Newcriteria for H-tensors and an application

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1 Wang and Sun Journal of Inequalities and Applications (016) 016:96 DOI /s R E S E A R C H Open Access Newcriteria for H-tensors and an application Feng Wang * and De-shu Sun * Correspondence: wangf991@163.com College of Science, Guizhou Minzu University, Guiyang, Guizhou 55005, P.R. China Abstract Some new criteria for H-tensors are obtained. As an application, some sufficient conditions of the positive definiteness for an even-order real symmetric tensor are given. The advantages of the results obtained are illustrated by numerical examples. MSC: 15A06; 15A4; 15A48 Keywords: H-tensor; symmetric tensor; positive definiteness; irreducible; nonzero elements chain 1 Introduction Let C(R)bethecomplex(real)fieldandN = {1,,...,n}.WecallA =(a i1 i i m )acomplex (real) order m dimension n tensor, if a i1 i i m C(R), where i j =1,...,n for j =1,...,m [1, ]. A tensor A =(a i1 i i m ) is called symmetric [3], if a i1 i i m = a π(i1 i i m ), π m, where m is the permutation group of m indices. Furthermore, an order m dimension n tensor I =(δ i1 i i m ) is called the unit tensor [4], if its entries δ i1 i i m = { 1, if i 1 = = i m, 0, otherwise. Let A =(a i1 i i m )beanorderm dimension n complex tensor. If there exist a complex number λ and a nonzero complex vector x =(x 1, x,...,x n ) T that are solutions of the following homogeneous polynomial equations: Ax m 1 = λx [m 1], then we call λ an eigenvalue of A and x the eigenvector of A associated with λ [5 7], Ax m 1,andλx [m 1] are vectors, whose ith components are ( Ax m 1 ) i = a ii i m x i x im i,...,i m N 016 Wang and Sun. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page of 1 and ( x [m 1] ) i = xm 1 i, respectively. If the eigenvalue λ and the eigenvector x are real, then λ is called an H-eigenvalue of A and x is its corresponding H-eigenvector [1]. Throughout this paper, we will use the following definitions. Definition 1 [8] LetA =(a i1 i i m ) be a tensor of order m dimension n. A is called a diagonally dominant tensor if a ii i a ii im i,...,i m N δ ii...im =0, i N. (1) If all inequalities in (1)hold,then we call A a strictly diagonally dominant tensor. Definition [9] LetA =(a i1 i m )beanorderm dimension n complex tensor. A is called an H-tensor if there is a positive vector x =(x 1, x,...,x n ) T R n such that a ii i x m 1 i > a x ii im i x im, i,...,i m N δ ii im =0 i =1,,...,n. Definition 3 [10] LetA =(a i1 i i m ) be a tensor of order m dimension n, X = diag(x 1, x,...,x n ). Denote B =(b i1 i m )=AX m 1, b i1 i i m = a i1 i i m x i x i3 x im, i j N, j N, we call B the product of the tensor A and the matrix X. Definition 4 [11] AcomplextensorA =(a i1 i m )oforderm dimension n is called reducible, if there exists a nonempty proper index subset I N such that a i1 i i m =0, i 1 I, i,...,i m / I. If A is not reducible, then we call A irreducible. Definition 5 Let A =(a i1 i m )beanordermdimension n complex tensor, for i, j N (i j), if there exist indices k 1, k,...,k r with a ks i i m 0, s =0,1,...,r, i,...,i m N δ ksi im =0 k s+1 {i,...,i m } where k 0 = i, k r+1 = j, we call there is a nonzero elements chain from i to j. For an mth-degree homogeneous polynomial of n variables f (x)canbedenoted f (x)= i 1,i,...,i m N a i1 i i m x i1 x i x im, ()

