# An SDP Method for Copositivity of Partially Symmetric Tensors

Authors: Chunyan Wang; Haibin Chen; Haitao Che
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In this paper, we consider the problem of detecting the copositivity of partially symmetric rectangular tensors. We first propose a semidefinite relaxation algorithm for detecting the copositivity of partially symmetric rectangular tensors. Then, the convergence of the proposed algorithm is given, and it shows that we can always catch the copositivity of given partially symmetric tensors. Several preliminary numerical results confirm our theoretical findings.


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## Body

## 1. Introduction

LetA=ai1i2,…,ipj1j2,…,jq be a real p,q-th order m×n-dimensional rectangular tensor, where ai1i2,…,ipj1j2,…,jq∈ℝ for ik∈m, k∈p, jl∈n, and l∈q. If the entries of the tensor are invariant under any permutation of i1,i2,…,ip and j1,j2,…,jq, A is called a partially symmetric tensor. For the sake of simplicity, let ℙSp,qm×n be the set of all partially symmetric rectangular tensors with order p,q and dimension m×n. By the relationship between partially symmetric tensors and homogeneous polynomials, we always use the following notation:(1)fx,y=Axpyq=∑i1,…,ip∈mj1,…,jq∈nai1i2,…,ipj1j2,…,jqxi1xi2,…,xipyj1yj2,…,yjq.By this notation, we know thatA=ai1,…,ipj1,…,jq∈ℙSp,qm×n is strictly copositive if and only if(2)Axpyq≥>0,for allx∈ℝ+m,y∈ℝ+nwithx=1,y=1.Particularly, ifm=n and x=y, then it reduces to the copositivity of symmetric tensors [1–10].The copositive tensor has attracted many researches’ attention since it plays an important role in polynomial optimization [11], hypergraph theory [1], vacuum stability of a general scalar potential [12], tensor complementarity problem [13, 14], tensor eigenvalue complementarity problem [15, 16], and so on [17–37]. Kannike proved the vacuum stability conditions for more complicated potentials with the help of the copositive tensor [12]. Ling et al. [16] proposed that the tensor generalized eigenvalue complementarity problem is solvable and has one solution at least under assumptions that the related square tensor is strictly copositive. During the process of application, a challenging problem is how to detect the copositivity of tensors numerically.Recently, several numerical algorithms are proposed to check the copositivity of symmetric tensors. To the best of our knowledge, the first numerical algorithm was proposed by Chen et al. in [2], where the algorithm is based on the representation of the multivariate form in barycentric coordinates with respect to the standard simplex. Then, by a suitable convex subcone of a copositive tensor cone, an updated numerical algorithm for copositivity of tensors was proposed in [1]. It must be pointed out that the methods of [1, 2] can only capture strictly copositive tensors and noncopositive tensors. To overcome this drawback, in [38], Li et al. proposed an SDP relaxation algorithm to test the copositivity of higher-order tensors. Very recently, Nie et al. gave a complete semidefinite relaxation algorithm for detecting the copositivity of a matrix or tensor [39]. If the potential tensor is copositive, the algorithm can get a certificate for the copositivity. Otherwise, the algorithm can get a point that refutes the copositivity. Furthermore, it is showed that the detection can be done by solving a finite number of semidefinite relaxations for all matrices and tensors.For the copositivity of partially symmetric tensors, Gu et al. gave the first two spectral properties in [40], and some necessary or sufficient conditions for a real partially symmetric rectangular tensor to be copositive are further established. Moreover, an equivalent notion of strictly copositive rectangular tensors is presented [40]. In [41], Wang et al. extended the simplicial partition method for symmetric tensors to check the copositivity of partially symmetric tensors. However, as we discussed above, it can only capture all strictly copositive rectangular tensors or noncopositive rectangular tensors. When the input tensor is copositive but not strictly copositive, the algorithm may not stop in general. To solve this, motivated by the algorithm of [38, 39], we propose a new algorithm to check the copositivity of partially symmetric tensors in this paper.The remainder of this paper is organized as follows. In Section2, we recall some preliminaries on polynomials. In Section 3, we formulate the potential problem as a proper polynomial optimization problem which can be efficiently solved by Lasserre-type semidefinite relaxations. Then, a numerical method is proposed to check whether a given partially symmetric tensor is copositive or not, and the convergence of this algorithm is established. Several numerical experiments are listed in Section 4, and final remarks are given in Section 5.


