# Pareto Z-eigenvalue inclusion theorems for tensor eigenvalue complementarity problems

## Abstract

This paper presents some sharp Pareto Z-eigenvalue inclusion intervals and discusses the relationships among different Pareto Z-eigenvalue inclusion intervals for tensor eigenvalue complementarity problems. As an application, we propose a sufficient condition for identifying the strict copositivity of tensors. Some examples are provided to illustrate the obtained results.

## Introduction

Let $$\mathcal{A}=(a_{{i_{1}}{i_{2}}\cdots{i_{m}}})\in \mathbb{R}^{[m,n]}$$ be an mth-order n-dimensional real tensor, x be a real n-vector and $$N=\{1,2,\dots,n\}$$. Denote by $$\mathcal{A} x^{m-1}$$ the vector in $$\mathbb{R}^{n}$$ with entries

\begin{aligned} \bigl(\mathcal{A} x^{m-1} \bigr)_{i}=\sum _{i_{2}, \ldots, i_{m}\in N}a_{{i}{i_{2}} \cdots{i_{m}}}x_{i_{2}}\cdots x_{i_{m}}. \end{aligned}

Consider the tensor eigenvalue complementarity problems of finding $$(\lambda,x)\in \mathbb{R}\times \mathbb{R}^{n}_{+}\backslash \{0\}$$ such that

\begin{aligned} 0\leq x\perp \bigl(\lambda x-\mathcal{A}x^{m-1} \bigr)\geq 0 \quad \text{and}\quad x^{\top}x=1, \end{aligned}

where $$a \bot b$$ means that vectors a and b are perpendicular to each other. For the problem, its solution $$(\lambda,x)\in \mathbb{R}\times \mathbb{R}^{n}_{+}\backslash \{0\}$$ is called a Pareto Z-eigenpair of tensor $$\mathcal{A}$$.

The Pareto Z-eigenpair of a tensor was introduced by Song , which is a natural generalization of that of a matrix . It is worth noting that Pareto Z-eigenvalues of $$\mathcal{A}$$ are closely related to Z ($$Z^{+}$$)-eigenvalues of $$\mathcal{A}$$ introduced by Lim  and Qi [7, 8], respectively.

### Definition 1

For a tensor $$\mathcal{A}=(a_{{i_{1}}{i_{2}}\cdots{i_{m}}})\in \mathbb{R}^{[m,n]}$$, if there exist $$(\lambda,x)\in \mathbb{R}\times \mathbb{R}^{n}\backslash \{0\}$$ such that

\begin{aligned} \mathcal{A}x^{m-1}=\lambda x, \quad x^{\top}x=1, \end{aligned}

then $$(\lambda,x)$$ is called a Z-eigenpair of tensor $$\mathcal{A}$$. Further, Z-eigenvalue λ of $$\mathcal{A}$$ is said to be a $$Z^{+}$$-eigenvalue, if its eigenvector $$x\in \mathbb{R}^{n}_{+}\backslash \{0\}$$.

Obviously, $$Z^{+}$$-eigenvalues of $$\mathcal{A}$$ are Pareto Z-eigenvalues. However, the converse may not hold as pointed by Zeng . Therefore, the tensor Pareto Z-eigenvalue received much attentions of researchers . For instance, Zeng  proposed a semidefinite relaxation algorithm to obtain Pareto Z-eigenvalues of tensor eigenvalue complementarity problems. Since it is not easy to find all Pareto Z-eigenvalues in practice [1, 9, 13], it is significant to make some characterizations to the distribution of Pareto Z-eigenvalues. Inspired by the results obtained in , we establish some Pareto Z-eigenvalues inclusion intervals, give comparisons among these Pareto Z-eigenvalue inclusion intervals, and propose a sufficient condition to identify the strict copositivity of real tensors in this paper.

The remainder of this paper is organized as follows. In Sect. 2, we recall some preliminary results and establish Pareto Z-eigenvalue inclusion intervals. Further, we give comparisons among these Pareto Z-eigenvalue inclusion intervals. In Sect. 3, we propose a sufficient condition to identify the strict copositivity of tensors.

To end this section, we give some notations needed. The set of all real numbers is denoted by $$\mathbb{R}$$, and the n-dimensional real Euclidean space is denoted by $$\mathbb{R}^{n}$$. For any $$a\in \mathbb{R}$$, we denote $$[a]_{+}:=\max \{0,a\}$$ and $$[a]_{-}:=\max \{0,-a\}$$. For any $$x\in \mathbb{R} ^{n}$$, $$x^{{\otimes}m}$$ denotes a tensor whose entries are defined by $$(x^{{\otimes}m})_{i_{1}i_{2}\cdots i_{m}}=x_{i_{1}}x_{i_{2}} \cdots x_{i_{m}}$$ for all $$i_{1},i_{2},\dots,i_{m}\in N$$. For any $$\mathcal{A}\in \mathbb{R} ^{[m,n]}$$ and $$x\in \mathbb{R} ^{n}$$, we define

\begin{aligned} &\mathcal{A}x^{m}:=x^{\top}\mathcal{A}x^{m-1}=\sum _{i_{1},i_{2}, \dots,i_{m}=1}^{n}a_{i_{1}i_{2}\cdots i_{m}}x_{i_{1}}x_{i_{2}} \cdots x_{i_{m}}, \\ & \Vert \mathcal{A} \Vert _{F}:= \Biggl(\sum _{i_{1},i_{2},\dots,i_{m}=1}^{n}a_{i_{1}i_{2} \cdots i_{m}}^{2} \Biggr)^{\frac{1}{2}}, \\ &[\mathcal{A}]_{+}:= \bigl([a_{i_{1}i_{2}\cdots i_{m}}]_{+} \bigr)\in \mathbb{R} ^{[m,n]},\qquad [\mathcal{A}]_{-}:= \bigl([a_{i_{1}i_{2}\cdots i_{m}}]_{-} \bigr)\in \mathbb{R} ^{[m,n]}. \end{aligned}

