$SDD_2$ tensors and $B_2$-tensors

1.   Introduction

In 2005, Qi studied the eigenvalues of a real supersymmetric tensor ^[1], and this work gave us a more profound understanding of the tensors. Indeed, in mathematics, tensors are a generalization of matrices; a first-order tensor is a vector, and a second-order tensor is a matrix. Tensors play a crucial role in numerous scientific fields, including signal and image processing ^[2], continuum physics, high-order statistics ^[3], and magnetic resonance imaging ^[4]. As multilinear functions, tensors can express linear relationships among vectors, scalars, and other tensors. Recent research on tensors has primarily focused on several key areas, for example, establishing criteria for identifying strong $\mathcal{H}$-tensors ^[5]; generalizing $\mathcal{H}$-tensors to $B$-tensors using matrix theory ^[6]; analyzing the positive definiteness of $\mathcal{H}$-tensors ^[7]; investigating whether newly defined tensors retain the properties of $\mathcal{H}$-tensors; and deriving bounds for the infinity norm of tensors. Consequently, the structural properties, identification criteria, and iterative algorithms for strong $\mathcal{H}$-tensors have garnered substantial attention from researchers recently. In 2015, Song et al. discussed relationships among higher-order tensors, positive semi-definite tensors, and some other structured tensors. They demonstrate that every principal sub-tensor of such a structured tensor is still a structured tensor in the same class, with a lower dimension^[8].

The positive definiteness of homogeneous polynomials plays a crucial role in numerous scientific fields, such as multivariate network realizability theory ^[9], a test for Lyapunov stability in multivariate filters ^[10], and polynomial problems ^[11]. And the $\mathcal{H}$-eigenvalues of tensors are widely used in data analysis, high-order Markov chains, and positive definiteness of even-order homogeneous polynomials^[7,12,13]. Given the broad applications of even-order homogeneous polynomials in areas such as medical imaging and the stability study of non-linear autonomous systems via Lyapunov's direct method in automatic control^[6,7,14,15]. Determining whether an even-order homogeneous polynomial is positive definite has become increasingly significant. In this paper, we investigate whether $SDD_2$ tensors retain the properties of strong $\mathcal{H}$-tensors and explore their application to the positive definiteness of even-order homogeneous polynomials. The definitions of homogeneous polynomials and positive definiteness are provided below.

For positive integers $n$ and $m$, $N = \{1, 2, \cdots, n\}$ and $\mathbb{C}(resp.\; \mathbb{R})$ denotes the set of all complex($resp$. real) numbers. Let ${\mathbb{C}}^{n\times n}(resp.\; {\mathbb{R}}^{n\times n})(n\geq 2)$ denotes the set of all $n$ by $n$ complex ($resp.$ real) matrices and let ${\mathbb{C}}^{\lbrack m, n\rbrack}$ $(resp.\; {\mathbb{R}}^{\lbrack m, n\rbrack})(m, n\geq 2)$ be the set of all complex ($resp.$ real) $m$th-order $n$-dimensional tensors. A tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})$ is called a complex($resp.$ real) $m$th-order $n$-dimensional tensor if $a_{i_{1}i_{2}\cdots i_{m}}\in \mathbb{C} (resp.\; \mathbb{R})$, where $i_j = 1, 2, \cdots, n$ for $j = 1, 2, \ldots, m$. A tensor $\mathcal{A}$ is called symmetric if its elements are invariant under any permutation of indices $\{i_1, i_2, \cdots, i_m\}$ ^[1]. An $m$th-degree homogeneous polynomial of $n$ variables, $f(x)$, can be usually denoted as

where $x = (x_1, x_2, \ldots, x_n)^T \in {\mathbb{R}^n}$ and $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ is a symmetric tensor ^[7]. An $m$th-order $n$-dimensional tensor is denoted by ${\mathcal{A}} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}(m, n\geq2)$, an $n$-dimensional vector is denoted by $x = (x_1, x_2, \ldots, x_n)^{T}$, and the $i$th of $\mathcal{A}x^{m-1}$ components are

and

If there exists a $\lambda$ such that the following homogeneous polynomial equation holds:

where $\mathcal{A}x^{m-1}$ and $\lambda x^{\lbrack m-1 \rbrack}$ are vectors, and $\lambda \in {\mathbb C}, \; x = (x_1, x_2, \ldots, x_n)^{T}$ being a nonzero complex vector, then $\lambda$ is referred to as an eigenvalue of $\mathcal{A}$, and $x$ is its corresponding eigenvector ^[1,16,17]. Specifically, if $\lambda$, $x$, and all entries of $\mathcal{A}$ are constrained to the real field, then $\lambda$ is termed an $\mathcal{H}$-eigenvalue of $\mathcal{A}$, and $x$ is its corresponding $\mathcal{H}$-eigenvector ^[1]. If $m$ is even, and

then we say that $f(x)$ is positive definite. The symmetric tensor $\mathcal{A}$ is called positive definite if $f(x)$ is positive definite^[7].

Definition 1.1. ^[18] Let $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$. If there is a positive vector $x = (x_1, x_2, \ldots, x_n)^T\in {\mathbb{R}^n}$ such that

where $|a|$ is the modulus of $a \in \mathbb{C}$, then $\mathcal{A}$ is called a strong $\mathcal{H}$-tensor.

Theorem 1.1. ^[19] Let $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ with $a_{kk\cdots k} > 0$ for all $k\in N$ and $m$ be even. If $\mathcal{A}$ is a strong $\mathcal{H}$-tensor, then $\mathcal{A}$ is positive definite.

Based on this theorem, to determine the positive definiteness of an even-order real symmetric tensor, one can first verify whether the given tensor is a strong $\mathcal{H}$-tensor. Numerous criteria for identifying strong $\mathcal{H}$-tensors have been extensively proposed in the literature; for example, using algorithmic criteria ^[20,21,22] and direct criteria to determine strong $\cal{H}$-tensor ^{[23,24,25,26,27]}. In the following sections, we will give the highlights of this article and present a new class of tensors, called $SDD_2$ tensors.

This paper is organized as follows: In Section 2, we introduce a new class of tensors, named $SDD_2$ tensors, which extend the concept of $SDD_2$ matrices. And we demonstrate that this new class of tensors is a subclass of strong $\mathcal{H}$-tensors. Furthermore, we use some numerical examples to illustrate these new results. In Section 3, we propose $B_2$-tensors inspired by $SDD_2$ tensors. Meanwhile, some properties of $B_2$-tensors are introduced. Finally, in Section 4, give a conclusion of this article.

2.   $SDD_2$ tensors

In this section, we proposed a new class of tensors, which was inspired by the $SDD_2$ matrices, and named it $SDD_2$ tensors. First, let us begin by reviewing the concept of $SDD_2$ matrix. For the convenience of discussion, now some notations, definitions, lemmas, and theorems are given, which will be used in the sequel.

The calligraphic letters $\mathcal{A}$, $\mathcal{B}$, $\cdots$, represent tensors; the capital letters $A$, $B$, $\cdots$, denote matrices; the lowercase letters $x$, $y$, $\cdots$, refer to vectors. A tensor $\mathcal{I} = (\delta_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ is called the unit tensor, where

For a given matrix $M = (m_{ij})\in {\mathbb{C}}^{n\times n}$, we denote

For a given tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$, we denote

where $r_{i}(\mathcal{A})$ denotes the weight of the off-diagonal entries in the $i$th row of the flattening $\mathcal{A}$. $N_A$ is the set of indices where the modulus of the diagonal entry is less than or equal to the corresponding off-diagonal weight. Conversely, $N_B$ is the set of indices where the modulus of the diagonal entry is greater than the corresponding off-diagonal weight. The set of $S^{m-1}$ includes indices where $i_2$ to $i_m$ belong to $S$ and $S\subseteq N$. The set $N^{m-1}\backslash S^{m-1}$ refers to the difference set between $N^{m-1}$ and $S^{m-1}$, where $i_2$ to $i_m$ belong to $N$ but not to $S$. And $N_C^{m-1}$ denotes the difference set between $N^{m-1}$ and $(N_A^{m-1}\cup N_B^{m-1})$, where $i_2$ to $i_m$ partially belong to $N_A$ and partially to $N_B$.