3 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page 3 of 1 where x =(x 1, x,...,x n ) R n. The homogeneous polynomial f (x) in() isequivalentto the tensor product of an order m dimensional n symmetric tensor A and x m defined by f (x) Ax m = i 1,i,...,i m N a i1 i i m x i1 x i x im, (3) where x =(x 1, x,...,x n ) R n [1]. The positive definiteness of homogeneous polynomials have applications in automatic control [1, 13],polynomial problems [14], magnetic resonance imaging [15, 16], and spectral hypergraph theory [17, 18]. However, for n >3andm > 4, it is a hard problem to identify the positive definiteness of such a multivariate form. For solving this problem, Qi [1] pointed out that f (x) definedby(3) is positive definite if and only if the real symmetric tensor A is positive definite, and Qi provided an eigenvalue method to verify the positive definiteness of A when m is even (see Theorem 1). Theorem 1 [1] 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. Although from Theorem 1 we can verify the positive definiteness of an even-order symmetric tensor A (the positive definiteness of the mth-degree homogeneous polynomial f (x)) by computing the H-eigenvalues of A, it is difficult to compute all these H-eigenvalues when m and n are large. Recently, by introducing the definition of H-tensor, Li et al. [9] provided a practical sufficient condition for identifying the positive definiteness of an even-order symmetric tensor (see Theorem ). Theorem [9] Let A =(a i1 i i m ) be an even-order real symmetric tensor of order m dimension n with a k k >0for all k N. If A is an H-tensor, then A is positive definite. Theorem provides a method for identifying the positive definiteness of an even-order symmetric tensor by determining H-tensors. Thus the identification of H-tensors is useful in checking the positive definiteness of homogeneous polynomials. In this paper, some new criteria for identifying H-tensors are presented, which is easy to calculatesince it only depends on the entries of tensors. As an application of these criteria, some sufficient conditions of the positive definiteness for an even-order real symmetric tensor are obtained. Numerical examples are also given to verify the corresponding results. Main results In this section, we give some new criteria for H-tensors. First of all, we give some notation and lemmas. Let S be a nonempty subset of N and let N \ S be the complement of S in N. Givenan order m dimension n complex tensor A =(a i1 i i m ), we denote R i (A)= a = ii im i,...,i m N δ ii...im =0 i,...,i m N a ii i m a ii i, N 1 = { i N :0<a ii i R i (A) }, N = { i N : a ii i > R i (A) }, s i = a ii i R i (A), t i = R { } i(a) a ii i, r = max max s i, max t i, i N 1 i N

4 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page 4 of 1 S m 1 = {i i 3 i m : i j S, j =,3,...,m}, N m 1 \ S m 1 = { i i 3 i m : i i 3 i m N m 1 and i i 3 i m / S m 1}. It is obvious that if N 1 =, thena is an H-tensor. It is known that, for an H-tensor A, N [9]. So we always assume that both N 1 and N are not empty. Otherwise, we assume that A satisfies: a ii i 0,R i (A) 0, i N. Lemma 1 [8] If A is a strictly diagonally dominant tensor, then A is an H-tensor. Lemma [10] Let A =(a i1 i m ) be a complex tensor of order m dimension n. If there exists a positive diagonal matrix X such that AX m 1 is an H-tensor, then A is an H-tensor. Lemma 3 [9] Let A =(a i1 i m ) be a complex tensor of order m dimension n. If A is irreducible, a i i R i (A), i N, and strictly inequality holds for at least one i, then A is an H-tensor. Lemma 4 Let A =(a i1 i m ) be an order m dimension n complex tensor. If (i) a ii i R i (A), i N, (ii) J(A)={i N : a ii i > R i (A)}, (iii) for any i / J(A), there exists a nonzero elements chain from i to j, such that j J(A), then A is an H-tensor. Proof It is evident that the result holds with J(A) =N. Next, we assume that J(A) N. Suppose J(A)={k +1,...,n}, N \ J(A)={1,...,k},1 k < n.byhypothesis, a kk k = R k (A). By the condition (iii), there exist indices k 1, k,...,k r such that i,...,i m N δ ksi im =0 k s+1 {i,...,i m } a ks i i m 0, s =0,1,...,r, where k 0 = k, k r+1 = j, j J(A). Then i,...,i m N δ kri im =0 j {i,...,i m } a kr i i m 0. Further, without loss of generality, we assume that k 1,...,k r / J(A), that is, 1 k 1,...,k r < k. From j J(A), we have a jj j > R j (A), so there exists 0 < ε <1suchthatεa jj j > R j (A). Construct a positive diagonal matrix X kr = diag(x 1,...,x n ), where x i = {ε m 1 1, i = j, 1, i j.

