**1. Introduction**

Tensor theory is widely used in digital signal processing, medical image processing, data mining, quantum entanglement, and other fields [1–7]. 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 differentialequationsandthedegreeofdiscretizationofnonlinearparabolicequations[8–13].

 Let C(R) be the complex (real) field and N = {1, 2, ... , *<sup>n</sup>*}. A complex (real)-order *m* dimension *n* tensor A = (*ai*1*i*2...*im*) consists of *n<sup>m</sup>* complex (real) entries:

$$a\_{i\_1 i\_2 \dots i\_m} \in \mathbb{C}(\mathbb{R})\_\*$$

where *ij* = 1, 2, ... , *n* and *j* = 1, 2, ... , *m*. A tensor A = (*ai*1*i*2...*im* ) is called symmetric [14], if:

$$a\_{i\_1 i\_2 \dots i\_m} = a\_{\pi(i\_1 i\_2 \dots i\_m)} \forall \pi \in \Pi\_{m \nu}$$

where Π*m* is the permutation group of *m* indices. Furthermore, a tensor I = (*<sup>δ</sup>i*1*i*2...*im* ) is called the unit tensor [15], if its entries:

$$
\delta\_{i\_1 i\_2 \dots i\_m} = \begin{cases} 1, \text{ if } \ i\_1 = i\_2 = \dots = i\_{m'} \\ 0, \text{ otherwise.} \end{cases}
$$

Let A = (*ai*1*i*2...*im* ) be a tensor with order *m* and dimension *n*. If there exist a complex number *λ* and a non-zero complex vector *x* = (*<sup>x</sup>*1, *x*2, ... , *xn*)*<sup>T</sup>* that are solutions of the following homogeneous polynomial equations:

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

then we call *λ* an eigenvalue of A and *x* an eigenvector of A associated with *λ* [14,16–20] , and A*xm*−<sup>1</sup> and *λx*[*<sup>m</sup>*−<sup>1</sup>] are vectors, whose *i*th components are:

$$(\mathcal{A}x^{\mathfrak{m}-1})\_i = \sum\_{i\_2, i\_3, \dots, i\_{\mathfrak{m}} \in \mathbb{N}} a\_{i i\_2 i\_3 \dots i\_{\mathfrak{m}}} \mathfrak{x}\_{i\_2} \mathfrak{x}\_{i\_3} \dotsm \mathfrak{x}\_{i\_{\mathfrak{m}}}$$

**Citation:** Li, M.; Sang, H.; Liu, P.; Huang, G. Practical Criteria for H-Tensors and Their Application. *Symmetry* **2022**, *14*, 155. https:// doi.org/10.3390/sym14010155

Academic Editor: Roman Orus

Received: 15 December 2021 Accepted: 11 January 2022 Published: 13 January 2022

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**Copyright:** © 2022 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

and:

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

In particular, if *λ* and *x* are restricted to the real field, then we call *λ* an *H*-eigenvalue of A and *x* an *H*-eigenvector of A associated with *λ* [14].

It is known that an *m*th-degree homogeneous polynomial of *n* variables *f*(*x*) can be denoted as:

$$f(\mathbf{x}) = \sum\_{\bar{i}\_1, \bar{i}\_2, \dots, \bar{i}\_m \in \mathbb{N}} a\_{\bar{i}\_1 \bar{i}\_2 \dots \bar{i}\_m} \mathbf{x}\_{\bar{i}\_1} \mathbf{x}\_{\bar{i}\_2} \cdot \dots \cdot \mathbf{x}\_{\bar{i}\_m \cdot \mathbf{x}}$$

where *x* = (*<sup>x</sup>*1, *x*2, ... , *xn*)*<sup>T</sup>* ∈ R*<sup>n</sup>*. The homogeneous polynomial *f*(*x*) can be expressed as the tensor product of a symmetric tensor A with order *m* and dimension *n* and *x<sup>m</sup>* defined by:

$$f(\mathbf{x}) \equiv \mathcal{A}\mathbf{x}^{\mathrm{m}} = \sum\_{\boldsymbol{i}\_{1}, \boldsymbol{i}\_{2}, \dots, \boldsymbol{i}\_{m} \in \mathbb{N}} a\_{\boldsymbol{i}\_{1}\boldsymbol{i}\_{2}, \dots, \boldsymbol{i}\_{m}} \mathbb{1}\_{\boldsymbol{i}\_{1}} \mathbb{1}\_{\boldsymbol{i}\_{2}} \cdot \cdots \cdot \mathbb{1}\_{\boldsymbol{i}\_{m}\boldsymbol{i}\_{m}}$$

where *x* = (*<sup>x</sup>*1, *x*2, ... , *xn*)*<sup>T</sup>* ∈ R*<sup>n</sup>*. When *m* is even, *f*(*x*) is called positive definite if *f*(*x*) > 0, for any *x* ∈ R*n* \ {0}. The symmetric tensor A is called positive definite if *f*(*x*) is positive definite [5].

