Practical Criteria for $\mathcal{H}$-Tensors and Their Application

Abstract

Identifying the positive definiteness of even-order real symmetric tensors is an important component in tensor analysis. $\mathcal{H}$-tensors have been utilized in identifying the positive definiteness of this kind of tensor. Some new practical criteria for identifying $\mathcal{H}$-tensors are given in the literature. As an application, several sufficient conditions of the positive definiteness for an even-order real symmetric tensor were obtained. Numerical examples are given to illustrate the effectiveness of the proposed method.

1. Introduction

Tensor theory is widely used in digital signal processing, medical image processing, data mining, quantum entanglement, and other fields [1,2,3,4,5,6,7]. $\mathcal{H}$-tensor theory is an integral part of tensor theory. It plays an important role in physics such as control theory and dynamic control systems and in mathematics such as the numerical solution of partial differential equations and the degree of discretization of nonlinear parabolic equations [8,9,10,11,12,13].

Let $\mathbb{C}\left(\mathbb{R}\right)$ be the complex (real) field and $\mathbb{N}=\{1,2,\dots ,n\}$. A complex (real)-order m dimension n tensor $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ consists of $n^m$ complex (real) entries:

$$a_{i_1{i}_2\dots i_m}\in \mathbb{C}\left(\mathbb{R}\right),$$

where $i_j=1,2,\dots ,n$ and $j=1,2,\dots ,m.$ A tensor $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ is called symmetric [14], if:

$$a_{i_1{i}_2\dots i_m}=a_{\pi (i_1{i}_2\dots i_m)},\forall \pi \in {\Pi }_m,$$

where ${\Pi }_m$ is the permutation group of m indices. Furthermore, a tensor $\mathcal{I}=\left({\delta }_{i_1{i}_2\dots i_m}\right)$ is called the unit tensor [15], if its entries:

$${\delta }_{i_1{i}_2\dots i_m}=\left\{\begin{array}{c}1,if{i}_1=i_2=\dots =i_m,\hfill \\ 0,otherwise.\hfill \end{array}\right.$$

Let $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ be a tensor with order m and dimension n. If there exist a complex number $\lambda$ and a non-zero complex vector $x={(x_1,x_2,\dots ,x_n)}^T$ that are solutions of the following homogeneous polynomial equations:

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

then we call $\lambda$ an eigenvalue of $\mathcal{A}$ and x an eigenvector of $\mathcal{A}$ associated with $\lambda$ [14,16,17,18,19,20], and $\mathcal{A}{x}^{m-1}$ and $\lambda x^{[m-1]}$ are vectors, whose ith components are:

$${\left(\mathcal{A}{x}^{m-1}\right)}_i={\displaystyle \sum _{i_2,i_3,\dots ,i_m\in \mathbb{N}}}{a}_{i{i}_2{i}_3\cdots i_m}{x}_{i_2}{x}_{i_3}\cdots x_{i_m}$$

and:

$${\left(x^{[m-1]}\right)}_i=x_i^{m-1}.$$

In particular, if $\lambda$ and x are restricted to the real field, then we call $\lambda$ an H-eigenvalue of $\mathcal{A}$ and x an H-eigenvector of $\mathcal{A}$ associated with $\lambda$ [14].

It is known that an mth-degree homogeneous polynomial of n variables $f\left(x\right)$ can be denoted as:

$$f\left(x\right)={\displaystyle \sum _{i_1,i_2,\dots ,i_m\in \mathbb{N}}}{a}_{i_1{i}_2\cdots i_m}{x}_{i_1}{x}_{i_2}\cdots x_{i_m},$$

where $x={(x_1,x_2,\dots ,x_n)}^T\in {\mathbb{R}}^n.$ The homogeneous polynomial $f\left(x\right)$ can be expressed as the tensor product of a symmetric tensor $\mathcal{A}$ with order m and dimension n and $x^m$ defined by:

$$f\left(x\right)\equiv \mathcal{A}{x}^m={\displaystyle \sum _{i_1,i_2,\dots ,i_m\in \mathbb{N}}}{a}_{i_1{i}_2\cdots i_m}{x}_{i_1}{x}_{i_2}\cdots x_{i_m},$$

where $x={(x_1,x_2,\dots ,x_n)}^T\in {\mathbb{R}}^n.$ When m is even, $f\left(x\right)$ is called positive definite if $f\left(x\right)>0$, for any $x\in {\mathbb{R}}^n\backslash \left\{0\right\}.$ The symmetric tensor $\mathcal{A}$ is called positive definite if $f\left(x\right)$ is positive definite [5].

It is well known that the positive definiteness of a multivariate polynomial $f\left(x\right)$ plays an important role in the stability study of non-linear autonomous systems [12,21]. However, for $n>3$ and $m>4$, it is a hard problem to identify the positive definiteness of such a multivariate form. To solve this problem, Qi [14] pointed out that $f\left(x\right)\equiv \mathcal{A}{x}^m$ is positive definite if and only if the real symmetric tensor $\mathcal{A}$ is positive definite and provided an eigenvalue method to verify the positive definiteness of $\mathcal{A}$ when m is even ([14], Theorem 1.1).

Let $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ be a complex tensor with order m and dimension n; we denote:

$$R_i\left(A\right)={\displaystyle \sum _{\begin{array}{c}{i}_2,\dots ,i_m\in \mathbb{N}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|={\displaystyle \sum _{i_2,\dots ,i_m\in \mathbb{N}}}|a_{i{i}_2\cdots i_m}|-|a_{ii\cdots i}|,\forall i\in \mathbb{N}.$$

Definition 1

([14]). Let $\mathcal{A}$ be a tensor with order m and dimension n. $\mathcal{A}$ is called a diagonally dominant tensor if $|a_{ii\cdots i}|\ge R_i\left(\mathcal{A}\right),\forall i\in \mathbb{N}.$$\mathcal{A}$ is called a strictly diagonally dominant tensor if $|a_{ii\cdots i}|>R_i\left(\mathcal{A}\right),\forall i\in \mathbb{N}.$

Definition 2

([22]). Let $\mathcal{A}$ be a complex tensor with order m and dimension n. $\mathcal{A}$ is called an $\mathcal{H}$-tensor if there is a positive vector $x={(x_1,x_2,\dots ,x_n)}^T\in {\mathbb{R}}^n,$ such that:

$$|a_{ii\cdots i}|x_i^{m-1}>{\displaystyle \sum _{\begin{array}{c}{i}_2,\dots ,i_m\in \mathbb{N}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|x_{i_2}\cdots x_{i_m},\forall i\in \mathbb{N}.$$

Definition 3

([23]). Let $\mathcal{A}$ be a complex tensor with order m and dimension n, $X=\mathrm{diag}(x_1,x_2,$$\dots ,x_n).$ Denote $\mathcal{B}=\left(b_{i_1{i}_2\cdots i_m}\right)=\mathcal{A}{X}^{m-1},$

$$b_{i_1{i}_2\cdots i_m}=a_{i_1{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_m},\forall i_j\in \mathbb{N},j=1,2,\dots ,m.$$

We call $\mathcal{B}$ the product of the tensor $\mathcal{A}$ and the matrix X.