## 2. Preliminaries

Letℝx be the ring of the polynomial with variables x=x1,x2,…,xn∈ℝn. Let ℝxd⊆ℝx denote the vector space of polynomials with degree at most d. The degree of a polynomial f is denoted as deg (f). Denote e as the vector of all entries which equals one. A polynomial p is called SOS if there exist p1,p2,…,pr∈ℝx such that p=p12+p22+,…,+pr2. Denote by ∑x the set of all SOS polynomials. For x∈ℝnandα∈Nn, let xα=x1α1x2α2,…,xnαn. Then, for any polynomial f∈ℝx, it can be denoted by fx=∑α∈Nnfαxα, and vecf:=fαα∈ℕn denotes its sequence of coefficients in the monomial basis of ℝx. For matrix A, its transpose is denoted by A⊤. For a symmetric matrix X,X≽0 means X is positive semidefinite. More details about polynomial optimization can be seen in [42–45].Forα=α1,α2,…,αn∈Nn, α=α1+,⋯,+αn, and denote Ndn=α∈Nnα≤d. For t∈ℝ, ⌈t⌉ denotes the smallest integer that is not smaller than t. If the subset I⊆ℝx satisfies that I+I⊆I and I⋅ℝx⊆ℝx, then I is called ideal. For a polynomial tuple h=h1,h2,…,hs, the ideal generated by h is defined such that(3)ℐh=h1ℝx+h2ℝx+,…,+hsℝx.Thek-th truncation ideal generated by h is(4)ℐhk=h1ℝxk−degh1+h2ℝxk−degh2+,⋯,+hsℝxk−deghs.For complex and real algebraic varieties of polynomial tupleh, define(5)VCh=x∈Cnhx=0,VRh=VCh∩ℝn.The quadratic module generated byg=g1,g2,…,gt is (denote g0=1)(6)Qgk=∑x+g1∑x+,…,+gt∑x.Fory=yα∈ℝNdn,α∈Ndn, where ℝNdn is the space of real vectors indexed by α∈Ndn, define(7)∑α∈Ndnpαx1α1x2α2,…,xnαn,y=∑α∈Ndnpαyα.For a polynomialq∈ℝx2k, the k-th localizing matrix of q is the symmetric matrix Lkqy satisfying(8)vecp1⊤Lqkyvecp2=qp1p2,y,for all p1,p2∈ℝx with degp1, degp2≤k−⌈degq/2⌉, where vecpi denotes the coefficient vector of the polynomial pi. When q=1,Lqky is the moment matrix Mky=L1ky. Let f=f1,f2,…,fr be a polynomial tuple; its localizing matrix is defined such that(9)Lfky=Lf1ky,Lf2ky,…,Lfrky.