It is well known that the positive definiteness of a multivariate polynomial *f*(*x*) 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*(*x*) ≡ A*x<sup>m</sup>* is positive definite if and only if the real symmetric tensor A is positive definite and provided an eigenvalue method to verify the positive definiteness of A when *m* is even ([14], Theorem 1.1).

Let A = (*ai*1*i*2...*im*) be a complex tensor with order *m* and dimension *n*; we denote:

$$R\_i(A) = \sum\_{\substack{i\_2, \ldots, i\_m \in \mathbb{N} \\ \delta i\_2 \cdots i\_m = 0}} |a\_{i i\_2 \cdots i\_m}| = \sum\_{i\_2, \ldots, i\_m \in \mathbb{N}} |a\_{i i\_2 \cdots i\_m}| - |a\_{i i \cdots i}|\_\prime \quad \forall i \in \mathbb{N}.$$

**Definition 1** ([14])**.** *Let* A *be a tensor with order m and dimension n.* A *is called a diagonally dominant tensor if* |*aii*···*<sup>i</sup>*| ≥ *Ri*(A), ∀*i* ∈ N. A *is called a strictly diagonally dominant tensor if* |*aii*···*<sup>i</sup>*| > *Ri*(A), ∀*i* ∈ N.

**Definition 2** ([22])**.** *Let* A *be a complex tensor with order m and dimension n.* A *is called an* H*-tensor if there is a positive vector x* = (*<sup>x</sup>*1, *x*2,..., *xn*)*<sup>T</sup>* ∈ R*<sup>n</sup>*, *such that:*

$$|a|\_{i i \cdots i} |\mathbf{x}\_i^{m-1}| > \sum\_{\substack{i\_2, \dots, i\_m \in \mathbb{N} \\ i\_{i i\_2 \cdots i\_m} = 0}} |a\_{i i\_2 \cdots i\_m}| \mathbf{x}\_{i\_2} \cdot \cdots \cdot \mathbf{x}\_{i\_{m'}} \quad \forall i \in \mathbb{N}.$$

**Definition 3** ([23])**.** *Let* A *be a complex tensor with order m and dimension n, X* = diag(*<sup>x</sup>*1, *x*2, ..., *xn*). *Denote* B = (*bi*1*i*2···*im* ) = A*Xm*−1,

$$b\_{i\_1 i\_2 \cdots i\_m} = a\_{i\_1 i\_2 \cdots i\_m} \mathbf{x}\_{i\_2} \cdots \mathbf{x}\_{i\_{m'}} \quad \forall i\_j \in \mathbb{N}, \ j = 1, 2, \dots, m.$$

*We call* B *the product of the tensor* A *and the matrix X.*

**Lemma 1** ([14])**.** *Let* A *be an even-order real symmetric tensor, then* 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 A (the positive definiteness of the *m*th-degree homogeneous polynomial *f*(*x*)) by computing the *H*-eigenvalues of 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 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* A = (*ai*1*i*2···*im* ) *is a strictly diagonally dominant tensor, then* A *is an* H*-tensor.*

**Lemma 3** ([27])**.** *Let* A = (*ai*1*i*2···*im* ) *be an even-order real symmetric tensor of order m and dimension n with aii*···*i* > 0 *for all i* ∈ N*. If* A *is an* H*-tensor, then* A *is positive definite.*

**Lemma 4** ([27])**.** *Let* A = (*ai*1*i*2...*im* ) *be a complex tensor with order m and dimension n. If there exists a positive diagonal matrix X such that* A*Xm*−<sup>1</sup> *is an* H*-tensor, then* A *is an* H*-tensor.*

**Lemma 5.** *Let* A = (*ai*1*i*2...*im* ) *be a complex tensor with order m and dimension n. If* A *is an* H*-tensor, then there exists at least one index i*0 ∈ N *such that* |*ai*0*i*0···*i*<sup>0</sup>| > *Ri*0(A)*.*

**Proof of Lemma 5.** According to Definition 2, there is a positive vector *x* = (*<sup>x</sup>*1, *x*2, ... , *xn*)*<sup>T</sup>* ∈ R*<sup>n</sup>*, such that:

$$\mathbb{P}\left|a\_{i\bar{i}\cdots i}\right|\mathbf{x}\_{\bar{i}}^{\text{nr}-1} > \sum\_{\substack{i\_{2},\ldots,i\_{m}\in\mathbb{N}\\\delta\_{i\bar{i}\_{2},\ldots i\_{m}}=0}} \left|a\_{i\bar{i}\_{2},\ldots,i\_{m}}\right|\mathbf{x}\_{\bar{i}\_{2}}\cdot\cdots\cdot\mathbf{x}\_{i\_{m}\prime} \quad \forall i\in\mathbb{N}.$$

Denote *xi*0 = min *i*∈N{*xi*}, then for the index *i*0 ∈ N, we have:

$$|a\_{i0:i\_0\cdots i\_0}| > \sum\_{\substack{i\_2,\ldots,i\_m \in \mathbb{N} \\ \delta\_{i i\_2\cdots i\_m} = 0}} |a\_{i0i\_2\cdots i\_m}| \frac{\varkappa\_{i\_2}}{\varkappa\_{i\_0}} \cdots \frac{\varkappa\_{i\_m}}{\varkappa\_{i\_0}} \ge \sum\_{\substack{i\_2,\ldots,i\_m \in \mathbb{N} \\ \delta\_{i i\_2\cdots i\_m} = 0}} |a\_{i0i\_2\cdots i\_m}| = R\_{i0}(\mathcal{A}).$$

The proof is complete.