For any $$i,j\in N$$, set

\begin{aligned} &R_{i}(\mathcal{A})_{+}:=\sum_{i_{2},\dots,i_{m}=1}^{n}[a_{ii_{2} \cdots i_{m}}]_{+}, \qquad R_{i}(\mathcal{A})_{-}:=\sum _{i_{2},\dots, i_{m}=1}^{n}[a_{ii_{2} \cdots i_{m}}]_{-}, \\ &R_{i}^{j}(\mathcal{A})_{+}:=R_{i}( \mathcal{A})_{+}-[a_{ij\cdots j}]_{+}, \qquad R_{i}^{j}(\mathcal{A})_{-}:=R_{i}( \mathcal{A})_{-}-[a_{ij \cdots j}]_{-}, \\ &P_{i}^{j}(\mathcal{A})_{+}:=\sum _{ \substack{ i_{2},\dots, i_{m} \in N\\j\notin \{{i_{2}},\dots,{i_{m}}\}}}[a_{ii_{2} \cdots i_{m}}]_{+}, \qquad P_{i}^{j}(\mathcal{A})_{-}:=\sum _{ \substack{ i_{2},\ldots, i_{m} \in N\\j\notin \{{i_{2}},\ldots,{i_{m}}\}}}[a_{ii_{2} \cdots i_{m}}]_{-}. \end{aligned}

## Pareto Z-eigenvalues inclusion intervals

First, we recall some results of strictly copositive tensors [19, 20], and then establish Pareto Z-eigenvalue inclusion theorems of tensor $$\mathcal{A}$$. Some comparisons among different Pareto Z-eigenvalue inclusion intervals are also made in this section.

### Definition 2

Tensor $$\mathcal{A} \in \mathbb{R}^{[m,n]}$$ is said to be:

(i) strictly copositive if $$\mathcal{A}x^{m}>0$$ for any $$x\in \mathbb{R}^{n}_{+}\backslash \{0\}$$;

(ii) symmetric if $$a_{{i_{1}}{i_{2}}\cdots{i_{m}}}= a_{i_{\pi (1)}\cdots{i}_{\pi (m)}}, \forall \pi \in \Gamma _{m}$$, where $$\Gamma _{m}$$ is the permutation group of m indices.

### Lemma 1

([1, Corollary 3.5])

Let $$\mathcal{A}=(a_{{i_{1}}{i_{2}}\cdots{i_{m}}})\in \mathbb{R}^{[m,n]}$$ be symmetric. Then $$\mathcal{A}$$ always has Pareto Z-eigenvalues; $$\mathcal{A}$$ is strictly copositive if and only if all of its Pareto Z-eigenvalues are positive.

### Lemma 2

([20, Proposition 2.1])

Let $$\mathcal{A}=(a_{{i_{1}}{i_{2}}\cdots{i_{m}}})\in \mathbb{R}^{[m,n]}$$. If $$\mathcal{A}$$ is strictly copositive, then $$a_{i\cdots i}>0,\forall i \in N$$.

Based on the above lemmas, we have the following conclusion.

### Theorem 1

Let $$\mathcal{A} \in \mathbb{R}^{[m,n]}$$. Denote the set of Pareto Z-eigenvalues by $$\sigma (\mathcal{A})$$ and assume $$\sigma (\mathcal{A})\neq \emptyset$$. Then,

\begin{aligned} \sigma (\mathcal{A})\subseteq \Psi (\mathcal{A}):= \bigl\{ \lambda \in \mathbb{R}:\max \bigl\{ -\bar{a}\cdot n^{\frac{m}{2}},- \bigl\Vert [ \mathcal{A}]_{-} \bigr\Vert _{F} \bigr\} \leq \lambda \leq \min \bigl\{ \bar{a}\cdot n^{\frac{m}{2}}, \bigl\Vert [\mathcal{A}]_{+} \bigr\Vert _{F} \bigr\} \bigr\} , \end{aligned}
(1)

where $$\bar{a}=\max_{i_{1},\dots,i_{m}\in N}|a_{i_{1}i_{2}\cdots i_{m}}|$$.

### Proof

Suppose that $$(\lambda,x)$$ is a Pareto Z-eigenpair of $$\mathcal{A}$$. Then

\begin{aligned} \lambda \sum_{i=1}^{n}x_{i}^{2}&= \mathcal{A}x^{m}\leq [\mathcal{A}]_{+}x^{m}\leq \bigl\Vert [ \mathcal{A}]_{+} \bigr\Vert _{F} \bigl\Vert x^{ \otimes m} \bigr\Vert _{F} \\ &= \bigl\Vert [\mathcal{A}]_{+} \bigr\Vert _{F} \Biggl( \sum_{i_{1},i_{2},\dots,i_{m}=1}^{n}x_{i_{1}}^{2}x_{i_{2}}^{2} \cdots x_{i_{m}}^{2} \Biggr)^{\frac{1}{2}} \\ &= \bigl\Vert [\mathcal{A}]_{+} \bigr\Vert _{F} \Biggl( \sum_{i=1}^{n}x_{i}^{2} \Biggr)^{ \frac{m}{2}}= \bigl\Vert [\mathcal{A}]_{+} \bigr\Vert _{F} \end{aligned}
(2)

and

\begin{aligned} -\lambda \sum_{i=1}^{n}x_{i}^{2}&= \mathcal{-A}x^{m}\leq [\mathcal{A}]_{-}x^{m}\leq \bigl\Vert [\mathcal{A}]_{-} \bigr\Vert _{F} \bigl\Vert x^{ \otimes m} \bigr\Vert _{F} \\ &= \bigl\Vert [\mathcal{A}]_{-} \bigr\Vert _{F} \Biggl(\sum_{i_{1},i_{2},\dots,i_{m}=1}^{n}x_{i_{1}}^{2}x_{i_{2}}^{2} \cdots x_{i_{m}}^{2} \Biggr)^{\frac{1}{2}} \\ &= \bigl\Vert [\mathcal{A}]_{-} \bigr\Vert _{F} \Biggl(\sum_{i=1}^{n}x_{i}^{2} \Biggr)^{ \frac{m}{2}}= \bigl\Vert [\mathcal{A}]_{-} \bigr\Vert _{F}. \end{aligned}
(3)

Combining (2) with (3) yields

\begin{aligned} - \bigl\Vert [\mathcal{A}]_{-} \bigr\Vert _{F}\leq \lambda \leq \bigl\Vert [\mathcal{A}]_{+} \bigr\Vert _{F}. \end{aligned}
(4)

Meanwhile, from the definition of Pareto Z-eigenpair, we obtain

\begin{aligned} \vert \lambda \vert &=\frac{ \vert \mathcal{A}x^{m} \vert }{\sum_{i=1}^{n} x_{i}^{2}}\leq \frac{\bar{a}(\sum_{i=1}^{n}x_{i})^{m}}{\sum_{i=1}^{n}x_{i}^{2}} \leq \bar{a} \Bigl(\sqrt{x_{1}^{2}+x_{2}^{2}+ \cdots +x_{n}^{2}}\cdot \sqrt{1+1+\cdots +1} \Bigr)^{m} \\ &=\bar{a} n^{\frac{m}{2}}, \end{aligned}
(5)

where the second inequality holds via Cauchy–Schwartz inequality. The desired result follows by combining (4) and (5). □

In the following, we will use some important elements of tensor to describe Pareto Z-eigenvalues inclusion intervals.

