# New result and some open problems on the primitive degree of nonnegative
tensors 111P. Yuan’s research is supported by the NSF of China (Grant No.
11271142) and the Guangdong Provincial Natural Science Foundation(Grant No.
S2012010009942), L. You’s research is supported by the Zhujiang Technology New
Star Foundation of Guangzhou (Grant No. 2011J2200090) and Program on
International Cooperation and Innovation, Department of Education, Guangdong
Province (Grant No.2012gjhz0007).

Pingzhi Yuan 222Corresponding author<EMAIL_ADDRESS>Zilong He333Email
address<EMAIL_ADDRESS>Lihua You444Email address<EMAIL_ADDRESS>
(School of Mathematical Sciences, South China Normal University,
Guangzhou, 510631, P.R. China
)

###### Abstract

In this paper, we show that the exponent set of nonnegative primitive tensors
with order $m(\geq 3)$ and dimension $n$ is $\\{1,2,\ldots,(n-1)^{2}+1\\},$
and propose some open problems for further research.

AMS classification: 05C50; 15A69 Keywords: tensor; the exponent set; primitive
tensor; primitive degree.

##  1 Introduction and a survey

A nonnegative square matrix $A=(a_{ij})$ of order $n$ is nonnegative primitive
(or simply, primitive) if $A^{k}>0$ for some positive integer $k$. The least
such $k$ is called the primitive exponent (or simply, exponent) of $A$ and is
denoted by $\gamma(A)$.

Since the work of Qi [9] and Lim [6], the study of tensors which regarded as
the generalization of matrices, the spectra of tensors (and hypergraphs) and
their various applications has attracted much attention and interest.

As is in [9], an order $m$ dimension $n$ tensor
$\mathbb{A}=(a_{i_{1}i_{2}\ldots i_{m}})_{1\leq i_{j}\leq n\hskip
5.69046pt(j=1,\ldots,m)}$ over the complex field $\mathbb{C}$ is a
multidimensional array with all entries $a_{i_{1}i_{2}\ldots
i_{m}}\in\mathbb{C}\,(i_{1},\ldots,i_{m}\in[n]=\\{1,\ldots,n\\})$. In [1] and
[2], Chang et al investigated the properties of the spectra of nonnegative
tensors, defined the irreducibility of tensors and the primitivity of
nonnegative tensors (as Definition 1.1), and extended many important
properties of primitive matrices to primitive tensors.

######  Definition 1.1.

([2, 8])  Let $\mathbb{A}$ be a nonnegative tensor with order $m$ and
dimension $n$, $x=(x_{1},x_{2},\ldots,x_{n})^{T}\in\mathbb{R}^{n}$ a vector
and $x^{[r]}=(x_{1}^{r},x_{2}^{r},\ldots,x_{n}^{r})^{T}$. Define the map
$T_{\mathbb{A}}$ from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$ as:
$T_{\mathbb{A}}(x)=(\mathbb{A}x)^{[\frac{1}{m-1}]}$. If there exists some
positive integer $r$ such that $T_{\mathbb{A}}^{r}(x)>0$ for all nonnegative
nonzero vectors $x\in\mathbb{R}^{n}$, then $\mathbb{A}$ is called primitive
and the smallest such integer $r$ is called the primitive degree of
$\mathbb{A}$, denoted by $\gamma(\mathbb{A})$.

Recently, Shao [12] defined the general product of two n-dimensional tensors
as follows.

######  Definition 1.2.

Let $\mathbb{A}$ (and $\mathbb{B}$) be an order $m\geq 2$ (and $k\geq 1$),
dimension $n$ tensor, respectively. Define the general product
$\mathbb{A}\mathbb{B}$ to be the following tensor $\mathbb{D}$ of order
$(m-1)(k-1)+1$ and dimension $n$:

$d_{i\alpha_{1}\ldots\alpha_{m-1}}=\sum\limits_{i_{2},\ldots,i_{m}=1}^{n}a_{ii_{2}\ldots
i_{m}}b_{i_{2}\alpha_{1}}\ldots
b_{i_{m}\alpha_{m-1}}\quad(i\in[n],\,\alpha_{1},\ldots,\alpha_{m-1}\in[n]^{k-1}).$

The tensor product is a generalization of the usual matrix product, and
satisfies a very useful property: the associative law ([12], Theorem 1.1).
With the general product, when $k=1$ and
$\mathbb{B}=x=(x_{1},\ldots,x_{n})^{T}\in\mathbb{C}^{n}$ is a vector of
dimension $n$, then $\mathbb{A}\mathbb{B}=\mathbb{A}x$ is still a vector of
dimension $n$, and for any $i\in[n],$

$(\mathbb{A}\mathbb{B})_{i}=(\mathbb{A}x)_{i}=\sum\limits_{i_{2},\ldots,i_{m}=1}^{n}a_{ii_{2}\ldots
i_{m}}x_{i_{2}}\ldots x_{i_{m}}.$

As an application of the general tensor product defined by Shao [12], Shao
presented a simple characterization of the primitive tensors. Now we give the
definition of “essentially positive” which introduced by Pearson.

######  Definition 1.3.

([7], Definition 3.1) A nonnegative tensor $\mathbb{A}$ is called essentially
positive, if for any nonnegative nonzero vector
$x\in\mathbb{R}^{n},\mathbb{A}x>0$ holds.

######  Proposition 1.4.

([12], Proposition 4.1) Let $\mathbb{A}$ be an order $m$ and dimension $n$
nonnegative tensor. Then the following three conditions are equivalent:

(i).  For any $i,j\in[n],a_{ij\cdots j}>0$ holds.

(ii).  For any $j\in[n],\mathbb{A}e_{j}>0$ holds (where $e_{j}$ is the
$j^{th}$ column of the identity matrix $I_{n}$).

(iii).  For any nonnegative nonzero vector $x\in\mathbb{R}^{n},\mathbb{A}x>0$
holds.

By Proposition 1.4, the following Definition 1.5 is equivalent to Definition
1.3.

######  Definition 1.5.

([12], Definition 4.1) A nonnegative tensor $\mathbb{A}$ is called essentially
positive, if it satisfies one of the three conditions in Proposition 1.4.

Let $Z(\mathbb{A})$ be the tensor obtained by replacing all the nonzero
entries of $\mathbb{A}$ by one. Then $Z(\mathbb{A})$ is called the zero-
nonzero pattern of A (or simply the zero pattern of $\mathbb{A}$). In [12],
Shao showed the following characterization and defined the primitive degree by
using the properties of tensor product and the zero patterns.

######  Proposition 1.6.

([12], Theorem 4.1) A nonnegative tensor $\mathbb{A}$ is primitive if and only
if there exists some positive integer $r$ such that $\mathbb{A}^{r}$ is
essentially positive. Furthermore, the smallest such $r$ is the primitive
degree of $\mathbb{A}$, $\gamma(\mathbb{A})$.

The concept of the majorization matrix of a tensor introduced by Pearson is
very useful.

######  Definition 1.7.

([7], Definition 2.1) The majorization matrix $M(\mathbb{A})$ of the tensor
$\mathbb{A}$ is defined as $(M(\mathbb{A}))_{ij}=a_{ij\ldots j},i,j\in[n]$.

By Definition 1.5, Proposition 1.6 and Definition 1.7, the following
characterization of the primitive tensors was easily obtained.

######  Proposition 1.8.

([13], Remark 2.6) Let $\mathbb{A}$ be a nonnegative tensor with order $m$ and
dimension $n$. Then $\mathbb{A}$ is primitive if and only if there exists some
positive integer $r$ such that $M(\mathbb{A}^{r})>0.$ Furthermore, the
smallest such $r$ is the primitive degree of $\mathbb{A}$,
$\gamma(\mathbb{A})$.

On the primitive degree $\gamma(\mathbb{A})$, Shao proposed the following
conjecture for further research.

######  Conjecture 1.9.

([12], Conjecture 1) When $m$ is fixed, then there exists some polynomial
$f(n)$ on $n$ such that $\gamma(\mathbb{A})\leq f(n)$ for all nonnegative
primitive tensors of order $m$ and dimension $n$.

In the case of $m=2$ ($\mathbb{A}$ is a matrix), the well-known Wielandt’s
upper bound tells us that we can take $f(n)=(n-1)^{2}+1.$

Recently, the authors [13] confirmed Conjecture 1.9 by proving Theorem 1.10.

######  Theorem 1.10.

([13], Theorem 1.2) Let $\mathbb{A}$ be a nonnegative primitive tensor with
order $m$ and dimension $n$. Then its primitive degree
$\gamma(\mathbb{A})\leq(n-1)^{2}+1$, and the upper bound is tight.

In fact, Theorem 1.10 not only confirmed the existence of $f(n)$, but also
showed that $f(n)$ is a quadratic function of $n$, independent of $m$,
furthermore the expression of $f(n)$ with the tensor case is the same with the
matrix case.

Let $m,n$ be positive integers with $m\geq 2,n\geq 2$, and the exponent set of
primitive tensors with order $m$ and dimension $n$ be

$E(m,n)=\\{k|\mbox{ there exists a primitive tensor }\mathbb{A}\mbox{ of order
}m\mbox{ dimension }n\mbox{ such that }k=\gamma(\mathbb{A})\\}.$

Now it is natural to consider the question to completely determine $E(m,n)$.