## 3. The SDP Algorithm for Copositivity of Partially Symmetric Tensors

In this section, we establish an equivalent condition for the copositivity of partially symmetric tensors. Then, the concerned problem can be reformulated as a polynomial optimization problem. To continue, recall that a partially symmetric tensorA∈ℙSp,qm×n is strictly copositive if and only if(10)Axpzq≥0>0,for allx∈ℝ+m,z∈ℝ+nwithx=1,z=1,which is equivalent with the following optimization problem:(11)f∗=minAxpzqs.t.e1⊤x=1,e2⊤z=1x∈ℝ+m,z∈ℝ+n.Clearly, tensorA is strictly copositive if and only if f∗≥0>0. Problem (11) can be solved by classical Lasserre relaxations [46]. Since the objection function is continuous and the feasible region is compact, problem (11) always has a solution. Without loss of generality, assume x∗,z∗ is one of the solutions of (11); then, it satisfies the following KKT-conditions with λ,μ∈ℝ,v∈ℝm,andw∈ℝn:(12)pAx∗p−1z∗q−λe1−v=0,qAx∗pz∗q−1−μe2−w=0,e1⊤x∗=1,e2⊤z∗=1,x∗≥0,z∗≥0,v≥0,w≥0,x∗⊤v=0,z∗⊤w=0.By (12), we obtain that λ=pAx∗pz∗q, μ=qAx∗pz∗q, and(13)pAx∗p−1zq−λe1≥0,qAx∗pz∗q−1−μe2≥0,x∗⊤pAx∗p−1z∗q−λx∗⊤e1=0,z∗⊤qAx∗pz∗q−1−μz∗⊤e2=0.Combining this with the fact thatx∗≤1,z∗≤1, we consider the following optimization problem:(14)minAxpzqs.t.x⊤Axp−1zq−Axpzqx⊤e1=0,z⊤Axpzq−1−Axpzqz⊤=0,Axp−1zq−Axpzqe1≥0,Axpzq−1−Axpzqe2≥0,e1⊤x=1,e2⊤z=1,1−x2≥0,1−z2≥0x∈ℝ+m,z∈ℝ+n.It is clear to see that problems (11) and (14) are equivalent in the sense that they have the same optimal solution. To solve (14), we introduce the following notations:(15)fx,z=Axpzq,gx,z=Axp−1zq−Axpzqe1,Axpzq−1−Axpzqe2,1−x2,1−z2,xi,zj,hx,z=e⊤x,0n+m=1,e⊤0,zn+m=1,xiAxp−1zqi−Axpzqxi,zjAxpzq−1j−Axpzqzj.So, the problem of (14) can be rewritten such that(16)f∗=minfx,zs.t.gx,z≥0,hx,z=0.By the Lasserre-type semidefinite relaxations of (16), consider the semidefinite program(17)ρk=min∑α∈Nn+mfαyαs.t.Lgky≽0,Lhky=0,y0=1,Mky≽0,y∈ℝN2kn+m,where k=k0,k0+1,…, with k0=max⌈p/2⌉,⌈q/2⌉. It is obvious that the feasible set is compact, and the Archimedean condition holds. Thus, the asymptotic convergence of (17) is always guaranteed. Moreover, A is copositive if ρk≥0 for some k, and ρk is a monotonically decreasing sequence, with the decreasing of order k, i.e.,(18)ρk0≤ρk0+1≤,…,≤ρk≤,…,≤f∗.Now, we present an algorithm to check the copositivity of a given partially symmetric rectangular tensor (Algorithm1).Algorithm 1: An SDP method for copositivity of a partially symmetric tensor A∈ℙSp,qm×n.
Step 0: given an arbitrary vectorξ∈ℝℕp+qn+m. Let k=max⌈p/2⌉,⌈q/2⌉.Step 1: solve the semidefinite relaxation (17). If ρk≥0, then stop, and A is copositive. If ρk<0, go to Step 2.Step 2: solve the following semidefinite program:for an optimizery∗ if it is feasible. If it is infeasible, let k=k+1 and go to Step 1.Step 3: letx∗,z∗=y∗e1,…,y∗em,y∗em+1,…,y∗em+n. If Ax∗pz∗q<0, then A is not copositive and stop. Otherwise, let k=k+1 and go to Step 1.(19)ρk∗=minξ⊤x,ym+n,ys.t.y0=1,Lgky≽0,Lhky=0Mky≽0,Lρk−fx,zk≽0,y∈ℝN2kn+m,The following theorem shows the convergence of Algorithm1 for any partially symmetric tensor.Theorem 1.
SupposeA∈ℙSp,qm×n is a partially symmetric tensor. Then, the following properties hold:(i)
For allk≥0, problem (16) is feasible and achieves its optimal value ρk=f∗ for all k sufficiently large(ii)
For allk≥0, problem (19) has an optimizer if it is feasible(iii)