### Theorem 2

Let $$\mathcal{A}\in \mathbb{R}^{[m,n]}$$ and $$\sigma (\mathcal{A})\neq \emptyset$$. Then,

\begin{aligned} \sigma (\mathcal{A})\subseteq \Omega (\mathcal{A})=\bigcup _{i\in N} \Omega _{i}(\mathcal{A}):= \bigl\{ \lambda \in \mathbb{R}: \vert \lambda \vert \leq \max \bigl\{ R_{i}( \mathcal{A})_{+},R_{i}( \mathcal{A})_{-} \bigr\} \bigr\} . \end{aligned}

### Proof

Suppose that $$(\lambda,x)$$ is a Pareto Z-eigenpair of $$\mathcal{A}$$. Then

\begin{aligned} \lambda x_{i}^{2}=\sum _{i_{2},\dots,i_{m}=1}^{n}a_{ii_{2}\cdots i_{m}}x_{i}x_{i_{2}} \cdots x_{i_{m}}. \end{aligned}
(6)

Denote $$x_{p}=\max_{i\in N}\{x_{i}\}$$. Then, $$0< x_{p}\leq 1$$ as $$x^{\top}x=1$$. Recalling the pth equation of (6), we get

\begin{aligned} \lambda x_{p}^{2}=\sum_{i_{2},\dots,i_{m}=1}^{n}a_{pi_{2}\cdots i_{m}}x_{p}x_{i_{2}} \cdots x_{i_{m}}. \end{aligned}

Taking the absolute value of the equation above, one has

\begin{aligned} \vert \lambda \vert x_{p}^{2}&= \Biggl\vert \sum_{i_{2},\dots,i_{m}=1}^{n}[a_{pi_{2} \cdots i_{m}}]_{+}x_{p}x_{i_{2}} \cdots x_{i_{m}}-\sum_{i_{2},\dots,i_{m}=1}^{n}[a_{pi_{2} \cdots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}} \Biggr\vert \\ &\leq \max \Biggl\{ \sum_{i_{2},\dots,i_{m}=1}^{n}[a_{pi_{2}\cdots i_{m}}]_{+}x_{p}x_{i_{2}} \cdots x_{i_{m}},\sum_{i_{2},\dots,i_{m}=1}^{n}[a_{pi_{2}\cdots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}} \Biggr\} \\ &\leq \max \Biggl\{ \sum_{i_{2},\dots,i_{m}=1}^{n}[a_{pi_{2}\cdots i_{m}}]_{+}, \sum_{i_{2},\dots,i_{m}=1}^{n}[a_{pi_{2}\cdots i_{m}}]_{-} \Biggr\} x_{p}^{2}. \end{aligned}
(7)

Dividing both sides by $$x_{p}^{2}$$, one has

\begin{aligned} \vert \lambda \vert \leq \max \bigl\{ R_{p}(\mathcal{A})_{+},R_{p}( \mathcal{A})_{-} \bigr\} , \end{aligned}

which implies $$\lambda \in \Omega _{p}(\mathcal{A})$$, and hence $$\sigma (\mathcal{A})\subseteq \Omega (\mathcal{A})$$. □

### Theorem 3

Let $$\mathcal{A}\in \mathbb{R}^{[m,n]}$$ and $$\sigma (\mathcal{A})\neq \emptyset$$. Then,

\begin{aligned} \sigma (\mathcal{A})\subseteq \Phi (\mathcal{A})=\bigcup _{i\in N} \bigcap_{j\in N,i\neq j}\Phi _{i,j}(\mathcal{A}), \end{aligned}

where

\begin{aligned} \Phi _{i,j}(\mathcal{A}):={}& \bigl\{ \lambda \in \mathbb{R}: \bigl\vert \lambda -R_{i}^{j}( \mathcal{A})_{+} \bigr\vert \vert \lambda \vert \leq [a_{ij\cdots j}]_{+} \max \bigl\{ R_{j}( \mathcal{A})_{+},R_{j}( \mathcal{A})_{-} \bigr\} \bigr\} \\ & {}\cup \bigl\{ \lambda \in \mathbb{R}: \bigl\vert \lambda -R_{i}^{j}( \mathcal{A})_{-} \bigr\vert \vert \lambda \vert \leq [a_{ij\cdots j}]_{-} \max \bigl\{ R_{j}( \mathcal{A})_{+},R_{j}( \mathcal{A})_{-} \bigr\} \bigr\} . \end{aligned}

### Proof

Suppose that $$(\lambda,x)$$ is a Pareto Z-eigenpair of $$\mathcal{A}$$. Setting $$0< x_{p}=\max_{i\in N}\{x_{i}\}$$ and referring to the pth equation of (6), for any $$q\in N, q\neq p$$, we obtain