Lemma 1

([14]). Let $\mathcal{A}$ be an even-order real symmetric tensor, then $\mathcal{A}$ is positive definite if and only if all of its H-eigenvalues are positive.

From Lemma 1, we can verify the positive definiteness of an even-order symmetric tensor $\mathcal{A}$ (the positive definiteness of the mth-degree homogeneous polynomial $f\left(x\right)$) by computing the H-eigenvalues of $\mathcal{A}$. In [6,24,25], for a non-negative tensor, some algorithms were provided to compute its largest eigenvalue. In [1,26], based on semi-definite programming approximation schemes, some algorithms were also given to compute the eigenvalues for general tensors with moderate sizes. However, it is not easy to compute all these H-eigenvalues when m and n are very large. Recently, by introducing the definition of $\mathcal{H}$-tensor, References [22,27] and Li et al. [27] provided a practical sufficient condition for identifying the positive definiteness of an even-order symmetric tensor (see Lemmas 2, 4, and 5).

Lemma 2

([22]). If $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)$ is a strictly diagonally dominant tensor, then $\mathcal{A}$ is an $\mathcal{H}$-tensor.

Lemma 3

([27]). Let $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)$ be an even-order real symmetric tensor of order m and dimension n with $a_{ii\cdots i}>0$ for all $i\in \mathbb{N}$. If $\mathcal{A}$ is an $\mathcal{H}$-tensor, then $\mathcal{A}$ is positive definite.

Lemma 4

([27]). Let $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ be a complex tensor with order m and dimension n. If there exists a positive diagonal matrix X such that $\mathcal{A}{X}^{m-1}$ is an $\mathcal{H}$-tensor, then $\mathcal{A}$ is an $\mathcal{H}$-tensor.

Lemma 5.

Let $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ be a complex tensor with order m and dimension n. If $\mathcal{A}$ is an $\mathcal{H}$-tensor, then there exists at least one index $i_0\in \mathbb{N}$ such that $|a_{i_0{i}_0\cdots i_0}|>R_{i_0}\left(\mathcal{A}\right)$.

Proof of Lemma 5. 

According to Definition 2, there is a positive vector $x=(x_1,x_2,\dots ,$$x_n{)}^T\in {\mathbb{R}}^n,$ such that:

$$|a_{ii\cdots i}|x_i^{m-1}>{\displaystyle \sum _{\begin{array}{c}{i}_2,\dots ,i_m\in \mathbb{N}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|x_{i_2}\cdots x_{i_m},\forall i\in \mathbb{N}.$$

Denote $x_{i_0}={\displaystyle \underset{i\in \mathbb{N}}{min}}\left\{x_i\right\}$, then for the index $i_0\in \mathbb{N}$, we have:

$$|a_{i_0{i}_0\cdots i_0}|>{\displaystyle \sum _{\begin{array}{c}{i}_2,\dots ,i_m\in \mathbb{N}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i_0{i}_2\cdots i_m}|\frac{x_{i_2}}{x_{i_0}}\cdots \frac{x_{i_m}}{x_{i_0}}\ge {\displaystyle \sum _{\begin{array}{c}{i}_2,\dots ,i_m\in \mathbb{N}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i_0{i}_2\cdots i_m}|=R_{i_0}\left(\mathcal{A}\right).$$

The proof is complete.    □

2. Practical Criteria for the $\mathcal{H}$-Tensor

Throughout this paper, we use the following definitions and notation.

$$\begin{array}{cc}\hfill & \mathbb{N}=\{1,2,\cdots ,n\}=\bigcup _{i=1}^k{\mathbb{N}}_i,{\mathbb{N}}_i\cap {\mathbb{N}}_j=\varnothing ,1\le i\ne j\le k.\hfill \\ \hfill & \forall t\in \{1,2,\cdots ,k\},{\mathbb{N}}_t^{m-1}=\{i_2{i}_3\cdots i_m|i_j\in {\mathbb{N}}_t,j=1,2,\cdots ,m\}.\hfill \\ \hfill & {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_t^{m-1}=\{i_2{i}_3\cdots i_m|i_2{i}_3\cdots i_m\in {\mathbb{N}}^{m-1},i_2{i}_3\cdots i_m\notin {\mathbb{N}}_t^{m-1}\}.\hfill \end{array}$$

For all $i\in \mathbb{N}$, there exists a constant $t\in \{1,2,\cdots ,k\}$ that satisfies $i\in {\mathbb{N}}_t$. Denote:

$$\begin{array}{cc}\hfill & {\alpha }_{{\mathbb{N}}_t}^{\left(i\right)}={\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_t^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|,{\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}={\displaystyle \sum _{i_2,\cdots ,i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_t^{m-1}}}|a_{i{i}_2\cdots i_m}|=R_i\left(\mathcal{A}\right)-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)}.\hfill \\ \hfill & {\mathbb{N}}^{*}=\{i||a_{ii\cdots i}|>R_i\left(\mathcal{A}\right),i\in \mathbb{N}\},\hfill \\ \hfill & {\mathbb{N}}^{+}=\{i||a_{ii\cdots i}|>{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)},i\in {\mathbb{N}}_t\subseteq \mathbb{N}\},{\mathbb{N}}^0=\{i||a_{ii\cdots i}|={\alpha }_{{\mathbb{N}}_t}^{\left(i\right)},i\in {\mathbb{N}}_t\subseteq \mathbb{N}\},\hfill \\ \hfill & {\mathbb{J}}^{+}=\{(i,j)|(|a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)})(|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_s}^{\left(j\right)})>{\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}·{\overline{\alpha }}_{{\mathbb{N}}_s}^{\left(j\right)},i\in {\mathbb{N}}_t,j\in {\mathbb{N}}_s,1\le t\ne s\le k\}\hfill \\ \hfill & {\mathbb{J}}^0=\{(i,j)|(|a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)})(|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_s}^{\left(j\right)})={\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}·{\overline{\alpha }}_{{\mathbb{N}}_s}^{\left(j\right)},i\in {\mathbb{N}}_t,j\in {\mathbb{N}}_s,1\le t\ne s\le k\}\hfill \end{array}$$

Definition 4.

