# Some Properties of the Strong Primitivity of Nonnegative Tensors

Authors: Lihua You; Yafei Chen; Pingzhi Yuan
Journal: Journal of Applied Mathematics
 (2018)

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## Abstract

We show that an orderm dimension 2 tensor is primitive if and only if its majorization matrix is primitive, and then we obtain the characterization of order m dimension 2 strongly primitive tensors and the bound of the strongly primitive degree. Furthermore, we study the properties of strongly primitive tensors with n≥3 and propose some problems for further research.


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## Body

## 1. Introduction

In recent years, the study of tensors and the spectra of tensors (and hypergraphs) with their various applications has attracted extensive attention and interest, since the work of L. Qi ([1]) and L.H. Lim ([2]) in 2005.As is in [1], an order m dimension n tensor A=(ai1i2…im)1≤ij≤n(j=1,…,m) over the complex field C is a multidimensional array with all entries (1)ai1i2⋯im∈Ci1,…,im∈n=1,…,n.A tensorA=(ai1i2…im) is called a nonnegative tensor if all of its entries ai1i2…im are nonnegative. Clearly, adjacency tensors and signless Laplacian tensors are nonnegative.In the theory of nonnegative matrices, the notion of primitivity plays an important role in the convergence of the Collatz method. For a nonnegative matrixA, the following are equivalent [3]:( 1 ) Let ρ(A) be the spectral radius of A. Then A is irreducible and ρ(A) is greater than any other eigenvalue in modulus.( 2 ) The only A-invariant nonempty subset of the boundary of the positive cone is {0}.( 3 ) There exists a natural number r such that Ar is positive.Matrices which satisfy any of the above conditions are called primitive. The least suchr such that Ar is positive is called the primitive exponent (or simply, exponent) of A and is denoted by exp⁡(A).In [4], K.C. Chang et al. defined the primitivity of nonnegative tensors (as Definition 1), extended the theory of nonnegative matrices to nonnegative tensors, and proved the convergence of the NQZ method which is an extension of the Collatz method and can be used to find the largest eigenvalue of any nonnegative irreducible tensor.Definition 1 (see [4]).
LetA be a nonnegative tensor with order m and dimension n, x=(x1,x2,…,xn)T∈Rn a vector, and x[r]=(x1r,x2r,…,xnr)T. Define the map TA from Rn to Rn as TA(x)=(Axm-1)[1/m-1]. If there exists some positive integer r such that TAr(x)>0 for all nonnegative nonzero vectors x∈Rn, then A is called primitive and the smallest such integer r is called the primitive degree of A, denoted by γ(A).As in [1], let A=(ai1i2…im)1≤ij≤n(j=1,…,m) be an order m dimension n tensor over the complex field C, x=(x1,…,xn)T∈Cn, x[r]=(x1r,x2r,…,xnr), and Axm-1 be a vector ∈Cn whose ith component is defined as follows:(2)Axm-1i=∑i2,…,im=1naii2⋯imxi2⋯xim.Then a number λ∈C is called an eigenvalue of A if there exists a nonzero vector x∈Cn such that(3)Axm-1=λxm-1.Recently, Shao [5] defined the general product of two n-dimensional tensors as follows, and one of the applications of the tensor product is that Axm-1 can be simply written as A·x.Definition 2 (see [5]).
LetA (and B) be an order m≥2 (and k≥1), dimension n tensor, respectively. Define the general product A·B (sometimes simplified as AB), to be the following tensor D of order (m-1)(k-1)+1 and dimension n: (4)diα1⋯αm-1=∑i2,…,im=1naii2⋯imbi2α1⋯bimαm-1i∈n,α1,…,αm-1∈nk-1.The tensor product is a generalization of the usual matrix product and satisfies a very useful property: the associative law ([5], Theorem 1.1). By the associative law, we can define Ak as the product of k many tensors A.With the general product, whenk=1 and B=x=(x1,…,xn)T∈Cn is a vector of dimension n, then A·B=A·x is still a vector of dimension n, and for any i∈[n], (A·B)i=(A·x)i=∑i2,…,im=1naii2…imxi2…xim=(Axm-1)i, where Axm-1 is defined in (2). Thus we have Axm-1=A·x.In order to study eigenvalue, Pearson defined “essentially positive” tensors as Definition3. By the general product of tensors, Shao obtained Proposition 4 and Definition 5 which is equivalent to Definition 3.Definition 3 (see [6], Definition 3.1).