\begin{aligned} \vert \lambda \vert x_{p}^{2}={}& \Biggl\vert \sum _{i_{2},\dots,i_{m}=1}^{n}a_{pi_{2} \cdots i_{m}}x_{p}x_{i_{2}} \cdots x_{i_{m}} \Biggr\vert \\ ={}& \Biggl\vert \sum_{i_{2},\dots,i_{m}=1}^{n}[a_{pi_{2}\cdots i_{m}}]_{+}x_{p}x_{i_{2}} \cdots x_{i_{m}}-\sum_{i_{2},\dots,i_{m}=1}^{n}[a_{pi_{2}\cdots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}} \Biggr\vert \\ \leq{}& \max \Biggl\{ \sum_{i_{2},\dots,i_{m}=1}^{n}[a_{pi_{2}\cdots i_{m}}]_{+}x_{p}x_{i_{2}} \cdots x_{i_{m}},\sum_{i_{2},\dots,i_{m}=1}^{n}[a_{pi_{2}\cdots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}} \Biggr\} \\ \leq{}& \max \biggl\{ [a_{pq\cdots q}]_{+}x_{p}x_{q}^{m-1}+ \sum_{\delta _{q{i_{2}} \ldots{i_{m}}}=0}[a_{pi_{2}\cdots i_{m}}]_{+}x_{p}x_{i_{2}} \cdots x_{i_{m}}, \\ &{} [a_{pq\cdots q}]_{-}x_{p}x_{q}^{m-1}+ \sum_{ \delta _{q{i_{2}}\ldots{i_{m}}}=0}[a_{pi_{2}\cdots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}} \biggr\} \\ \leq{}& \max \biggl\{ [a_{pq\cdots q}]_{+}x_{p}x_{q}+ \sum_{\delta _{q{i_{2}} \ldots{i_{m}}}=0}[a_{pi_{2}\cdots i_{m}}]_{+}x_{p}^{2}, \\ &{} [a_{pq\cdots q}]_{-}x_{p}x_{q}+\sum _{ \delta _{q{i_{2}}\ldots{i_{m}}}=0}[a_{pi_{2}\cdots i_{m}}]_{-}x_{p}^{2} \biggr\} \\ ={}&\max \bigl\{ R_{p}^{q}(\mathcal{A})_{+}x_{p}^{2}+[a_{pq\cdots q}]_{+}x_{p}x_{q},R_{p}^{q}( \mathcal{A})_{-}x_{p}^{2}+[a_{pq\cdots q}]_{-}x_{p}x_{q} \bigr\} , \end{aligned}

which implies

\begin{aligned} \vert \lambda \vert x_{p}^{2}\leq \max \bigl\{ R_{p}^{q}(\mathcal{A})_{+}x_{p}^{2}+[a_{pq \cdots q}]_{+}x_{p}x_{q},R_{p}^{q}( \mathcal{A})_{-}x_{p}^{2}+[a_{pq \cdots q}]_{-}x_{p}x_{q} \bigr\} . \end{aligned}
(8)

Recalling the qth equation of (6), one has

\begin{aligned} \vert \lambda \vert x_{q}^{2}&= \Biggl\vert \sum _{i_{2},\dots,i_{m}=1}^{n}a_{qi_{2} \cdots i_{m}}x_{q}x_{i_{2}} \cdots x_{i_{m}} \Biggr\vert \\ &\leq \max \Biggl\{ \sum_{i_{2},\dots,i_{m}=1}^{n}[a_{qi_{2}\cdots i_{m}}]_{+}x_{q}x_{i_{2}} \cdots x_{i_{m}},\sum_{i_{2},\dots,i_{m}=1}^{n}[a_{qi_{2}\cdots i_{m}}]_{-}x_{q}x_{i_{2}} \cdots x_{i_{m}} \Biggr\} \\ &\leq \max \Biggl\{ \sum_{i_{2},\dots,i_{m}=1}^{n}[a_{qi_{2}\cdots i_{m}}]_{+}x_{q}x_{p}, \sum_{i_{2},\dots,i_{m}=1}^{n}[a_{qi_{2}\cdots i_{m}}]_{-}x_{q}x_{p} \Biggr\} \\ &=\max \bigl\{ R_{q}(\mathcal{A})_{+}x_{p}x_{q},R_{q}( \mathcal{A})_{-}x_{p}x_{q} \bigr\} , \end{aligned}

which shows

\begin{aligned} \vert \lambda \vert x_{q}^{2}\leq \max \bigl\{ R_{q}(\mathcal{A})_{+},R_{q}( \mathcal{A})_{-} \bigr\} x_{p}x_{q}. \end{aligned}
(9)

We now break up the argument into two cases for (8).

Case I. $$|\lambda | x_{p}^{2}\leq R_{p}^{q}(\mathcal{A})_{+}x_{p}^{2}+[a_{pq \cdots q}]_{+}x_{p}x_{q}$$. In this case, if $$x_{q}>0$$, multiplying (8) with (9) and dividing $$x_{p}^{2}x_{q}^{2}$$ yields

\begin{aligned} \bigl( \vert \lambda \vert -R_{p}^{q}( \mathcal{A})_{+} \bigr) \vert \lambda \vert \leq [a_{pq\cdots q}]_{+} \max \bigl\{ R_{q}( \mathcal{A})_{+},R_{q}( \mathcal{A})_{-} \bigr\} , \end{aligned}

which implies $$\lambda \in \Phi _{p,q}(\mathcal{A})$$.

Otherwise, $$x_{q}=0$$. From (8), it holds that

\begin{aligned} \bigl( \vert \lambda \vert -R_{p}^{q}( \mathcal{A})_{+} \bigr) \vert \lambda \vert \leq 0\leq [a_{pq \cdots q}]_{+}\max \bigl\{ R_{q}( \mathcal{A})_{+},R_{q}( \mathcal{A})_{-} \bigr\} , \end{aligned}

which shows that $$\lambda \in \Phi _{p,q}(\mathcal{A})$$.

Case II. $$|\lambda | x_{p}^{2}\leq R_{p}^{q}(\mathcal{A})_{-}x_{p}^{2}+[a_{pq \cdots q}]_{-}x_{p}x_{q}$$. Following similar arguments as in the proof of Case I, we obtain $$\lambda \in \Phi _{p,q}(\mathcal{A})$$.

Combining Cases I and II, we obtain the desired results. □

Compared with Theorem 2, the result of Theorem 3 requires relatively many calculations but has accurate results. Detailed investigation is given in Corollary 1.

### Corollary 1

For a tensor $$\mathcal{A}\in \mathbb{R}^{[m,n]}$$, it holds that

\begin{aligned} \Phi (\mathcal{A})\subseteq \Omega (\mathcal{A}), \end{aligned}

where $$\Phi (\mathcal{A})$$ and $$\Omega (\mathcal{A})$$ are defined in Theorems 2and 3.

### Proof

For any $$\lambda \in \Phi (\mathcal{A})$$, there exist $$p,q\in N$$ with $$p\neq q$$ such that

\begin{aligned} \bigl( \vert \lambda \vert -R_{p}^{q}( \mathcal{A})_{+} \bigr) \vert \lambda \vert \leq [a_{pq\cdots q}]_{+} \max \bigl\{ R_{q}( \mathcal{A})_{+},R_{q}( \mathcal{A})_{-} \bigr\} , \end{aligned}

or

\begin{aligned} \bigl( \vert \lambda \vert -R_{p}^{q}( \mathcal{A})_{-} \bigr) \vert \lambda \vert \leq [a_{pq\cdots q}]_{-} \max \bigl\{ R_{q}( \mathcal{A})_{+},R_{q}( \mathcal{A})_{-} \bigr\} . \end{aligned}

We now break up the argument into two cases.