#### **2. Practical Criteria for the** H**-Tensor**

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

$$\mathbb{N} = \{1, 2, \cdots, m\} = \bigcup\_{i=1}^{k} \mathbb{N}\_i, \mathbb{N}\_i \cap \mathbb{N}\_j = \mathcal{O}, 1 \le i \ne j \le k.$$

$$\forall t \in \{1, 2, \cdots, \cdot, k\}, \mathbb{N}\_t^{m-1} = \{i\_2 i\_3 \cdots i\_m | i\_j \in \mathbb{N}\_t, j = 1, 2, \cdots, m\}.$$

$$\mathbb{N}^{m-1} \backslash \mathbb{N}\_t^{m-1} = \{i\_2 i\_3 \cdots i\_m | i\_2 i\_3 \cdots i\_m \in \mathbb{N}^{m-1}, i\_2 i\_3 \cdots i\_m \notin \mathbb{N}\_t^{m-1}\}.$$

For all *i* ∈ N, there exists a constant *t* ∈ {1, 2, ··· , *k*} that satisfies *i* ∈ N*<sup>t</sup>*. Denote:

*α*(*i*) N*t* = ∑ *<sup>i</sup>*2···*im*∈N*<sup>m</sup>*−<sup>1</sup> *t <sup>δ</sup>ii*2···*im* =0 | *aii*<sup>2</sup>···*im* |, *α*(*i*) N*t* = ∑ *i*2,··· ,*im*∈N*<sup>m</sup>*−<sup>1</sup>\N*<sup>m</sup>*−<sup>1</sup> *t* | *aii*<sup>2</sup>···*im* |= *Ri*(A) − *α*(*i*) N*t* . N∗ = {*i*| | *aii*···*i* |> *Ri*(A), *i* ∈ N}, N<sup>+</sup> = {*i*| | *aii*···*i* |> *α*(*i*) N*t* , *i* ∈ N*t* ⊆ N}, N<sup>0</sup> = {*i*| | *aii*···*i* |= *α*(*i*) N*t* , *i* ∈ N*t* ⊆ N}, J<sup>+</sup> = {(*<sup>i</sup>*, *j*)|(| *aii*···*i* | −*<sup>α</sup>*(*i*) <sup>N</sup>*t*)(| *ajj*···*j* | −*<sup>α</sup>*(*j*) N*s* ) > *α*(*i*) N*t* · *α*(*j*) N*s* , *i* ∈ N*<sup>t</sup>*, *j* ∈ N*<sup>s</sup>*, 1 ≤ *t* = *s* ≤ *k*} J0 = {(*<sup>i</sup>*, *j*)|(| *aii*···*i* | −*<sup>α</sup>*(*i*) <sup>N</sup>*t*)(| *ajj*···*j* | −*<sup>α</sup>*(*j*) N*s* ) = *α*(*i*) N*t* · *α*(*j*) N*s* , *i* ∈ N*<sup>t</sup>*, *j* ∈ N*<sup>s</sup>*, 1 ≤ *t* = *s* ≤ *k*}

**Definition 4.** *Let* A = (*ai*1*i*2···*im* ) *be a complex tensor with order m and dimension n. For all i* ∈ N*t, j* ∈ N*<sup>s</sup>*, 1 ≤ *t* = *s* ≤ *k, if:*

$$(|\ |\ a\_{i\bar{i}l,\cdots\bar{i}\ }|\ -a\_{\mathbb{N}\_{\ell}}^{(i)})(|\ |\ a\_{j\bar{j},\cdots\bar{j}\ }|\ -a\_{\mathbb{N}\_{\nu}}^{(j)})\ \geq \overline{a}\_{\mathbb{N}\_{\ell}}^{(i)}\cdot\overline{a}\_{\mathbb{N}\_{\nu}\prime}^{(j)}\tag{1}$$

*we call* A *a locally double-diagonally dominant tensor and denote* A ∈ *LDD*0*T. If:*

$$(|\ |\ a\_{i\bar{i}\cdots\bar{i}\ }|\ -a\_{\mathbb{N}\_t}^{(i)})(|\ |\ a\_{j\bar{j}\cdots j}\ |\ -a\_{\mathbb{N}\_s}^{(j)}\ )\ > \overline{a}\_{\mathbb{N}\_t}^{(i)}\cdot\overline{a}\_{\mathbb{N}\_s\prime}^{(j)}\tag{2}$$

