. (2)

由式(1)得，

$\left|a_{ii\cdots i}\right|{\beta }_i-M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}-{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}-{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\right\}\left|a_{i{i}_2\cdots i_m}\right|}>0$. (3)

而 $R_i\left(A\right)<r_i\left(A\right)$ 且 $M\ge \frac{r_i\left(A\right)}{\left|a_{ii\cdots i}\right|}\left(\forall i\in N_3\right)$，所以

$M>\frac{R_i\left(A\right)}{\left|a_{ii\cdots i}\right|}$， $\forall i\in N_3$. (4)

由(3)式和(4)式得，一定存在足够小的正数 $\epsilon$，使得

$M>\frac{R_i\left(A\right)}{\left|a_{ii\cdots i}\right|}+\epsilon$， $\forall i\in N_3$， (5)

$\begin{array}{l}\left|a_{ii\cdots i}\right|{\beta }_i-M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}-{\displaystyle \underset{\underset{{\delta }_{i{i}_2\cdots i_m=0}}{i_2\cdots i_m\in N_2^{m-1}}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ -{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\right\}\left|a_{i{i}_2\cdots i_m}\right|}>\epsilon {\displaystyle \underset{i_2\cdots i_m\in N_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\end{array}$， $\forall i\in N_2$. (6)

构造正对角阵 $X=diag\left(x_1,x_2,\cdots ,x_n\right)$，记，其中

$x_i=M$， $i\in N_1$ ； $x_i={\beta }_i^{\frac{1}{m-1}}$， $i\in N_2$ ； $x_i={\left(\frac{R_i\left(A\right)}{\left|a_{ii\cdots i}\right|}+\epsilon \right)}^{\frac{1}{m-1}}$， $i\in N_3$.

而 $M\ge \frac{r_i\left(A\right)-\left|a_{ii\cdots i}\right|}{r_i\left(A\right)}\left(i\in N_2\right)$ 且 $M>\frac{r_i\left(A\right)}{\left|a_{ii\cdots i}\right|}+\epsilon \left(i\in N_3\right)$，故对 $\forall i\in N_1$，

$\begin{array}{c}{r}_i\left(B\right)={\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_0^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|x_{i_2}}\cdots x_{i_m}+{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\beta }_{i_2}^{\frac{1}{m-1}}\cdots }{\beta }_{i_m}^{\frac{1}{m-1}}\\ +{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\left(\frac{R_{i_2}\left(A\right)}{\left|a_{i_2{i}_2\cdots i_2}\right|}+\epsilon \right)}^{\frac{1}{m-1}}\cdots }{\left(\frac{R_{i_m}\left(A\right)}{a_{i_m{i}_m\cdots i_m}}+\epsilon \right)}^{\frac{1}{m-1}}\\ \le M{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_0^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|+}{\displaystyle \underset{i_2\cdots i_m\in N_2^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}+\epsilon \right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ \le M{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_0^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+M{\displaystyle \underset{i_2\cdots i_m\in N_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+M{\displaystyle \underset{i_2\cdots i_m\in N_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\\ =M{r}_j\left(A\right)=M\left|a_{ii\cdots i}\right|=\left|b_{ii\cdots i}\right|.\end{array}$

根据(6)式，对 $\forall i\in N_2$，

$\begin{array}{c}{r}_i\left(B\right)={\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|x_{i_2}}\cdots x_{i_m}+{\displaystyle \underset{\begin{array}{c}{i}_2{i}_3\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\beta }_{i_2}^{\frac{1}{m-1}}\cdots }{\beta }_{i_m}^{\frac{1}{m-1}}\\ +{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\left(\frac{R_{i_2}\left(A\right)}{\left|a_{i_2{i}_2\cdots i_2}\right|}+\epsilon \right)}^{\frac{1}{m-1}}\cdots }{\left(\frac{R_{i_m}\left(A\right)}{a_{i_m{i}_m\cdots i_m}}+\epsilon \right)}^{\frac{1}{m-1}}\\ \le M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|+}{\displaystyle \underset{\begin{array}{c}{i}_2{i}_3\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}+\epsilon {\displaystyle \underset{i_2\cdots i_m\in N_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\end{array}$