By Theorem 1.10, we know $E(m,n)\subseteq[(n-1)^{2}+1]$. Clearly,
$E(2,n)=E_{n}$, where $E_{n}=\\{k|\mbox{ there exists a primitive matrix
}A\mbox{ of order }n\mbox{ such that }$ $\gamma(A)=k\\}.$

Let $A=(a_{ij})$ be a nonnegative primitive matrix of order $n$. In 1950, H.
Wielandt [10] first stated the sharp upper bound for $\gamma(A)$, that is,
$\gamma(A)\leq w_{n}=(n-1)^{2}+1$ for all primitive matrices of order $n$ and
thus $E_{n}\subseteq[1,w_{n}]^{o}$, where $a,b$ are positive integers with
$b>a$ and $[a,b]^{o}=\\{k|k\mbox{ is an integer and }a\leq k\leq b\\}$. In
1964, A. L. Dulmage and N. S. Mendelsohn [4] revealed the existence of the so-
called gaps in the exponent set of primitive matrices, that is,
$E_{n}\subset[1,w_{n}]^{o}$, where “gap” is a set of consecutive integers
$[a,b]^{o}(\subset[1,w_{n}]^{o})$, such that no matrix $A$ of order $n$
satisfying $\gamma(A)\in[a,b]^{o}$. In 1981, M. Lewin and Y. Vitek [5] found
all gaps in $[\lfloor\frac{1}{2}w_{n}\rfloor+1,w_{n}]^{o}$, and conjectured
that $[1,\lfloor\frac{1}{2}w_{n}\rfloor]^{o}$ has no gaps, where $\lfloor
x\rfloor$ denotes the greatest integer $\leq x$. In 1985, Shao [11] proved
that this Lewin-Vitek Conjecture is true for all sufficiently large $n$, and
the conjecture has one counterexample when $n=11$ since $48\not\in E_{11}$.
Finally, in 1987, Zhang [14] continued and completed the work. He showed that
the Lewin-Vitek Conjecture holds for all $n$ except $n=11$. Thus the exponent
set $E_{n}$ for primitive matrices of order $n$ is completely determined.

For general tensors (the case of $m\geq 3$), in [3], the authors showed that
there are no gaps in tensor case when $m\geq n\geq 3$. but the tensor cases
$n>m\geq 3$ and $m>n=2$ are still open.

In this paper, we show that there are no gaps in tensor case $m\geq 3$ in
Section 3, that is, $E(m,n)=[(n-1)^{2}+1]=\\{1,2,\ldots,(n-1)^{2}+1\\}$ when
$m\geq 3$, and propose some open problems for further research in Section 4.

##  2 Preliminaries

For proving Conjecture 1.9, the authors [13] defined $j$-primitive and
$j$-primitive degree for a nonnegative tensor and obtained the following
result.

######  Definition 2.1.

([13], Definition 2.13) Let $\mathbb{A}$ be a nonnegative tensor with order
$m$ and dimension $n$. For a fixed integer $j\in[n]$, if there exists a
positive integer $k$ such that

$(M(\mathbb{A}^{k}))_{uj}>0,\hskip 5.69046pt{\mbox{f}or\,\,all}\,u\in[n],$

then $\mathbb{A}$ is called $j$-primitive and the smallest such integer $k$ is
called the $j$-primitive degree of $\mathbb{A}$, denoted by
$\gamma_{j}(\mathbb{A})$.

######  Proposition 2.2.

([13], Proposition 2.14) Let $\mathbb{A}$ be a nonnegative primitive tensor
with order $m$ and dimension $n$. Then $\gamma(\mathbb{A})=\max\limits_{1\leq
j\leq n}\\{\gamma_{j}(\mathbb{A})\\}.$

Let $\mathbb{A}$ be a nonnegative tensor with order $m$ and dimension $n$. For
positive integers $k$ and $j\in[n]$, a notation $S_{k}(\mathbb{A},j)$ was
introduced in [3] as follows:

$S_{k}(\mathbb{A},j)=\\{u\in[n]\hskip 2.27626pt|\hskip
2.27626pt(M(\mathbb{A}^{k}))_{uj}>0\\},\hskip 5.69046ptk=1,2,\ldots.$ (2.1)

and by equation

$(M(\mathbb{A}^{k+1}))_{uj}=\sum\limits_{i_{2},\ldots,i_{m}=1}^{n}a_{ui_{2}\cdots
i_{m}}(M(\mathbb{A}^{k}))_{i_{2}j}\cdots(M(\mathbb{A}^{k}))_{i_{m}j},$

which investigated in [13], the following recurrence relation (2.2) was
obtained as follows:

$S_{k+1}(\mathbb{A},j)=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there
exist }i_{2},\ldots,i_{m}\in S_{k}(\mathbb{A},j)\mbox{ and }a_{ui_{2}\cdots
i_{m}}>0\\}.$ (2.2)

According to this relation (2.2), some good properties were obtained.

######  Proposition 2.3.

([3], Lemma 3.1 and Remark 3.2) Let $\mathbb{A}$ be a nonnegative tensor with
order $m$ and dimension $n$.

(i)  Let $k,l,i,j$ be positive integers such that $1\leq i,j\leq n$. Suppose
that $S_{k}(\mathbb{A},i)=S_{l}(\mathbb{A},j)$, then
$S_{k+r}(\mathbb{A},i)=S_{l+r}(\mathbb{A},j)$ holds for every positive integer
$r$.

(ii)  For any $j\in[n]$, let $k$ be the least positive integer such that
$S_{k}(\mathbb{A},j)=[n]$. Then for any integer $l\geq k$,
$S_{l}(\mathbb{A},j)=[n]$.

(iii). For any $j\in[n]$, $\gamma_{j}(\mathbb{A})$ is the least positive
integer $k$ satisfying $S_{k}(\mathbb{A},j)=[n]$.

######  Remark 2.4.

For convenience, we replace $S_{t}(\mathbb{A},n)$ by the notation
$S_{t}(\mathbb{A},0)$, and replace $\gamma_{n}(\mathbb{A})$ by
$\gamma_{0}(\mathbb{A})$, respectively. Under these notations, we have
$\gamma(\mathbb{A})=\max\limits_{0\leq j\leq n-1}\\{\gamma_{j}(\mathbb{A})\\}$
by Proposition 2.2.

In [13], the authors introduce some theoretical concepts of digraphs and
matrices.

Let $D=(V,E)$ denote a digraph on $n$ vertices with vertex set $V(D)=V$ and
arc set $E(D)=E$. Loops are permitted, but no multiple arcs. A $u\rightarrow
v$ walk in $D$ is a sequence of vertices $u,u_{1},\ldots,u_{k}=v$ and a
sequence of arcs
$e_{1}=(u,u_{1}),e_{2}=(u_{1},u_{2}),\ldots,e_{k}=(u_{k-1},v)$, where the
vertices and the arcs are not necessarily distinct. We use the notation
$u\rightarrow u_{1}\rightarrow u_{2}\rightarrow\cdots\rightarrow
u_{k-1}\rightarrow v$ to refer to this $u\rightarrow v$ walk. A closed walk is
a $u\rightarrow v$ walk where $u=v$. A path is a walk with distinct vertices.
A cycle is a closed $u\rightarrow v$ walk with distinct vertices except for
$u=v$. The length of a walk $W$ is the number of arcs in $W$, denoted by
$l(W)$.

######  Definition 2.5.

([13], Definition 2.9) Let $D=(V,E)$ denote a digraph on $n$ vertices. A
digraph $D^{\prime}=(V,E^{\prime})$ is called the reversed digraph of $D$
where $(j,i)\in E^{\prime}$ if and only if $(i,j)\in E$ for any $i,j\in V$,
denoted by $\overleftarrow{D}$.

Let $A=(a_{ij})$ be a square nonnegative matrix of order $n$. The associated
digraph $D(A)=(V,E)$ of $A$ (possibly with loops) is defined to be the digraph
with vertex set $V=\\{1,2,\ldots,n\\}$ and arc set $E=\\{(i,j)|a_{ij}\neq
0\\}$. The associated reversed digraph $\overleftarrow{D(A)}=(V,E^{\prime})$
of $A$ (possibly with loops) is defined to be the digraph with vertex set
$V=\\{1,2,\ldots,n\\}$ and arc set $E^{\prime}=\\{(j,i)|a_{ij}\neq 0\\}$.
Clearly, the associated reversed digraph of $A$ is the reversed digraph of the
associated digraph of $A$.

######  Proposition 2.6.

([3], Proposition 4.1) Let $A$ be a nonnegative matrix of order $n$,
$j(\in[n]),k$ be positive integers, $S_{k}(A,j)=\\{u\in[n]\hskip
2.27626pt|\hskip 2.27626pt(A^{k})_{uj}>0\\}$. Then

$S_{k}(A,j)=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there exists a
walk of length $k$ from $j$ to $u$ in the digraph }\overleftarrow{D(A)}\\}.$

######  Proposition 2.7.

([3], Lemma 4.2) Let $\mathbb{A}$ be a nonnegative tensor with order $m$ and
dimension $n$ such that $a_{ii_{2}\cdots i_{m}}=0$ if $i_{2}\cdots
i_{m}\not=i_{2}\cdots i_{2}$ for any $i\in[n]$. Then for any positive integers
$j(\in[n])$ and $k$,

$S_{k}(\mathbb{A},j)=S_{k}(M(\mathbb{A}),j).$ (2.3)

The relations between a tensor $\mathbb{A}$ and the majorization matrix
$M(\mathbb{A})$ are important and useful.