*we call* A *a strictly locally double-diagonally dominant tensor and denote* A ∈ *LDDT.*

**Remark 1.** *If* A ∈ *LDD*0*T and* N∗ = ∅*, then* N = N<sup>0</sup> ∪ N+*; if* A ∈ *LDDT and* N∗ = ∅*, then* N = N+*.*

**Lemma 6.** *Let* A = (*ai*1*i*2...*im* ) *be a complex tensor with order m and dimension n. If* A ∈ *LDDT and* N∗ = ∅*, then there exists at most one l*0 ∈ {1, 2, ··· , *k*} *such that,*

$$|\
u\_{i i \cdots i}| \ -\mathfrak{a}^{(i)}\_{\mathbb{N}\_{\mathbb{I}\_0}} \leq \overline{\mathfrak{a}}^{(i)}\_{\mathbb{N}\_{\mathbb{N}\_0}}, \ \forall i \in \mathbb{N}\_{\mathbb{I}\_0}.$$

**Proof of Lemma 6.** If there exists *l*10 = *l*20 ∈ {1, 2, ··· , *k*}, such that for all *i* ∈ N*l*10, *j* ∈ N*l*20,

$$|\ a\_{i\bar{i}\cdots\bar{i}\ }| \ -a\_{\mathbb{N}\_{l\_0^1}}^{(i)} \leq \overline{a}\_{\mathbb{N}\_{l\_0^1}\prime}^{(i)} \quad |\ a\_{j\bar{j}\cdots\bar{j}\ }| \ -a\_{\mathbb{N}\_{l\_0^2}}^{(j)} \leq \overline{a}\_{\mathbb{N}\_{l\_0^2}}^{(j)}.$$

Notice that A ∈ *LDDT* and N∗ = ∅. By Remark 1, we have:

$$|\left|\left|a\_{i\bar{i}\cdots\bar{i}\ }\right|\right|-\mathfrak{a}^{(i)}\_{\mathbb{N}\_{l\_0^1}} > 0, \quad |\left|a\_{j\bar{j}\cdots\bar{j}\ }\right|\ -\mathfrak{a}^{(j)}\_{\mathbb{N}\_{l\_0^2}} > 0.$$

These imply:

$$(\mid a\_{ii\cdots i} \mid -a\_{\mathbb{N}\_{\{\frac{1}{6}\}}}^{(i)}) \cdot (\mid a\_{j\dot{j}\cdots j} \mid -a\_{\mathbb{N}\_{\{\frac{2}{6}\}}}^{(j)}) \leq \overline{\pi}\_{\mathbb{N}\_{\{\frac{3}{6}\}}}^{(i)} \cdot \overline{\pi}\_{\mathbb{N}\_{\{\frac{2}{6}\}}}^{(j)}$$

which contradicts A ∈ *LDDT*. The proof is complete.

**Remark 2.** *Based on Lemma 6, when* A ∈ *LDDT and* N∗ = ∅*, we always assume that* N1 = {*i*| | *aii*···*i* | −*<sup>α</sup>*(*i*) N*t*≤ *α*(*i*) N*t*, *i* ∈ N}.

**Lemma 7.** *Let* A = (*ai*1*i*2...*im* ) *be a complex tensor with order m and dimension n,* A ∈ *LDDT and* N∗ = ∅*. If* N1 = ∅*, then* A *is an* H*-tensor.*

**Proof of Lemma 7.** Since A ∈ *LDDT*, N∗ = ∅ and N1 = ∅. For all *i* ∈ N = N2 ∪ N3 ∪···∪ N*k*, we have:

$$\mid a\_{ii\cdots i} \mid -\alpha\_{\mathbb{N}\_t}^{(i)} > \overline{\alpha}\_{\mathbb{N}\_t}^{(i)} \ t \in \{2, 3, \cdots, k\} \dots$$

This implies:

$$|\ a\_{ii\cdots i}| \geqslant |\alpha\_{\mathbb{N}\_t}^{(i)} + \overline{\alpha}\_{\mathbb{N}\_t}^{(i)} = \mathcal{R}\_i(\mathcal{A}), i \in \mathbb{N}.$$

By Lemma 2, A is an H-tensor.