Let $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)$ be a complex tensor with order m and dimension n. For all $i\in {\mathbb{N}}_t$, $j\in {\mathbb{N}}_s,1\le t\ne s\le k$, if: 

$$\begin{array}{cc}\hfill & (|a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)})(|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_s}^{\left(j\right)})\ge {\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}·{\overline{\alpha }}_{{\mathbb{N}}_s}^{\left(j\right)},\hfill \end{array}$$

(1)

we call $\mathcal{A}$ a locally double-diagonally dominant tensor and denote $\mathcal{A}\in LD{D}_{0}T$. If: 

$$\begin{array}{cc}\hfill & (|a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)})(|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_s}^{\left(j\right)})>{\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}·{\overline{\alpha }}_{{\mathbb{N}}_s}^{\left(j\right)},\hfill \end{array}$$

(2)

we call $\mathcal{A}$ a strictly locally double-diagonally dominant tensor and denote $\mathcal{A}\in LDDT$.

Remark 1.

If $\mathcal{A}\in LD{D}_{0}T$ and ${\mathbb{N}}^{*}\ne \varnothing$, then $\mathbb{N}={\mathbb{N}}^0\cup {\mathbb{N}}^{+}$; if $\mathcal{A}\in LDDT$ and ${\mathbb{N}}^{*}\ne \varnothing$, then $\mathbb{N}={\mathbb{N}}^{+}$.

Lemma 6.

Let $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ be a complex tensor with order m and dimension n. If $\mathcal{A}\in LDDT$ and ${\mathbb{N}}^{*}\ne \varnothing$, then there exists at most one $l_0\in \{1,2,\cdots ,k\}$ such that,

$$\begin{array}{cc}\hfill & |a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_{l_0}}^{\left(i\right)}\le {\overline{\alpha }}_{{\mathbb{N}}_{l_0}}^{\left(i\right)},\forall i\in {\mathbb{N}}_{l_0}.\hfill \end{array}$$

Proof of Lemma 6. 

If there exists $l_0^1\ne l_0^2\in \{1,2,\cdots ,k\}$, such that for all $i\in {\mathbb{N}}_{l_0^1},j\in {\mathbb{N}}_{l_0^2}$,

$$\begin{array}{cc}\hfill & |a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_{l_0^1}}^{\left(i\right)}\le {\overline{\alpha }}_{{\mathbb{N}}_{l_0^1}}^{\left(i\right)},|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_{l_0^2}}^{\left(j\right)}\le {\overline{\alpha }}_{{\mathbb{N}}_{l_0^2}}^{\left(j\right)}.\hfill \end{array}$$

Notice that $\mathcal{A}\in LDDT$ and ${\mathbb{N}}^{*}\ne \varnothing$. By Remark 1, we have:

$$\begin{array}{c}\hfill |a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_{l_0^1}}^{\left(i\right)}>0,|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_{l_0^2}}^{\left(j\right)}>0.\end{array}$$

These imply:

$$\begin{array}{cc}\hfill & (|a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_{l_0^1}}^{\left(i\right)})·(|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_{l_0^2}}^{\left(j\right)})\le {\overline{\alpha }}_{{\mathbb{N}}_{l_0^1}}^{\left(i\right)}·{\overline{\alpha }}_{{\mathbb{N}}_{l_0^2}}^{\left(j\right)},\hfill \end{array}$$

which contradicts $\mathcal{A}\in LDDT$. The proof is complete.    □

Remark 2.

Based on Lemma 6, when $\mathcal{A}\in LDDT$ and ${\mathbb{N}}^{*}\ne \varnothing$, we always assume that ${\mathbb{N}}_1=\{i||a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)}\le {\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)},i\in \mathbb{N}\}.$

Lemma 7.

Let $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ be a complex tensor with order m and dimension n, $\mathcal{A}\in LDDT$ and ${\mathbb{N}}^{*}\ne \varnothing$. If ${\mathbb{N}}_1=\varnothing$, then $\mathcal{A}$ is an $\mathcal{H}$-tensor.

Proof of Lemma 7. 

Since $\mathcal{A}\in LDDT$, ${\mathbb{N}}^{*}\ne \varnothing$ and ${\mathbb{N}}_1=\varnothing$. For all $i\in \mathbb{N}={\mathbb{N}}_2\cup {\mathbb{N}}_3\cup \cdots \cup {\mathbb{N}}_k$, we have:

$$\begin{array}{cc}\hfill & |a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)}>{\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)},t\in \{2,3,\cdots ,k\}.\hfill \end{array}$$

This implies:

$$\begin{array}{c}\hfill |a_{ii\cdots i}|>{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)}+{\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}=R_i\left(\mathcal{A}\right),i\in \mathbb{N}.\end{array}$$

By Lemma 2, $\mathcal{A}$ is an $\mathcal{H}$-tensor.    □

Theorem 1.

Let $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ be a complex tensor with order m and dimension n, $\mathcal{A}\in LDDT$ and ${\mathbb{N}}^{*}\ne \varnothing$. If for all $i\in {\mathbb{N}}_t,t\in \{2,3,\cdots ,k\},j\in {\mathbb{N}}_1\ne \varnothing$, we have:

$$\begin{array}{cc}\hfill & (|a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)})(|a_{jj\cdots j}|-{\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{m-1}\\ {\delta }_{j{i}_2\cdots i_m}=0\end{array}}}|a_{j{i}_2\cdots i_m}|)\hfill \\ \hfill & >{\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}·({\displaystyle \sum _{i_2{i}_3\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{m-1}}}|a_{j{i}_2{i}_3\cdots i_m}|),\hfill \end{array}$$

(3)

then $\mathcal{A}$ is an $\mathcal{H}$-tensor.

Proof of Theorem 1. 

From the inequality (3), there exists a positive constant d such that:

$$\begin{array}{cc}\hfill & {\displaystyle \underset{\begin{array}{c}i\in {\mathbb{N}}_t\\ t\in \{2,3,\cdots ,k\}\end{array}}{min}}\{\frac{|a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)}}{{\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}}\}>d>{\displaystyle \underset{j\in {\mathbb{N}}_1}{max}}\{\frac{{\displaystyle \sum _{i_2{i}_3\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{m-1}}}|a_{j{i}_2{i}_3\cdots i_m}|}{|a_{jj\cdots j}|-{\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{m-1}\\ {\delta }_{j{i}_2\cdots i_m}=0\end{array}}}|a_{j{i}_2\cdots i_m}|}\}.\hfill \end{array}$$

(4)