IfA is copositive, Algorithm 1 must stop with ρk≥0 when k is sufficiently large(iv)
IfA is not copositive, Algorithm 1 must stop with fx,z<0 for almost all ξ∈ℝℕp+qn+m when k is sufficiently largeProof.
(i)
Since the feasible set of (11) is compact, it must have a minimizer x∗,z∗. On the contrary, x∗,z∗ is a feasible point for (16), which implies that the semidefinite relaxation (17) is always feasible. Since L1−x2k≽0, let X=x,0,0,zx∈ℝm,z∈ℝn⊆ℝm+n; then, it holds that L1−x2k≽0. We now show that the feasible set of (17) is compact as follows. First of all, we have(20)1≥y2e1+y2e2+,…,+y2en+m.Then,0≤y2ei≤1; since Mky≽0, i∈m+n. Furthermore, for all 0<α≤k−1, the α,α-th diagonal entry of L1−x2k is nonnegative, which implies that(21)y2α≥y2e1+2α+y2e2+2α+,…,+y2en+m+2α.Takeα=e1,e2,…,em+n in the following analysis. By the same argument as (21) and repeating k−1 times, we can show that 0≤y2β≤1 for all β≤k. By the definition of Mky, we know that the diagonal entries Mky are precisely y2β, β≤k. Since Mky≽0, all the entries of Mky must be between −1 and 1. So, y is bounded, and the feasible set of (17) is compact. Hence, the optimal value can always be achieved. In the following, we will show that ρk=f∗ for all k sufficiently large.By direct computation, the optimization (16) is equivalent with the following problem:(22)minAxpzqs.t.xiAxp−1zqi−Axpzqxi=0,zjAxpzq−1j−Axpzqzj=0,Axp−1zq−Axpzqe1≥0,Axpzq−1−Axpzqe2≥0,e1⊤x=1,e2⊤z=1,x∈ℝ+m,z∈ℝ+n.For simplicity, denote(23)fx,z=Axpzq,gx,z=x,z,Axp−1zq−Axpzqe1,Axpzq−1−Axpzqe2hx,z=e⊤x,0m+n−1,e⊤0,zm+n−1,xiAxp−1zqi−Axpzqxi,zjAxpzq−1j−Axpzqzj.Corresponding Lasserre’s relaxations for (22) are(24)ρk′=min∑α∈Nnfαyαs.t.Lgky≽0,Lhky=0,y0=1,Mky≽0,y∈ℝN2kn+m.Fork=k0,k0+1,…, where k0=max⌈p/2⌉,⌈q/2⌉, any feasible solution of (17) is also a feasible solution of (24), so(25)ρk′≤ρk≤f∗,k=k0,k0+1,….Next, we show that the set of polynomials(26)F=1−eTxϕ+∑i=1nxi∑lsl2+1−eTzψ+∑i=1nzj∑tst2.is Archimedean, i.e., there existsf∈F such that the inequality fx≥0 defines a compact set in ℝm+n. Let f=2−X2 and X=x,zm+n; we have(27)2−x2=1−eTx1+x2+∑i=1nxi1−xi2+∑i≠j=1mxi2xj+1−eTz1+z2+∑j=1nzj1−zj2+∑i≠j=1nzi2zj.So,F is Archimedean by Theorem 3.3 of [47]; we know that ρk′=f∗ when k is sufficiently large. Hence, ρk=f∗ when all k values are sufficiently large.(ii)
The proof is the same with (i).(iii)
Clearly,A is copositive if and only if f∗≥0. By item (i), ρk=f∗ for all k big enough. Therefore, if A is copositive, we must have ρk≥0 for all k large enough.(iv)
IfA is not copositive, then f∗<0. By (i), there exists k1∈N such that ρk=f∗ for all k≥k1. Hence, for all k≥k1, problem (19) is equivalent with the following problem:(28)ρk^=minξ⊤x,ym+n,ys.t.y0=1,L1−eTxky≽0,L1−eTyk≽0,LXky≽0,Mky≽0,Lf∗−fXk≽0,X=x,zm+n,y∈ℝN2kn+m.
It isk-th Lasserre’s relaxation for the polynomial optimization(29)minξ⊤x,zm+ns.t.1−e⊤x,0m+n≥0,1−e⊤0,zm+n≥0,x≥0,z≥0,f∗−fX≥0.
The feasible region of (29) is clearly compact. When ξ∈ℝℕp+qn+m is arbitrary, (29) has a unique optimizer X∗=x∗,z∗. Hence, for almost all ξ∈ℝℕp+qn+m, X∗ is the unique optimizer. For notation convenience, denote by y^k the optimizer of (19) with the relaxation order k. Let Xk=y^ke1,…,y^ken+m. By Corollary 3.5 of [48] or Theorem 3.3 of [49], the sequence Xkk=k0∞ must converge to X∗. Since f∗≤ρk∗<0, we must have fXk<0 when k is sufficiently large.