$\begin{array}{l}<M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|+}{\displaystyle \underset{\begin{array}{c}{i}_2{i}_3\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}+\left|a_{ii\cdots i}\right|{\beta }_i-M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\\ -{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}-{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ =\left|a_{ii\cdots i}\right|{\beta }_i=\left|b_{ii\cdots i}\right|.\end{array}$

对 $\forall i\in N_3$，由(2)式得

$\begin{array}{c}{r}_i\left(B\right)={\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|x_{i_2}}\cdots x_{i_m}+{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\beta }_{i_2}^{\frac{1}{m-1}}\cdots {\beta }_{i_m}^{\frac{1}{m-1}}}\\ +{\displaystyle \underset{\begin{array}{c}{i}_2{i}_3\cdots i_m\in N_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\left(\frac{R_{i_2}\left(A\right)}{\left|a_{i_2{i}_2\cdots i_2}\right|}+\epsilon \right)}^{\frac{1}{m-1}}\cdots }{\left(\frac{R_{i_m}\left(A\right)}{a_{i_m{i}_m\cdots i_m}}+\epsilon \right)}^{\frac{1}{m-1}}\\ \le M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|+}{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_2^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_{\text{3}}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}+\epsilon {\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\end{array}$

$\begin{array}{l}\le M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|+}{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_2^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_{\text{3}}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }K\left|a_{i{i}_2\cdots i_m}\right|}+\epsilon {\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\\ =R_i\left(A\right)+\epsilon {\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}<R_i\left(A\right)+\epsilon {\displaystyle \underset{i_2\cdots i_m\in N_3^{m-1}}{\sum }\left|a_{ii\cdots i}\right|}\\ =\left|a_{ii\cdots i}\right|\left(\frac{R_i\left(A\right)}{\left|a_{ii\cdots i}\right|}+\epsilon \right)=\left|b_{ii\cdots i}\right|.\end{array}$

综上可得 $\left|b_{ii\cdots i}\right|>r_i\left(B\right)\left(\forall i\in N\right)$。由引理1知 $B$ 是H-张量，故由引理2知 $A$ 是H-张量。

定理2 设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维张量且不可约。若 $A$ 满足

$\begin{array}{c}\left|a_{ii\cdots i}\right|{\beta }_i\ge M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\right\}\left|a_{i{i}_2\cdots i_m}\right|}\end{array}$， $\forall i\in N_2$， (7)

且(7)式中至少有一个严格不等式成立，则 $A$ 是H-张量。

证明：构造正对角阵 $X=diag\left(x_1,x_2,\cdots ,x_n\right)$，记 $B=A{X}^{m-1}=\left(b_{i_1{i}_2\cdots i_m}\right)$，其中

$x_i=M$， $i\in N_1$ ； $x_i={\beta }_i^{\frac{1}{m-1}}$， $i\in N_2$ ； $x_i={\left(\frac{R_i\left(A\right)}{\left|a_{ii\cdots i}\right|}\right)}^{\frac{1}{m-1}}$， $i\in N_3$.

由M的定义得，对 $\forall i\in N_1$，

$\begin{array}{c}{r}_i\left(B\right)={\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_0^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|x_{i_2}}\cdots x_{i_m}+{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\beta }_{i_2}^{\frac{1}{m-1}}\cdots {\beta }_{i_m}^{\frac{1}{m-1}}}\\ +{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\left(\frac{R_{i_2}\left(A\right)}{\left|a_{i_2{i}_2\cdots i_2}\right|}\right)}^{\frac{1}{m-1}}\cdots }{\left(\frac{R_{i_m}\left(A\right)}{a_{i_m{i}_m\cdots i_m}}\right)}^{\frac{1}{m-1}}\\ \le M{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_0^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|+}{\displaystyle \underset{i_2\cdots i_m\in N_2^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ \le M{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_0^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|+}M{\displaystyle \underset{i_2\cdots i_m\in N_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+M{\displaystyle \underset{i_2\cdots i_m\in N_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\\ =M{r}_j\left(A\right)=M\left|a_{ii\cdots i}\right|=\left|b_{ii\cdots i}\right|.\end{array}$