Definition 2.1. ^[28] Given a matrix $M = \left({{m_{ij}}} \right) \in {\mathbb C^{n \times n}}\left({n \ge 2} \right)$ is called an $SDD_2$ matrix, if

where

Definition 2.2. ^[1] Let $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$. $\mathcal{A}$ is called a diagonally dominant tensor if

$\mathcal{A}$ is called a strictly diagonally dominant ($SDD$) tensor if all inequalities hold strictly.

Definition 2.3. ^[29] Let $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ and $X = diag(x_1, x_2, \ldots, x_n)$. If

where

then $\mathcal{B}$ is referred to as the product of the tensor $\mathcal{A}$ and the matrix $X$.

Definition 2.4. ^[6] A tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ is called an $SD{{D}_{1}}$ tensor if

where

Through ^[6], it is established that $SDD_1$ matrices can be extended to $SDD_1$ tensors. Furthermore, we further attempt to generalize $SDD_2$ matrices to $SDD_2$ tensors. Specifically, we will demonstrate that $SDD_2$ tensors are a subclass of strong $\mathcal{H}$-tensors.

Definition 2.5. A tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ is called an $SD{{D}_{2}}$ tensor if

where

and ${p_i}(\mathcal{A})$ is defined as the Definition 2.1.

Lemma 2.1. ^[18] If $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ is an $SDD$ tensor, then $\mathcal{A}$ is a strong $\mathcal{H}$-tensor.

Lemma 2.2. ^[29] Let $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$. If there exists a positive diagonal matrix $X$ such that $\mathcal{A}X^{m-1}$ is a strong $\mathcal{H}$-tensor, then $\mathcal{A}$ is a strong $\mathcal{H}$-tensor.

In the following, we will give some properties of the $SDD_2$ tensor.

Theorem 2.1. If a tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ is an $SD{{D}_{2}}$ tensor and $N_A\neq\varnothing$, then we have $\sum\limits_{\substack{i_{2}\cdots i_{m} \in N_B^{m-1}\\ \delta_{ii_{2}\cdots i_{m}} = 0}}|a_{ii_{2}\cdots i_{m}}|\neq0, \; i\in N_A$.

Proof. For a tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$, if $\sum\limits_{\substack{i_{2}\cdots i_{m} \in N_B^{m-1}\\ \delta_{ii_{2}\cdots i_{m}} = 0}}|a_{ii_{2}\cdots i_{m}}| = 0, \; i\in N_A$, then there is ${{r}_{i}}(\mathcal{A}) = {{q}_{i}}(\mathcal{A})$. Since $\mathcal{A}$ is an $SDD_2$ tensor, by definition we have $\left| {{a}_{i\cdots i}} \right| > {{q}_{i}}(\mathcal{A}) = {{r}_{i}}(\mathcal{A})$ for all $i\in N_A$, it contradicts the definition of $N_A$. The proof is complete. □

Theorem 2.2. A tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ is an $SD{{D}_{2}}$ tensor if and only if $\left| {{a}_{i\cdots i}} \right| > {{q}_{i}}(\mathcal{A})$ for all $i\in N$.

Proof. Let $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ be an $SD{{D}_{2}}$ tensor. From Definition 2.5, we have $\left| {{a}_{i\cdots i}} \right| > {{q}_{i}}(\mathcal{A})$ for any $i\in N_A$. For any $i\in N_B$, from the definition of $N_B$ and $q_i(\mathcal{A})$, we have $\left| {{a}_{i\cdots i}} \right| > {{r}_{i}}(\mathcal{A})\geq{{q}_{i}}(\mathcal{A})$. Therefore, we obtain $\left| {{a}_{i\cdots i}} \right| > {{q}_{i}}(\mathcal{A})$ for all $i\in N$. □

Next, we will prove the $SDD_2$ tensor is a strong $\mathcal{H}$ tensor.

Theorem 2.3. If a tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ is an $SD{{D}_{2}}$ tensor, then $\mathcal{A}$ is a strong $\mathcal{H}$-tensor.