5 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page 5 of 1 Let A kr =[a (k r) a (k r) ii i a (k r ) i 1 i i m ]=AXk m 1 r.then = a ii i = R i (A) R i (A kr ), 1 i k, i k r, k r k r k r = akr k r k r = R kr (A)>R kr (A kr ), a (k r ) ii i = aii i > R i (A) R kr (A kr ), i J(A), i j, a (k r) = εa jj j > R j (A) R kr (A kr ). jj j Obviously, A kr is also a diagonally dominant tensor, and J(A kr )=J(A) {k r }. If J(A kr )=N,thenA kr is strictly diagonally dominant. By Lemma, A is an H-tensor. If N \ J(A kr ), thena kr also satisfies the conditions of the lemma, that is, for any i N \ J(A kr ), there exist indices l 1, l,...,l s,suchthat i,...,i m N, δ l ti im =0, l t+1 {i,...,i m } a lt i i m 0, t =0,1,...,s, where l 0 = i, l s+1 = j, j J(A kr ). Then i,...,i m N, δ lsi im =0, j {i,...,i m } a ls i i m 0. Similar to the above argument, for A kr, there exists a positive diagonal matrix X ls such that A ls = A kr Xl m 1 s is diagonally dominant, and J(A ls )=J(A kr ) {l s }. If J(A ls )=N,thenA ls is strictly diagonally dominant. By Lemma, A is an H-tensor. If N \ J(A ls ), thena ls also satisfies the conditions of the lemma. Similarly as the above argument, for A ls,thereexistatmostk positive diagonal matrices X kr, X ls,...,x pq such that B is strictly diagonally dominant, where B = A(X kr X ls X pq ) m 1.Hence,B is an H-tensor, and by Lemma, A is an H-tensor. The proof is completed. Theorem 3 Let A =(a i1 i m ) be an order m dimension n complex tensor. If a ii i s i > r i i m N m 1 \N m 1 δ ii...im =0 a ii i m + i i m N m 1 max {t j}a ii i m, i N 1, (4) j {i,...,i m } then A is an H-tensor. Proof Let M i = a ii i s i r i i m N m 1 \N m 1 a ii i m i i m N m 1 δ ii...im =0 i i m N m 1 a ii i m max j {i,...,i m }{t j }a ii i m i N 1. (5),

6 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page 6 of 1 If i i 3 i m N m 1 a ii i m =0,wedenoteM i =+. Frominequality(4), we obtain M i >0 (i N 1 ). Hence, there exists a positive number ε >0suchthat { } 0<ε < min min M i,1 max t i. (6) i N 1 i N Let the matrix X = diag(x 1, x,...,x n ), where x i = {(s i ) m 1 1, i N 1, (ε + t i ) m 1 1, i N. By inequality (6), we have (ε + t i ) m 1 1 <1(i N ). As ε +, sox i +, which implies that X is a diagonal matrix with positive entries. Let B =(b i1 i i m )=AX m 1.Next,wewill prove that B is strictly diagonally dominant. For all i N 1,if i i 3 i m N m 1 a ii i m =0,thenbyinequality(4), we have R i (B) = = r i i m N m 1 \N m 1 δ ii im =0 i i m N m 1 \N m 1 δ ii im =0 i i m N m 1 \N m 1 δ ii im =0 b ii i m + i i m N m 1 a ii i m x i x im + b ii i m i i m N m 1 a ii i m x i x im a ii i m < a ii i s i = b ii i. (7) If i i 3 i m N m 1 a ii i m 0, then by inequalities (5)and(6), we obtain R i (B) = = r = r i i m N m 1 \N m 1 δ ii im =0 i i m N m 1 \N m 1 δ ii im =0 + i i m N m 1 i i m N m 1 \N m 1 δ ii im =0 i i m N m 1 \N m 1 δ ii im =0 + ε i i m N m 1 a ii i m x i x im + a ii i m x i x im i i m N m 1 a ii i m (ε + t i ) 1 m 1 (ε + t im ) 1 m 1 a ii i m + a ii i m + a ii i m i i m N m 1 i i m N m 1 ( a ii i m ε + a ii i m x i x im ) max {t j} j {i,...,i m } max {t j}a ii i m j {i,...,i m } < a ii i s i = b ii i. (8)