根据(7)式知，对 $\forall i\in N_2$，

$\begin{array}{c}{r}_i\left(B\right)={\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|x_{i_2}}\cdots x_{i_m}+{\displaystyle \underset{\begin{array}{c}{i}_2{i}_3\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\beta }_{i_2}{}^{\frac{1}{m-1}}\cdots }{\beta }_{i_m}{}^{\frac{1}{m-1}}\\ +{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_3^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\left(\frac{R_{i_2}\left(A\right)}{\left|a_{i_2{i}_2\cdots i_2}\right|}\right)}^{\frac{1}{m-1}}\cdots }{\left(\frac{R_{i_m}\left(A\right)}{a_{i_m{i}_m\cdots i_m}}\right)}^{\frac{1}{m-1}}\\ \le M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|+}{\displaystyle \underset{\begin{array}{c}{i}_2{i}_3\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}\\ \le \left|a_{ii\cdots i}\right|{\beta }_i=\left|b_{ii\cdots i}\right|.\end{array}$

又对 $\forall i\in N_3$，由K的定义知

$\begin{array}{c}{r}_i\left(B\right)={\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|x_{i_2}}\cdots x_{i_m}+{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_2^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\beta }_{i_2}^{\frac{1}{m-1}}\cdots {\beta }_{i_m}^{\frac{1}{m-1}}}\\ +{\displaystyle \underset{\begin{array}{c}{i}_2{i}_3\cdots i_m\in N_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|{\left(\frac{R_{i_2}\left(A\right)}{\left|a_{i_2{i}_2\cdots i_2}\right|}\right)}^{\frac{1}{m-1}}\cdots }{\left(\frac{R_{i_m}\left(A\right)}{a_{i_m{i}_m\cdots i_m}}\right)}^{\frac{1}{m-1}}\\ \le M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|+}{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_2^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_{\text{3}}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}\end{array}$

$\begin{array}{l}+{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_{\text{3}}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\left|a_{i{i}_2\cdots i_m}\right|}\\ \le M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|+}{\displaystyle \underset{i_2{i}_3\cdots i_m\in N_2^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_{\text{3}}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }K\left|a_{i{i}_2\cdots i_m}\right|}+\epsilon {\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_3^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}\\ =R_i\left(A\right)=\left|a_{ii\cdots i}\right|\frac{R_i\left(A\right)}{\left|a_{ii\cdots i}\right|}=\left|b_{ii\cdots i}\right|.\end{array}$

因此， $\left|b_{ii\cdots i}\right|\ge r_i\left(B\right)\left(\forall i\in N\right)$。因(7)式中至少有一个严格不等式成立，所以存在指标 $i_0$ 满足 $\left|b_{i_0{i}_0\cdots i_0}\right|>r_{i_0}\left(B\right)$，且由 $A$ 不可约知 $B$ 不可约，于是由引理3知 $B$ 是H-张量。从而，由引理2知 $A$ 是H-张量。

记

$\begin{array}{c}\Omega \left(A\right)=\{\forall i\in N_1:\left|a_{ii\cdots i}\right|{\beta }_i\ge M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{\frac{R_j(A)}{\left|a_{jj\cdots j}\right|}\right\}\left|a_{i{i}_2\cdots i_m}\right|}\}.\end{array}$

定理3 设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维张量。若 $A$ 满足

$\begin{array}{l}\left|a_{ii\cdots i}\right|{\beta }_i\ge M{\displaystyle \underset{i_2\cdots i_m\in N_0^{m-1}}{\sum }\left|a_{i{i}_2\cdots i_m}\right|}+{\displaystyle \underset{\begin{array}{c}{i}_2\cdots i_m\in N_2^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{i{i}_2\cdots i_m}\right|}\\ +{\displaystyle \underset{i_2\cdots i_m\in N_{\text{3}}^{m-1}}{\sum }\underset{j\in \left\{i_2,i_3,\cdots ,i_m\right\}}{\max }\left\{\frac{R_j\left(A\right)}{\left|a_{jj\cdots j}\right|}\right\}\left|a_{i{i}_2\cdots i_m}\right|}\end{array}$， $\forall i\in N_2$， (8)