Proof. Let a tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ be an $SD{{D}_{2}}$ tensor; according to Theorems 2.1 and 2.2, $\left| {{a}_{i\cdots i}} \right| > {{q}_{i}}(\mathcal{A})$ for all $i\in N$. Hence, we have

and

Then there exists a positive number $\varepsilon > 0$ such that

if $\sum\limits_{\substack{i_{2}\cdots i_{m} \in N_B^{m-1}\\ \delta_{ii_{2}\cdots i_{m}} = 0}}|a_{ii_{2}\cdots i_{m}}| = 0, \; i\in N_B$, then the corresponding fraction is defined to be $\infty$. Next we construct a diagonal matrix $X = \mathrm{diag}(x_1, x_2, \ldots, x_n)$, where

From inequality (2.1) we can obtain that$\frac{q_i(\mathcal{A})}{|a_{ii\cdots i}|}+\varepsilon < \frac{q_i(\mathcal{A})}{|a_{ii\cdots i}|}+(1-\frac{q_i(\mathcal{A})}{|a_{ii\cdots i}|}) = 1$, so $x_{i}\neq{+\infty}$, which shows that $X$ is a positive diagonal matrix. Let $\mathcal{B} = (b_{ i_{1}i_{2}\cdots i_{m}}) = \mathcal{A}X^{m-1}$; then we have $b_{ i_{1}i_{2}\cdots i_{m}} = a_{ i_{1}i_{2}\cdots i_{m}}x_{i_2}\cdots x_{i_m}$, for any $i_j\in N, \; j\in\{1, 2, \ldots, m\}$.

Next, we will prove that $\mathcal{B}$ is an $SDD$ tensor.

For any $i\in N_A$, according to inequality (2.1) and Theorem 2.1, we have

And for any $i\in N_B$, from definition of $N_B$, we have

Thus, we obtain $|b_{ii\cdots i}| > r_{i}(\mathcal{B})$ for any $i\in N$. This indicates that $\mathcal{B}$ is a strictly diagonally dominant ($SDD$) tensor, and by Lemma 2.1 we conclude that $\mathcal{B}$ is a strong $\mathcal{H}$-tensor. Furthermore, applying Lemma 2.2, it is straightforward to deduce that $\mathcal{A}$ is also a strong $\mathcal{H}$-tensor. The proof is completed. □

Remark 2.1. From the Definitions 2.2, 2.4, and 2.5, it can be readily deduced that $q_{i}(\mathcal{A}) \leq p_{i}(\mathcal{A}) \leq r_{i}(\mathcal{A})$. Consequently, $SDD$ tensors constitute a subclass of $SDD_1$ tensors, and $SDD_1$ tensors, in turn, form a subclass of $SDD_2$ tensors. Furthermore, from Theorem 2.3, we conclude that $SDD_2$ tensors are strong $\mathcal{H}$-tensor. This establishes the following inclusion relationship:

Utilizing the following chart, we illustrate the relationships among these tensors.

Example 2.1. Let us consider tensor $\mathcal{A} = (a_{ijk}) = [A(1, :, :), A(2, :, :), A(3, :, :)]\in {\mathbb{C}}^{[3, 3]}$, where

Obviously,

so $N_A = \{3\}$, $N_B = \{1, 2\}$. By calculation, we obtain ${p_1}({\cal A}) = \frac{{37}}{{9}}$ and ${p_2}({\cal A}) = \frac{{26}}{{9}}$, then

when $i\in N_A$, we obtain

By Definition 2.5, $\mathcal{A}$ is not an $SDD_1$ tensor. However, there exists a positive diagonal matrix $D = diag(d_1, d_2, d_3)$, where ${d_1} = {0.7^{\frac{1}{2}}}, {d_2} = {0.6^{\frac{1}{2}}}, {d_3} = {2.2^{\frac{1}{2}}}$ such that $\mathcal{A}D$ is an $SDD$ tensor, by Lemma 2.2, that tensor $\cal{A}$ is a strong $H$ tensor.

Example 2.2. Let us consider tensor $\mathcal{A} = (a_{ijk}) = [A(1, :, :), A(2, :, :), A(3, :, :)]\in {\mathbb{C}}^{[3, 3]}$, where

Obviously,

so $N_A = \{3\}$, $N_B = \{1, 2\}$. By calculation, we obtain

when $i\in N_A$, we obtain

By Definition 2.4, we obtain that $\cal{A}$ is not an $SDD_2$ tensor. Moreover, through computation we find ${p_1}({\cal A}) = \frac{{37}}{{9}}, {p_2}({\cal A}) = \frac{{26}}{{9}}$, then

when $i\in N_A$, we obtain

From Definition 2.5, we conclude that $\cal{A}$ is an $SDD_2$ tensor.