**Theorem 1.** *Let* A = (*ai*1*i*2...*im* ) *be a complex tensor with order m and dimension n,* A ∈ *LDDT and* N∗ = ∅*. If for all i* ∈ N*<sup>t</sup>*, *t* ∈ {2, 3, ··· , *k*}, *j* ∈ N1 = ∅*, we have:*

$$\begin{aligned} \left( \mid a\_{\bar{n}\bar{\cdot}\cdots\bar{\cdot}} \mid -a\_{\mathbb{N}\_{\mathbb{N}\_{\ell}}}^{(i)} \right) \left( \mid a\_{\bar{j}\bar{\cdot}\cdots\bar{\cdot}} \mid -\sum\_{\substack{\bar{i}\_{2}\bar{i}\_{3}\cdots\bar{i}\_{m}\in\mathbb{N}^{m-1}\backslash\{\mathbb{N}\nmid\mathbb{N}\_{1}\}^{m-1} \\ \delta\_{\bar{j}\bar{\cdot}\_{2}\cdots\bar{i}\_{m}} = 0 \\ > \overline{a}\_{\mathbb{N}\_{\mathbb{N}\_{\mathbb{N}\_{\mathbb{N}\_{\mathbb{N}\_{\mathbb{N}}}}}} \mid a\_{\bar{j}\bar{\cdot}\_{2}\cdots\bar{i}\_{m}} \mid} \mid a\_{\bar{j}\bar{\cdot}\_{2}\cdots\bar{i}\_{m}} \mid) \right) \end{aligned} \tag{3}$$

*then* A *is an* H*-tensor.*

**Proof of Theorem 1.** From the inequality (3), there exists a positive constant *d* such that:

$$\min\_{\begin{subarray}{c}\boldsymbol{i}\in\mathbb{N}\_{\boldsymbol{t}}\\\boldsymbol{\alpha}\in\{2,3,\cdots,k\}\end{subarray}}\{\frac{|\boldsymbol{a}\_{\boldsymbol{i}\boldsymbol{i}\cdots\boldsymbol{i}}\mid-\boldsymbol{a}\_{\boldsymbol{\mathcal{N}}\_{\boldsymbol{t}}}^{({\boldsymbol{i}})}\}>\boldsymbol{d}>\boldsymbol{d}>\max\_{\boldsymbol{j}\in\mathbb{N}\_{\boldsymbol{t}}}\{\frac{{}\_{j\_{2}\boldsymbol{i}\_{3}\cdots\boldsymbol{i}\_{m}\in\{\boldsymbol{N}\}\mid\boldsymbol{N}\_{1}}\mid\boldsymbol{a}\_{\boldsymbol{j}\boldsymbol{i}\_{2}\cdots\boldsymbol{i}\_{m}}}{\|\boldsymbol{a}\_{\boldsymbol{j}\boldsymbol{j}\cdots\boldsymbol{j}}\|-\sum\_{\begin{subarray}{c}\boldsymbol{i}\_{2}\boldsymbol{i}\_{3}\cdots\boldsymbol{i}\_{m}\in\{\boldsymbol{N}\}\mid\boldsymbol{N}\_{1}\end{subarray}}\mid\boldsymbol{a}\_{\boldsymbol{j}\boldsymbol{i}\_{2}\cdots\boldsymbol{i}\_{m}}|\}.\tag{4}$$

If *α*(*i*) N*t* = 0, we denote |*aii*···*<sup>i</sup>*|−*α*(*i*) N*t α*(*i*) N*t* = +<sup>∞</sup>. It is obvious that *d* > 1. Construct a positive diagonal matrix *X* = diag(*<sup>x</sup>*1, *x*2, ··· , *xn*), where:

$$\mathfrak{x}\_{i} = \begin{cases} \mathbf{1}\_{\prime} & i \in \mathbb{N} \backslash \mathbb{N}\_{1}, \\ d^{\frac{1}{\mathfrak{m}-1}}, & i \in \mathbb{N}\_{1}. \end{cases}$$

Let B = A*Xm*−<sup>1</sup> = (*bi*1*i*2···*im* ). For *i* ∈ N\N1 = {N2, N3, ···, N*k*}, by the first inequality in (4),

$$\begin{split} |b\_{\bar{i}\bar{i}\cdots i}| &= |a\_{\bar{i}\bar{i}\cdots i}| \cdot \overbrace{1\cdot1\cdots1\cdot1}^{m-1} \\ &= |a\_{\bar{i}\bar{i}\cdots i}| \\ &> a\_{\bar{i}\bar{i}\cdots i}^{(i)} + \overline{a}\_{\bar{i}\bar{i}\bar{i}}^{(i)}\cdot d \\ &= \sum\_{\substack{i\_{1}\cdots i\_{m}\in\mathbb{N}\_{0}^{m-1} \\ \delta\_{\bar{i}\bar{i}\cdots i\bar{n}}=0}} |\; a\_{\bar{i}\bar{i}\cdots i\_{m}}| + \sum\_{\substack{i\_{2}\cdots i\_{m}\in\mathbb{N}^{m-1}\mathbb{N}\_{0}^{m-1} \\ \delta\_{\bar{i}\bar{i}\cdots i\bar{n}}=0}} |\; a\_{\bar{i}\bar{i}\cdots i\_{m}}| \; \sum\_{\substack{i\_{2}\cdots i\_{m}\in\mathbb{N}^{m-1}\mathbb{N}\_{0}^{m-1} \\ \delta\_{\bar{i}\bar{i}\cdots i\bar{n}}=0}}^{m-1} |\; a\_{\bar{i}\bar{i}\cdots i\_{m}}| \; \sum\_{\substack{i\_{2}\cdots i\_{m}\in\mathbb{N}^{m-1}\mathbb{N}\_{0}^{m-1} \\ \delta\_{\bar{i}\bar{i}}\cdots i\bar{i} = 0 \\ \delta\_{\bar{i}\bar{i}\cdots i}\leq 0}}^{m-1} |\; \delta\_{\bar{i}\bar{i}}\cdots\; \delta\_{m}| + \sum\_{\substack{i\_{2}\cdots i\_{m}\in\mathbb{N}^{m-1}\mathbb{N}\_{0}^{m-1} \\ \delta\_{\bar{i}\bar{i}\cdots i}=0}} |\; \delta\_{\bar{i}\bar{i}\cdots i\_{m}}| \\ = R\_{\bar{i}}(\mathcal{B}). \end{split}$$