######  Proposition 2.8.

([12], Corollary 4.1 ) Let $\mathbb{A}$ be a nonnegative tensor with order $m$
and dimension $n$. If $M(\mathbb{A})$ is primitive, then $\mathbb{A}$ is also
primitive.

######  Proposition 2.9.

([13], Corollary 3.4) Let $\mathbb{A}$ be a nonnegative primitive tensor with
order $m$ and dimension $n$ such that $a_{ii_{2}\cdots i_{m}}=0$ if
$i_{2}\cdots i_{m}\not=i_{2}\cdots i_{2}$ for any $i\in[n]$. If
$M(\mathbb{A})$ is primitive, then $\gamma(\mathbb{A})=\gamma(M(\mathbb{A})).$

Now we define the nonnegative tensor $\mathbb{A}_{0}=(a_{i_{1}i_{2}\ldots
i_{m}})_{1\leq i_{j}\leq n\hskip 5.69046pt(j=1,\ldots,m)}$ with order $m$ and
dimension $n$ such that

(i). The majorization matrix $M(\mathbb{A}_{0})$ is given by

$M_{1}=\left(\begin{array}[]{cccccc}0&0&\cdots&0&1&1\\\ 1&0&\cdots&0&0&0\\\
0&1&\cdots&0&0&0\\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\
0&0&\cdots&1&0&0\\\ 0&0&\cdots&0&1&0\\\ \end{array}\right).$

(ii). $a_{ii_{2}\cdots i_{m}}=0$, if $i_{2}\cdots i_{m}\not=i_{2}\cdots i_{2}$
for any $i\in[n]$.

It is well-known that $M_{1}$ is primitive and the primitive exponent
$\gamma(M_{1})=(n-1)^{2}+1$. Then by Propositions 2.8 $\sim$ 2.9, it can
easily be seen that the tensor $\mathbb{A}_{0}$ is primitive and its primitive
degree $\gamma(\mathbb{A}_{0})=(n-1)^{2}+1$. Furthermore, there are more good
properties on tensor $\mathbb{A}_{0}$ as follows.

######  Proposition 2.10.

([3], Proposition 4.5) Let $\mathbb{A}_{0}$ be the nonnegative primitive
tensor with order $m$ and dimension $n$ defined as above. Then
$\gamma_{n-1}(\mathbb{A}_{0})=n^{2}-3n+3$.

######  Proposition 2.11.

([3], Remark 4.6 and Proposition 4.9) Let $k$ be positive integer with $1\leq
k\leq n^{2}-3n+2$. Then
$S_{1}(\mathbb{A}_{0},n-1),S_{2}(\mathbb{A}_{0},n-1),\ldots,S_{k}(\mathbb{A}_{0},n-1)$
are pairwise distinct proper subsets of $[n]$, and
$S_{n^{2}-3n+2}(\mathbb{A}_{0},n-1)=[n-1]$.

For convenience, we let $|a|_{n}$ denote the least positive integer $t$ with
$t\equiv a\pmod{n}$, and let a set $S=\\{a_{1},\cdots,a_{s}\\}\pmod{n}$ denote
the set $S=\\{|a_{1}|_{n},\cdots,|a_{s}|_{n}\\}$.

######  Proposition 2.12.

Let $k$ be a positive integer and $1\leq k\leq n^{2}-3n+2$.

(i). If $k=(n-1)q+r$ with $q\geq 0$ and $1\leq r\leq n-1$, then

$S_{k}(\mathbb{A}_{0},n-1)=\\{r-q-1,\cdots,r-1,r\\}\pmod{n}$ (2.4)

and $|S_{k}(\mathbb{A}_{0},n-1)|=q+2$.

(ii).  Let $t(\geq 1)$, $j(\geq 0)$ be integers. Then for any positive integer
$t$ and any integer $j\in\\{0,1,\ldots,n-2\\}$, we have
$S_{t+(n-1-j)}(\mathbb{A}_{0},j)=S_{t}(\mathbb{A}_{0},n-1)$.

###### Proof.

(i). By the definition of $\mathbb{A}_{0}$ and Propositions 2.6$\sim$2.7, we
have

$S_{k}(\mathbb{A}_{0},n-1)$

$=S_{k}(M(\mathbb{A}_{0}),n-1)$

$=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{ there exists a walk of
length }k\mbox{ from }n-1\mbox{ to }u\mbox{ in the digraph
}\overleftarrow{D(M(\mathbb{A}_{0}))}\\}.$

Now we can obtain the conclusion (i) from the digraph
$\overleftarrow{D(M(\mathbb{A}_{0}))}=\overleftarrow{D(M_{1})}$ (See Figure
$1$) by simple calculation.

$\textstyle{n}$$\textstyle{\hskip
3.0pt{n-1}}$$\textstyle{\mathrm{1}}$$\textstyle{2}$$\textstyle{\hskip
3.0pt{n-2}}$ $\mbox{ \qquad Figure 1. digraph
}\overleftarrow{D(M(\mathbb{A}_{0}))}$

(ii). By the definition of $\mathbb{A}_{0}$, an easy computation shows that

$\\{n,1\\}=S_{1}(\mathbb{A}_{0},n-1)=S_{2}(\mathbb{A}_{0},n-2)=\cdots=S_{n-1}(\mathbb{A}_{0},1)=S_{n}(\mathbb{A}_{0},0).$

It implies that

$S_{1}(\mathbb{A}_{0},n-1)=S_{n-j}(\mathbb{A}_{0},j)\mbox{ for any
}j\in\\{0,1,\ldots,n-2\\}.$ (2.5)

Then for any integer $r\geq 1$, by (i) of Proposition 2.3, we have

$S_{1+r}(\mathbb{A}_{0},n-1)=S_{n-j+r}(\mathbb{A}_{0},j)\mbox{ for any
}j\in\\{0,1,\ldots,n-2\\}.$ (2.6)

Combining (2.5) and (2.6), (ii) holds. ∎

######  Remark 2.13.

For convenience, we regard a set $\\{n,1\\}$ as a consecutive positive
integers subset of $\\{1,2,\ldots,n\\}$. Then it implies that $\\{n-1,n,1\\}$,
$\\{n,1,2\\}$, $\\{n,1,2,3\\}$ and so on, and thus $S_{k}(\mathbb{A}_{0},n-1)$
are consecutive positive integers subsets of $\\{1,2,\cdots,n\\}$ for all
$k\in[n^{2}-3n+2]$ by the result (i)  of Proposition 2.12.

##  3 The exponent set of nonnegative primitive tensors

In this section, we will show that there are no gaps in tensor case $m\geq 3$,
that is, $E(m,n)=[(n-1)^{2}+1]$ when $m\geq 3$. It implies that the results of
the case $m\geq 3$ is totally different from the case $m=2$.

Let $s_{1},s_{2},\ldots,s_{l}$ be $l$ elements which are not necessarily to be
distinct. For convenience, we let $\\{s_{1},s_{2},\ldots,s_{l}\\}$ denote the
set of all different elements from $s_{1},s_{2},\ldots,s_{l}$.

Let $k(\geq 1),n(\geq 2),q(\geq 0),r(\geq 1)$ be integers,
$\mathbb{A}_{0}=(a_{i_{1}i_{2}\ldots i_{m}})_{1\leq i_{j}\leq n\hskip
5.69046pt(j=1,\ldots,m)}$ be a nonnegative tensor with order $m$ and dimension
$n$ defined in Section 2, $k=(n-1)q+r$ with $1\leq r\leq n-1$, then $0\leq
q\leq n-3$. We define the nonnegative tensor
$\mathbb{A}_{k}=(a^{(k)}_{i_{1}i_{2}\cdots i_{m}})_{1\leq i_{j}\leq n\hskip
5.69046pt(j=1,\ldots,m)}$ with order $m$ and dimension $n$ such that (i).
$M(\mathbb{A}_{k})=M(\mathbb{A}_{0})=M_{1};$

(ii). $a^{(k)}_{i\alpha}=1$, if
$i\in[n]\backslash\\{r-q,r-q+1,\ldots,r,r+1\\}\pmod{n}$ and
$\alpha=i_{2}\cdots i_{m}\in[n]^{m-1}$ with
$\\{i_{2},\cdots,i_{m}\\}=\\{r-q-1,r\\}\pmod{n};$

(iii). $a^{(k)}_{i_{1}i_{2}\cdots i_{m}}=0$, except for (i) and (ii). By
Proposition 2.8, it’s easy to see the tensor $\mathbb{A}_{k}$ is primitive
from the fact that $M(\mathbb{A}_{k})=M(\mathbb{A}_{0})=M_{1}$ is primitive.
Now we investigate the relations between the tensors $\mathbb{A}_{k}$ and
$\mathbb{A}_{0}$.