A nonnegative tensorA is called essentially positive, if, for any nonnegative nonzero vector x∈Rn,A·x>0 holds.Proposition 4 (see [5], Proposition 4.1).
LetA be an order m and dimension n nonnegative tensor. Then the following three conditions are equivalent:
(1) For anyi,j∈[n],aij…j>0 holds.
(2) For anyj∈[n],A·ej>0 holds (where ej is the j-th column of the identity matrix In).
(3) For any nonnegative nonzero vectorx∈Rn,A·x>0 holds.Definition 5 (see [5], Definition 4.1).
A nonnegative tensorA is called essentially positive, if it satisfies one of the three conditions in Proposition 4.Based on the above arguments and the zero patterns defined by Shao in [5], Shao showed a characterization of primitive tensors and defined the primitive degree as follows.Proposition 6 (see [5], Theorem 4.1).
A nonnegative tensorA is primitive if and only if there exists some positive integer r such that Ar is essentially positive. Furthermore, the smallest such r is the primitive degree of A, γ(A).The concept of the majorization matrix of a tensor introduced by Pearson is very useful.Definition 7 (see [6], Definition 2.1).
The majorization matrixM(A) of the tensor A is defined as (M(A))ij=aij…j for i,j∈[n].By Definition5, Proposition 6, and Definition 7, the following characterization of the primitive tensors was easily obtained.Proposition 8 (see [7], Remark 2.6).
LetA be a nonnegative tensor with order m and dimension n. Then A is primitive if and only if there exists some positive integer r such that M(Ar)>0. Furthermore, the smallest such r is the primitive degree of A, γ(A).On the primitive degreeγ(A), Shao proposed the following conjecture for further research.Conjecture 9 (see [5], Conjecture 1).
Whenm is fixed, then there exists some polynomial f(n) on n such that γ(A)≤f(n) for all nonnegative primitive tensors of order m and dimension n.In the case ofm=2 (A is a matrix), the well-known Wielandt upper bound tells us that we can take f(n)=(n-1)2+1. Recently, the authors [7] confirmed Conjecture 9 by proving Theorem 10.Theorem 10 (see [7], Theorem 1.2).
LetA be a nonnegative primitive tensor with order m and dimension n. Then its primitive degree γ(A)≤(n-1)2+1, and the upper bound is tight.They also showed that there are no gaps in the tensor case in [8], which implies that the result of the case m≥3 is totally different from the case m=2 (A is a matrix). In [5], Shao also proposed the concept of strongly primitive tensor for further research.Definition 11 (see [5], Definition 4.3).
LetA be a nonnegative tensor with order m and dimension n. If there exists some positive integer k such that Ak>0 is a positive tensor, then A is called strongly primitive, and the smallest such k is called the strongly primitive degree of A.LetA=(ai1i2…im) be a nonnegative tensor with order m and dimension n. It is clear that if A is strongly primitive, then A is primitive. For convenience, let η(A) be the strongly primitive degree of A. Clearly, γ(A)≤η(A). In fact, it is obvious that, in the matrix case (m=2), a nonnegative matrix A is primitive if and only if A is strongly primitive, and γ(A)=η(A)=exp⁡(A). But in the case m≥3 Shao gave an example to show that these two concepts are not equivalent. In [8], the authors proposed the following question.Question 12 ([8], Question 4.18).
Can we define and study the strongly primitive degree, the strongly primitive degree set, thej-strongly primitive degree  of strongly primitive tensors and so on?Based on Question12, we study primitive tensors and strongly primitive tensors in this paper, show that an order m dimension 2 tensor is primitive if and only if its majorization matrix is primitive, and obtain the characterization of order m dimension 2 strongly primitive tensors and the bound of the strongly primitive degree. Furthermore, we study the properties of strongly primitive tensors with n≥3 and propose some problems for further research.


## 2. Preliminaries

In [8], the authors obtained the following Proposition 13 and gave Example 15 by computing the strongly primitive degree.Proposition 13 ([8], Proposition 4.16).