If ${\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}=0$, we denote $\frac{|a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)}}{{\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}}=+\infty$. It is obvious that $d>1$. Construct a positive diagonal matrix $X=\mathrm{diag}(x_1,x_2,\cdots ,x_n)$, where:

$$x_i=\left\{\begin{array}{c}1,i\in \mathbb{N}\backslash {\mathbb{N}}_1,\hfill \\ d^{\frac{1}{m-1}},i\in {\mathbb{N}}_1.\hfill \end{array}\right.$$

Let $\mathcal{B}=\mathcal{A}{X}^{m-1}=\left(b_{i_1{i}_2\cdots i_m}\right)$. For $i\in \mathbb{N}\backslash {\mathbb{N}}_1=\{{\mathbb{N}}_2,{\mathbb{N}}_3,\cdots ,{\mathbb{N}}_k\}$, by the first inequality in (4),

$$\begin{array}{cc}\hfill |b_{ii\cdots i}|& =|a_{ii\cdots i}|·\stackrel{m-1}{\overbrace{1·1\cdots 1}}\hfill \\ \hfill & =|a_{ii\cdots i}|\hfill \\ \hfill & >{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)}+{\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}·d\hfill \\ \hfill & ={\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_t^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{i_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_t^{m-1}}}|a_{i{i}_2\cdots i_m}|\stackrel{m-1}{\overbrace{d^{\frac{1}{m-1}}\cdots d^{\frac{1}{m-1}}}}\hfill \\ \hfill & \ge {\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_t^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{i_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_t^{m-1}}}|a_{i{i}_2\cdots i_m}|x_{i_2}\cdots x_{i_m}\hfill \\ \hfill & ={\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_t^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|b_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{i_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_t^{m-1}}}|b_{i{i}_2\cdots i_m}|\hfill \\ \hfill & =R_i\left(\mathcal{B}\right).\hfill \end{array}$$

For $i\in {\mathbb{N}}_1$, by the second inequality in (4),

$$\begin{array}{cc}\hfill |b_{ii\cdots i}|& =|a_{ii\cdots i}|·d\hfill \\ \hfill & =|a_{ii\cdots i}|·\stackrel{m-1}{\overbrace{d^{\frac{1}{m-1}}\cdots d^{\frac{1}{m-1}}}}\hfill \\ \hfill & >d·{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{i_2\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{m-1}}}|a_{i{i}_2\cdots i_m}|\hfill \\ \hfill & =d({\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_1^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{i_2\cdots i_m\in [{\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{m-1}]\backslash {\mathbb{N}}_1^{m-1}}}|a_{i{i}_2\cdots i_m}|)+{\displaystyle \sum _{i_2\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{m-1}}}|a_{i{i}_2\cdots i_m}|\hfill \\ \hfill & \ge {\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_1^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|·\stackrel{m-1}{\overbrace{d^{\frac{1}{m-1}}\cdots d^{\frac{1}{m-1}}}}\hfill \\ \hfill & +{\displaystyle \sum _{i_2\cdots i_m\in [{\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{m-1}]\backslash {\mathbb{N}}_1^{m-1}}}|a_{i{i}_2\cdots i_m}|x_{i_2}\cdots x_{i_m}+{\displaystyle \sum _{i_2\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{m-1}}}|a_{i{i}_2\cdots i_m}|\hfill \\ \hfill & ={\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_1^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|b_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{i_2{i}_3\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_1^{m-1}}}|b_{i{i}_2\cdots i_m}|\hfill \\ \hfill & =R_i\left(\mathcal{B}\right).\hfill \end{array}$$

Thus, we have proven that:

$$\begin{array}{cc}\hfill & |b_{ii\cdots i}|>{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|b_{i{i}_2\cdots i_m}|=R_i\left(\mathcal{B}\right),i\in \mathbb{N}.\hfill \end{array}$$

i.e., $\mathcal{B}$ is a strictly diagonally dominant tensor. By Lemmas 2 and 4, $\mathcal{A}$ is an $\mathcal{H}$-tensor. The proof is complete.    □

Lemma 8.

Let $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ be a complex tensor with order m and dimension n. If ${\mathbb{N}}^{*}\ne \varnothing$ and for all $i\in {\mathbb{N}}_t,j\in {\mathbb{N}}_s,1\le t\ne s\le k$,

$$\begin{array}{cc}\hfill & (|a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)})(|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_s}^{\left(j\right)})\le {\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)}·{\overline{\alpha }}_{{\mathbb{N}}_s}^{\left(j\right)},\hfill \end{array}$$

(5)

then there exists only one natural number $k_0\in \{1,2,\cdots ,k\}$ such that ${\mathbb{N}}^{*}\subseteq {\mathbb{N}}_{k_0}$.

Proof of Lemma 8. 

Since ${\mathbb{N}}^{*}\ne \varnothing$, then there exists one $k_0\in \{1,2,\cdots ,k\}$ such that ${\mathbb{N}}^{*}\cap {\mathbb{N}}_{k_0}\ne \varnothing$. If there exists another $k_1\in \{1,2,\cdots ,k\}\backslash \left\{k_0\right\}$ such that ${\mathbb{N}}^{*}\cap {\mathbb{N}}_{k_1}\ne \varnothing$, then we have:

$$\begin{array}{cc}\hfill & (|a_{ii\cdots i}|-{\alpha }_{{\mathbb{N}}_{k_0}\cap {\mathbb{N}}^{*}}^{\left(i\right)})(|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_{k_1}\cap {\mathbb{N}}^{*}}^{\left(j\right)})>{\overline{\alpha }}_{{\mathbb{N}}_{k_0}\cap {\mathbb{N}}^{*}}^{\left(i\right)}·{\overline{\alpha }}_{{\mathbb{N}}_{k_1}\cap {\mathbb{N}}^{*}}^{\left(j\right)}\hfill \end{array}$$

which contradicts the inequality (5). Therefore, there exists only one natural number $k_0\in \{1,2,\cdots ,k\}$ that satisfies ${\mathbb{N}}^{*}\subseteq {\mathbb{N}}_{k_0}$.    □

Remark 3.

Based on Lemma 8, when $\mathcal{A}$ satisfies (5) and ${\mathbb{N}}^{*}\ne \varnothing$, then we always assume that ${\mathbb{N}}^{*}\subseteq {\mathbb{N}}_k$ (where $\mathbb{N}={\mathbb{N}}_1\cup {\mathbb{N}}_2\cup \cdots \cup {\mathbb{N}}_k$).

Theorem 2.

Let $\mathcal{A}=\left(a_{i_1{i}_2\dots i_m}\right)$ be a complex tensor with order m and dimension n and ${\mathbb{N}}^{*}\ne \varnothing$. For all $i\in {\mathbb{N}}_k,j\in {\mathbb{N}}_s,s\in \{1,2,\cdots ,k-1\},$ if:

$$\begin{array}{cc}\hfill 0& <(|a_{ii\cdots i}|-{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|)(|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_s}^{\left(j\right)})\hfill \\ \hfill & \le ({\displaystyle \sum _{i_2\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}}}|a_{i{i}_2\cdots i_m}|)·{\overline{\alpha }}_{{\mathbb{N}}_s}^{\left(j\right)},\hfill \end{array}$$

(6)

then $\mathcal{A}$ is not an $\mathcal{H}$-tensor.

