Pareto Z-eigenvalue inclusion theorems for tensor eigenvalue complementarity problems

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 A = (a i 1 i 2 ···i m ) ∈ R [m,n] be an mth-order n-dimensional real tensor, x be a real n-vector and N = {1, 2, . . . , n}. Denote by Ax m-1 the vector in R n with entries Ax m-1 i = i 2 ,...,i m ∈N a ii 2 ···i m x i 2 · · · x i m .
Consider the tensor eigenvalue complementarity problems of finding (λ, x) ∈ R×R n + \{0} such that where a⊥b means that vectors a and b are perpendicular to each other. For the problem, its solution (λ, x) ∈ R × R n + \{0} is called a Pareto Z-eigenpair of tensor A. The Pareto Z-eigenpair of a tensor was introduced by Song [1], which is a natural generalization of that of a matrix [2][3][4][5]. It is worth noting that Pareto Z-eigenvalues of A are closely related to Z (Z + )-eigenvalues of A introduced by Lim [6] and Qi [7,8], respectively. Definition 1 For a tensor A = (a i 1 i 2 ···i m ) ∈ R [m,n] , if there exist (λ, x) ∈ R × R n \{0} such that Ax m-1 = λx, x x = 1, then (λ, x) is called a Z-eigenpair of tensor A. Further, Z-eigenvalue λ of A is said to be a Z + -eigenvalue, if its eigenvector x ∈ R n + \{0}.
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 [14][15][16][17][18], 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 R, and the n-dimensional real Euclidean space is denoted by R n . For any a ∈ R, we denote [a] + := max{0, a} and [a] -:= max{0, -a}. For any x ∈ R n , x ⊗m denotes a tensor whose entries are defined by (x ⊗m ) i 1 i 2 ···i m = x i 1 x i 2 · · · x i m for all i 1 , i 2 , . . . , i m ∈ N . For any A ∈ R [m,n] and x ∈ R n , we define For any i, j ∈ N , set

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 A. Some comparisons among different Pareto Z-eigenvalue inclusion intervals are also made in this section.
Based on the above lemmas, we have the following conclusion. and Combining (2) with (3) yields Meanwhile, from the definition of Pareto Z-eigenpair, we obtain 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 Zeigenvalues inclusion intervals. Denote Recalling the pth equation of (6), we get Taking the absolute value of the equation above, one has Dividing both sides by x 2 p , one has which implies λ ∈ p (A), and hence σ (A) ⊆ (A).
referring to the pth equation of (6), for any q ∈ N, q = p, we obtain Recalling the qth equation of (6), one has We now break up the argument into two cases for (8).
Otherwise, x q = 0. From (8), it holds that Following similar arguments as in the proof of Case I, we obtain λ ∈ p,q (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. Proof For any λ ∈ (A), there exist p, q ∈ N with p = q such that

Corollary 1 For a tensor
We now break up the argument into two cases.
Following similar arguments as in the proof of Case I, we can prove that λ ∈ (A).
Combining Case I with Case II, we conclude that (A) ⊆ (A).
To get accurate results, we divide precisely the index set of A and establish Theorem 4. Proof Suppose that (λ, x) is a Pareto Z-eigenpair of A. Setting 0 < x p = max i∈N {x i } and referring to the pth equation of (6), for any q ∈ N, q = p, one has

Theorem 4 Let
Taking the absolute value of the equation above, we obtain where the third inequality holds from 0 < x m-1 p ≤ x p ≤ 1 and 0 ≤ x q < 1. Further, In view of the qth equation of (6), we deduce We now break up the argument into two cases. Case I: x q > 0. Multiplying (11) with (12) and dividing x 2 p x 2 q , we obtain Case II: x q = 0. It follows from (11) In what follows, we now test the efficiency of the obtained results.
Example 1 Consider a 3rd order 3-dimensional tensor A = (a ijk ) defined by  According to Theorem 1, we obtain Referring to Theorem 2, we deduce Recalling Theorem 3, one has It follows from Theorem 4 that

Judging strict copositivity of tensors
In this section, we mainly propose a sufficient condition for judging strict copositivity of A.
problem, we may symmetrize the tensors A = (a i 1 i 2 ···i m ) ∈ R [m,n] as follows: where A = ( a i 1 i 2 ···i m ) ∈ R [m,n] is the symmetrization tensor under permutation group m .
The following example shows that the result given in Theorem 5 can verify the strict copositivity of tensors. It is easy to see that A is symmetric with R 1 (A) -= 2, R 2 (A) -= 1.
According to Theorem 5, we have which means that A is strictly copositive. When A is asymmetric, we still identify the strict copositivity by Theorem 5. Since a 112 = -1, a 121 = -2, and a 211 = -1, we know that A is asymmetric. Therefore, we cannot directly use Theorem 5 to judge whether A is strictly copositive. Symmetrizing A, we obtain A with It is easy to see that A is symmetric with