## 4. Numerical Examples

In this section, we give several numerical examples to show the efficiency of Algorithm1. Let Sπi1i2,…,im denote the set of all permutations of i1i2,…,im, and let ρk∗=0 when ρk∗<1e−5. All experiments are done in Matlab2014b on a desktop computer with Intel (R) Core (TM)i7-6500 CPU @ 2.50 GHz 2.60 GHz and 16 GB of RAM.Example 1.
Suppose thatA∈ℙS2,22×4 is given by(30)a1111=1,a1122=1,a1133=1,a1144=1,a2211=1,a2222=1,a2233=1,a2244=1,∑i1i2j1j2∈Sπ1112ai1i2j1j2=2,∑i1i2j1j2∈Sπ1134ai1i2j1j2=2,∑i1i2j1j2∈Sπ1113ai1i2j1j2=−2,∑i1i2j1j2∈Sπ1114ai1i2j1j2=−2,∑i1i2j1j2∈Sπ1123ai1i2j1j2=−2,∑i1i2j1j2∈Sπ1124ai1i2j1j2=−2,∑i1i2j1j2∈Sπ2212ai1i2j1j2=2,∑i1i2j1j2∈Sπ2234ai1i2j1j2=2,∑i1i2j1j2∈Sπ2213ai1i2j1j2=−2,∑i1i2j1j2∈Sπ2214ai1i2j1j2=−2,∑i1i2j1j2∈Sπ2223ai1i2j1j2=−2,∑i1i2j1j2∈Sπ2224ai1i2j1j2=−2,∑i1i2j1j2∈Sπ1211ai1i2j1j2=−2,∑i1i2j1j2∈Sπ1222ai1i2j1j2=−2,∑i1i2j1j2∈Sπ1212ai1i2j1j2=−4,∑i1i2j1j2∈Sπ1244ai1i2j1j2=−2,∑i1i2j1j2∈Sπ1234ai1i2j1j2=−4,∑i1i2j1j2∈Sπ1213ai1i2j1j2=4,∑i1i2j1j2∈Sπ1214ai1i2j1j2=4,∑i1i2j1j2∈Sπ1223ai1i2j1j2=4,∑i1i2j1j2∈Sπ1224ai1i2j1j2=4,∑i1i2j1j2∈Sπ1233ai1i2j1j2=−2.
The corresponding polynomial for tensorA is(31)fx,y=x1−x22y1+y2−y3−y42,x=x1,x2,y=y1,y2,y3,y4.
By Algorithm1, we know that f∗=0 with x=0.5000,0.5000, y=0.2500,0.2500,0.2500,0.2500, which implies that rectangular tensor A is copositive.Example 2.
Suppose thatA∈ℙS2,21×2 with entries such that(32)a1111=1,a1122=1,∑i1i2j1j2∈Sπ11,12ai1i2j1j2=−2.
The corresponding polynomial ofA is(33)fx,y=x12y12−2x12y1y2+x12y22,x=x1,y=y1,y2.
By Algorithm1, we obtain that f∗=0 with optimal solution x,y=1.0000,0.7071,0.7071, which implies that A is copositive but not strictly copositive.Example 3.
Suppose thatA∈ℙS2,22×2 is given by(34)a1111=1,a1122=1,a2211=1,a2222=1,∑i1i2j1j2∈Sπ11,12ai1i2j1j2=−2,∑i1i2j1j2∈Sπ22,12ai1i2j1j2=−2,∑i1i2j1j2∈Sπ12,11ai1i2j1j2=2,∑i1i2j1j2∈Sπ12,22ai1i2j1j2=2,∑i1i2j1j2∈Sπ12,12ai1i2j1j2=−4.
So, the corresponding polynomial ofA is that(35)fx,y=x12y12+x12y22−2x12y1y2+x22y12+x22y22−2x22y1y2+2x1x2y12+2x1x2y22−4x1x2y1y2,where x=x1,x2,y=y1,y2. By Algorithm 1, we have f∗=0 with x∗=0.5126,0.4874, y∗=0.5000,.5000, which implies that the rectangular tensor is copositive.Example 4.
Suppose thatA∈ℙS3,23×2 is given by(36)a11122=1,a22222=1,a33311=1,∑i1i2i3j1j2∈Sπ123,12ai1i2i3j1j2=−3.
The corresponding polynomial of the partially symmetric rectangular tensorA is(37)fx,y=x13y22+x23y22+x33y12−4x1x2x3y1y2,where x=x1,x2,x3,y=y1,y2. By Algorithm 1, we know that f∗=−0.0639 with x∗=0.7652,0.4702,0.7652, y∗=0.3572,0.8724, which implies that the rectangular tensor is not copositive.Example 5.
SupposeA∈ℙS2,22×2 is a tensor with entries such that(38)a1111=1,a1122=−1,a2211=1,a2222=1,∑i1i2j1j2∈Sπ11,12ai1i2j1j2=2,∑i1i2j1j2∈Sπ22,12ai1i2j1j2=2,∑i1i2j1j2∈Sπ12,11ai1i2j1j2=2,∑i1i2j1j2∈Sπ12,22ai1i2j1j2=2,∑i1i2j1j2∈Sπ12,12ai1i2j1j2=−4.
The corresponding polynomial of the partially symmetric rectangular tensorA is(39)fx,y=x12y12−x12y22+2x12y1y2+x22y12+x22y22+2x22y1y2+2x1x2y12+2x1x2y22−4x1x2y1y2,where x=x1,x2,y=y1,y2. By Algorithm 1, we know that f∗=0.3333 with x∗=0.6666,0.3334, y∗=0.5000,.5000, which implies that the rectangular tensor is strictly copositive.


## 5. Conclusions

In this paper, based on Lasserre’s hierarchy of semidefinite relaxations, we propose a new criterion to judge whether a given partially symmetric rectangular tensor is copositive or not. The convergence for the proposed algorithm is established. Furthermore, numerical examples demonstrate that the proposed algorithm is effective when the input rectangular tensor has lower dimension and orders, and it is difficult for the case with higher order or higher dimension. We will continue to study this problem in the future.



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