######  Proposition 3.1.

Let $k(\geq 1),n(\geq 2),q(\geq 0),r$ be integers, and
$k=(n-1)q+r\in[n^{2}-3n+2]$ with $1\leq r\leq n-1$, $\mathbb{A}_{0}$ and
$\mathbb{A}_{k}$ defined as above. Then for any integer $t\in[k]$, we have

$S_{t}(\mathbb{A}_{k},n-1)=S_{t}(\mathbb{A}_{0},n-1).$ (3.1)

###### Proof.

Now we prove (3.1) holds for any $t\in[k]$ by induction on $t$.

It’s obvious that
$S_{1}(\mathbb{A}_{k},n-1)=S_{1}(\mathbb{A}_{0},n-1)=\\{1,n\\}$ from (2.1) and
$M(\mathbb{A}_{k})=M(\mathbb{A}_{0})$. Assume that
$S_{t}(\mathbb{A}_{k},n-1)=S_{t}(\mathbb{A}_{0},n-1)$ holds for $t<k$. Now we
only need show that $S_{t+1}(\mathbb{A}_{k},n-1)=S_{t+1}(\mathbb{A}_{0},n-1)$
holds.

Case 1:  $\\{r-q-1,r\\}\pmod{n}\not\subseteq S_{t}(\mathbb{A}_{0},n-1)$.

We note that the recurrence relation (2.2) and the definition of
$\mathbb{A}_{k}$, then

$S_{t+1}(\mathbb{A}_{k},n-1)=\\{u\in[n]\hskip 2.27626pt|\hskip
2.27626pt\mbox{there exist }i_{2},\ldots,i_{m}\in
S_{t}(\mathbb{A}_{k},n-1)\mbox{ and }a^{(k)}_{ui_{2}\cdots i_{m}}>0\\}$

$=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there exist
}i_{2},\ldots,i_{m}\in S_{t}(\mathbb{A}_{0},n-1)\mbox{ and
}a^{(k)}_{ui_{2}\cdots i_{m}}>0\\}$

$=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there exist }v\in
S_{t}(\mathbb{A}_{0},n-1)\mbox{ and }a_{uv\cdots v}>0\\}$

$\cup\hskip 2.27626pt\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there
exist }i_{2},\ldots,i_{m}\in S_{t}(\mathbb{A}_{0},n-1)\mbox{ such that }$

$i_{2}\cdots i_{m}\not=i_{2}\cdots i_{2}\mbox{ and }a^{(k)}_{ui_{2}\cdots
i_{m}}>0\\}$

$=S_{t+1}(\mathbb{A}_{0},n-1)\hskip 2.27626pt\cup\varnothing$

$=S_{t+1}(\mathbb{A}_{0},n-1).$

Case 2:  $\\{r-q-1,r\\}\pmod{n}\subseteq S_{t}(\mathbb{A}_{0},n-1)$.

From the Remark 2.13, we know that $S_{t}(\mathbb{A}_{0},n-1)$ is a
consecutive positive integers subset of $\\{1,2,\cdots,n\\}$. Thus

$\\{r-q-1,r-q,\cdots,r\\}\pmod{n}\subseteq S_{t}(\mathbb{A}_{0},n-1)$

or

$\\{r,r+1,\cdots,r-q-1\\}\pmod{n}\subseteq S_{t}(\mathbb{A}_{0},n-1)$

holds.

If $\\{r-q-1,r-q,\cdots,r\\}\pmod{n}\subseteq S_{t}(\mathbb{A}_{0},n-1)$, then
$S_{k}(\mathbb{A}_{0},n-1)\subseteq S_{t}(\mathbb{A}_{0},n-1)$ and
$|S_{t}(\mathbb{A}_{0},n-1)|=\lfloor\frac{t-1}{n-1}\rfloor+2\leq|S_{k}(\mathbb{A}_{0},n-1)|=q+2$
by the result (i) of Proposition 2.12 and $t<k$. Then
$S_{t}(\mathbb{A}_{0},n-1)=S_{k}(\mathbb{A}_{0},n-1)$, it implies a
contradiction since $S_{t}(\mathbb{A}_{0},n-1)\not=S_{k}(\mathbb{A}_{0},n-1)$
by $t<k$ and Proposition 2.11. It follows that

$\\{r,r+1,\cdots,r-q-1\\}\pmod{n}\subseteq S_{t}(\mathbb{A}_{0},n-1).$

Thus by the definition of $\mathbb{A}_{0}$ and Propositions 2.6 $\sim$ 2.7, we
have

$\\{r+1,r+2,\cdots,r-q\\}\pmod{n}\subseteq S_{t+1}(\mathbb{A}_{0},n-1).$ (3.2)

Then by the definition of $\mathbb{A}_{k}$, the assumption and (3.2), we can
see that

$S_{t+1}(\mathbb{A}_{k},n-1)=\\{u\in[n]\hskip 2.27626pt|\hskip
2.27626pt\mbox{there exist }i_{2},\ldots,i_{m}\in
S_{t}(\mathbb{A}_{k},n-1)\mbox{ and }a^{(k)}_{ui_{2}\cdots i_{m}}>0\\}$

$=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there exist
}i_{2},\ldots,i_{m}\in S_{t}(\mathbb{A}_{0},n-1)\mbox{ and
}a^{(k)}_{ui_{2}\cdots i_{m}}>0\\}$

$=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there exist }v\in
S_{t}(\mathbb{A}_{0},n-1)\mbox{ and }a_{uv\cdots v}>0\\}$

$\cup\hskip 2.27626pt\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there
exist }i_{2},\ldots,i_{m}\in S_{t}(\mathbb{A}_{0},n-1)\mbox{ such that }$

$i_{2}\cdots i_{m}\not=i_{2}\cdots i_{2}\mbox{ and }a^{(k)}_{ui_{2}\cdots
i_{m}}>0\\}$

$=S_{t+1}(\mathbb{A}_{0},n-1)\hskip 2.27626pt\cup\hskip
2.27626pt([n]\backslash\\{r-q,r-q+1,\ldots,r,r+1\\}\pmod{n})$

$=S_{t+1}(\mathbb{A}_{0},n-1)$.

Combining the above two cases, the equation (3.1) is proved for all $t\in[k]$.
∎

######  Proposition 3.2.

Let $k(\geq 1),n(\geq 2),q(\geq 0),r$ be integers, and
$k=(n-1)q+r\in[n^{2}-3n+2]$ with $1\leq r\leq n-1$, $\mathbb{A}_{0}$ and
$\mathbb{A}_{k}$ defined as above. Then for any integer $t\in[k]$, we have

$S_{t}(\mathbb{A}_{k},j)=S_{t}(\mathbb{A}_{0},j),\mbox{where
}j\in\\{0,1,2,\ldots,n-2\\}$ (3.3)

###### Proof.

Now we show (3.3) holds for any $t\in[k]$ by the following two cases.

Case 1:  $1\leq t\leq n-1$.

Now we prove (3.3) holds for any $t\hskip 2.84544pt(1\leq t\leq n-1)$ by
induction on $t$.

Clearly, $S_{1}(\mathbb{A}_{k},j)=\\{j+1\\}=S_{1}(\mathbb{A}_{0},j)$ for any
$j\in\\{0,1,2,\ldots,n-2\\}$. Now we assume that
$S_{t}(\mathbb{A}_{k},j)=S_{t}(\mathbb{A}_{0},j)$ holds for any
$j\in\\{0,1,2,\ldots,n-2\\}$ and $t<n-1$. Then we show that
$S_{t+1}(\mathbb{A}_{k},j)=S_{t+1}(\mathbb{A}_{0},j)$ holds for any
$j\in\\{0,1,2,\ldots,n-2\\}$.

By observing and the result (ii) of Proposition 2.12, we know for any $t$ with
$2\leq t\leq n-1$,

$S_{t}(\mathbb{A}_{0},j)$

$=\left\\{\begin{array}[]{ll}\\{j+t\\},&j\in\\{0,1,\ldots,n-t-1\\};\\\
S_{t-(n-1-j)}(\mathbb{A}_{0},n-1)=\\{t+j-n,t+j-n+1\\}\pmod{n},&j\in\\{n-t,\ldots,n-2\\}.\end{array}\right.$

We note that $(t+j-n+1)-(t+j-n)\equiv 1\pmod{n}$, but $(r-q-1)-r\not\equiv
1\pmod{n}$ by $0\leq q\leq n-3$. Thus $\\{r-q-1,r\\}\pmod{n}\subseteq
S_{t}(\mathbb{A}_{0},j)$ if and only if $q=0$ and $r=k$, then
$\\{r-q-1,r\\}\pmod{n}=\\{k-1,k\\}\pmod{n}$. It is a contradiction by the fact
$1\leq t+j-n+1<t\leq k$ for $j\in\\{n-t,\ldots,n-2\\}$.