LetA=(ai1i2…im) be a nonnegative strongly primitive tensor with order m and dimension n. Then for any α∈[n]m-1, there exists some i∈[n] such that aiα>0.Letk(≥0), n(≥2), q(≥0), r(≥1) be integers and k=(n-1)q+r with 1≤r≤n-1 when k≥1. In [7–9], the authors defined some nonnegative tensors with order m and dimension n as follows: (5)Ak=ai1i2⋯imk1≤ij≤nj=1,…,m,where one has the following:(1)(6)MAk=M1=00⋯01110⋯00001⋯000⋮⋮⋱⋮⋮⋮00⋯10000⋯010.(2)aii2…im[0]=0, if i2…im≠i2…i2 for any i∈[n].(3)aiα[k]=1, if i∈[n]\{r-q,r-q+1,…,r,r+1}(modn) and α=i2…im∈[n]m-1 with {i2,…,im}={r-q-1,r}(modn).(4)ai1i2…im[k]=0, except for (1) and (3).The authors showed the tensorsAk(k≥0) are primitive, the primitive degree γ(A0)=(n-1)2+1([7]) and γ(Ak)=k+n ([8], Theorem 3.3) for 1≤k≤n2-3n+2.Remark 14.
It is clear that, for anyAk(0≤k≤n2-3n+2), there exists some α∈[n]m-1, for any i∈[n], aiα[k]=0. Thus, for each 0≤k≤n2-3n+2, Ak is not a strongly primitive tensor by Proposition 13.Example 15 ([8], Example 4.17).
Letm=n=3, and let A=(ai1i2i3) be a nonnegative tensor with order m and dimension n, where a111=a222=a333=a233=a311=0 and other ai1i2i3=1. Then η(A)=4.Remark 16.
In fact, we can obtainγ(A)=η(A)=4 because of a233333333(3)=0, where A3=(ai1i2…i9(3)).In the computation of Example15, we note that the following equation is useful and will be used repeatedly. It is not difficult to obtain the equation by the general product of two n-dimensional tensors which is defined in Definition 1.2 in [5].LetA be a nonnegative primitive tensor with order m and dimension n, α2,…,αm∈[n](m-1)k-1. Then we have(7)Akiα2⋯αm=∑i2,i3,…,im=1naii2i3⋯imAk-1i2α2⋯Ak-1imαm.Proposition 17 (see [7], Proposition 2.7).
LetA be a nonnegative primitive tensor with order m and dimension n and M(A) be the majorization matrix of A. Then we have the following:
(1) For eachj∈[n], there exists an integer i∈[n]\{j} such that (M(A))ij>0.
(2) There exist somej∈[n] and integers u,v with 1≤u<v≤n such that (M(A))uj>0 and (M(A))vj>0.Letα=jj…j∈[n]m-1; then M(A)ij=aiα. We can see that Proposition 13 is the generalization of result (1) of Proposition 17 from a primitive tensor to a strongly primitive tensor. We note that Proposition 17 played an important role in [7], and if A is a nonnegative strongly primitive tensor, then A must be a nonnegative primitive tensor; thus result (2) of Proposition 17 also holds for nonnegative strongly primitive tensors.Proposition 18.
LetA=(ai1i2…im) be a nonnegative strongly primitive tensor with order m and dimension n. Then there exists at least one j∈[n] and integers u,v with 1≤u<v≤n such that (M(A))uj>0 and (M(A))vj>0.Proposition 19.
LetA=(ai1i2…im)1≤ij≤n(j=1,…,m) be a nonnegative tensor with order m and dimension n and A≠J. For given i∈[n], if aiα=ajii…i=1 for any α∈[n]m-1 and any j∈[n]\{i}, then A is strongly primitive with η(A)=2.Proof.
By (7), for any k∈[n] and α2,…,αm∈[n]m-1, we have
( A 2 ) k α 2 … α m = ∑ k 2 , k 3 , … , k m = 1 n a k k 2 … k m a k 2 α 2 … a k m α m ≥ a k i i … i a i α 2 … a i α m = 1, which implies A is strongly primitive and η(A)=2.Remark 20.
From Proposition19, we can see the following:
(1) There exist at leastn(2(n-1)(nm-1-1)-1) strongly primitive tensors such that their strongly primitive degree is equal to 2.