Next we give the application of the $SDD_2$ tensor from Theorems 1.1 and 2.3 as follows.

Theorem 2.4. Let $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{R}}^{[m, n]}$ be an even-order symmetric tensor with $a_{kk\cdots k} > 0$ for all $k\in N$. If $\mathcal{A}$ is an $SDD_2$ tensor, then $\mathcal{A}$ is positive definite.

We give an example to illustrate how the definition of an $SDD_2$ tensor can be applied to determine whether a given tensor is a strong $\mathcal{H}$-tensor.

Example 2.3. Let us consider tensor $\mathcal{A} = (a_{ijk}) = [A(1, :, :), A(2, :, :), A(3, :, :)]\in {\mathbb{C}}^{[3, 3]}$, where

Obviously,

so $N_A = \{1\}$, $N_B = \{2, 3\}$. By calculation, we obtain

when $i\in N$, we obtain

Furthermore, we obtain

when $i\in N_A$, we obtain

Hence, $\mathcal{A}$ satisfies the conditions of the $SDD_2$ tensor. By Theorem 2.3, we can get that $\mathcal{A}$ is a strong $\mathcal{H}$-tensor.

Additionally, another example is provided to demonstrate the positive definiteness of an even-degree homogeneous polynomial.

Example 2.4. Consider the following 4th-degree homogeneous polynomial

where $x = (x_1, x_2, x_3, x_4)^T$. Then we can obtain a symmetric tensor $\mathcal{A} = (a_{ijkl})\in {\mathbb{R}}^{[4, 4]}$, where

and others are zeros. Then,

hence $N_A = \{1\}$, $N_B = \{2, 3, 4\}$. By calculation, we obtain

when $i\in N_B$, we obtain

Furthermore, we obtain

when $i\in N_A$,

Therefore, by the definition of an $SDD_2$ tensor, $\mathcal{A}$ is an $SDD_2$ tensor. According to Theorem 2.3, $\mathcal{A}$ also is a strong $\mathcal{H}$-tensor. Moreover, all its diagonal elements are positive. Furthermore, by applying Theorem 2.4, we conclude that $\mathcal{A}$ is positive definite, and consequently, $f(x)$ is positive definite.

3.   $B_2$-tensor and its properties

In this section, we first introduce a new class of tensors, termed $B_2$-tensor, which is based on the $SDD_2$ tensor. This new class of tensors encompasses $B$-tensors and $B_1$-tensors as its subclass. Subsequently, we present several properties of $B_2$-tensors. For convenience, some notations, definitions, theorems, and lemmas are provided as follows:

Given a tensor $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$, for each row $i$, we denote

Let $\mathcal{B}^+ = (b_{i_1i_2\cdots i_m})$ be the tensor defined as

clearly, $\mathcal{B}^+$ is a $Z$-tensor. A tensor ${\mathcal{A}} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{C}}^{\lbrack m, n\rbrack}$ is called a $Z$-tensor if and only if $a_{i_{1}i_{2}\cdots i_{m}}\leq0$ for all $(i_1, \ldots, i_m)\neq(i, \ldots, i)$^[15].

Definition 3.1. ^[8] A tensor $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is called a $B$-tensor if and only if for all $i\in N$,

and

In this section, we reviewed the concept of $B$-tensor. Furthermore, in^[8], it was also proved that a $B$-tensor can be characterized by the following equivalent definition.

Definition 3.2. ^[8] A tensor $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is called a $B$-tensor if and only if for each $i\in N$,

i.e.,

Definition 3.3. ^[6] A tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{R}}^{[m, n]}$ is a $B_1$-tensor if for all $i\in N$,

Lemma 3.1. ^[6] If a tensor $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is a $B$-tensor, then $\mathcal{A}$ is a $B_1$-tensor.

Now, we give the definition of a $B_2$-tensor.

Definition 3.4. A tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})\in {\mathbb{R}}^{[m, n]}$ is a $B_2$-tensor if for all $i\in N$,

Next, we introduce some useful properties of a $B_2$-tensors.

Proposition 3.1. If a tensor $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is a $B_1$-tensor, then $\mathcal{A}$ is a $B_2$-tensor.