The above arguments imply that $\\{r-q-1,r\\}\pmod{n}\not\subseteq
S_{t}(\mathbb{A}_{0},j)$ for $j=0,1,\cdots,n-2$. Then by the definition of
$\mathbb{A}_{k}$, and the assumption, we can see that

$S_{t+1}(\mathbb{A}_{k},j)=\\{u\in[n]\hskip 2.27626pt|\hskip
2.27626pt\mbox{there exist }i_{2},\ldots,i_{m}\in
S_{t}(\mathbb{A}_{k},j)\mbox{ and }a^{(k)}_{ui_{2}\cdots i_{m}}>0\\}$

$=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there exist
}i_{2},\ldots,i_{m}\in S_{t}(\mathbb{A}_{0},j)\mbox{ and
}a^{(k)}_{ui_{2}\cdots i_{m}}>0\\}$

$=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there exist }v\in
S_{t}(\mathbb{A}_{0},j)\mbox{ and }a_{uv\cdots v}>0\\}$

$\cup\hskip 2.27626pt\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there
exist }i_{2},\ldots,i_{m}\in S_{t}(\mathbb{A}_{0},j)\mbox{ such that }$

$i_{2}\cdots i_{m}\not=i_{2}\cdots i_{2}\mbox{ and }a^{(k)}_{ui_{2}\cdots
i_{m}}>0\\}$

$=S_{t+1}(\mathbb{A}_{0},j)\hskip 2.27626pt\cup\varnothing$

$=S_{t+1}(\mathbb{A}_{0},j)$.

Case 2:  $t\geq n$.

Now we prove (3.3) holds for any $n\leq t\leq k$ by induction on $t$.

By (ii) of Proposition 2.12,
$S_{n}(\mathbb{A}_{0},j)=S_{j+1}(\mathbb{A}_{0},n-1)=\\{j,j+1\\}\pmod{n}$ for
any $j\in\\{0,1,\ldots,n-2\\}$. On the other hand, by Proposition 2.7, the
result of Case 1 and (2.2), we have

$S_{n-1}(\mathbb{A}_{0},j)=\left\\{\begin{array}[]{ll}\\{n-1\\},&j=0;\\\
\\{j-1,j\\}\pmod{n},&1\leq j\leq n-2.\end{array}\right.$

and

$S_{n}(\mathbb{A}_{k},j)=\\{u\in[n]\hskip 2.27626pt|\hskip
2.27626pt\mbox{there exist }i_{2},\ldots,i_{m}\in
S_{n-1}(\mathbb{A}_{k},j)\mbox{ and }a^{(k)}_{ui_{2}\cdots i_{m}}>0\\}$

$=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there exist
}i_{2},\ldots,i_{m}\in S_{n-1}(\mathbb{A}_{0},j)\mbox{ and
}a^{(k)}_{ui_{2}\cdots i_{m}}>0\\}$

$=\left\\{\begin{array}[]{ll}\\{1,n\\},&j=0;\\\ \\{j,j+1\\}\pmod{n},&1\leq
j\leq n-2.\end{array}\right.$

$=\\{j,j+1\\}\pmod{n}.$

Thus $S_{n}(\mathbb{A}_{k},j)=S_{n}(\mathbb{A}_{0},j)$.

Now we assume that $S_{t}(\mathbb{A}_{k},j)=S_{t}(\mathbb{A}_{0},j)$ holds for
$n\leq t<k$ for any $j\in\\{0,1,\ldots,n-2\\}$. Then we only need show that
$S_{t+1}(\mathbb{A}_{k},j)=S_{t+1}(\mathbb{A}_{0},j)$ holds for any
$j\in\\{0,1,\ldots,n-2\\}$. By the assumption, the fact that $1\leq
t-(n-1-j)<k$, (ii) of Proposition 2.12 and Proposition 3.1, we have

$S_{t}(\mathbb{A}_{k},j)=S_{t}(\mathbb{A}_{0},j)=S_{t-(n-1-j)}(\mathbb{A}_{0},n-1)=S_{t-(n-1-j)}(\mathbb{A}_{k},n-1).$

Thus by (i) of Proposition 2.3, Proposition 3.1 and (ii) of Proposition 2.12,
we have

$\displaystyle S_{t+1}(\mathbb{A}_{k},j)$
$\displaystyle=S_{t+1-(n-1-j)}(\mathbb{A}_{k},n-1)$
$\displaystyle=S_{t+1-(n-1-j)}(\mathbb{A}_{0},n-1)$
$\displaystyle=S_{t+1}(\mathbb{A}_{0},j).$

Combining the above two cases, we complete the proof. ∎

######  Theorem 3.3.

Let $k$ be a positive integer with $1\leq k\leq n^{2}-3n+2$. Then

(i).  $\gamma_{j}({\mathbb{A}_{k}})=k+n-j$ where $j=0,1,\cdots,n-1.$

(ii).  $\gamma(\mathbb{A}_{k})=k+n.$

###### Proof.

Firstly, we show $\gamma_{n-1}({\mathbb{A}_{k}})=k+1.$ By Proposition 3.1 and
(i) of Proposition 2.12, we have

$S_{k}(\mathbb{A}_{k},n-1)=S_{k}(\mathbb{A}_{0},n-1)=\\{r-q-1,r-q,\cdots,r\\}\pmod{n}.$

And by Proposition 2.7 and (i) of Proposition 2.12, we have

$\\{r-q,r-q+1,\cdots,r+1\\}\pmod{n}\subseteq
S_{k+1}(M(\mathbb{A}_{0}),n-1)=S_{k+1}(\mathbb{A}_{0},n-1).$

Thus by Proposition 3.1, we have

$S_{k+1}(\mathbb{A}_{k},n-1)=\\{u\in[n]\hskip 2.27626pt|\hskip
2.27626pt\mbox{there exist }i_{2},\ldots,i_{m}\in
S_{k}(\mathbb{A}_{k},n-1)\mbox{ and }a^{(k)}_{ui_{2}\cdots i_{m}}>0\\}$

$=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there exist
}i_{2},\ldots,i_{m}\in S_{k}(\mathbb{A}_{0},n-1)\mbox{ and
}a^{(k)}_{ui_{2}\cdots i_{m}}>0\\}$

$=\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there exist }v\in
S_{k}(\mathbb{A}_{0},n-1)\mbox{ and }a_{uv\cdots v}>0\\}$

$\cup\hskip 2.27626pt\\{u\in[n]\hskip 2.27626pt|\hskip 2.27626pt\mbox{there
exist }i_{2},\ldots,i_{m}\in S_{k}(\mathbb{A}_{0},n-1)\mbox{ such that }$

$i_{2}\cdots i_{m}\not=i_{2}\cdots i_{2}\mbox{ and }a^{(k)}_{ui_{2}\cdots
i_{m}}>0\\}$

$=S_{k+1}(\mathbb{A}_{0},n-1)\hskip 2.27626pt\cup\hskip
2.27626pt([n]\backslash\\{r-q,r-q+1,\ldots,r,r+1\\}\pmod{n})$

$=[n]$

On the other hand, we have
$S_{t}(\mathbb{A}_{k},n-1)=S_{t}(\mathbb{A}_{0},n-1)\subsetneqq[n]$ holds for
all $1\leq t\leq k$ by Proposition 3.1 and Proposition 2.11. Thus
$\gamma_{n-1}(\mathbb{A}_{k})=k+1$ by (iii) of Proposition 2.3.

Now we show $\gamma_{j}({\mathbb{A}_{k}})=k+n-j$ for any
$j\in\\{0,1,\ldots,n-2\\}$.

By Propositions 3.1$\sim$ 3.2, we have

$S_{1}(\mathbb{A}_{k},n-1)=S_{2}(\mathbb{A}_{k},n-2)=S_{3}(\mathbb{A}_{k},n-3)=\cdots=S_{n-1}(\mathbb{A}_{k},1)=S_{n}(\mathbb{A}_{k},n)=\\{1,n\\}.$

Thus by (i) of Proposition 2.3 and the definition of $j$-primitive degree, we
have

$S_{1}(\mathbb{A}_{k},n-1)=S_{2}(\mathbb{A}_{k},n-2)$

$\Longrightarrow\left\\{\begin{array}[]{ll}([n]\not=)S_{1+r}(\mathbb{A}_{k},n-1)=S_{2+r}(\mathbb{A}_{k},n-2),&\mbox{
if }1\leq r\leq k-1;\\\
([n]=)S_{1+k}(\mathbb{A}_{k},n-1)=S_{2+k}(\mathbb{A}_{k},n-2),&\mbox{ if
}r=k.\end{array}\right.$

$\Longrightarrow\gamma_{n-2}(\mathbb{A}_{k})=k+2$.

Similarly, we can prove
$\gamma_{n-3}(\mathbb{A}_{k})=k+3,\ldots,\gamma_{1}(\mathbb{A}_{k})=k+n-1,\gamma_{0}(\mathbb{A}_{k})=k+n.$

Combining the above arguments, we obtain $\gamma_{j}({\mathbb{A}_{k}})=k+n-j$
for any $j\in\\{0,1,\ldots,n-1\\}.$

By Proposition 2.2 or Remark 2.4, we have

$\gamma(\mathbb{A}_{k})=\max\limits_{0\leq j\leq
n-1}\\{\gamma_{j}(\mathbb{A}_{k})\\}=\gamma_{0}(\mathbb{A}_{k})=k+n,$

and we complete the proof of (ii) immediately. ∎

######  Theorem 3.4.

Let $m,n$ be positive integers with $m\geq 3,n\geq 2$. Then
$E(m,\,n)=[(n-1)^{2}+1]$.

###### Proof.

Let $t$ be any positive integer with $1\leq t\leq(n-1)^{2}+1$, we will
complete the proof by constructing primitive tensor $\mathbb{B}_{t}$ with
order $m$ and dimension $n$ such that $\gamma(\mathbb{B}_{t})=t$. We consider
the following two cases.