For *i* ∈ N1, by the second inequality in (4),


*bii*···*<sup>i</sup>* | =| *aii*···*i* | · *d* =| *aii*···*i* | · *<sup>m</sup>*−1 8 9: ; *d* 1 *<sup>m</sup>*−1 ··· *d* 1 *<sup>m</sup>*−1 > *d* · ∑ *<sup>i</sup>*<sup>2</sup>···*im*∈N*<sup>m</sup>*−<sup>1</sup>\(N\N1)*<sup>m</sup>*−<sup>1</sup> *<sup>δ</sup>ii*2···*im* =0 | *aii*<sup>2</sup>···*im* | + ∑ *<sup>i</sup>*<sup>2</sup>···*im*∈(N\N1)*<sup>m</sup>*−<sup>1</sup> | *aii*<sup>2</sup>···*im* | = *d*( ∑ *<sup>i</sup>*<sup>2</sup>···*im*∈N*<sup>m</sup>*−<sup>1</sup> 1 *<sup>δ</sup>ii*2···*im* =0 | *aii*<sup>2</sup>···*im* | + ∑ *<sup>i</sup>*<sup>2</sup>···*im*∈[N*<sup>m</sup>*−<sup>1</sup>\(N\N1)*<sup>m</sup>*−<sup>1</sup>]\N*<sup>m</sup>*−<sup>1</sup> 1 | *aii*<sup>2</sup>···*im* |) + ∑ *<sup>i</sup>*<sup>2</sup>···*im*∈(N\N1)*<sup>m</sup>*−<sup>1</sup> | *aii*<sup>2</sup>···*im* | ≥ ∑ *<sup>i</sup>*<sup>2</sup>···*im*∈N*<sup>m</sup>*−<sup>1</sup> 1 *<sup>δ</sup>ii*2···*im* =0 | *aii*<sup>2</sup>···*im* | · *<sup>m</sup>*−1 8 9: ; *d* 1 *<sup>m</sup>*−1 ··· *d* 1 *<sup>m</sup>*−1 + ∑ *<sup>i</sup>*<sup>2</sup>···*im*∈[N*<sup>m</sup>*−<sup>1</sup>\(N\N1)*<sup>m</sup>*−<sup>1</sup>]\N*<sup>m</sup>*−<sup>1</sup> 1 | *aii*<sup>2</sup>···*im* | *xi*2 ··· *xim* + ∑ *<sup>i</sup>*<sup>2</sup>···*im*∈(N\N1)*<sup>m</sup>*−<sup>1</sup> | *aii*<sup>2</sup>···*im* | = ∑ *<sup>i</sup>*<sup>2</sup>···*im*∈N*<sup>m</sup>*−<sup>1</sup> 1 *<sup>δ</sup>ii*2···*im* =0 | *bii*<sup>2</sup>···*im* | + ∑ *<sup>i</sup>*2*i*3···*im*∈N*<sup>m</sup>*−<sup>1</sup>\N*<sup>m</sup>*−<sup>1</sup> 1 | *bii*<sup>2</sup>···*im* | = *Ri*(B).

Thus, we have proven that:

$$|\ b\_{ii\cdots i}| > \sum\_{\substack{i\_2 \cdots i\_{\text{lw}} \in \mathbb{N}^{\text{w}-1} \\ \delta\_{ii\_2\cdots i\_{\text{lw}}} = 0}} |\ b\_{ii\_2\cdots i\_{\text{lw}}}| = \mathcal{R}\_i(\mathcal{B}), \ i \in \mathbb{N}.$$

i.e., B is a strictly diagonally dominant tensor. By Lemmas 2 and 4, A is an H-tensor. The proof is complete.