(2) We cannot improve the result of Proposition13 any more by the fact that there exists i∈[n] such that aiα=1>0 for any α∈[n]m-1 and there is exactly one i such that aiα>0 for any α≠ii…i.
(3) Similarly, we cannot improve the result of Proposition18 any more by the fact that there is exactly one i∈[n] such that (M(A))ui>0 for any u∈[n] and for any other j∈[n]\{i}, there exists only i∈[n] such that (M(A))ij>0.
(4) What is more, combining the above arguments, we know whether a nonnegative tensor is a nonnegative strongly primitive tensor or not, and the values of the strongly primitive degree of a nonnegative strongly primitive tensor do not depend on the number of nonzero entries but the positions of the nonzero entries.Proposition 21.
LetA=(ai1i2…im) be a nonnegative strongly primitive tensor with order m and dimension n. Then for any i∈[n], there exists some α∈[n]m-1 such that aiα>0.Proof.
SinceA is strongly primitive, there exists some k>0 such that Ak>0 by Definition 1. Assume that there exists some i∈[n] such that aiα=0 for any α∈[n]m-1. Then by (7), we have (8)Akiα2⋯αm=∑i2,i3,…,im=1naii2i3⋯imAk-1i2α2⋯Ak-1imαm=0,which leads to a contraction.Remark 22.
LetA=(ai1i2…im)1≤ij≤n(j=1,…,m) be a nonnegative tensor with order m and dimension n. For given i∈[n], we take aiα=ajii…i=1 for any α∈[n]m-1 and any j∈[n]\{i} and any other entry ai1i2…im=0. Then A is strongly primitive with η(A)=2 by Proposition 19. This implies that we cannot improve the result of Proposition 21 anymore, and it indicates the importance of the positions of the nonzero entries again.Proposition 23.
LetA be a nonnegative strongly primitive tensor and k=η(A). Then, for any integer t>k>0, we have At>0.Proof.
It is clear thatAk>0 by k=η(A). We only need to show Ak+1>0; say, for any i∈[n] and any α2,…,αm∈[n](m-1)k, we show (Ak+1)iα2…αm>0.
By Proposition21, there exists some α=j2j3…jm∈[n]m-1 such that aiα=aij2…jm>0. By Ak>0 we have (Ak)j2α2>0,…,(Ak)jmαm>0; then by (7), we have(9)Ak+1iα2⋯αm=∑i2,i3,…,im=1naii2i3⋯imAki2α2Aki3α3⋯Akimαm≥aij2j3⋯jmAkj2α2…Akjmαm>0.Proposition 24.
LetA be a nonnegative tensor with order m and dimension n and t be a positive integer. Then A is strongly primitive if and only if At is strongly primitive.Proof.
Firstly, the sufficiency is obvious. Now we show the necessity. Letk=η(A). Then Ak>0 by A is strongly primitive. Let s be a positive integer such that st≥k; then Ast>0 by Proposition 23. Thus (At)s=Ast>0, which implies At is strongly primitive.


## 3. A Characterization of the (Strongly) Primitive Tensor with Orderm and Dimension 2

In this section, we study primitive tensors and strongly primitive tensors in this paper, show that an orderm dimension 2 tensor is primitive if and only if its majorization matrix is primitive, and obtain the characterization of order m dimension 2 strongly primitive tensors and the bound of the strongly primitive degree.Lemma 25 (see [5], Corollary 4.1).
LetA be a nonnegative tensor with order m and dimension n. If M(A) is primitive, then A is also primitive and in this case we have γ(A)≤γ(M(A))≤(n-1)2+1.Theorem 26.
LetA be a nonnegative tensor with order m and dimension n=2. Then A is primitive if and only if M(A) is primitive.Proof.
Firstly, the sufficiency is obvious by Lemma25. Now we only show the necessity. Clearly, all primitive (0,1) matrices of order 2 are listed as follows:(10)1110,0111,1111.
LetA be primitive. Then γ(A)≤2 by Theorem 10 and M(A2)>0 by Proposition 8. Now we assume that M(A) is not primitive; we will show A is also not primitive.