7 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page 7 of 1 Now, we consider i N.Sincea ii i > R i (A), we have a ii i i i 3 i m N m 1 δ ii im =0 a ii i m >0 (9) and r i i m N m 1 \N m 1 a ii i m + i i m N m 1 δ ii im =0 max {t j}a ii i m R i (A) 0. (10) j {i,...,i m } By inequalities (9), (10), and ε >0,we get ε > r i i m N m 1 \N m 1 a ii i m + i i m N m 1 max j {i,...,i m }{t j }a ii i m R i (A) δ ii im =0 a ii i i i m N m 1 a ii i m δ ii im =0 From inequality (11), for any i N,weobtain b ii i R i (B) =a ii i (ε + t i ) i i m N m 1 δ ii im =0 a ii i (ε + t i ) r i i m N m 1 δ ii im =0 ( = ε a ii i r i i m N m 1 \N m 1 a ii i m x i x im a ii i m (ε + t i ) 1 m 1 (ε + t im ) 1 m 1 i i m N m 1 \N m 1 a ii i m ( ) a ii i m ε + max {t j} j {i,...,i m } i i m N m 1 δ ii im =0 i i m N m 1 \N m 1 ) a ii i m + R i (A) a ii i m i i m N m 1 δ ii im =0 max {t j}a ii i m j {i,...,i m }. (11) >0. (1) Therefore, from inequalities (7), (8), and (1), we obtain b ii i > R i (B) for all i N,thatis, B is strictly diagonally dominant. By Lemma 1 and Lemma, A is an H-tensor. The proof is completed. Theorem 4 Let A =(a i1 i m ) be an order m dimension n complex tensor. If A is irreducible and a ii i s i r i i m N m 1 \N m 1 δ ii...im =0 a ii i m + i i m N m 1 max {t j}a ii i m, i N 1, (13) j {i,...,i m } and a strict inequality holds for at least one i N 1, then A is an H-tensor.

8 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page 8 of 1 Proof Let the matrix X = diag(x 1, x,...,x n ), where x i = {(s i ) m 1 1, i N 1, (t i ) m 1 1, i N. By the irreducibility of A, wehavex i +, thenx is a diagonal matrix with positive diagonal entries. Let B =[b i1 i m ]=AX m 1. Adopting the same procedure as in the proof of Theorem 3, we can obtain b ii i R i (B) ( i N), and there exists at least an i 0 N 1 such that b i0 i 0 i 0 > R i0 (B). On the other hand, since A is irreducible and so is B. Then by Lemma 3,weseethatB is an H-tensor. By Lemma, A is an H-tensor. The proof is completed. Let { J 1 = i N 1 : a ii i s i > r { J = i N : a ii i t i > r i i m N m 1 \N m 1 δ ii...im =0 i i m N m 1 \N m 1 a ii i m + a ii i m + i i m N m 1 i i m N m 1 δ ii...im =0 Theorem 5 Let A =(a i1 i m ) beanordermdimensionncomplextensor. If a ii i s i r i i m N m 1 \N m 1 δ ii...im =0 a ii i m + i i m N m 1 } max {t j}a ii i m, j {i,...,i m } } max {t j}a ii i m. j {i,...,i m } max {t j}a ii i m, (14) j {i,...,i m } J 1 J, and for i (N 1 \ J 1 ) (N \ J ), there exists a nonzero elements chain from i to jsuchthatj J 1 J, then A is an H-tensor. Proof Let the matrix X = diag(x 1, x,...,x n ), where x i = {(s i ) m 1 1, i N 1, (t i ) m 1 1, i N. Obviously x i +, thenx is a diagonal matrix with positive diagonal entries. Let B = [b i1 i m ]=AX m 1. Similarly as in the proof of Theorem 3, we can obtain b ii i R i (B) ( i N). From J 1 J,thereexistsatleastani 0 N such that b i0 i 0 i 0 > R i0 (B). On the other hand, if b ii i = R i (B), then i (N 1 \ J 1 ) (N \ J ), by the assumption, we know that there exists a nonzero elements chain from i to j of A such that j J 1 J.Then there exists a nonzero elements chain from i to j of B with j satisfying b jj j > R j (B). Based on above analysis, we conclude that the tensor B satisfies the conditions of Lemma 4, sob is an H-tensor. By Lemma, A is an H-tensor. The proof is completed. Theorem 6 Let A =(a i1 i i m ) be a complex tensor of order m dimension n. If a ii i s i > r i i m N m 1 \N m 1 δ ii...im =0 a ii i m, i N 1 (15)