Case 1:  $1\leq t\leq n$.

It is well known that there exists a primitive matrix $A_{t}$ of order $n$
such that $\gamma(A_{t})=t$. We define the tensor $\mathbb{B}_{t}$ to be the
nonnegative primitive tensor with order $m$ and dimension $n$ such that
$(\mathbb{B}_{t})_{ii_{2}\cdots i_{m}}=0$ if $i_{2}\cdots
i_{m}\not=i_{2}\cdots i_{2}$ for any $i\in[n]$, and $M(\mathbb{B}_{t})=A_{t}$.
Then $\gamma(\mathbb{B}_{t})=\gamma(A_{t})=t$ by Proposition 2.9.

Case 2:  $n+1\leq t\leq(n-1)^{2}+1$.

We choose $\mathbb{B}_{t}=\mathbb{A}_{t-n}$. Then
$\gamma(\mathbb{B}_{t})=(t-n)+n=t$ by Theorem 3.3. ∎

##  4 Some open problems for further research

In this section, we will propose some interesting open problems for further
research.

Let $\mathbb{A}$ be a nonnegative primitive tensor with order $m$ and
dimension $n$. By Theorem 1.10, we have $\gamma(\mathbb{A})\leq(n-1)^{2}+1$.
It is well-known that when $m=2$, $\gamma(\mathbb{A})=(n-1)^{2}+1$ if and only
if $\mathbb{A}\cong M_{1}$. But the case of $m\geq 3$ is totally different. In
fact, we know $\mathbb{A}_{n^{2}-3n+2}\not\cong\mathbb{A}_{0}$, but
$\gamma(\mathbb{A}_{n^{2}-3n+2})=\gamma(\mathbb{A}_{0})=(n-1)^{2}+1$ by
Proposition 2.9 and Theorem 3.3.

######  Question 4.1.

Let $\mathbb{A}$ be a nonnegative primitive tensor with order $m$ and
dimension $n$. If $\gamma(\mathbb{A})=(n-1)^{2}+1$, can we give a
characterization of such $\mathbb{A}$?

Let $k$ be a positive integer with $1\leq k\leq n^{2}-3n+2$, $\mathbb{A}_{k}$
defined as Section 3. Then by Theorem 3.3, we have

$\\{\gamma_{1}(\mathbb{A}_{k}),\gamma_{2}(\mathbb{A}_{k}),\ldots,\gamma_{n-1}(\mathbb{A}_{k}),\gamma_{n}(\mathbb{A}_{k})\\}=\\{k+1,k+2,\ldots,k+n\\}=[k+1,k+n]^{o}.$
(4.1)

Based on (4.1), we propose the following question.

######  Question 4.2.

Let $\mathbb{A}$ be a nonnegative primitive tensor with order $m$ and
dimension $n$. Does there exist two positive integers $a,b$ with $a<b$ such
that
$\\{\gamma_{1}(\mathbb{A}),\gamma_{2}(\mathbb{A}),\ldots,\gamma_{n-1}(\mathbb{A}),\gamma_{n}(\mathbb{A})\\}$
$=[a,b]^{o}?$

Now we recall the definition of reducibility.

######  Definition 4.3.

([1], Definition 2.1) A tensor $\mathbb{C}=(c_{i_{1}\ldots i_{m}})$ of order
$m$ dimension $n$ is called reducible, if there exists a nonempty proper index
subset $I\subset\\{1,\ldots,n\\}$ such that

$c_{i_{1}\ldots i_{m}}=0\hskip 8.5359pt\forall i_{1}\in I,\forall i_{2},\ldots
i_{m}\not\in I.$

If $\mathbb{C}$ is not reducible, then we call $\mathbb{C}$ irreducible.

When $m=2$, Definition 4.3 give the definition of reducible matrices and
irreducible matrices. It is well-known from the basic relations between
matrices and digraphs that a square matrix $A$ is irreducible if and only if
its associated digraph $D(A)$ is strongly connected, a matrix $A$ is primitive
if and only if $A$ is irreducible and the period of $A$ is 1. It is also well-
known that $A$ is primitive if and only if $D(A)$ is primitive, and a digraph
$D$ is primitive if and only if $D$ is strongly connected and the greatest
common divisor of the lengths of all the cycles of $D$ is 1. Then by the
definitions of primitive matrices, irreducible matrices and $j$-primitive, we
obtain the following characterization.

######  Proposition 4.4.

Let $A$ be a nonnegative matrix of order $n$. Then $A$ is primitive if and
only if $A$ is irreducible and there exists some $j\in[n]$ such that $A$ is
$j$-primitive.

###### Proof.

Necessity is obvious. So we omit it. Now we only show sufficiency. Let
$\gamma_{j}(A)=k$, it implies that $(A^{k})_{uj}>0$ for any $u\in[n]$, thus
there exists a walk of length $k$ from $u$ to $j$ in the associated digraph
$D(A)$ by the relation between matrices and digraphs. Now we show there exists
a positive integer $l$ such that $(A^{l})_{uv}>0$ for any $u,v\in[n]$, that
is, we need show there exists a walk of length $l$ from $u$ to $v$ in the
associated digraph $D(A)$ for any $u,v\in V(D(A))$.

Since $A$ is irreducible, $D(A)$ is strong connected. Then for any $v$, there
exists a path $P$ with length $l_{1}(0\leq l_{1}\leq n-1)$ from $j$ to $v$.
Similarly, there exists $w$ such that there exists a walk $Q$ with length
$n-1-l_{1}$ from $u$ to $w$. Since $\gamma_{j}(A)=k$, then there exists a walk
$R$ with length $k$ from $w$ to $j$. Thus the walk $Q+R+P$ is a walk of length
$(n-1-l_{1})+k+l_{1}=k+n-1$ from $u$ to $v$. Let $l=k+n-1$, we complete the
proof. ∎

Based on Proposition 4.4, we propose the following conjecture.

######  Conjecture 4.5.

Let $\mathbb{A}$ be a nonnegative tensor with order $m$ and dimension $n$.
Then $\mathbb{A}$ is primitive if and only if $\mathbb{A}$ is irreducible and
there exists some $j\in[n]$ such that $\mathbb{A}$ is $j$-primitive.

Let $m(\geq 2),n(\geq 2),j\in[n],k(\geq 1)$ be positive integers, $\mathbb{A}$
be a nonnegative tensor with order $m$ and dimension $n$. Recall the
definitions of $\gamma_{j}(\mathbb{A})$, $S_{k}(\mathbb{A},j)$ and Proposition
2.3, we know $\gamma_{j}(\mathbb{A})$ is the least positive integer satisfying
$S_{k}(\mathbb{A},j)=[n]$. If $\mathbb{A}$ is primitive, then
$1\leq\gamma_{j}(\mathbb{A})\leq(n-1)^{2}+1$ and the upper and lower bounds
are tight by Theorem 1.10 and Proposition 2.2. But if $\mathbb{A}$ is not
primitive, and there exists some $j\in[n]$ and positive integer $k$ such that
$S_{k}(\mathbb{A},j)=[n]$, that is, there exists some $j\in[n]$ such that
$\mathbb{A}$ is $j$-primitive, does $\gamma_{j}(\mathbb{A})$ have a tight
lower bound and a tight upper bound? It is obvious that
$\gamma_{j}(\mathbb{A})\geq 1$. Now we only consider the upper bound of
$\gamma_{j}(\mathbb{A})$.

Let $m,n,j(\in[n])$ be positive integers with $m\geq 2,n\geq 2$, and notations

$r(m,n)=\max\\{\gamma_{j}(\mathbb{A})\hskip 2.84544pt|\hskip
2.84544pt\mathbb{A}\mbox{ is a nonnegative but not primitive tensor of order
}m\mbox{ dimension }n$

$\mbox{ and there exists some }j\in[n]\mbox{ such that }\mathbb{A}\mbox{ is
}j\mbox{-primitive}\\}.$

$R_{j}(m,n)=\\{k\hskip 2.84544pt|\hskip 2.84544pt\mbox{there exists a
nonnegative but not primitive tensor }\mathbb{A}\mbox{ of order }m\mbox{
dimension }$

$n\mbox{ such that }\mathbb{A}\mbox{ is }j\mbox{-primitive and
}\gamma_{j}(\mathbb{A})=k\\}.$

Now we investigate the properties of $r(m,n)$ and $R_{j}(m,n)$.

######  Proposition 4.6.

$r(2,n)=n^{2}-4n+6.$

###### Proof.

Let $A$ be a nonnegative matrix of order $n$, but $A$ is not primitive and
there exists some $j\in[n]$ such that $A$ is $j$-primitive. Let
$\gamma_{j}(A)=k$, it implies that $(A^{k})_{uj}>0$ for any $u\in[n]$, thus
there exists a walk of length $k$ from $u$ to $j$ in the associated digraph
$D=D(A)$ for any $u\in[n]$ by the relation between matrices and digraphs.