It is not difficult to find that(11)A2ijj⋯j=∑i2,i3,…,im=12aii2i3⋯imai2jj⋯j⋯aimjj⋯j(12)=ai22⋯2a2jj⋯jm-1+∑i2,i3,…,im=1,i2i3⋯im≠22⋯22aii2⋯imai2jj⋯j⋯aimjj⋯j(13)=ai11⋯1a1jj⋯jm-1+∑i2,i3,…,im=1,i2i3⋯im≠11⋯12aii2⋯imai2jj⋯j⋯aimjj⋯j.
In (12), we note thati2i3…im≠22…2, which implies that there exists at least one entry, say, is=1, where 2≤s≤m; then a1jj…j∈{ai2jj…j,…,aimjj…j}.
Similarly, in (13), we note thati2i3…im≠11…1, which implies that there exists at least one entry, say, is=2, where 2≤s≤m; then a2jj…j∈{ai2jj…j,…,aimjj…j}.
Thus, by (12), (13), and the above arguments, we have(14)MA2ij=A2ijj⋯j=ai22⋯2a2jj⋯jm-1+a1jj⋯jP(15)=ai11⋯1a1jj⋯jm-1+a2jj⋯jQ.
SinceM(A) is not primitive; by (10), we can complete the proof by the following two cases.
Case 1. M(A)=∗0∗∗or∗∗0∗.
Subcase  1.1. M ( A ) 12 = 0.
Thena12…2=0. By (14), we have M(A2)12=(A2)12…2=0, which implies A2 is not essentially positive.
Subcase  1.2. M ( A ) 21 = 0.
Thena21…1=0. By (15), we have M(A2)21=(A2)21…1=0, which implies A2 is not essentially positive.
Case  2. M(A)=0110.
Then we haveM(A)11=M(A)22=0; that is, a11…1=a22…2=0; by (15) we have M(A2)12=(A2)12…2=0, which implies A2 is not essentially positive.
Based on the above two cases and Proposition6, we complete the proof of the necessity.A nature question is whether the result of Theorem26 is true for n≥3 or not. The following Example 27 shows that the necessity of Theorem 26 is false with n≥3.Example 27.
LetA=(ai1i2…im) be a nonnegative tensor of order m and dimension n≥3, where (16)ai1i2⋯im=0,ifi1=1,i2=i3=⋯=im≠1;1,otherwise.Then A is (strongly) primitive, but M(A) is not primitive.Proof.
By direct calculation and Definition2, we know that A2 is the tensor of order (m-1)2+1 and dimension n, and for any 1≤i≤n, we have (17)A2iα2α3⋯αm=∑i2,i3,…,im=12aii2i3⋯imai2α2⋯aimαm≥a12⋯ma2α2⋯amαm=1,ifi=1;aii⋯iaiα2⋯aiαm=1,otherwise.Obviously, A2 is positive; then A is strongly primitive with η(A)=2 and thus A is primitive with γ(A)=2.
On the other hand, by the definition ofA, we have(18)MA=10⋯011⋯1⋮11⋯1.Since the associated digraph of M(A) is not strongly connected, thus M(A) is not primitive.Next, we will study the strongly primitive degree of orderm and dimension 2 tensors. Firstly, we discuss an example with order m=5 and dimension n=2 tensor as follows.Definition 28 (see [10]).
LetA be a tensor with order m and dimension n. The i-th slice of A, denoted by A[i], is the subtensor of A with order m-1 and dimension n such that (A[i])i2…im=aii2…im.Example 29.
LetA=(ai1i2i3i4i5)1≤ij≤2(j=1,…,5) be a nonnegative tensor with order m=5 and dimension n=2, where a12122=a21121=0 and other ai1i2i3i4i5=1. Then there exists at least one zero element in each slice of A2.Proof.
Letα1=2122, α2=1121, and denote β2=β4=β5=α1, β3=α2. Then we have(19)
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Similarly, we letγ2=γ3=γ5=α2 and γ4=α1, we can show (A2)2γ2γ3γ4γ5=0, and we omit it.
Combining the above arguments, we know there exists at least one zero element in each slice ofA2 by (A2)1β2β3β4β5=(A2)2γ2γ3γ4γ5=0.Similarly, the result of Example29 can be generalized to any nonnegative tensor with order m and dimension n=2.Lemma 30.