Case I. $$(|\lambda |-R_{p}^{q}(\mathcal{A})_{+})|\lambda |\leq [a_{pq\cdots q}]_{+} \max \{R_{q}(\mathcal{A})_{+},R_{q}(\mathcal{A})_{-}\}$$.

If $$[a_{pq\cdots q}]_{+}\max \{R_{q}(\mathcal{A})_{+},R_{q}(\mathcal{A})_{-} \}=0$$, it holds that

\begin{aligned} \vert \lambda \vert \leq R_{p}^{q}( \mathcal{A})_{+}\leq R_{p}(\mathcal{A})_{+} \leq \max \bigl\{ R_{p}( \mathcal{A})_{+},R_{p}( \mathcal{A})_{-} \bigr\} , \end{aligned}

or

\begin{aligned} \vert \lambda \vert =0\leq \max \bigl\{ R_{q}( \mathcal{A})_{+},R_{q}( \mathcal{A})_{-} \bigr\} , \end{aligned}

which indicates that

\begin{aligned} \lambda \in \Omega _{p}(\mathcal{A})\subseteq \Omega ( \mathcal{A})\quad \text{or}\quad \lambda \in \Omega _{q}(\mathcal{A})\subseteq \Omega (\mathcal{A}). \end{aligned}
(10)

Otherwise, $$[a_{pq\cdots q}]_{+}\max \{R_{q}(\mathcal{A})_{+},R_{q}(\mathcal{A})_{-} \}> 0$$. Then,

\begin{aligned} \frac{ \vert \lambda \vert -R_{p}^{q}(\mathcal{A})_{+}}{[a_{pq\cdots q}]_{+}} \cdot \frac{ \vert \lambda \vert }{\max \{R_{q}(\mathcal{A})_{+},R_{q}(\mathcal{A})_{-}\}} \leq 1, \end{aligned}

which implies

\begin{aligned} \frac{ \vert \lambda \vert -R_{p}^{q}(\mathcal{A})_{+}}{[a_{pq\cdots q}]_{+}} \leq 1 \quad\text{or} \quad\frac{ \vert \lambda \vert }{\max \{R_{q}(\mathcal{A})_{+},R_{q}(\mathcal{A})_{-}\}} \leq 1. \end{aligned}

Consequently, (10) holds.

Case II. $$(|\lambda |-R_{p}^{q}(\mathcal{A})_{-})|\lambda |\leq [a_{pq\cdots q}]_{-} \max \{R_{q}(\mathcal{A})_{+},R_{q}(\mathcal{A})_{-}\}$$. Following similar arguments as in the proof of Case I, we can prove that $$\lambda \in \Omega (\mathcal{A})$$.

Combining Case I with Case II, we conclude that $$\Phi (\mathcal{A})\subseteq \Omega (\mathcal{A})$$. □

To get accurate results, we divide precisely the index set of $$\mathcal{A}$$ and establish Theorem 4.

### Theorem 4

Let $$\mathcal{A}\in \mathbb{R}^{[m,n]}$$ and $$\sigma (\mathcal{A})\neq \emptyset$$. Then,

\begin{aligned} \sigma (\mathcal{A})\subseteq \mathcal{N}(\mathcal{A})=\bigcup _{i \in N}\bigcap_{j\in N,j\neq i} \mathcal{N}_{i,j}(\mathcal{A}), \end{aligned}

where $$\mathcal{N}_{i,j}(\mathcal{A})=\{\lambda \in \mathbb{R}:(|\lambda |- \max \{P_{i}^{j}(\mathcal{A})_{+},P_{i}^{j}(\mathcal{A})_{-}\})| \lambda |\leq \max \{R_{i}(\mathcal{A})_{+}-P_{i}^{j}(\mathcal{A})_{+},R_{i}( \mathcal{A})_{-}-P_{i}^{j}(\mathcal{A})_{-}\}\cdot \max \{R_{j}( \mathcal{A})_{+},R_{j}(\mathcal{A})_{-}\}\}$$.

### Proof

Suppose that $$(\lambda,x)$$ is a Pareto Z-eigenpair of $$\mathcal{A}$$. Setting $$0< x_{p}=\max_{i\in N}\{x_{i}\}$$ and referring to the pth equation of (6), for any $$q\in N, q\neq p$$, one has

\begin{aligned} \lambda x_{p}^{2}&= \sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\in \{i_{2},\ldots, i_{m}\}}}a_{p i_{2} \ldots i_{m}}x_{p}x_{i_{2}} \cdots x_{i_{m}}+\sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}a_{p i_{2} \ldots i_{m}}x_{p}x_{i_{2}} \cdots x_{i_{m}}. \end{aligned}

Taking the absolute value of the equation above, we obtain

\begin{aligned} \vert \lambda \vert x_{p}^{2} \leq{}& \biggl\vert \sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\in \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{+}x_{p}x_{i_{2}} \cdots x_{i_{m}} -\sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\in \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}} \biggr\vert \\ & {}+ \biggl\vert \sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{+}x_{p}x_{i_{2}} \cdots x_{i_{m}} -\sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}} \biggr\vert \\ \leq{}& \max \biggl\{ \sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\in \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{+}x_{p}x_{i_{2}} \cdots x_{i_{m}},\sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\in \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}} \biggr\} \\ & {}+\max \biggl\{ \sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{+}x_{p}x_{i_{2}} \cdots x_{i_{m}},\sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}} \biggr\} \\ \leq{}& x_{p}x_{q}\max \biggl\{ \sum _{ \substack{ i_{2},\ldots, i_{m} \in N\\q\in \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{+},\sum _{ \substack{ i_{2},\ldots, i_{m} \in N\\q\in \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{-} \biggr\} \\ &{} +x_{p}^{2}\max \biggl\{ \sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{+}, \sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{-} \biggr\} , \end{aligned}

where the third inequality holds from $$0< x_{p}^{m-1}\leq x_{p}\leq 1$$ and $$0\leq x_{q}<1$$. Further,

\begin{aligned} & \biggl[ \vert \lambda \vert -\max \biggl\{ \sum _{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{+},\sum _{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{-} \biggr\} \biggr]x_{p}^{2} \\ & \quad\leq x_{p}x_{q}\max \biggl\{ \sum _{ \substack{ i_{2},\ldots, i_{m} \in N\\q\in \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{+},\sum _{ \substack{ i_{2},\ldots, i_{m} \in N\\q\in \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{-} \biggr\} . \end{aligned}
(11)