且对 $\forall i\in N/\Omega \left(A\right)\ne \varnothing$， $A$ 中存在从i到j的非零元素链，满足 $j\in \Omega \left(A\right)\ne \varnothing$，则 $A$ 为H-张量。

证明：构造正对角阵 $X=diag\left(x_1,x_2,\cdots ,x_n\right)$，记，其中

$x_i=M$， $i\in N_1$ ； $x_i={\beta }_i^{\frac{1}{m-1}}$， $i\in N_2$ ； $x_i={\left(\frac{R_i\left(A\right)}{\left|a_{ii\cdots i}\right|}\right)}^{\frac{1}{m-1}}$， $i\in N_3$.

类似于定理2的证明，对任意的 $i\in N$，有 $\left|b_{ii\cdots i}\right|\ge r_i\left(B\right)$，且至少有一个严格不等式成立。

另一方面，若 $\left|b_{ii\cdots i}\right|=r_i\left(B\right)$，则 $\forall i\in N/\Omega \left(A\right)$。设 $A$ 中有从i到j的一条非零元素链，满足 $j\in \Omega \left(A\right)$，则 $B$ 中也有从i到j一条非零元素链，满足 $\left|b_{jj\cdots j}\right|>r_j\left(B\right)$。于是，由引理4知 $B$ 是H-张量，再由引理2知 $A$ 是H-张量。

例1 设是一个3阶3维张量，其中

$A\left(1,:,:\right)=\left(\begin{array}{ccc}12& 1& 0\\ 0& 6& 0\\ 1& 0& 16\end{array}\right)$， $A\left(2,:,:\right)=\left(\begin{array}{ccc}1& \text{0}& 0\\ 0& \text{6}& 0\\ 0& 0& \text{3}\end{array}\right)$， $A\left(3,:,:\right)=\left(\begin{array}{ccc}1& 0& \text{0}\\ 0& 2& 0\\ 1& 0& 20\end{array}\right)$.

则

$\left|a_{111}\right|=12$， $r_1\left(A\right)=24$， $\left|a_{222}\right|=\text{6}$， $r_2\left(A\right)=4$， $\left|a_{333}\right|=20$， $r_3\left(A\right)=\text{4}$.

所以 $N_1=\varnothing ,N_2=\left\{1\right\},N_2=\left\{2,3\right\}$。计算得

${\beta }_1=\frac{1}{2}$， $M=\frac{1}{2}$， $K=\frac{1}{\text{6}}$， $R_2\left(A\right)=\text{1}$， $R_3\left(A\right)=\frac{\text{4}}{\text{3}}$.

因为

${\displaystyle \underset{jk\in N_0^2}{\sum }\left|a_{1jk}\right|}+{\displaystyle \underset{\underset{{\delta }_{1jk=0}}{jk\in N_2^2}}{\sum }\underset{l\in \left\{j,k\right\}}{\max }\left\{{\beta }_l\right\}\left|a_{1jk}\right|}+{\displaystyle \underset{jk\in N_3^2}{\sum }\underset{l\in \left\{j,k\right\}}{\max }\frac{R_l\left(A\right)}{\left|a_{ll}\right|}\left|a_{1jk}\right|}=\frac{1}{2}\times 2+\text{0}+22\times \frac{\text{1}}{\text{6}}=\frac{\text{14}}{\text{3}}<6=a_{111}{\beta }_1$，

所以张量 $A$ 满足本文定理1的条件，故张量 $A$ 为H-张量。但

${\displaystyle \underset{\begin{array}{c}jk\in N^2/N_3^2\\ {\delta }_{1jk}=0\end{array}}{\sum }\left|a_{1jk}\right|}+{\displaystyle \underset{jk\in N_3^2}{\sum }\underset{l\in \left\{j,k\right\}}{\max }\frac{r_l\left(A\right)}{\left|a_{ll}\right|}\left|a_{1jk}\right|}=2+\frac{1}{2}\times 22=13>12=\left|a_{111}\right|$，