LetA=(ai1i2…im)1≤ij≤2(j=1,…,m) be a nonnegative tensor with order m and dimension n=2. If there exist α1=j2j3…jm∈[2]m-1 and α2=k2k3…km∈[2]m-1 such that a1α1=a2α2=0, for any 2≤t≤m, let βt=α2,ifjt=1;α1,ifjt=2, and γt=α2,ifkt=1;α1,ifkt=2. Then(20)A21β2⋯βm=0,A22γ2⋯γm=0.Proof.
We first show(A2)1β2…βm=0. For any 2≤t≤m, jt∈{1,2}, we denote jt¯∈{1,2}\{jt}. Then we have ajt¯βt=0 by a1α1=a2α2=0, and(21)A21β2⋯βm=ajn¯β2=0∑i2,i3,…,im=12a1i2i3⋯imai2β2⋯aimβm=ajn¯β3=0∑i3,…,im=12a11i3⋯ima1β2ai3β3⋯aimβm+∑i3,…,im=12a12i3⋯ima2β2ai3β3⋯aimβm=aj2¯β2=0∑i3,…,im=12a1j2i3⋯imaj2βj2ai3β3⋯aimβm=ajn¯β6=0∑i4,…,im=12a1j21i4⋯imaj2βj2a1β3ai4β4⋯aimβm+∑i4,…,im=12a1j22i4⋯imaj2βj2a2β3ai4β4⋯aimβm=aj3¯β3=0∑i4,…,im=12a1j2j3i4⋯imaj2βj2aj3βj3ai4β4⋯aimβm=aji¯β3=0⋯⋯=ajp¯β3=0∑im=12a1j2j3⋯jm-1imaj2βj2aj3βj3⋯ajm-1βjm-1aimβm=ajp¯β5=0a1j2⋯jm-11aj2βj2⋯ajm-1βjm-1a1βm+a1j2⋯jm-12aj2βj2⋯ajm-1βjm-1a2βm=ajm¯βm=0a1j2⋯jmaj2βj2⋯ajmβjm=a1α1=00.
Similarly, for any2≤t≤m, kt∈{1,2}, we denote kt¯∈{1,2}\{kt}. Then we have akt¯γt=0 and we can show (A2)2γ2…γm=0 by (22)A22γ2⋯γm=∑i2,i3,…,im=12a2i2i3⋯imai2γ2⋯aimγmand similar process of the above arguments. Thus we complete the proof of (20).Theorem 31.
LetA=(ai1i2…im)1≤ij≤2(j=1,…,m) be a nonnegative tensor with order m and dimension n=2. If there exist α1=j2j3…jm∈[2]m-1 and α2=k2k3…km∈[2]m-1 such that a1α1=a2α2=0, then A is not strongly primitive.Proof.
Now we show that there exists at least one zero element in each slice ofAr by induction on r(≥2).
Firstly, by Lemma30, we know there exists at least one zero element in each slice of A2. Now we assume that there exists at least one zero element in each slice of Ar-1; say, there exist δ1,δ2∈[2](m-1)r-1 such that (Ar-1)1δ1=(Ar-1)2δ2=0. For any 2≤t≤m, let βt=δ2ifjt=1δ1ifjt=2 and γt=δ2ifkt=1δ1ifkt=2. Then by (7) and the similar proof of Lemma 30, we have(23)Ar1β2⋯βm=∑i2,i3,…,im=1na1i2i3⋯imAr-1i2β2⋯Ar-1imβm=0,and(24)Ar2γ2⋯γm=∑i2,i3,…,im=1na2i2i3⋯imAr-1i2γ2⋯Ar-1imγm=0.
By (23) and (24), we obtain that there exists at least one zero element in each slice of Ar, and thus we complete the proof.Now we give the characterization of the strongly primitive tensor with orderm and dimension 2.Theorem 32.
LetA=(ai1i2…im)1≤ij≤2(j=1,…,m) be a nonnegative tensor with order m and dimension n=2. Then(1)
A is strongly primitive if and only if one of the following holds:(a)
A=J;(b)
A≠J and a1i2…im=a211…1=1(1≤ij≤2,j=2,…,m);(c)
A≠J and a2i2…im=a122…2=1(1≤ij≤2,j=2,…,m);(2)
ifA is strongly primitive, then η(A)≤2.Proof.