In view of the qth equation of (6), we deduce

\begin{aligned} \vert \lambda \vert x_{q}^{2}&= \biggl\vert \sum_{i_{2},\dots,i_{m}\in N}a_{q i_{2} \cdots i_{m}}x_{q}x_{i_{2}} \cdots x_{i_{m}} \biggr\vert \\ &= \biggl\vert \sum_{i_{2},\dots,i_{m}\in N}[a_{q i_{2}\cdots i_{m}}]_{+}x_{q}x_{i_{2}} \cdots x_{i_{m}}-\sum_{i_{2},\dots,i_{m}\in N}[a_{q i_{2}\cdots i_{m}}]_{-}x_{q}x_{i_{2}} \cdots x_{i_{m}} \biggr\vert \\ &\leq \max \biggl\{ \sum_{i_{2},\dots,i_{m}\in N}[a_{q i_{2}\cdots i_{m}}]_{+}x_{q}x_{i_{2}} \cdots x_{i_{m}},\sum_{i_{2},\dots,i_{m}\in N}[a_{q i_{2}\cdots i_{m}}]_{-}x_{q}x_{i_{2}} \cdots x_{i_{m}} \biggr\} \\ &\leq x_{p}x_{q} \max \biggl\{ \sum _{i_{2},\dots,i_{m}\in N}[a_{q i_{2} \cdots i_{m}}]_{+},\sum _{i_{2},\dots,i_{m}\in N}[a_{q i_{2}\cdots i_{m}}]_{-} \biggr\} . \end{aligned}
(12)

We now break up the argument into two cases.

Case I: $$x_{q} >0$$. Multiplying (11) with (12) and dividing $$x_{p}^{2}x_{q}^{2}$$, we obtain

\begin{aligned} \bigl( \vert \lambda \vert -\max \bigl\{ P_{p}^{q}( \mathcal{A})_{+},P_{p}^{q}( \mathcal{A})_{-} \bigr\} \bigr) \vert \lambda \vert \leq {}&\max \bigl\{ R_{p}(\mathcal{A})_{+}-P_{p}^{q}( \mathcal{A})_{+},R_{p}(\mathcal{A})_{-}-P_{p}^{q}( \mathcal{A})_{-} \bigr\} \\ &{}\times \max \bigl\{ R_{q}(\mathcal{A})_{+},R_{q}( \mathcal{A})_{-} \bigr\} , \end{aligned}

which implies $$\lambda \in \mathcal{N}_{p,q}(\mathcal{A})\subseteq \mathcal{N}( \mathcal{A})$$.

Case II: $$x_{q}=0$$. It follows from (11) that

\begin{aligned} \vert \lambda \vert \leq \max \biggl\{ \sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{+}, \sum_{ \substack{ i_{2},\ldots, i_{m} \in N\\q\notin \{i_{2},\ldots, i_{m}\}}}[a_{p i_{2} \ldots i_{m}}]_{-} \biggr\} , \end{aligned}

that is,

\begin{aligned} \bigl[ \vert \lambda \vert -\max \bigl\{ P_{p}^{q}( \mathcal{A})_{+},P_{p}^{q}( \mathcal{A})_{-} \bigr\} \bigr] \vert \lambda \vert \leq{}& \max \bigl\{ R_{p}(\mathcal{A})_{+}-P_{p}^{q}( \mathcal{A})_{+},R_{p}(\mathcal{A})_{-}-P_{p}^{q}( \mathcal{A})_{-} \bigr\} \\ &{} \times \max \bigl\{ R_{q}(\mathcal{A})_{+},R_{q}( \mathcal{A})_{-} \bigr\} , \end{aligned}

which implies $$\lambda \in \mathcal{N}_{p,q}(\mathcal{A})\subseteq \mathcal{N}( \mathcal{A})$$. □

In what follows, we now test the efficiency of the obtained results.

### Example 1

Consider a 3rd order 3-dimensional tensor $$\mathcal{A}=(a_{ijk})$$ defined by

\begin{aligned} a_{ijk}=\textstyle\begin{cases} a_{111}=1; \qquad a_{112}=-1; \qquad a_{131}=1;\qquad a_{133}=1; \\ a_{211}=-1;\qquad a_{222}=2;\qquad a_{232}=1; \\ a_{311}=1;\qquad a_{322}=3;\qquad a_{323}=1; \\ a_{ijk}=0, \quad \text{otherwise}. \end{cases}\displaystyle \end{aligned}

By calculating, we have

\begin{aligned} &\bigl\Vert [\mathcal{A}]_{-} \bigr\Vert _{F}=\sqrt{2},\qquad \bigl\Vert [ \mathcal{A}]_{+} \bigr\Vert _{F}= \sqrt{19},\qquad \bar{a}\cdot n^{\frac{m}{2}}=9\sqrt{3},\\ &R_{1}( \mathcal{A})_{+}=3,\qquad R_{1}(\mathcal{A})_{-}=1,\qquad R_{2}( \mathcal{A})_{+}=3,\qquad R_{2}( \mathcal{A})_{-}=1,\\ & R_{3}(\mathcal{A})_{+}=5,\qquad R_{3}( \mathcal{A})_{-}=0,\\ &R_{1}^{2}( \mathcal{A})_{+}=3,\qquad R_{1}^{2}( \mathcal{A})_{-}=1, \qquad R_{1}^{3}( \mathcal{A})_{+}=2,\qquad R_{1}^{3}( \mathcal{A})_{-}=1,\\ &R_{2}^{1}( \mathcal{A})_{+}=3,\qquad R_{2}^{1}(\mathcal{A})_{-}=0,\qquad R_{2}^{3}(\mathcal{A})_{+}=3, \qquad R_{2}^{3}( \mathcal{A})_{-}=1,\\ &R_{3}^{1}( \mathcal{A})_{+}=4, \qquad R_{3}^{1}( \mathcal{A})_{-}=0,\qquad R_{3}^{2}( \mathcal{A})_{+}=2,\qquad R_{3}^{2}(\mathcal{A})_{-}=0,\\ &P_{1}^{2}( \mathcal{A})_{+}=3,\qquad P_{1}^{2}( \mathcal{A})_{-}=0,\qquad P_{1}^{3}( \mathcal{A})_{+}=1,\qquad P_{1}^{3}( \mathcal{A})_{-}=1,\\ &P_{2}^{1}( \mathcal{A})_{+}=3,\qquad P_{2}^{1}(\mathcal{A})_{-}=0,\qquad P_{2}^{3}(\mathcal{A})_{+}=2, \qquad P_{2}^{3}( \mathcal{A})_{-}=1,\\ &P_{3}^{1}( \mathcal{A})_{+}=4, \qquad P_{3}^{1}( \mathcal{A})_{-}=0,\qquad P_{3}^{2}( \mathcal{A})_{+}=1,\qquad P_{3}^{2}(\mathcal{A})_{-}=0. \end{aligned}