9 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page 9 of 1 and i i m N m 1 \N m 1 then A is an H-tensor. a ii i m =0, i N, (16) Proof By inequality (15), for each i N 1, there exists a positive number K i >1,suchthat a ii i s i > r i i m N m 1 \N m 1 δ ii...im =0 a ii i m + 1 K i ( Let K max i N1 {K i }.Byinequality(17), we obtain a ii i s i > r i i m N m 1 \N m 1 δ ii...im =0 a ii i m + 1 K i i m N m 1 ( i i m N m 1 ) max {t j}a ii i m. (17) j {i,...,i m } ) max {t j}a ii i m, j {i,...,i m } i N 1. (18) Since a ii i R i (A)(i N 1 ) and inequality (15), so i i m N m 1 For any i N 1,denote T i = a ii i m >0, i N 1. (19) a ii i s i r i i m N m 1 \N m 1 a ii i m 1 K ( i i m N m 1 δ ii...im =0 i i 3 i m N m 1 a ii i m max j {i,...,i m }{t j }a ii i m ) From inequalities (18) and(19), we have T i > 0. Therefore there exists a positive number ε >0suchthat { } t i 0<ε < min min T i,1 max. i N 1 i N K Let the matrix X = diag(x 1, x,...,x n ), where x i = {(s i ) m 1 1, i N1, (ε + t i K ) m 1 1, i N. Mark B = AX m 1. Similarly as in the proof of Theorem 3, we can prove that B is strictly diagonally dominant. By Lemma 1 and Lemma, A is an H-tensor. The proof is completed. There is no inclusion relation between the conditions of Theorem 3 and the conditions of Theorem 6. This can be seen from the following examples..

10 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page 10 of 1 Example 1 Consider a tensor A =(a ijk ) of order 3 dimension 3 defined as follows: A = [ A(1,:,:),A(,:,:),A(3,:,:) ], A(1,:,:)= , A(,:,:)= 0 1 0, A(3,:,:)= Obviously, a 111 =15, R 1 (A)=4, a =1, R (A)=4, a 333 =15, R 3 (A)=5, so N 1 = {1}, N = {, 3}. By calculation, we have Since s 1 = a 111 R 1 (A) = 15 4, t = R (A) a = 1 3, t 3 = R 3(A) a 333 = 1 3, r = r jk N \N δ 1jk =0 a 1jk + jk N max {t l}a 1jk = 15 l {j,k} 4 ( )+ 1 3 ( ) = 13 4 < 5 4 = a 111s 1, we know that A satisfies the conditions of Theorem 3,then A is an H-tensor. But a jk =3 0, a 3jk = 0. jk N \N jk N \N so A does not satisfy the conditions of Theorem 6. Example Consider a tensor A =(a ijk ) of order 3 dimension 3 defined as follows: A = [ A(1,:,:),A(,:,:),A(3,:,:) ], A(1,:,:)= , A(,:,:)= 0 8, A(3,:,:)=

11 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page 11 of 1 Obviously, a 111 =8, R 1 (A)=4, a =8, R (A)=4, a 333 =10, R 3 (A)=5, so N 1 = {1}, N = {, 3}. By calculation, we have s 1 = a 111 R 1 (A) = 1 3, t = R (A) a = 1, t 3 = R 3(A) a 333 = 1, r = 1. Since r a 1jk = 1 ( )= 3 < 8 3 = a 111s 1 jk N \N δ 1jk =0 and a jk =0, a 3jk =0, jk N \N jk N \N we see that A satisfies the conditions of Theorem 6,then A is an H-tensor. But r jk N \N δ 1jk =0 a 1jk + jk N max l {j,k} {t l}a 1jk = 1 ( )+ 1 ( ) =1> 8 3 = a 111s 1, so A does not satisfy the conditions of Theorem 3. 3 An application In this section, based on the criteria for H-tensors in Section, we present some criteria for identifying the positive definiteness of an even-order real symmetric tensor (the positive definiteness of a multivariate form). From Theorems -6, we obtain easily the followingresult. Theorem 7 Let A =(a i1 i m ) be an even-order real symmetric tensor of order m dimension nwitha kk k >0for all k N. If A satisfies one of the following conditions, then A is positive definite: (i) all the conditions of Theorem 3; (ii) all the conditions of Theorem 4; (iii) all the conditions of Theorem 5; (iv) all the conditions of Theorem 6. Example 3 Let f (x)=ax 4 =11x x4 +18x4 3 +1x4 4 +1x 1 x x 3 4x 1 x x 3 x 4 be a 4thdegree homogeneous polynomial. We can get an order 4 dimension 4 real symmetric tensor A =(a i1 i i 3 i 4 ), where a 1111 =11, a =18, a 3333 =18, a 4444 =1,