且

$\begin{array}{l}\frac{r_1\left(A\right)}{r_1\left(A\right)-\left|a_{111}\right|}\left[q\left({\displaystyle \underset{i_2{i}_3\cdots i_m\in N_0^2}{\sum }\left|a_{1i_2{i}_3}\right|}+{\displaystyle \underset{\begin{array}{c}{i}_2{i}_3\cdots i_m\in N_2^2\\ {\delta }_{1i_2{i}_3}=0\end{array}}{\sum }\left|a_{1i_2{i}_3}\right|}\right)+{\displaystyle \underset{i_2{i}_3\in N_3^2}{\sum }\underset{j\in \left\{i_2,i_3\right\}}{\max }}\frac{t{P}_j\left(A\right)}{\left|a_{jjj}\right|}\left|a_{1i_2{i}_3}\right|\right]\\ =\frac{24}{24-12}\left[\frac{1}{2}\left(1+0+1+0\right)+\frac{1}{4}\left(16+6\right)\right]=\frac{26}{2}>12=\left|a_{111}\right|.\end{array}$

因此 $A$ 不满足文献 [11] 中定理1的条件且 $A$ 不满足文献 [14] 中定理2的条件。

4. 应用

基于H-张量的新判定条件，下面给出判定高次多元偶次齐次多项式正定性的新结论。

引理5 [6] 设 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 是m阶n维的实对称张量，m是偶数， $a_i{{}_{i\cdots }}_i>0\left(\forall i\in N\right)$。若 $A$ 是H-张量，则 $A$ 是正定的。

根据引理5，定理1，定理2和定理3，可得到以下结论。

定理4设m阶n维张量 $A=\left(a_{i_1{i}_2\cdots i_m}\right)$ 为偶数阶实对称张量， $a_i{{}_{i\cdots }}_i>0\left(\forall i\in N\right)$。若 $A$ 满足下列条件之一：定理1的条件；或定理2的条件；或定理3的条件，则 $A$ 是正定的。

例2 设6次齐次多项式

$\begin{array}{c}f\left(x\right)=A{x}^6=8x_1^6+17x_2^6+1\text{6}4x_3^6+93x_4^6+10x_5^6+15x_6^6-6x_1{x}_2^5-30x_1{x}_3^5\\ -6x_1{x}_4^5-6x_2{x}_3^5-6x_2{x}_4^5-24x_3{x}_4^5-20x_2^3{x}_3^3+6x_1^5{x}_4,\end{array}$

其中 $A=\left(a_{i_1}{{}_{i_2}}_{i_3{i}_4}{}_{i_5{i}_6}\right)$ 是一个6阶6维实对称张量，且

$\begin{array}{l}{a}_{111111}=8,a_{222222}=17,a_{333333}=164,a_{444444}=93,a_{555555}=10,a_{666666}=15,\\ a_{122222}=a_{212222}=a_{221222}=a_{222122}=a_{222212}=a_{222221}=-1,\\ a_{133333}=a_{313333}=a_{331333}=a_{333133}=a_{333313}=a_{333331}=-5,\\ a_{144444}=a_{414444}=a_{441444}=a_{444144}=a_{444414}=a_{444441}=-1,\end{array}$

$\begin{array}{l}{a}_{233333}=a_{323333}=a_{332333}=a_{333233}=a_{333323}=a_{3333332}=-1,\\ a_{244444}=a_{424444}=a_{442444}=a_{444244}=a_{444424}=a_{444442}=-1,\\ a_{344444}=a_{434444}=a_{444344}=a_{444344}=a_{444434}=a_{444443}=-4,\\ a_{222333}=a_{223233}=a_{223323}=a_{223332}=a_{232233}=a_{232323}=a_{232332}=-1,\end{array}$

$\begin{array}{l}{a}_{233223}=a_{233232}=a_{233322}=a_{333222}=a_{332322}=a_{332232}=a_{332223}=-1,\\ a_{323322}=a_{323232}=a_{323223}=a_{322332}=a_{322323}=a_{322233}=-1,\\ a_{411111}=a_{141111}=a_{114111}=a_{111411}=a_{111141}=a_{111141}=1.\end{array}$