Proof. If $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})$ is a $B_1$-tensor, and we have $r_i(\mathcal{B}^+)\geq p_i(\mathcal{B}^+)\geq q_i(\mathcal{B}^+)$, then by Definition 3.3,

that is, $\mathcal{A}$ is a $B_2$-tensor. □

Remark 3.1. From Lemma 3.1 and Proposition 3.1, it is evident that the $B_2$-tensors encompass the $B_1$-tensors, and the $B_1$-tensors encompass the $B$-tensors; that is,

Proposition 3.2. Let tensor $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ be a $B_2$-tensor; then $\mathcal{B}^+$ is an $SDD_2$ tensor.

Proof. Since ${b_{{i_1}{i_2} \cdots {i_m}}} = {a_{{i_1}{i_2} \cdots {i_m}}} - r_{{i_1}}^ + ({\mathcal A}) > {q_{{i_1}}}\left({{{\mathcal B}^ + }} \right),$ we have $\left| {{b_{{i_1}{i_2} \cdots {i_m}}}} \right| > {q_{{i_1}}}\left({{{\cal B}^ + }} \right)$, so $\mathcal{B}^+$ is an $SDD_2$ tensor. □

Proposition 3.3. If tensor $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is a $B_2$-tensor, then $\mathcal{B}^+$ has positive diagonal elements.

Proof. If tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})$ is a $B_2$-tensor, then it follows that $a_{i\cdots i}-r_{i}^+(\mathcal{A}) > q_i(\mathcal{B}^+)\geq0$, which implies $a_{i\cdots i}-r_{i}^+(\mathcal{A}) > 0$. The proof is complete. □

From Proposition 3.3, we can easily obtain the following corollary.

Corollary 3.1. If tensor $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is a $B_2$-tensor, then there must be $a_{i\cdots i} > r_{i}^+(\mathcal{A})$.

Proposition 3.4. A tensor $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is a $B_2$-tensor if and only if $\mathcal{B}^+$ is an $SDD_2$ tensor with positive diagonal entries.

Proof. If the tensor $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})$ is a $B_2$-tensor, then by Propositions 3.2 and 3.3, $\mathcal{B}^+$ is an $SDD_2$ tensor with positive diagonal entries. Conversely, if $\mathcal{B}^+$ is an $SDD_2$ tensor, then $|a_{i\cdots i}-r_{i}^+(\mathcal{A})| > q_i(\mathcal{B}^+)$. Since $\mathcal{B}^+$ has positive diagonal entries, i.e., $a_{i\cdots i}-r_{i}^+(\mathcal{A}) > 0$, it follows that $a_{i\cdots i}-r_{i}^+(\mathcal{A}) > q_i(\mathcal{B}^+)$. Therefore, $\mathcal{A}$ is $B_2$-tensor. □

Proposition 3.5. Let $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ be a $Z$-tensor with positive diagonal entries. Then $\mathcal{A}$ is a $B_2$-tensor if and only if $\mathcal{A}$ is an $SDD_2$ tensor.

Proof. Since $\mathcal{A} = (a_{i_{1}i_{2}\cdots i_{m}})$ is a $Z$-tensor, we have $r_{i}^+(\mathcal{A}) = 0$ for all $i\in N$, and $\mathcal{B}^+ = \mathcal{A}$. Consequently, $\mathcal{A}$ is a $B_2$-tensor if and only if $a_{i\cdots i}-r_{i}^+(\mathcal{A}) > q_i(\mathcal{B}^+)$, which simplifies to $a_{i\cdots i} > q_i(\mathcal{A})$. Therefore, $\mathcal{A}$ is an $SDD_2$ tensor. Conversely, if $\mathcal{A}$ is an $SDD_2$ tensor, we have $|a_{i\cdots i}| > q_i(\mathcal{A})$. Since $\mathcal{A}$ has positive diagonal entries, it follows that $a_{i\cdots i} > q_i(\mathcal{A})$. Thus, $a_{i\cdots i}-r_{i}^+(\mathcal{A}) > q_i(\mathcal{B}^+)$ holds immediately, which implies that $\mathcal{A}$ is a $B_2$-tensor. □

The following corollary can be obtained from Proposition 3.5 and Theorem 2.3.