12 Wang and Sun Journal of Inequalities and Applications (016) 016:96 Page 1 of 1 a 113 = a 113 = a 113 = a 131 = a 131 = a 131 =1, a 113 = a 131 = a 311 = a 311 = a 311 = a 311 =1, a 134 = a 143 = a 134 = a 134 = a 143 = a 143 = 1, a 134 = a 143 = a 314 = a 341 = a 413 = a 431 = 1, a 314 = a 314 = a 314 = a 341 = a 341 = a 341 = 1, a 413 = a 413 = a 413 = a 431 = a 431 = a 431 = 1, and other a i1 i i 3 i 4 = 0. It can be verified that A satisfies all the conditions of Theorem 3. Thus, from Theorem 7,we see that A is positive definite, that is, f (x) is positive definite. Competing interests The authors declare that they have no competing interests. Authors contributions The two authors contributed equally to this work. Both authors read and approved the final manuscript. Acknowledgements The authors are very indebted to the referees for their valuable comments and corrections, which improved the original manuscript of this paper. This work was supported by the National Natural Science Foundation of China ( ), the Foundation of Science and Technology Department of Guizhou Province ([015]073, [015]706), the Natural Science Programs of Education Department of Guizhou Province ([015]40), and the Research Foundation of Guizhou Minzu University (15XRY004). Received: 5 March 015 Accepted: 10 March 016 References 1. Qi, LQ: Eigenvalues of a real supersymmetric tensor. J. Symb. Comput. 40, (005). Lim, LH: Singular values and eigenvalues of tensors: a variational approach. In: CAMSAP 05: Proceeding of the IEEE International Workshop on Computational Advances in Multi-Sensor Adaptive Processing, pp (005) 3. Li,CQ,Qi,LQ,Li,YT:MB-Tensors and MB 0 -tensors. Linear Algebra Appl. 484, (015) 4. Yang, YN, Yang, QZ: Further results for Perron-Frobenius theorem for nonnegative tensors. SIAM J. Matrix Anal. Appl. 31, (010) 5. Qi, LQ,Wang, YJ, Wu, EX: D-Eigenvalues of diffusion kurtosis tensors. J. Comput. Appl. Math. 1, (008) 6. Cartwright, D, Sturmfels, B: The number of eigenvalues of a tensor. Linear Algebra Appl. 438, (013) 7. Kolda, TG, Mayo, JR: Shifted power method for computing tensor eigenpairs. SIAM J. Matrix Anal. Appl. 3, (011) 8. Ding,WY, Qi,LQ,Wei, YM: M-Tensors and nonsingular M-tensors. Linear Algebra Appl. 439, (013) 9. Li, CQ, Wang, F, Zhao, JX, Zhu, Y, Li, YT: Criterions for the positive definiteness of real supersymmetric tensors. J. Comput. Appl. Math. 55, 1-14 (014) 10. Kannana, MR, Mondererb, NS, Bermana, A: Some properties of strong H-tensors and general H-tensors. Linear Algebra Appl. 476,4-55 (015) 11. Chang, KC, Pearson, K, Zhang, T: Perron-Frobenius theorem for nonnegative tensors. Commun. Math. Sci. 6, (008) 1. Zhang, LP, Qi, LQ, Zhou, GL: M-Tensors and some applications. SIAM J. Matrix Anal. Appl. 35, (014) 13. Wang, F, Qi, LQ: Comments on explicit criterion for the positive definiteness of a general quartic form. IEEE Trans. Autom. Control 50, (005) 14. Reznick, B: Some concrete aspects of Hilbert s 17th problem. In: Real Algebraic Geometry and Ordered Structures. Contemp. Math., vol. 53, pp Am. Math. Soc., Providence (000) 15. Qi, LQ, Yu, GH, Wu, EX: Higher order positive semi-definite diffusion tensor imaging. SIAM J. Imaging Sci. 3, (010) 16. Qi, LQ, Yu, GH, Xu, Y: Nonnegative diffusion orientation distribution function. J. Math. Imaging Vis. 45, (013) 17. Hu, SL, Qi, LQ, Shao, JY: Cored hypergraphs, power hypergraphs and their Laplacian eigenvalues. Linear Algebra Appl. 439, (013) 18. Li,GY,Qi,LQ,Yu,GH:The Z-eigenvalues of a symmetric tensor and its application to spectral hyper graph theory. Numer. Linear Algebra Appl. 0, (013)