Proof of Theorem 2. 

From the inequality (6), there exists a positive constant $d_0$ such that:

$$\begin{array}{cc}\hfill & {\displaystyle \underset{i\in {\mathbb{N}}_k}{max}}\{\frac{|a_{ii\cdots i}|-{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|}{{\displaystyle \sum _{i_2\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}}}|a_{i{i}_2\cdots i_m}|}\}\le d_0\le {\displaystyle \underset{j\in \mathbb{N}\backslash {\mathbb{N}}_k}{min}}\{\frac{{\overline{\alpha }}_{{\mathbb{N}}_s}^{\left(j\right)}}{|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_s}^{\left(j\right)}}\}.\hfill \end{array}$$

(7)

If $|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_s}^{\left(j\right)}=0,j\in \mathbb{N}\backslash {\mathbb{N}}_k$, we denote $\frac{{\overline{\alpha }}_{{\mathbb{N}}_s}^{\left(j\right)}}{|a_{jj\cdots j}|-{\alpha }_{{\mathbb{N}}_s}^{\left(j\right)}}=+\infty$. Obviously,

$$\begin{array}{cc}\hfill & {\displaystyle \underset{i\in {\mathbb{N}}_k}{max}}\{\frac{|a_{ii\cdots i}|-{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|}{{\displaystyle \sum _{i_2\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}}}|a_{i{i}_2\cdots i_m}|}\}\ge {\displaystyle \underset{i\in {\mathbb{N}}^{*}}{max}}\{\frac{|a_{ii\cdots i}|-{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|}{{\displaystyle \sum _{i_2\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}}}|a_{i{i}_2\cdots i_m}|}\}>1,\hfill \end{array}$$

so $d_0>1$. Construct a positive diagonal matrix $X=\mathrm{diag}(x_1,x_2,\cdots ,x_n)$, where:

$$x_i=\left\{\begin{array}{c}1,i\in {\mathbb{N}}_k,\hfill \\ d_0^{\frac{1}{m-1}},i\in \mathbb{N}\backslash {\mathbb{N}}_k.\hfill \end{array}\right.$$

Let $\mathcal{B}=\mathcal{A}{X}^{m-1}=\left(b_{i_1{i}_2\cdots i_m}\right)$. For $i\in {\mathbb{N}}^{*}$, by the first inequality in (7),

$$\begin{array}{cc}\hfill |b_{ii\cdots i}|& =|a_{ii\cdots i}|·\stackrel{m-1}{\overbrace{1·1\cdots 1}}=|a_{ii\cdots i}|\hfill \\ \hfill & \le {\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|+d_0·{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}\end{array}}}|a_{i{i}_2\cdots i_m}|\hfill \\ \hfill & \le {\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_k^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in [{\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}]\backslash {\mathbb{N}}_k^{m-1}\end{array}}}|a_{i{i}_2\cdots i_m}|x_{i_2}\cdots x_{i_m}\hfill \\ \hfill & +{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}\end{array}}}|a_{i{i}_2\cdots i_m}|·\stackrel{m-1}{\overbrace{d^{\frac{1}{m-1}}\cdots d^{\frac{1}{m-1}}}}\hfill \\ \hfill & ={\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_k^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|b_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_k^{m-1}\end{array}}}|b_{i{i}_2\cdots i_m}|\hfill \\ \hfill & =R_i\left(B\right).\hfill \end{array}$$

For $i\in {\mathbb{N}}_k\backslash {\mathbb{N}}^{*}$,

$$\begin{array}{cc}\hfill |b_{ii\cdots i}|& =|a_{ii\cdots i}|·\stackrel{m-1}{\overbrace{1·1\cdots 1}}=|a_{ii\cdots i}|\hfill \\ \hfill & \le {\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_k^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_k^{m-1}\end{array}}}|a_{i{i}_2\cdots i_m}|\hfill \\ \hfill & \le {\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_k^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash ({\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_k^{m-1})\end{array}}}|a_{i{i}_2\cdots i_m}|x_{i_2}\cdots x_{i_m}\hfill \\ \hfill & +{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_k^{m-1}\end{array}}}|a_{i{i}_2\cdots i_m}|\stackrel{m-1}{\overbrace{d_0^{\frac{1}{m-1}}\cdots d_0^{\frac{1}{m-1}}}}\hfill \\ \hfill & ={\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_k^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|b_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_k^{m-1}\end{array}}}|b_{i{i}_2\cdots i_m}|\hfill \\ \hfill & =R_i\left(B\right).\hfill \end{array}$$

For $i\in \mathbb{N}\backslash {\mathbb{N}}_k={\mathbb{N}}_1\bigcup {\mathbb{N}}_2\bigcup \cdots \bigcup {\mathbb{N}}_{k-1}$, by the second inequality in (7),    

$$\begin{array}{cc}\hfill |b_{ii\cdots i}|& =d_0|a_{ii\cdots i}|=|a_{ii\cdots i}|\stackrel{m-1}{\overbrace{d_0^{\frac{1}{m-1}}\cdots d_0^{\frac{1}{m-1}}}}\hfill \\ \hfill & \le d_0·{\alpha }_{{\mathbb{N}}_s}^{\left(i\right)}+{\overline{\alpha }}_{{\mathbb{N}}_s}^{\left(i\right)}\hfill \\ \hfill & ={\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_s^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|\stackrel{m-1}{\overbrace{d_0^{\frac{1}{m-1}}\cdots d_0^{\frac{1}{m-1}}}}+{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_s^{m-1}\end{array}}}|a_{i{i}_2\cdots i_m}|\hfill \\ \hfill & \le {\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_s^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|\stackrel{m-1}{\overbrace{d_0^{\frac{1}{m-1}}\cdots d_0^{\frac{1}{m-1}}}}+{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_s^{m-1}\end{array}}}|a_{i{i}_2\cdots i_m}|x_{i_2}\cdots x_{i_m}\hfill \\ \hfill & ={\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}_s^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|b_{i{i}_2\cdots i_m}|+{\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {\mathbb{N}}_s^{m-1}\end{array}}}|b_{i{i}_2\cdots i_m}|\hfill \\ \hfill & =R_i\left(B\right).\hfill \end{array}$$

Thus we have proved that:

$$\begin{array}{cc}\hfill & |b_{ii\cdots i}|\le {\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|b_{i{i}_2\cdots i_m}|=R_i\left(\mathcal{B}\right),i\in \mathbb{N}.\hfill \end{array}$$

i.e., $\mathcal{B}$ is not strictly diagonally dominant for all $i\in \mathbb{N}$. By Lemma 5, $\mathcal{B}$ is not an $\mathcal{H}$-tensor. By Lemma 4, $\mathcal{A}$ is not an $\mathcal{H}$-tensor. The proof is complete.    □

3. An Algorithm for Identifying $\mathcal{H}$-Tensors

Based on the results in the above section, an algorithm for identifying $\mathcal{H}$-tensors is put forward in this section:

Remark 4. 