According to Theorem 1, we obtain

\begin{aligned} \Psi (\mathcal{A})=\{\lambda \in \mathbb{R}:-\sqrt{2}\leq \lambda \leq \sqrt{19} \}. \end{aligned}

Referring to Theorem 2, we deduce

\begin{aligned} \Omega (\mathcal{A})=\bigcup_{i\in N}\Omega _{i}(\mathcal{A})=\bigl\{ \lambda \in \mathbb{R}: \vert \lambda \vert \leq 5 \bigr\} . \end{aligned}

Recalling Theorem 3, one has

\begin{aligned} \Phi (\mathcal{A})=\bigcup_{i\in N}\bigcap _{j\in N,i\neq j}\Phi _{i,j}( \mathcal{A})=\bigl\{ \lambda \in \mathbb{R}: \vert \lambda \vert \leq 1+\sqrt{10}\bigr\} , \end{aligned}

where

 $$\Phi _{1,2}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq 3\}$$ $$\Phi _{1,3}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq 1+\sqrt{6}\}$$ $$\Phi _{2,1}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq 3\}$$ $$\Phi _{2,3}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq 3\}$$ $$\Phi _{3,1}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq 2+\sqrt{7}\}$$ $$\Phi _{3,2}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq 1+\sqrt{10} \}$$.

It follows from Theorem 4 that

\begin{aligned} \mathcal{N}(\mathcal{A})=\bigcup_{i\in N}\bigcap _{j\in N,i\neq j} \mathcal{N}_{i,j}(\mathcal{A})= \biggl\{ \lambda \in \mathbb{R}: \vert \lambda \vert \leq \frac{1+\sqrt{41}}{2} \biggr\} , \end{aligned}

where

 $$\mathcal{N}_{1,2}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq \frac{3+\sqrt{21}}{2}\}$$ $$\mathcal{N}_{1,3}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq \frac{1+\sqrt{41}}{2}$$ $$\mathcal{N}_{2,1}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq \frac{3+\sqrt{21}}{2}\}$$ $$\mathcal{N}_{2,3}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq 1+\sqrt{6}$$ $$\mathcal{N}_{3,1}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq 2+\sqrt{7}\}$$ $$\mathcal{N}_{3,2}(\mathcal{A})=\{\lambda \in \mathbb{R}: |\lambda |\leq 3\}$$.

## Judging strict copositivity of tensors

In this section, we mainly propose a sufficient condition for judging strict copositivity of $$\mathcal{A}$$.

### Theorem 5

Let $$\mathcal{A}=(a_{{i_{1}}{i_{2}}\cdots{i_{m}}})\in \mathbb{R}^{[m,n]}$$ be symmetric with $$a_{i\cdots i}>0$$ for $$i\in N$$. Then $$\mathcal{A}$$ is strictly copositive provided that

\begin{aligned} a_{i\cdots i} \biggl(\frac{1}{\sqrt{n}} \biggr)^{m-2}-R_{i}( \mathcal{A})_{-}>0. \end{aligned}
(13)

### Proof

Suppose that $$(\lambda,x)$$ is a Pareto Z-eigenpair of $$\mathcal{A}$$. Setting $$0< x_{p}=\max_{i\in N}\{x_{i}\}$$ and referring to the pth equation of (6), we obtain

\begin{aligned} \lambda x_{p}^{2}&=\sum _{i_{2},\dots,i_{m}=1}^{n}a_{pi_{2} \cdots i_{m}}x_{p}x_{i_{2}} \cdots x_{i_{m}} \\ &=a_{p\cdots p}x_{p}^{m}+\sum _{\delta _{pi_{2}\cdots i_{m}}=0}[a_{pi_{2} \cdots i_{m}}]_{+}x_{p}x_{i_{2}} \cdots x_{i_{m}}-\sum_{\delta _{pi_{2} \cdots i_{m}}=0}[a_{pi_{2}\cdots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}}. \end{aligned}

Further,

\begin{aligned} \lambda x_{p}^{2}&\geq a_{p\cdots p}x_{p}^{m}- \sum_{\delta _{pi_{2} \cdots i_{m}}=0}[a_{pi_{2}\cdots i_{m}}]_{-}x_{p}x_{i_{2}} \cdots x_{i_{m}} \\ &\geq a_{p\cdots p}x_{p}^{m} -\sum _{\delta _{pi_{2}\cdots i_{m}}=0}[a_{pi_{2} \cdots i_{m}}]_{-}x_{p}^{2} . \end{aligned}

Dividing both sides by $$x_{p}^{2}$$, we have

\begin{aligned} \lambda &\geq a_{p\cdots p}x_{p}^{m-2}- \sum_{\delta _{pi_{2}\cdots i_{m}}=0}[a_{pi_{2} \cdots i_{m}}]_{-}=a_{p\cdots p}x_{p}^{m-2}-R_{p}( \mathcal{A})_{-}. \end{aligned}
(14)

Since $$x_{p}=\max_{i\in N}\{x_{i}\}$$ and $$x^{\top}x=1$$, we deduce $$x_{p}\geq \frac{1}{\sqrt{n}}$$. It follows from $$a_{i\cdots i}>0$$ and (14) that

\begin{aligned} \lambda \geq a_{p\cdots p} \biggl(\frac{1}{\sqrt{n}} \biggr)^{m-2}-\sum_{ \delta _{pi_{2}\cdots i_{m}}=0}[a_{pi_{2}\cdots i_{m}}]_{-}= a_{p \cdots p} \biggl(\frac{1}{\sqrt{n}} \biggr)^{m-2}-R_{p}( \mathcal{A})_{-}. \end{aligned}
(15)