Take $u=j$, we know there exists a closed walk of length $k$ from $j$ to $j$,
it implies that $j$ belongs in some strong connected subdigraph of $D$.
Without loss of generality, we assume a set $V_{1}$ is the maximal subset of
$V(D)$ such that $j\in V_{1}\subsetneqq V(D)$ and the induced subdigraph
$D_{1}=D[V_{1}]$ is strong connected. Thus for any $v\not\in V_{1}$, there
does not exist a walk from $j$ to $v$ in $D$ since there exists a walk of
length $k$ from $v$ to $j$ by $A$ is $j$-primitive.

Now we show the adjacent matrix $A(D_{1})$ is $j$-primitive. That is, for any
$u\in V_{1}$, let $W$ be a walk of length $k$ from $u$ to $j$ in $D$, we only
need show $V(W)\subseteq V_{1}$. Otherwise, if there exists vertex
$v\in(V(D)\backslash V_{1})\cap V(W)$, then there exists a walk from $j$ to
$v$ since $u,j\in V_{1}$ and $D_{1}=D[V_{1}]$ is strong connected, and thus
$v\in V_{1}$, it is a contradiction.

Therefore $A(D_{1})$ is primitive by Proposition 4.4 and thus there exists a
positive integer $l\leq(|V_{1}|-1)^{2}+1$ such that there exists a walk of
length $l$ from $u$ to $j$ for any $u\in V_{1}$ by the well-known Wieland’s
upper bound.

Then $k\leq n-|V_{1}|+l\leq|V_{1}|^{2}-3|V_{1}|+n+2\leq n^{2}-4n+6$ by
$|V_{1}|\leq n-1$, and thus $r(2,n)\leq n^{2}-4n+6$ by the definition of
$r(2,n)$.

On the other hand, let $M_{2}$ be a nonnegative matrix of order $n$ and
$D(M_{2})$ be the associated digragh as follows. Clearly, $M_{2}$ and
$D(M_{2})$ are not primitive.

$M_{2}=\left(\begin{array}[]{cccccccc}0&1&0&\cdots&0&0&0\\\
0&0&1&\cdots&0&0&0\\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\
0&0&0&\cdots&1&0&0\\\ 1&0&0&\cdots&0&1&0\\\ 1&0&0&\cdots&0&0&0\\\
0&0&0&\cdots&0&1&0\end{array}\right),$

$\textstyle{n}$$\textstyle{\hskip 9.0pt{n-1}}$$\textstyle{\hskip
3.0pt{n-2}}$$\textstyle{\mathrm{1}}$$\textstyle{2}$$\textstyle{\hskip
3.0pt{n-3}}$ $\mbox{ \qquad Figure 2. digraph }D(M_{2})$

It is easy to show there does not exist a walk of length $n^{2}-4n+5$ from $n$
to $n-1$. Then $\gamma_{n-1}(M_{2})\geq n^{2}-4n+6.$

Combining the above two inequalities, we obtain $r(2,n)=n^{2}-4n+6$. ∎

The following necessary condition for a tensor to be primitive is useful.

######  Proposition 4.7.

([13], (i) of Proposition 2.7) Let $\mathbb{A}$ be a nonnegative primitive
tensor with order $m$ and dimension $n$, $M(\mathbb{A})$ the majorization
matrix of $\mathbb{A}$. Then for each $j\in[n]$, there exists an integer
$i\in[n]\backslash\\{j\\}$ such that $(M(\mathbb{A}))_{ij}>0$.

######  Proposition 4.8.

If $m\geq\lfloor\frac{n-1}{2}\rfloor+1$, then $\left(\begin{array}[]{c}n-1\\\
\lfloor\frac{n-1}{2}\rfloor\end{array}\right)+1\leq r(m,n)\leq 2^{n}-1.$

###### Proof.

Let $S_{1},S_{2},\ldots,S_{k}$ be all the subset of $[n]\backslash\\{1\\}$
with $|S_{1}|=|S_{2}|=\ldots=|S_{k}|=\lfloor\frac{n-1}{2}\rfloor$. It is clear
that $k=\left(\begin{array}[]{c}n-1\\\
\lfloor\frac{n-1}{2}\rfloor\end{array}\right)$. Let
$\mathbb{A}=(a_{i_{1}\ldots i_{m}})$ be a nonnegative tensor with order $m$
and dimension $n$, where

$a_{i_{1}i_{2}\ldots i_{m}}=\left\\{\begin{array}[]{ll}1,&\mbox{ if }i_{1}\in
S_{1},\mbox{ and }i_{2}=\ldots=i_{m}=1;\\\ 1,&\mbox{ if }i_{1}\in
S_{j+1},\mbox{ and }\\{i_{2},\ldots,i_{m}\\}=S_{j}\mbox{ for
}j=1,2,\ldots,k-1;\\\ 1,&\mbox{ if }i_{1}\in[n],\mbox{ and
}\\{i_{2},\ldots,i_{m}\\}=S_{k};\\\ 0,&\mbox{ otherwise. }\end{array}\right.$

By the above definition, we know for any $j\in[n]\backslash\\{1\\}$,
$(M(\mathbb{A}))_{ij}=0$ for any $i\in[n]$. Then $\mathbb{A}$ is not primitive
by Proposition 4.7.

Now we show $\gamma_{1}(\mathbb{A})=k+1$. By (2.1) and (2.2), we have

$S_{1}(\mathbb{A},1)=\\{u\in[n]\hskip 2.84544pt|\hskip
2.84544ptM(\mathbb{A})_{u1}=a_{u1\ldots 1}>0\\}=S_{1},$
$S_{2}(\mathbb{A},1)=\\{u\in[n]\hskip 2.84544pt|\hskip 2.84544pt\mbox{there
exist }i_{2},\ldots,i_{m}\in S_{1}(\mathbb{A},1)\mbox{ and }a_{ui_{2}\ldots
i_{m}}>0\\}=S_{2},$ $\cdots,$ $S_{l}(\mathbb{A},1)=\\{u\in[n]\hskip
2.84544pt|\hskip 2.84544pt\mbox{there exist }i_{2},\ldots,i_{m}\in
S_{l-1}(\mathbb{A},1)\mbox{ and }a_{ui_{2}\ldots i_{m}}>0\\}=S_{l},$ $\cdots,$
$S_{k}(\mathbb{A},1)=\\{u\in[n]\hskip 2.84544pt|\hskip 2.84544pt\mbox{there
exist }i_{2},\ldots,i_{m}\in S_{k-1}(\mathbb{A},1)\mbox{ and }a_{ui_{2}\ldots
i_{m}}>0\\}=S_{k},$ $S_{k+1}(\mathbb{A},1)=\\{u\in[n]\hskip 2.84544pt|\hskip
2.84544pt\mbox{there exist }i_{2},\ldots,i_{m}\in S_{k}(\mathbb{A},1)\mbox{
and }a_{ui_{2}\ldots i_{m}}>0\\}=[n].$

Then $\gamma_{1}(\mathbb{A})=k+1$ by Proposition 2.3 and thus
$r(m,n)\geq\left(\begin{array}[]{c}n-1\\\
\lfloor\frac{n-1}{2}\rfloor\end{array}\right)+1$.

Next, we show $r(m,n)\leq 2^{n}-1.$ Let $\mathbb{B}=(b_{i_{1}\ldots i_{m}})$
be a nonnegative tensor with order $m$ and dimension $n$, and $\mathbb{B}$ be
not primitive. If $\mathbb{B}$ is $j$-primitive with some $j\in[n]$, then we
denote $K=\gamma_{j}(\mathbb{A})$. Then
$S_{1}(\mathbb{B},j),S_{2}(\mathbb{B},j),\ldots,S_{K-1}(\mathbb{B},j)$ are
proper subsets of $[n]$ and $S_{K}(\mathbb{B},j)=[n]$.

Now we show
$S_{1}(\mathbb{B},j),S_{2}(\mathbb{B},j),\ldots,S_{K}(\mathbb{B},j)$ are
pairwise distinct. Otherwise, there exist $u,v$ such that $1\leq u<v\leq K$
and
$S_{1}(\mathbb{B},j),S_{2}(\mathbb{B},j),\ldots,S_{u}(\mathbb{B},j),\ldots,S_{v-1}(\mathbb{B},j)$
are pairwise distinct, but $S_{u}(\mathbb{B},j)=S_{v}(\mathbb{B},j)$. Then by
(2.2), we have $S_{u+t+s(v-u)}(\mathbb{B},j)=S_{u+t}(\mathbb{B},j)$ for any
$t$ with $0\leq t<v-u$ and nonnegative integer $s$, and thus there exist
nonnegative integers $s_{1},t_{1}$ with $0\leq t_{1}<v-u$ such that
$K=u+t_{1}+s_{1}(v-u)$ and
$S_{K}(\mathbb{B},j)=S_{u+t_{1}}(\mathbb{B},j)\not=[n],$ it is a
contradiction. We note that
$S_{1}(\mathbb{B},j),S_{2}(\mathbb{B},j),\ldots,S_{K}(\mathbb{B},j)$ are
subset of $[n]$ and they are not empty set, so $K\leq 2^{n}-1.$ Thus
$r(m,n)\leq 2^{n}-1.$ ∎

######  Proposition 4.9.

Let $m\geq 2$, and $\mathbb{A}=(a_{i_{1}\ldots i_{m}})$ be a nonnegative
tensor with order $m$ and dimension $n$, $\mathbb{B}=(b_{i_{1}\ldots
i_{m}i_{m+1}})$ be a nonnegative tensor with order $m+1$ and dimension $n$
where