Corollary 3.2. Let $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ be a $Z$-tensor with positive diagonal entries. If $\mathcal{A}$ is a $B_2$-tensor, then $\mathcal{A}$ is a strong $\mathcal{H}$-tensor.

Proposition 3.6. $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is a $B_2$-tensor if and only if $\mathcal{B}^+$ is a $B_2$-tensor.

Proof. By Proposition 3.4, $\mathcal{A} = (a_{i_1i_2\cdots i_m})$ is a $B_2$-tensor if and only if $\mathcal{B}^+$ is an $SDD_2$ tensor with positive diagonal entries. Since $\mathcal{B}^+$ is a $Z$-tensor, according to Proposition 3.5, the conclusion follows immediately. □

Proposition 3.7. If $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is a $B_2$-tensor, and $\mathcal{D}\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is a nonnegative diagonal tensor of the same order and dimension, then $\mathcal{A+D}$ is a $B_2$-tensor.

Proof. Let $\mathcal{D} = (d_{{i_1}{i_2}\cdots{i_m}})$, where

and $d_i\geq0$. Let $\mathcal{C} = \mathcal{A+D}$, where

Then, $\mathcal{A}$ and $\mathcal{C}$ have the same nondiagonal elements, so that $r_{i}^+(\mathcal{A}) = r_{i}^+(\mathcal{C})$.

Next, let us prove $c_{i\cdots i}-r_{i}^+(\mathcal{C}) > q_i(\mathcal{C}^+)$ for all $i\in N$, where $\mathcal{C}^+ = (v_{{i_1}{i_2}\cdots{i_m}})$ with $v_{i{i_2}\cdots{i_m}} = c_{i{i_2}\cdots{i_m}}-r_{i}^+(\mathcal{C})$.

Since $\mathcal{A}$ is a $B_2$-tensor, for all $i\in N$, we have

Meanwhile, we have the following equation that holds:

Hence, for any $i\in N_A(\mathcal{C}^+)$, we have $|c_{i\cdots i}-r_{i}^+(\mathcal{C})|\leq r_i(\mathcal{C}^+)$, i.e., $|a_{i\cdots i}+d_i-r_{i}^+(\mathcal{A})|\leq r_i(\mathcal{B}^+)$, we can immediately obtain that $|a_{i\cdots i}-r_{i}^+(\mathcal{A})|\leq r_i(\mathcal{B}^+)$. Therefore, $i\in N_A(\mathcal{B}^+)$. This indicates that $N_A(\mathcal{C}^+)\subset N_A(\mathcal{B}^+)$; the same method can illustrate that $N_B(\mathcal{B}^+)\subset N_B(\mathcal{C}^+)$.

For any $i\in N$,

Combining Eq (3.3), there is $c_{i\cdots i}-r_{i}^+(\mathcal{C}) > q_i(\mathcal{C}^+)$, and this proof is completed. □

Proposition 3.8. If $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is a $B_2$-tensor, then we can write $\mathcal{A}$ as $\mathcal{A} = \mathcal{B+C}$, where $\mathcal{B}$ is a $Z$-tensor with positive diagonal entries, and $\mathcal{C}$ is a nonnegative tensor with $c_{ii_2\cdots i_m} = c_i, \; c_i\geq0$, for any $i\in N$. In particular, if $\mathcal{A}$ is both a $B_2$-tensor and a $Z$-tensor, then $\mathcal{C}$ is a zero tensor.

Proof. Suppose $\mathcal{B} = (b_{i_{1}i_{2}\cdots i_{m}})$, where $b_{ii_{2}\cdots i_{m}} = a_{ii_{2}\cdots i_{m}}-r_{i}^+(\mathcal{A})$. According to the definition of $r_{i}^+(\mathcal{A})$, it can be inferred that $b_{ii_{2}\cdots i_{m}}\leq0, \; (i_2, \ldots, i_m)\neq(i, \ldots i)$. Since $\mathcal{A}$ is a $B_2$-tensor, $b_{i\cdots i} = a_{i\cdots i}-r_{i}^+(\mathcal{A}) > q_i(\mathcal{B})\geq0$. Then $\mathcal{B}$ is a $Z$-tensor with positive diagonal entries.