**Lemma 8.** *Let* A = (*ai*1*i*2...*im* ) *be a complex tensor with order m and dimension n. If* N∗ = ∅ *and for all i* ∈ N*<sup>t</sup>*, *j* ∈ N*<sup>s</sup>*, 1 ≤ *t* = *s* ≤ *k,*

$$(|\ |\ a\_{i\bar{i}\cdots\bar{i}\ }| - \mathfrak{a}^{(i)}\_{\mathbb{N}\_{\ell}})(|\ |\ a\_{j\bar{j}\cdots\bar{j}\ }| - \mathfrak{a}^{(j)}\_{\mathbb{N}\_{\kappa}}) \leq \overline{\mathfrak{a}}^{(i)}\_{\mathbb{N}\_{\ell}} \cdot \overline{\mathfrak{a}}^{(j)}\_{\mathbb{N}\_{\kappa}\prime} \tag{5}$$

*then there exists only one natural number k*0 ∈ {1, 2, ··· , *k*} *such that* N∗ ⊆ <sup>N</sup>*k*0*.*

**Proof of Lemma 8.** Since N∗ = ∅, then there exists one *k*0 ∈ {1, 2, ··· , *k*} such that N∗ ∩ <sup>N</sup>*k*0 = ∅. If there exists another *k*1 ∈ {1, 2, ··· , *k*}\{*k*0} such that N∗ ∩ <sup>N</sup>*k*1 = ∅, then we have:

$$(\mid a\_{i i \cdots i} \mid -a\_{\mathbb{N}\_{k\_0} \cap \mathbb{N}^\*}^{(i)}) (\mid a\_{j j \cdots j} \mid -a\_{\mathbb{N}\_{k\_1} \cap \mathbb{N}^\*}^{(j)}) > \overline{\mathfrak{a}}\_{\mathbb{N}\_{k\_0} \cap \mathbb{N}^\*}^{(i)} \cdot \overline{\mathfrak{a}}\_{\mathbb{N}\_{k\_1} \cap \mathbb{N}^\*}^{(j)}$$

which contradicts the inequality (5). Therefore, there exists only one natural number *k*0 ∈ {1, 2, ··· , *k*} that satisfies N∗ ⊆ <sup>N</sup>*k*0.

**Remark 3.** *Based on Lemma 8, when* A *satisfies* (5) *and* N∗ = ∅*, then we always assume that* N∗ ⊆ N*k (where* N = N1 ∪ N2 ∪···∪ N*k).*

**Theorem 2.** *Let* A = (*ai*1*i*2...*im* ) *be a complex tensor with order m and dimension n and* N∗ = ∅*. For all i* ∈ N*k*, *j* ∈ N*<sup>s</sup>*,*<sup>s</sup>* ∈ {1, 2, ··· , *k* − <sup>1</sup>}, *if:*

$$\begin{split} 0 &< \left( \mid a\_{\overline{i}\cdots i} \mid - \sum\_{\substack{i\_2 \cdots i\_m \in \mathbb{N}^{m-1} \backslash \{\mathbb{N} \backslash \mathbb{N}\_k\}^{m-1} \\ \delta\_{i\overline{i}2\cdots i\_m = 0} = 0 \end{split} \mid a\_{\overline{i}\overline{i}2\cdots i\_m} \mid) \left( \mid a\_{\overline{j}\overline{i}\cdots j} \mid -a\_{\overline{i}\overline{\mathbb{N}}\_k}^{(j)} \right) \\ &\leq \left( \sum\_{\substack{i\_2 \cdots i\_m \in \mathbb{N} \backslash \mathbb{N}\_k\Vert^{m-1}}} \mid a\_{\overline{i}\overline{i}2\cdots i\_m} \mid \right) \cdot \overline{a}\_{\mathbb{N}\_k\prime}^{(j)} \end{split} \tag{6}$$

*then* A *is not an* H*-tensor.*

**Proof of Theorem 2.** From the inequality (6), there exists a positive constant *d*0 such that:

$$\max\_{i \in \mathbb{N}\_k} \{ \frac{a\_{i\bar{\imath}\cdots i} \mid -\sum\_{\substack{i\_2 \cdots i\_m \in \mathbb{N}^{m-1} \ \langle \langle \mathcal{N} \rangle N\_k \rangle^{m-1} \\ \delta\_{\bar{i}\bar{\jmath}\cdots i\_m} = 0}}{\sum\_{\substack{i\_2 \cdots i\_m \in \langle \mathcal{N} \rangle N\_k \rangle^{m-1}}} \lfloor a\_{i\bar{\imath}\bar{\imath}\_2 \cdots i\_m} \rfloor} \} \le d\_0 \le \min\_{j \in \mathbb{N} \cup \{ \bar{\imath}\_k \; | \; i\_{\bar{j}\bar{\jmath}} \cdots i\_{\bar{j}} \mid -a\_{\bar{N}\_s}^{(j)} \}} \}. \tag{7}$$