Combining (13) with (15), we have $$\lambda >0$$ and $$\mathcal{A}$$ is strictly copositive. □

From the conclusion, identifying the strict copositivity of tensor $$\mathcal{A}$$ requires that it is symmetric. For general tensors, symmetry is a relatively strict condition. To tackle this problem, we may symmetrize the tensors $$\mathcal{A}=(a_{{i_{1}}{i_{2}}\cdots{i_{m}}})\in \mathbb{R}^{[m,n]}$$ as follows:

\begin{aligned} \widetilde{a}_{{i_{1}}{i_{2}}\cdots{i_{m}}}= \textstyle\begin{cases} a_{{i_{1}}{i_{2}}\cdots{i_{m}}}&\text{if } {i_{1}}={i_{2}}=\cdots ={i_{m}}, \\ \frac{1}{m!}\sum_{{i_{2}}\cdots{i_{m}}\in \Gamma _{m}}a_{{i_{1}}{i_{2}} \cdots{i_{m}}}&\text{otherwise}, \end{cases}\displaystyle \end{aligned}

where $$\widetilde{\mathcal{A}}=(\widetilde{a}_{{i_{1}}{i_{2}}\cdots{i_{m}}}) \in \mathbb{R}^{[m,n]}$$ is the symmetrization tensor under permutation group $$\Gamma _{m}$$.

The following example shows that the result given in Theorem 5 can verify the strict copositivity of tensors.

### Example 2

Consider a 3rd order 2-dimensional tensor $$\mathcal{A}=(a_{ijk})$$ defined by

\begin{aligned} a_{ijk}=\textstyle\begin{cases} a_{111}=4;\qquad a_{112}=-1; \qquad a_{121}=-1;\qquad a_{122}=0; \\ a_{222}=2;\qquad a_{211}=-1;\qquad a_{212}=0; \qquad a_{221}=0. \end{cases}\displaystyle \end{aligned}

It is easy to see that $$\mathcal{A}$$ is symmetric with

\begin{aligned} R_{1}(\mathcal{A})_{-}=2, \qquad R_{2}( \mathcal{A})_{-}=1. \end{aligned}

According to Theorem 5, we have

\begin{aligned} &a_{111} \biggl(\frac{1}{\sqrt{2}} \biggr)^{3-2}- R_{1}(\mathcal{A})_{-}=2( \sqrt{2}-1)>0, \\ &a_{222} \biggl(\frac{1}{\sqrt{2}} \biggr)^{3-2}- R_{2}(\mathcal{A})_{-}= \sqrt{2}-1>0, \end{aligned}

which means that $$\mathcal{A}$$ is strictly copositive.

When $$\mathcal{A}$$ is asymmetric, we still identify the strict copositivity by Theorem 5.

### Example 3

Consider a 3rd order 2-dimensional tensor $$\mathcal{A}=(a_{ijk})$$ defined by

\begin{aligned} a_{ijk}=\textstyle\begin{cases} a_{111}=4;\qquad a_{112}=-1;\qquad a_{121}=-2;\qquad a_{122}=0; \\ a_{222}=2;\qquad a_{211}=-1;\qquad a_{212}=0;\qquad a_{221}=0. \end{cases}\displaystyle \end{aligned}

Since $$a_{112}=-1, a_{121}=-2$$, and $$a_{211}=-1$$, we know that $$\mathcal{A}$$ is asymmetric. Therefore, we cannot directly use Theorem 5 to judge whether $$\mathcal{A}$$ is strictly copositive. Symmetrizing $$\mathcal{A}$$, we obtain $$\widetilde{\mathcal{A}}$$ with

\begin{aligned} \widetilde{a}_{ijk}=\textstyle\begin{cases} \widetilde{a}_{111}=4;\qquad \widetilde{a}_{112}=-\frac{4}{3}; \qquad\widetilde{a}_{121}=-\frac{4}{3}; \qquad\widetilde{a}_{122}=0; \\ \widetilde{a}_{222}=2;\qquad \widetilde{a}_{211}=-\frac{4}{3}; \qquad\widetilde{a}_{212}=0; \qquad\widetilde{a}_{221}=0. \end{cases}\displaystyle \end{aligned}

It is easy to see that $$\widetilde{\mathcal{A}}$$ is symmetric with

\begin{aligned} R_{1}(\widetilde{\mathcal{A}})_{-}=\frac{8}{3},\qquad R_{2}( \widetilde{\mathcal{A}})_{-}=\frac{4}{3}. \end{aligned}

According to Theorem 5, we have

\begin{aligned} &a_{111} \biggl(\frac{1}{\sqrt{2}} \biggr)^{3-2}-R_{1}( \widetilde{\mathcal{A}})_{-}=\frac{6\sqrt{2}-8}{3}>0, \\ &a_{222} \biggl(\frac{1}{\sqrt{2}} \biggr)^{3-2}-R_{2}( \widetilde{\mathcal{A}})_{-}=\frac{3\sqrt{2}-4}{3}>0, \end{aligned}

which implies that $$\widetilde{\mathcal{A}}$$ is strictly copositive. Taking into account that $$\mathcal{A}x^{3}=\widetilde{\mathcal{A}}x^{3}>0$$, we deduce that $$\mathcal{A}$$ is strictly copositive.

## Conclusion

In this paper, we proposed sharp Pareto Z-eigenvalue inclusion intervals and established comparisons among different Pareto Z-eigenvalue inclusion intervals for tensor eigenvalue complementarity problems. Meanwhile, we gave a sufficient condition to check strict copositivity of real tensors. Further studies can be considered to develop some algorithms by Pareto Z-eigenvalue inclusion intervals for tensor eigenvalue complementarity problems, as done in  for solving the matrix eigenvalue complementarity problems.

## Availability of data and materials

Data sharing not applicable to this article as no data were generated or analyzed during the current study.

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## Acknowledgements

The authors would like to thank anonymous referees and editors for their helpful comments and suggestions, which greatly improved the quality of this paper.

## Funding

This research is supported by the Natural Science Foundation of Shandong Province (ZR2020MA025, ZR2019PA016) and the National Natural Science Foundation of China (12071250, 11901343).

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### Contributions

PY: original draft writing, review writing, and editing. YJW: conceptualization, supervision, and funding acquisition. GW: computation and review writing. QLH: computation and review writing. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Yiju Wang.

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