其余的 $a_{i_1}{{}_{i_2}}_{i_3{i}_4}{}_{i_5{i}_6}=0$。则

$r_1\left(A\right)=12$， $r_2\left(A\right)=17$， $r_3\left(A\right)=44$， $r_4\left(A\right)=31$， $r_5\left(A\right)=0$， $r_6\left(A\right)=0$，

且 $N_1=\left\{2\right\},N_2=\left\{1\right\},N_3=\left\{3,4,5,6\right\}$。计算得

${\beta }_1=\frac{1}{3}$， $M=\frac{1}{3}$， $K=\frac{1}{12}$， $R_3\left(A\right)=\frac{41}{3}$， $R_4\left(A\right)=\frac{16}{3}$.

当 $i=1$ 时，计算得

$\begin{array}{l}M{\displaystyle \underset{\begin{array}{c}{i}_2{i}_3{i}_4{i}_5{i}_6\in N_0^5\\ {\delta }_{1i_2{i}_3{i}_4{i}_5{i}_6}=0\end{array}}{\sum }\left|a_{1i_2{i}_3{i}_4{i}_5{i}_6}\right|}+{\displaystyle \underset{i_2{i}_3{i}_4{i}_5{i}_6\in N_2^5}{\sum }\underset{j\in \left\{i_2,i_3,i_4,i_5,i_6\right\}}{\max }\left\{{\beta }_j\right\}\left|a_{1i_2{i}_3{i}_4{i}_5{i}_6}\right|}+{\displaystyle \underset{i_2{i}_3{i}_4{i}_5{i}_6\in N_3^5}{\sum }\underset{j\in \left\{i_2,i_3,i_4,i_5,i_6\right\}}{\max }\frac{R_j\left(A\right)}{\left|a_{jjjjjj}\right|}\left|a_{1i_2{i}_3{i}_4{i}_5{i}_6}\right|}\\ =\frac{1}{3}\times 6+0+\frac{1}{12}\times 6=\frac{5}{2}<\frac{8}{3}=\left|a_{111111}\right|{\beta }_1,\end{array}$

因此 $A$ 满足本文定理1的条件，由定理4知 $A$ 是正定的，即 $f\left(x\right)$ 是正定的。但

$a_{111111}=8<12=r_1\left(A\right)$，

$a_{444444}\left(a_{111111}-r_1\left(A\right)+\left|a_{144444}\right|\right)=-465<-31=r_4\left(A\right)\left|a_{144444}\right|$，

且

$\left|a_{111111}\right|=8=8={\displaystyle \underset{\begin{array}{c}{i}_2{i}_3{i}_4{i}_5{i}_6\in N^5/N_1^5\\ {\delta }_{1i_2{i}_3{i}_4{i}_5{i}_6}=0\end{array}}{\sum }\left|a_{1i_2{i}_3{i}_4{i}_5{i}_6}\right|}+{\displaystyle \underset{i_2{i}_3{i}_4{i}_5{i}_6\in N_1^5}{\sum }\underset{j\in \left\{i_2,i_3,i_4,i_5,i_6\right\}}{\max }\frac{R_j\left(A\right)}{\left|a_{jjjjjj}\right|}\left|a_{1i_2{i}_3{i}_4{i}_5{i}_6}\right|}$，

因此不能用 [15] 中的定理3， [16] 中的定理4和 [11] 中的定理1判断 $A$ 的正定性。

5. 结论

本文通过构建不同的正对角矩阵，结合不等式的放缩技巧，得到了判别H-张量的新不等式，且这些不等式只涉及到张量的元素关系，因此它们是容易计算的。作为应用，给出了偶数阶实对称张量，即高次多元偶次齐次多项式正定性的判定新方法，数值例子表明了新结论的有效性。下一步，H-张量的高效数值迭代判定算法将是研究的重点。

致谢

感谢编辑老师和审稿老师提出了宝贵修改意见。

基金项目

贵州省科学技术基金(20181079，20191161)，贵州民族大学自然科学基金(GZMU[2019]YB08)。