(a)

$s=$ the total number of tensors, $k_1=$ the number of $\mathcal{H}$-tensors, $k_2=$ the number of tensors which are not $\mathcal{H}$-tensor, and $s-k_1-k_2=$ the number of tensors, which are not checkable by using Algorithm 1;

(b)

The calculations of Algorithm 1 only depend on the elements of the tensor, so Algorithm 1 stops after a finite amount of steps.

Algorithm 1 An algorithm for identifying $\mathcal{H}$-tensors

Step 1.  Set $k_1:=0,k_2:=0,k_3:=0$ and $s:=50$. Step 2. Given a complex tensor $\mathcal{A}=\left(a_{i_1\cdots i_m}\right)$. If $k_3=s$, then output $k_1$ and $k_2$, and stop. Otherwise, Step 3. Compute $|a_{i\cdots i}|$ and $R_i\left(\mathcal{A}\right)$ for all $i\in \mathbb{N}$. Step 4. If $N^{*}=N$, then print “$\mathcal{A}$ is a $\mathcal{H}$-tensor”, and go to Step 5. Otherwise, go to Step 6. Step 5. Replace $k_1$ by $k_1+1$, and replace $k_3$ by $k_3+1$, then go to Step 2. Step 6. If $N^{*}=\varnothing$, then print “$\mathcal{A}$ is not a $\mathcal{H}$-tensor”, and go to Step 7. Otherwise, go to Step 8. Step 7. Replace $k_2$ by $k_2+1$, and replace $k_3$ by $k_3+1$, then go to Step 2. Step 8. Compute $|a_{i\cdots i}|,{\alpha }_{{\mathbb{N}}_t}^{\left(i\right)},{\overline{\alpha }}_{{\mathbb{N}}_t}^{\left(i\right)},|a_{j\cdots j}|,{\alpha }_{{\mathbb{N}}_s}^{\left(j\right)},{\overline{\alpha }}_{{\mathbb{N}}_s}^{\left(j\right)}$ for all $i\in {\mathbb{N}}_t$, $j\in {\mathbb{N}}_s,1\le t\ne s\le k$. Step 9. If Inequality (3) holds, then print “$\mathcal{A}$ is a $\mathcal{H}$-tensor”, and go to Step 5. Otherwise, Step 10. Compute: $${\displaystyle \sum _{\begin{array}{c}{i}_2\cdots i_m\in {\mathbb{N}}^{m-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}\\ {\delta }_{i{i}_2\cdots i_m}=0\end{array}}}|a_{i{i}_2\cdots i_m}|\mathrm{and}{\displaystyle \sum _{i_2\cdots i_m\in {(\mathbb{N}\backslash {\mathbb{N}}_k)}^{m-1}}}|a_{i{i}_2\cdots i_m}|.$$ Step 11. If Inequality (6) holds, then print “$\mathcal{A}$ is not a $\mathcal{H}$-tensor”, and go to Step 5. Otherwise, Step 12. Print “Whether A is a strong $\mathcal{H}$-tensor is not checkable”, and replace $k_3$ by $k_3+1$. Go to Step 2.

4. Numerical Examples

Example 1.

Consider a tensor $\mathcal{A}=\left(a_{i_1{i}_2{i}_3}\right)$, with order three and dimension six, defined as follows:

$$\begin{array}{cc}\hfill & a_{111}=3.9,a_{112}=a_{121}=a_{122}=1,a_{113}=a_{131}=a_{133}=a_{144}=a_{155}=0.2,\hfill \\ \hfill & a_{222}=5,a_{211}=a_{212}=a_{221}=a_{235}=a_{236}=1,a_{214}=a_{215}=a_{232}=a_{256}=a_{266}=0.2,\hfill \\ \hfill & a_{333}=4,a_{334}=a_{343}=0.5,a_{344}=1,a_{311}=a_{322}=a_{331}=a_{326}=a_{352}=0.2,\hfill \\ \hfill & a_{444}=6.5,a_{433}=a_{434}=a_{443}=1,a_{412}=a_{414}=a_{421}=a_{441}=0.5,\hfill \\ \hfill & a_{555}=6.3,a_{556}=1,a_{512}=a_{515}=a_{521}=a_{523}=a_{551}=a_{565}=a_{566}=0.5,\hfill \\ \hfill & a_{666}=8,a_{655}=a_{656}=a_{665}=1,a_{625}=a_{626}=a_{652}=a_{662}=0.5\hfill \end{array}$$

and other $a_{i_1{i}_2{i}_3}=0.$ By calculations, we have:

$$\begin{array}{c}\hfill R_1\left(\mathcal{A}\right)=4,R_2\left(\mathcal{A}\right)=6,R_3\left(\mathcal{A}\right)=3,R_4\left(\mathcal{A}\right)=5,R_5\left(\mathcal{A}\right)=4.5,R_6\left(\mathcal{A}\right)=5,\end{array}$$

$$\begin{array}{c}\hfill {\mathbb{N}}_1=\{1,2\},{\mathbb{N}}_2=\{3,4\},{\mathbb{N}}_3=\{5,6\},\mathbb{N}\backslash {\mathbb{N}}_1=\{3,4,5,6\}={\mathbb{N}}_2\bigcup {\mathbb{N}}_3,\end{array}$$

$$\begin{array}{c}{\mathbb{N}}_1^{3-1}={\mathbb{N}}_1^2=\{11,12,21,22\},{\mathbb{N}}_2^{3-1}={\mathbb{N}}_2^2=\{33,34,43,44\},{\mathbb{N}}_3^{3-1}={\mathbb{N}}_3^2=\{55,56,65,66\},\hfill \\ {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{3-1}={(\mathbb{N}\backslash {\mathbb{N}}_1)}^2=\{33,34,35,36,43,44,45,46,53,54,55,56,63,64,65,66\},\hfill \\ {\mathbb{N}}^{3-1}\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^{3-1}={\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2=\{11,12,13,14,15,16,21,22,23,24,25,26,31,32,41,42,51,52,61,62\}.\hfill \end{array}$$