Firstly, we show that (1) is sufficient. It is easy to see thatA=J is strongly primitive with η(J)=1, and if A satisfies (b) or (c), A is strongly primitive with η(A)=2 by Proposition 19 immediately.
Now we show the necessity of (1); that is, ifA does not satisfy the conditions of (a), (b), or (c), then we will show that A is not strongly primitive. We complete the proof by the following three cases.
Case  1. a1α=1 for any α∈[2]m-1 and a211…1=0.
It is not difficult to find thatM(A)=110∗. Then A is not primitive by Theorem 26, and thus A is not strongly primitive.
Case  2. a2α=1 for any α∈[2]m-1 and a122…2=0.
Similarly, we can find thatM(A)=∗011. Then A is not primitive by Theorem 26, and thus A is not strongly primitive.
Case  3. There is at least one zero element in each slice of A.
Then there existα1=j2j3…jm∈[2]m-1 and α2=k2k3…km∈[2]m-1 such that a1α1=a2α2=0. Thus A is not strongly primitive by Theorem 31.
(2) IfA is strongly primitive, by Definition 11 and the proof of (1), we obtain η(A)≤2 immediately.Remark 33.
By Theorem32, we can see that the strongly primitive degree η(A) of an nonnegative tensor with order m and dimension n=2 is irrelevant to its order m.


## 4. Some Properties and Problems of Orderm Dimension n(≥3) Strongly Primitive Tensors

In this section, we will study some properties of the strongly primitive tensors with orderm and dimension n≥3 and propose some questions for further research.Proposition 34.
LetA=(ai1i2…im)1≤ij≤n(j=1,…,m) be a nonnegative tensor with order m and dimension n. Let s∈[n], 2≤t≤m, jt(s)∈[n], and αs=j2(s)j3(s)…jm(s). If aiαs=0 for any i∈[n] and any s∈[n]\{i}, then there exist γ1,γ2,…,γn∈[n](m-1)2 such that (A2)iγk=0 for any i∈[n] and any k∈[n]\{i}.Proof.
For eachi∈[n], let βt(k)=αjt(k)∈[n]m-1 for any k∈[n] and 2≤t≤m; then ajt(k)¯βt(k)=0 for any jt(k)¯∈[n]\{jt(k)} by aiαs=0 for any i∈[n] and any s∈[n]\{i}. Let γk=β2(k)β3(k)…βm(k)∈[n](m-1)2 for any k∈[n]. Now we show (A2)iγk=0 for any k∈[n]\{i}.(25)
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We note thatk∈[n]\{i} which means there are n-1 zero elements in i-th slice of A2; thus we complete the proof by i∈[n].We note that Proposition34 is the generalization of Lemma 30; now we will obtain the generalization of Theorem 31.Theorem 35.
LetA=(ai1i2…im)1≤ij≤n(j=1,…,m) be a nonnegative tensor with order m and dimension n. Let s∈[n], 2≤t≤m, jt(s)∈[n], and αs=j2(s)j3(s)…jm(s). If aiαs=0 for any i∈[n] and any s∈[n]\{i}, then A is not strongly primitive.Proof.
Now we show that there existε1,ε2,…,εn∈[n](m-1)r such that (Ar)iεk=0 for any i∈[n] and any k∈[n]\{i} by induction on r(≥2); say, there exist at least n-1 zero elements in each slice of Ar and thus A is not strongly primitive.
Firstly, by Proposition34, we know there exist γ1,γ2,…,γn∈[n](m-1)2 such that (A2)iγk=0 for any i∈[n] and any k∈[n]\{i}; say, there exist at least n-1 zero elements in each slice of A2. Now we assume that there exist δ1,δ2,…,δn∈[n](m-1)r-1 such that (Ar-1)iδk=0 for any i∈[n] and any k∈[n]\{i}; say, there exist at least n-1 zero elements in each slice of Ar-1.
Letηt(s)=δjt(s) for any s∈[n] and 2≤t≤m; then (Ar-1)jt(s)¯ηt(s)=0 for any jt(s)¯∈[n]\{jt(s)} by (Ar-1)iδk=0 for any i∈[n] and any k∈[n]\{i}. Let εk=η2(k)η3(k)…ηm(k)∈[n](m-1)r for any k∈[n]. Now we show (Ar)iεk=0 for any i∈[n] and any k∈[n]\{i}.