$b_{i_{1}\ldots i_{m}i_{m+1}}=\left\\{\begin{array}[]{ll}a_{i_{1}\ldots
i_{m}},&\mbox{ if }i_{m+1}=i_{m};\\\ 0,&\mbox{ otherwise. }\end{array}\right.$

Then we have

(i).  There exists some $j\in[n]$ satisfying $\mathbb{A}$ is $j$-primitive if
and only if there exists some $j\in[n]$ satisfying $\mathbb{B}$ is
$j$-primitive, and $\gamma_{j}(\mathbb{B})=\gamma_{j}(\mathbb{A}).$

(ii).  $\mathbb{A}$ is primitive if and only if $\mathbb{B}$ is primitive, and
thus $\gamma(\mathbb{B})=\gamma(\mathbb{A}).$

###### Proof.

Firstly, we show $S_{k}(\mathbb{B},j)=S_{k}(\mathbb{A},j)$ for $k\geq 1$ by
induction on $k$. Clearly,

$S_{1}(\mathbb{B},j)=\\{u\in[n]|b_{uj\ldots j}>0\\}=\\{u\in[n]|a_{uj\ldots
j}>0\\}=S_{1}(\mathbb{A},j).$

Now we assume that $S_{k}(\mathbb{B},j)=S_{k}(\mathbb{A},j)$ for $k\geq 1$.
Then by (2.2), we have

$S_{k+1}(\mathbb{B},j)=\\{u\in[n]|\mbox{ there exist
}i_{2},\ldots,i_{m},i_{m+1}\in S_{k}(\mathbb{B},j)\mbox{ and }b_{ui_{2}\ldots
i_{m}i_{m+1}}>0\\}$

$=\\{u\in[n]|\mbox{ there exist }i_{2},\ldots,i_{m},i_{m+1}\in
S_{k}(\mathbb{A},j)\mbox{ and }b_{ui_{2}\ldots i_{m}i_{m+1}}>0\\}$

$=\\{u\in[n]|\mbox{ there exist }i_{2},\ldots,i_{m}\in
S_{k}(\mathbb{A},j)\mbox{ and }b_{ui_{2}\ldots i_{m}i_{m}}>0\\}$

$=\\{u\in[n]|\mbox{ there exist }i_{2},\ldots,i_{m}\in
S_{k}(\mathbb{A},j)\mbox{ and }a_{ui_{2}\ldots i_{m}}>0\\}$

$=S_{k+1}(\mathbb{A},j).$

Then $\mathbb{B}$ is $j$-primitive if and only if $\mathbb{A}$ is
$j$-primitive, and $\gamma_{j}(\mathbb{B})=\gamma_{j}(\mathbb{A}).$

It is obvious that $\mathbb{A}$ is primitive if and only if for all
$j(\in[n])$, $\mathbb{A}$ is $j$-primitive. Thus (ii) holds by (i). ∎

By Proposition 4.9, we obtain the following results and propose some questions
immediately.

######  Proposition 4.10.

Let $l$ be a large positive integer. Then

(i).  $r(2,n)\leq r(3,n)\leq r(4,n)\leq\ldots\leq r(l,n)\leq r(l+1,n)$.

(ii).  $R_{j}(2,n)\subseteq R_{j}(3,n)\subseteq
R_{j}(4,n)\subseteq\ldots\subseteq R_{j}(l,n)\subseteq R_{j}(l+1,n)$.

######  Question 4.11.

Let $m(\geq 3),n(\geq 2)$ be positive integers. Then $r(m,n)=?$ Can you
characterize the extremal tensor?

######  Question 4.12.

Does there exist gaps in $R_{j}(m,n)$?

In [12], Shao proposed the concept of strongly primitive.

######  Definition 4.13.

([12], Definition 4.3) Let $\mathbb{A}$ be a nonnegative tensor with order $m$
and dimension $n$. If there exists some positive integer $k$ such that
$\mathbb{A}^{k}>0$ is a positive tensor, then $\mathbb{A}$ is called strongly
primitive, and the smallest such $k$ is called the strongly primitive degree
of $\mathbb{A}$.

Let $\mathbb{A}=(a_{i_{1}i_{2}\ldots i_{m}})$ be a nonnegative tensor with
order $m$ and dimension $n$. It is clear that if $\mathbb{A}$ is strongly
primitive, then $\mathbb{A}$ is primitive. 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, but in the case $m\geq 3$, Shao gave an example to show
these two concepts are not equivalent([12]). Till now, there are no further
research on strongly primitive. For convenience, let $\eta(\mathbb{A})$ be the
strongly primitive degree of $\mathbb{A}$.

######  Proposition 4.14.

Let $\mathbb{A}=(a_{i_{1}i_{2}\ldots i_{m}})$ be a nonnegative strongly
primitive tensor with order $m$ and dimension $n$. Then for any
$\alpha\in[n]^{m-1}$, there exists some $i\in[n]$ such that $a_{i\alpha}>0$.

###### Proof.

Without loss of generality, we assume the strongly primitive degree of
$\mathbb{A}$, denoted by $\eta(\mathbb{A})$, is $k$. Then $A^{k}$ be a
positive tensor with order $(m-1)^{k}+1$ and dimension $n$. Therefore for any
$i_{1}\in[n]$ and any $\alpha_{2},\ldots,\alpha_{t}\in[n]^{m-1}$ where
$t=(m-1)^{k-1}$, we can complete the proof by the following equation.

$(A^{k})_{i_{1}\alpha_{2}\ldots\alpha_{t}}=\sum\limits_{i_{2},\ldots,i_{t}=1}^{n}(A^{k-1})_{i_{1}i_{2}\ldots
i_{t}}a_{i_{2}\alpha_{2}}\ldots a_{i_{t}\alpha_{t}}>0.$ (4.2)

∎

######  Example 4.15.

Let $m=n=3$, $\mathbb{A}=(a_{i_{1}i_{2}i_{3}})$ be a nonnegative tensor with
order $m$ and dimension $n$, where $a_{111}=a_{222}=a_{333}=a_{233}=a_{311}=0$
and other $a_{i_{1}i_{2}i_{3}}=1$. Then $\eta(\mathbb{A})=4.$

###### Proof.

Firstly, we show that $\mathbb{A}^{2}=(a_{i_{1}i_{2}i_{3}i_{4}i_{5}}^{(2)})$
is a nonnegative tensor with order $5$ and dimension $3$, and
$a_{i_{1}j_{2}j_{3}j_{4}j_{5}}>0$ except for
$a_{13333}=a_{21111}=a_{33333}=0$. We complete the proof by the following four
cases and
$a_{i_{1}j_{2}j_{3}j_{4}j_{5}}^{(2)}=\sum\limits_{i_{2},i_{3}=1}^{3}a_{i_{1}i_{2}i_{3}}a_{i_{2}j_{2}j_{3}}a_{i_{3}j_{4}j_{5}}.$

Case 1:  $j_{2}\not=j_{3}$ and $j_{4}\not=j_{5}$.

Then $a_{i_{2}j_{2}j_{3}}=a_{i_{3}j_{4}j_{5}}=1$ for any
$i_{2},i_{3}\in\\{1,2,3\\}$. Thus
$a_{i_{1}j_{2}j_{3}j_{4}j_{5}}^{(2)}=\sum\limits_{i_{2},i_{3}=1}^{3}a_{i_{1}i_{2}i_{3}}>0.$

Case 2:  $j_{2}=j_{3}$ and $j_{4}\not=j_{5}$.

Then $a_{i_{3}j_{4}j_{5}}=1$ for any $i_{3}\in\\{1,2,3\\}$. Thus

$a_{i_{1}j_{2}j_{3}j_{4}j_{5}}^{(2)}=\sum\limits_{i_{2},i_{3}=1}^{3}a_{i_{1}i_{2}i_{3}}a_{i_{2}j_{2}j_{3}}\geq\left\\{\begin{array}[]{ll}a_{i_{1}23}a_{211},&\mbox{if
}j_{2}j_{3}=11;\\\ a_{i_{1}13}a_{122}+a_{i_{1}31}a_{322},&\mbox{if
}j_{2}j_{3}=22;\\\ a_{i_{1}13}a_{133},&\mbox{if }j_{2}j_{3}=33\\\
\end{array}\right.$

$>0.$

Case 3:  $j_{2}\not=j_{3}$ and $j_{4}=j_{5}$.

The proof is similar to the proof of Case 2.

Case 4:  $j_{2}=j_{3}$ and $j_{4}=j_{5}$.

In this case, we know only $a_{13333}=a_{21111}=a_{33333}=0$ by direct
computation.

Similarly, we can show that $\mathbb{A}^{3}=(a_{i_{1}i_{2}\ldots
i_{9}}^{(3)})$ is a almost positive tensor with order $9$ and dimension $3$
except for $a_{233333333}^{(3)}=0$ by (4.2), but
$\mathbb{A}^{4},\mathbb{A}^{5},\ldots$ are positive tensors. Thus combining
the above cases, we know $\eta(\mathbb{A})=4.$ ∎

######  Question 4.16.

Can we define and study the strongly primitive degree, the strongly exponent
set, the $j$-strongly primitive of strongly primitive tensors and so on?

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