$$\begin{array}{c}\hfill {\alpha }_{{\mathbb{N}}_1}^{\left(1\right)}={\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}_1^2\\ {\delta }_{1i_2{i}_3}=0\end{array}}}|a_{1i_2{i}_3}|=|a_{112}|+|a_{121}|+|a_{122}|=1+1+1=3.\end{array}$$

$$\begin{array}{c}\hfill {\alpha }_{{\mathbb{N}}_1}^{\left(2\right)}={\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}_1^2\\ {\delta }_{2i_2{i}_3}=0\end{array}}}|a_{2i_2{i}_3}|=|a_{211}|+|a_{212}|+|a_{221}|=1+1+1=3.\end{array}$$

$$\begin{array}{c}\hfill {\alpha }_{{\mathbb{N}}_2}^{\left(3\right)}={\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}_2^2\\ {\delta }_{3i_2{i}_3}=0\end{array}}}|a_{3i_2{i}_3}|=|a_{334}|+|a_{343}|+|a_{344}|=0.5+0.5+1=2.\end{array}$$

$$\begin{array}{c}\hfill {\alpha }_{{\mathbb{N}}_2}^{\left(4\right)}={\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}_2^2\\ {\delta }_{4i_2{i}_3}=0\end{array}}}|a_{4i_2{i}_3}|=|a_{433}|+|a_{434}|+|a_{443}|=1+1+1=3.\end{array}$$

$$\begin{array}{c}\hfill {\alpha }_{{\mathbb{N}}_3}^{\left(5\right)}={\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}_3^2\\ {\delta }_{5i_2{i}_3}=0\end{array}}}|a_{5i_2{i}_3}|=|a_{556}|+|a_{565}|+|a_{566}|=1+0.5+0.5=2.\end{array}$$

$$\begin{array}{c}\hfill {\alpha }_{{\mathbb{N}}_3}^{\left(6\right)}={\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}_3^2\\ {\delta }_{6i_2{i}_3}=0\end{array}}}|a_{6i_2{i}_3}|=|a_{655}|+|a_{656}|+|a_{665}|=1+1+1=3.\end{array}$$

$$\begin{array}{c}\hfill {\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\\ {\delta }_{1i_2{i}_3}=0\end{array}}}|a_{1i_2{i}_3}|=|a_{112}|+|a_{121}|+|a_{122}|+|a_{113}|+|a_{131}|+0=1+1+1+0.2+0.2=3.4.\end{array}$$

$$\begin{array}{c}\hfill {\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\\ {\delta }_{2i_2{i}_3}=0\end{array}}}|a_{2i_2{i}_3}|=|a_{211}|+|a_{212}|+|a_{221}|+|a_{214}|+|a_{215}|+|a_{232}|+0=1+1+1+0.2+0.2+0.2=3.6.\end{array}$$

$$\begin{array}{c}\hfill {\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in \backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\end{array}}}|a_{1i_2{i}_3}|=|a_{133}|+|a_{144}|+|a_{155}|+0=0.2+0.2+0.2=0.6.\end{array}$$

$$\begin{array}{c}\hfill {\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\end{array}}}|a_{2i_2{i}_3}|=|a_{235}|+|a_{236}|+|a_{256}|+|a_{266}|+0=1+1+0.2+0.2=2.4.\end{array}$$

$$\begin{array}{c}\hfill {\overline{\alpha }}_{{\mathbb{N}}_1}^{\left(1\right)}=1,{\overline{\alpha }}_{{\mathbb{N}}_1}^{\left(2\right)}=3,{\overline{\alpha }}_{{\mathbb{N}}_2}^{\left(3\right)}=1,{\overline{\alpha }}_{{\mathbb{N}}_2}^{\left(4\right)}=2,{\overline{\alpha }}_{{\mathbb{N}}_3}^{\left(5\right)}=2.5,{\overline{\alpha }}_{{\mathbb{N}}_3}^{\left(6\right)}=2.\end{array}$$

$$\begin{array}{c}\hfill (|a_{111}|-{\alpha }_{{\mathbb{N}}_1}^{\left(1\right)})·(|a_{333}|-{\alpha }_{{\mathbb{N}}_2}^{\left(3\right)})=(3.9-3)\times (4-2)=1.8>{\overline{\alpha }}_{{\mathbb{N}}_1}^{\left(1\right)}·{\overline{\alpha }}_{{\mathbb{N}}_2}^{\left(3\right)}=1\times 1=1.\end{array}$$

$$\begin{array}{c}\hfill (|a_{111}|-{\alpha }_{{\mathbb{N}}_1}^{\left(1\right)})·(|a_{444}|-{\alpha }_{{\mathbb{N}}_2}^{\left(4\right)})=(3.9-3)\times (6.5-3)=3.15>{\overline{\alpha }}_{{\mathbb{N}}_1}^{\left(1\right)}·{\overline{\alpha }}_{{\mathbb{N}}_2}^{\left(4\right)}=1\times 2=2.\end{array}$$

$$\begin{array}{c}\hfill (|a_{111}|-{\alpha }_{{\mathbb{N}}_1}^{\left(1\right)})·(|a_{555}|-{\alpha }_{{\mathbb{N}}_3}^{\left(5\right)})=(3.9-3)\times (6.3-2)=3.87>{\overline{\alpha }}_{{\mathbb{N}}_1}^{\left(1\right)}·{\overline{\alpha }}_{{\mathbb{N}}_3}^{\left(5\right)}=1\times 2.5=2.5.\end{array}$$

$$\begin{array}{c}\hfill (|a_{111}|-{\alpha }_{{\mathbb{N}}_1}^{\left(1\right)})·(|a_{666}|-{\alpha }_{{\mathbb{N}}_3}^{\left(6\right)})=(3.9-3)\times (8-3)=4.5>{\overline{\alpha }}_{{\mathbb{N}}_1}^{\left(1\right)}·{\overline{\alpha }}_{{\mathbb{N}}_3}^{\left(6\right)}=1\times 2=2.\end{array}$$

$$\begin{array}{c}\hfill (|a_{222}|-{\alpha }_{{\mathbb{N}}_1}^{\left(2\right)})·(|a_{333}|-{\alpha }_{{\mathbb{N}}_2}^{\left(3\right)})=(5-3)\times (4-2)=4>{\overline{\alpha }}_{{\mathbb{N}}_1}^{\left(2\right)}·{\overline{\alpha }}_{{\mathbb{N}}_2}^{\left(3\right)}=3\times 1=3.\end{array}$$