Let $\mathcal{C} = c_{ii_2\cdots i_m}$ with $c_{ii_2\cdots i_m} = r_{i}^+(\mathcal{A})$; obviously $\mathcal{C}$ is a nonnegative tensor with $c_i = r_{i}^+(\mathcal{A})$. In particular, if $\mathcal{A}$ is both a $B_2$-tensor and $Z$-tensor, we have $r_{i}^+(\mathcal{A}) = 0$, then $\mathcal{C}$ is a zero tensor. □

Proposition 3.9. Let $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ be a $B_2$-tensor, and let $\mathcal{C} = (c_{i_{1}i_{2}\cdots i_{m}})$ be a nonnegative tensor of the form $c_{ii_2\cdots i_m} = c_i$; then $\mathcal{A+C}$ is a $B_2$-tensor.

Proof. Let $\mathcal{P} = \mathcal{A+C}$, where $\mathcal{P} = (p_{i_{1}i_{2}\cdots i_{m}})$. By Proposition 3.6, we need to prove that $\mathcal{P}^+$ is a $B_2$-tensor. By definition, we can see that for all $i\in N$,

Hence, for all $i, i_2, \ldots, i_m\in N$, we have

then we obtain that $\mathcal{P}^+ = \mathcal{B}^+$. Since $\mathcal{A}$ is a $B_2$-tensor, by Proposition 3.6, $\mathcal{B}^+$ is a $B_2$-tensor. The conclusion follows immediately. □

Proposition 3.10. If $\mathcal{A} = (a_{i_1i_2\cdots i_m})\in {\mathbb{R}}^{\lbrack m, n\rbrack}$ is a $B_2$-tensor, then every principal sub-tensor of $\mathcal{A}$ is also a $B_2$-tensor.

Proof. Let $T$ be a nonempty subset of $N$ with $|T| = r$, and let $\mathcal{C} = \mathcal{A}_r^T\in \mathbb{R}^{[m, r]}$ be the principal sub-tensor of $\mathcal{A}$. Since $\mathcal{A}$ is a $B_2$-tensor, we have $a_{i\cdots i}-r_{i}^+(\mathcal{A}) > q_i(\mathcal{B}^+)\geq0$. Consequently, $a_{i\cdots i} > r_{i}^+(\mathcal{A})\geq r_{i}^+(\mathcal{C})$. Next, we prove that $a_{i\cdots i}-r_{i}^+(\mathcal{C}) > q_i(\mathcal{C}^+)$ for all $i\in T$. We have

For any $i\in N_A(\mathcal{C}^+)$, we have $0\leq a_{i\cdots i}-r_{i}^+(\mathcal{C})\leq r_i(\mathcal{C}^+)\leq r_i(\mathcal{B}^+)$, hence $0\leq a_{i\cdots i}-r_{i}^+(\mathcal{A})\leq r_i(\mathcal{B}^+)$. Therefore, $i\in N_A(\mathcal{B}^+)$. This indicates that $N_A(\mathcal{C}^+)\subset N_A(\mathcal{B}^+)$. The same method can illustrate that $N_B(\mathcal{B}^+)\subset N_B(\mathcal{C}^+)$.

For any $i\in T$,

So,

In summary, we have $a_{i\cdots i}-r_{i}^+(\mathcal{C}) > q_i(\mathcal{C}^+)$, and this proof is completed. □

4.   Conclusions

In this paper, we mainly introduce a novel class of tensors, which we named $SDD_2$ tensors, inspired by the $SDD_1$ tensors. We demonstrate that the $SDD_2$ tensor is a subclass of strong $\mathcal{H}$-tensor and give an application of the $SDD_2$ tensor to the determination of the positive definiteness of even-order real symmetric tensors. The validity of our results is supported by illustrated examples. Furthermore, we extend the concept of the $SDD_2$ tensor to the $B$-tensor, thereby introducing the $B_2$-tensor. Finally, we proposed several properties of the $B_2$-tensor. Regarding the error bounds for the linear complementarity problems based on strong $\mathcal{H}$-tensors. In the future, we can conduct similar research on $SDD_2$ tensors and $B_2$-tensors.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors are grateful to the referee for carefully reading of the paper and valuable suggestions and comments. This work is partly supported by the Natural Science Basic Research Program of Shaanxi, China (2020JM-622).

Conflict of interest

The authors declare there are no conflicts of interest.