If | *ajj*···*j* | −*<sup>α</sup>*(*j*) N*s* = 0, *j* ∈ N \ N*k*, we denote *α*(*j*) N*s* <sup>|</sup>*ajj*···*j*|−*α*(*j*) N*s* = +<sup>∞</sup>. Obviously,

max *i*∈N*k* { | *aii*···*i* | − ∑ *<sup>i</sup>*2···*im*∈N*<sup>m</sup>*−<sup>1</sup>\(N\N*<sup>k</sup>* )*<sup>m</sup>*−<sup>1</sup> *<sup>δ</sup>ii*2···*im* =0 | *aii*<sup>2</sup>···*im* | ∑ *<sup>i</sup>*2···*im*∈(N\N*<sup>k</sup>* )*<sup>m</sup>*−<sup>1</sup> | *aii*<sup>2</sup>···*im* | } ≥ max *i*∈N<sup>∗</sup> { | *aii*···*i* | − ∑ *<sup>i</sup>*2···*im*∈N*<sup>m</sup>*−<sup>1</sup>\(N\N*<sup>k</sup>* )*<sup>m</sup>*−<sup>1</sup> *<sup>δ</sup>ii*2···*im* =0 | *aii*<sup>2</sup>···*im* | ∑ *<sup>i</sup>*2···*im*∈(N\N*<sup>k</sup>* )*<sup>m</sup>*−<sup>1</sup> | *aii*<sup>2</sup>···*im* | } > 1,

so *d*0 > 1. Construct a positive diagonal matrix *X* = diag(*<sup>x</sup>*1, *x*2, ··· , *xn*), where:

$$x\_i = \begin{cases} 1, & i \in \mathbb{N}\_{k'} \\ d\_0^{\frac{1}{m-1}}, & i \in \mathbb{N} \backslash \mathbb{N}\_{k-1} \end{cases}$$

Let B = A*Xm*−<sup>1</sup> = (*bi*1*i*2···*im* ). For *i* ∈ N<sup>∗</sup>, by the first inequality in (7),


For *i* ∈ N*k*\N<sup>∗</sup>,


For *i* ∈ N\N*k* = N1 = N2 = ··· = N*k*−1, by the second inequality in (7),

$$\begin{split} |b\_{\vec{i}\vec{i}\cdots\vec{i}}| &= d\_{0}|a\_{\vec{i}\vec{i}\cdots\vec{i}}| = |a\_{\vec{i}\vec{i}\cdots\vec{i}} \xleftarrow{m^{-1}} \xleftarrow{m^{-1}} \\ &\leq d\_{0} \cdot a\_{\vec{i}\vec{i}\cdots\vec{i}}^{(i)} + \overline{a}\_{\vec{i}\vec{i}\cdots\vec{i}}^{(i)} \\ &= \sum\_{\begin{subarray}{c}i\_{2}\cdots i\_{m}\in\mathbb{N}\_{0}^{m-1} \\ i\_{2}\cdots i\_{m}=0 \end{subarray}} |a\_{\vec{i}i\_{2}\cdots i\_{m}} \xleftarrow{m^{-1}} \xleftarrow{m^{-1}} \sum\_{\begin{subarray}{c}i\_{2}\cdots i\_{m}\in\mathbb{N}\_{0}^{m-1} \\ i\_{2}\cdots i\_{m}=0 \end{subarray}} |a\_{\vec{i}i\_{2}\cdots i\_{m}}| \\ &\leq \sum\_{\begin{subarray}{c}i\_{2}\cdots i\_{m}\in\mathbb{N}\_{0}^{m-1} \\ i\_{2}\cdots i\_{m}=0 \end{subarray}} |a\_{\vec{i}i\_{2}\cdots i\_{m}} \xleftarrow{m^{-1}} \xleftarrow{m^{-1}} |a\_{\vec{i}i\_{2}\cdots i\_{m}}| \\ &= \sum\_{\begin{subarray}{c}i\_{2}\cdots i\_{m}\in\mathbb{N}\_{0}^{m-1} \\ i\_{2}\cdots i\_{m}=0 \end{subarray}} |a\_{\vec{i}i\_{2}\cdots i\_{m}}| \cdot \sum\_{\begin{subarray}{c}i\_{2}\cdots i\_{m}\in\mathbb{N}\_{0}^{m-1} \\ i\_{2}\cdots i\_{m}=0 \end{subarray}} |a\_{\vec{i}i\_{2}\cdots i\_{m}}| \\ &= R\_{i}(\mathcal{B}). \end{split}$$

Thus we have proved that:

$$|\ b\_{ii\cdots i}| \le \sum\_{\substack{i\_2 \cdots i\_m \in \mathbb{N}^{m-1} \\ \delta\_{ii\_2\cdots i\_m} = 0}} |\ b\_{ii\_2\cdots i\_m}| = R\_i(\mathcal{B}), \ i \in \mathbb{N}.$$

i.e., B is not strictly diagonally dominant for all *i* ∈ N. By Lemma 5, B is not an H-tensor. By Lemma 4, A is not an H-tensor. The proof is complete.

#### **3. An Algorithm for Identifying** H**-Tensors**

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