$$\begin{array}{c}\hfill (|a_{222}|-{\alpha }_{{\mathbb{N}}_1}^{\left(2\right)})·(|a_{444}|-{\alpha }_{{\mathbb{N}}_2}^{\left(4\right)})=(5-3)\times (6.5-3)=7>{\overline{\alpha }}_{{\mathbb{N}}_1}^{\left(2\right)}·{\overline{\alpha }}_{{\mathbb{N}}_2}^{\left(4\right)}=3\times 2=6.\end{array}$$

$$\begin{array}{c}\hfill (|a_{222}|-{\alpha }_{{\mathbb{N}}_1}^{\left(2\right)})·(|a_{555}|-{\alpha }_{{\mathbb{N}}_3}^{\left(5\right)})=(5-3)\times (6.3-2)=8.6>{\overline{\alpha }}_{{\mathbb{N}}_1}^{\left(2\right)}·{\overline{\alpha }}_{{\mathbb{N}}_3}^{\left(5\right)}=3\times 2.5=7.5.\end{array}$$

$$\begin{array}{c}\hfill (|a_{222}|-{\alpha }_{{\mathbb{N}}_1}^{\left(2\right)})·(|a_{666}|-{\alpha }_{{\mathbb{N}}_3}^{\left(6\right)})=(5-3)\times (8-3)=10>{\overline{\alpha }}_{{\mathbb{N}}_1}^{\left(2\right)}·{\overline{\alpha }}_{{\mathbb{N}}_3}^{\left(6\right)}=3\times 2=6.\end{array}$$

From Definition 4, we have $\mathcal{A}\in LDDT$. Furthermore,

$$\begin{array}{cc}\hfill & (|a_{333}|-{\alpha }_{{\mathbb{N}}_2}^{\left(3\right)})·(|a_{111}|-{\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\\ {\delta }_{1i_2{i}_3}=0\end{array}}}|a_{1i_2{i}_3}|)=(4-2)\times (3.9-3.4)=1\hfill \\ \hfill & >{\overline{\alpha }}_{{\mathbb{N}}_2}^{\left(3\right)}·({\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\end{array}}}|a_{1i_2{i}_3}|)=1\times 0.6=0.6,\hfill \end{array}$$

$$\begin{array}{cc}\hfill & (|a_{333}|-{\alpha }_{{\mathbb{N}}_2}^{\left(3\right)})·(|a_{222}|-{\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\\ {\delta }_{2i_2{i}_3}=0\end{array}}}|a_{2i_2{i}_3}|)=(4-2)\times (5-3.6)=2.8\hfill \\ \hfill & >{\overline{\alpha }}_{{\mathbb{N}}_2}^{\left(3\right)}·({\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\end{array}}}|a_{2i_2{i}_3}|)=1\times 2.4=2.4,\hfill \end{array}$$

$$\begin{array}{cc}\hfill & (|a_{444}|-{\alpha }_{{\mathbb{N}}_2}^{\left(4\right)})·(|a_{111}|-{\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\\ {\delta }_{1i_2{i}_3}=0\end{array}}}|a_{1i_2{i}_3}|)=(6.5-3)\times (3.9-3.4)=1.75\hfill \\ \hfill & >{\overline{\alpha }}_{{\mathbb{N}}_2}^{\left(4\right)}·({\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\end{array}}}|a_{1i_2{i}_3}|)=2\times 0.6=1.2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill & (|a_{444}|-{\alpha }_{{\mathbb{N}}_2}^{\left(4\right)})·(|a_{222}|-{\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\\ {\delta }_{2i_2{i}_3}=0\end{array}}}|a_{2i_2{i}_3}|)=(6.5-3)\times (5-3.6)=4.9\hfill \\ \hfill & >{\overline{\alpha }}_{{\mathbb{N}}_2}^{\left(4\right)}·({\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\end{array}}}|a_{2i_2{i}_3}|)=2\times 2.4=4.8,\hfill \end{array}$$

$$\begin{array}{cc}\hfill & (|a_{555}|-{\alpha }_{{\mathbb{N}}_3}^{\left(5\right)})·(|a_{111}|-{\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\\ {\delta }_{1i_2{i}_3}=0\end{array}}}|a_{1i_2{i}_3}|)=(6.3-2)\times (3.9-3.4)=2.15\hfill \\ \hfill & >{\overline{\alpha }}_{{\mathbb{N}}_3}^{\left(5\right)}·({\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\end{array}}}|a_{1i_2{i}_3}|)=2.5\times 0.6=1.5,\hfill \end{array}$$

$$\begin{array}{cc}\hfill & (|a_{555}|-{\alpha }_{{\mathbb{N}}_3}^{\left(5\right)})·(|a_{222}|-{\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\\ {\delta }_{2i_2{i}_3}=0\end{array}}}|a_{2i_2{i}_3}|)=(6.3-2)\times (5-3.6)=6.02\hfill \\ \hfill & >{\overline{\alpha }}_{{\mathbb{N}}_3}^{\left(5\right)}·({\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\end{array}}}|a_{2i_2{i}_3}|)=2.5\times 2.4=6,\hfill \end{array}$$

$$\begin{array}{cc}\hfill & (|a_{666}|-{\alpha }_{{\mathbb{N}}_3}^{\left(6\right)})·(|a_{111}|-{\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\\ {\delta }_{1i_2{i}_3}=0\end{array}}}|a_{1i_2{i}_3}|)=(8-3)\times (3.9-3.4)=2.5\hfill \\ \hfill & >{\overline{\alpha }}_{{\mathbb{N}}_3}^{\left(6\right)}·({\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\end{array}}}|a_{1i_2{i}_3}|)=2\times 0.6=1.2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill & (|a_{666}|-{\alpha }_{{\mathbb{N}}_3}^{\left(6\right)})·(|a_{222}|-{\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {\mathbb{N}}^2\backslash {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\\ {\delta }_{2i_2{i}_3}=0\end{array}}}|a_{2i_2{i}_3}|)=(8-3)\times (5-3.6)=7\hfill \\ \hfill & >{\overline{\alpha }}_{{\mathbb{N}}_3}^{\left(6\right)}·({\displaystyle \sum _{\begin{array}{c}{i}_2{i}_3\in {(\mathbb{N}\backslash {\mathbb{N}}_1)}^2\end{array}}}|a_{2i_2{i}_3}|)=2\times 2.4=4.8.\hfill \end{array}$$

These imply that $\mathcal{A}$ satisfies all the conditions of Theorem 1. Therefore, $\mathcal{A}$ is an $\mathcal{H}$-tensor.

5. Conclusions

In this paper, several practical criteria for identifying $\mathcal{H}$-tensors were given. These criteria only depend on the self-constituent elements of the tensors. Therefore, they are easy to implement. Based on the above criteria, we also gave the corresponding algorithm to determine the $\mathcal{H}$-tensor.