By (7) and similar proof of Proposition 34, we have (26)Ariεk=Ariη2kη3k⋯ηmk=∑i2,i3,…,im=1naii2i3⋯imAr-1i2η2k⋯Ar-1imηmk=0,and then we complete the proof.Proposition 36.
LetA=(ai1i2⋯im)1≤ij≤n(j=1,⋯,m) be a nonnegative tensor with order m and dimension n, and M(A) be the majorization matrix of A. If there exist i,j∈[n], such that (M(A))ij>0,(M(A))uj=0 for any u∈[n]\{i} and (M(A))ji>0,(M(A))vi=0 for any v∈[n]\{j}, then A is not primitive, and thus A is not strongly primitive.Proof.
Firstly, we show the following assertion:
If  kis odd, then  (M(Ak))ij>0,  (M(Ak))ji>0,  (M(Ak))uj=0for any  u∈[n]\{i},  (M(Ak))vi=0for any  v∈[n]\{j}.
If  kis even, then  (M(Ak))ii>0,  (M(Ak))jj>0,  (M(Ak))ui=0for any  u∈[n]\{i},  (M(Ak))vj=0for any  v∈[n]\{j}.
Whenk=1, the above result holds which is obvious. When k=2, by Definition 7 and (7), we have(27)MA2ii=A2ii⋯i=∑i2,i3,…,im=1naii2i3⋯imai2i⋯i⋯aimi⋯i=aij⋯jaji⋯im-1=MAijMAjim-1>0,and for any u∈[n]\{i}, we have(28)MA2ui=A2ui⋯i=∑i2,i3,…,im=1naui2i3⋯imai2i⋯i⋯aimi⋯i=auj⋯jaji⋯im-1=MAujMAjim-1=0.
Similarly, we can show(M(A2))jj>0 and (M(A2))vj=0 for any v∈[n]\{j}.
Now we assume that, for anyk, the above assertion holds. Then, for k+1, we consider the following two cases.
Case  1. k is odd.
Then, by (7), we have(29)MAk+1ii=Ak+1ii⋯i=∑i2,i3,…,im=1naii2i3⋯imAki2i⋯i⋯Akimi⋯i=aij⋯jAkji⋯im-1=MAijMAkjim-1>0,and, for any u∈[n]\{i}, we have(30)MAk+1ui=Ak+1ui⋯i=∑i2,i3,…,im=1naui2i3⋯imAki2i⋯i⋯Akimi⋯i=auj⋯jAkji⋯im-1=MAujMAkjim-1=0.
Similarly, we can show(M(Ak+1))jj>0 and (M(Ak+1))vj=0 for any v∈[n]\{j}.
Case  2. k is even.
By (7) and similar proof of Case 1, we can show (M(Ak+1))ij>0, (M(Ak+1))ji>0, (M(Ak+1))uj=0 for any u∈[n]\{i}, and (M(Ak+1))vi=0 for any v∈[n]\{j}.
By Proposition8 and the above assertion, we know A is not primitive, and thus A is not strongly primitive.LetA=(ai1i2⋯im)1≤ij≤n(j=1,⋯,m) be a nonnegative strongly primitive tensor with order m and dimension n. When n=2, we know η(A)≤2 by Theorem 32. When n≥3, we do not know the value or bound of η(A). Even n=3, we do not find out all strongly primitive tensors. Thus we think it is not easy to obtain the value or bound of η(A). Based on the computation of the case n=3, we propose the following problem for further research.Question 37.
Letn≥3, A=(ai1i2⋯im)1≤ij≤n(j=1,⋯,m) be a nonnegative strongly primitive tensor with order m and dimension n. Then ηA<n-12+1.In [7, 9], the authors gave some algebraic characterizations of a nonnegative primitive tensor, and in [11] the authors showed that a nonnegative tensor is primitive if and only if the greatest common divisor of all the cycles in the associated directed hypergraph is equal to 1. It is natural for us to consider the following.Question 38.
Study the algebraic or graphic characterization of a nonnegative strongly primitive tensor.We are sure the above two questions are interesting and not easy.



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