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# Some inequalities for nonnegative tensors

Journal of Inequalities and Applications20142014:340

https://doi.org/10.1186/1029-242X-2014-340

• Received: 26 March 2014
• Accepted: 14 August 2014
• Published:

## Abstract

Let $\mathcal{A}$ be a nonnegative tensor and $x=\left({x}_{i}\right)>0$ its Perron vector. We give lower bounds for ${x}_{t}^{m-1}/\sum {x}_{{i}_{2}}\cdots {x}_{{i}_{m}}$ and upper bounds for ${x}_{s}^{m-1}/\sum {x}_{{i}_{2}}\cdots {x}_{{i}_{m}}$, where ${x}_{s}={max}_{1\le i\le n}{x}_{i}$ and ${x}_{t}={min}_{1\le i\le n}{x}_{i}$.

MSC:15A18, 15A69, 65F15, 65F10.

## Keywords

• Perron vector
• nonnegative tensor
• eigenvalues

## 1 Introduction

Eigenvalue problems of higher order tensors have become an important topic of study in a new applied mathematics branch, numerical multilinear algebra, and they have a wide range of practical applications . The main difficulty in tensor problems is that they are generally nonlinear. Therefore, large amounts of results for matrices are never in force for higher order tensors. However, there are still some results preserved in the case of higher order tensors.

Throughout this paper we consider an m th order n-dimensional tensor $\mathcal{A}$ consisting of ${n}^{m}$ entries in :
$\mathcal{A}=\left({a}_{{i}_{1},{i}_{2},\dots ,{i}_{m}}\right),\phantom{\rule{1em}{0ex}}{a}_{{i}_{1},{i}_{2},\dots ,{i}_{m}}\in \mathbb{R},1\le {i}_{1},{i}_{2},\dots ,{i}_{m}\le n.$

The tensor $\mathcal{A}$ is called nonnegative (or positive) if all the entries ${a}_{{i}_{1},{i}_{2},\dots ,{i}_{m}}\ge 0$ (or ${a}_{{i}_{1},{i}_{2},\dots ,{i}_{m}}>0$). We also denote by the field of complex numbers.

Definition 1.1 A pair $\left(\lambda ,x\right)\in \mathbb{C}×{\mathbb{C}}^{n}/\left\{0\right\}$ is called an eigenvalue-eigenvector pair of $\mathcal{A}$, if they satisfy
$\mathcal{A}{x}^{m-1}=\lambda {x}^{\left[m-1\right]},$
(1)
where n-dimensional column vectors $\mathcal{A}{x}^{m-1}$ and ${x}^{\left[m-1\right]}$ are defined as
$\mathcal{A}{x}^{m-1}:={\left({a}_{i{i}_{2}\cdots {i}_{n}}{x}_{{i}_{2}}\cdots {x}_{{i}_{n}}\right)}_{1\le i\le n}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}^{\left[m-1\right]}:={\left({x}_{i}^{m-1}\right)}_{1\le i\le n}.$
This definition was introduced by Qi  when m is even and $\mathcal{A}$ is symmetric. Independently, Lim  gave such a definition but restricted x to be a real vector and λ to be a real number. Let

where $|\lambda |$ denotes the modulus of λ. We call $\rho \left(\mathcal{A}\right)$ the spectral radius of tensor $\mathcal{A}$.

In , Chang et al. generalized the Perron-Frobenius theorem from nonnegative matrices to irreducible nonnegative tensors. Some further results based on this theorem are discussed by Yang [10, 11].

Definition 1.2 The tensor $\mathcal{A}$ is called reducible if there exists a nonempty proper index subset $\mathbb{J}\subset \left\{1,2,\dots ,n\right\}$ such that
${a}_{{i}_{1},{i}_{2},\dots ,{i}_{m}}=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }{i}_{1}\in \mathbb{J},\mathrm{\forall }{i}_{2},\dots ,{i}_{m}\notin \mathbb{J}.$

If $\mathcal{A}$ is not reducible, then we call $\mathcal{A}$ irreducible.

Theorem 1.3 

If $\mathcal{A}$ is irreducible and nonnegative, then there exist a number $\rho \left(\mathcal{A}\right)>0$ and a vector ${x}_{0}>0$, such that
$\mathcal{A}{x}_{0}^{m-1}=\rho \left(\mathcal{A}\right){x}_{0}^{\left[m-1\right]}.$

Moreover, if λ is an eigenvalue with a nonnegative eigenvector, then $\lambda =\rho \left(\mathcal{A}\right)$. If λ is an eigenvalue of $\mathcal{A}$, then $|\lambda |\le \rho \left(\mathcal{A}\right)$.

We call ${x}_{0}$ a Perron vector of $\mathcal{A}$ corresponding to its largest nonnegative eigenvalue $\rho \left(\mathcal{A}\right)$.

In this paper, we are interested in studying some bounds for the Perron vector of $\mathcal{A}$. For this purpose, we define
${x}_{s}=\underset{1\le i\le n}{max}{x}_{i},\phantom{\rule{2em}{0ex}}{x}_{t}=\underset{1\le i\le n}{min}{x}_{i}.$

In the following we first give a new and simple bound for ${x}_{s}/{x}_{t}$. Then we give some lower bounds for ${x}_{t}^{m-1}/\sum {x}_{{i}_{2}}\cdots {x}_{{i}_{m}}$ and upper bounds for ${x}_{s}^{m-1}/\sum {x}_{{i}_{2}}\cdots {x}_{{i}_{m}}$, which can be used to get another bound for ${x}_{s}/{x}_{t}$.

The paper is organized as follows. In Section 2, some efforts of establishing the bounds of the nonnegative tensor are made. An application of these bounds is studied in Section 3.

## 2 Bounds

In , the authors have studied the perturbation bound for the spectral radius of $\mathcal{A}$, and they show that a Perron vector x must be known in advance so that the perturbation bound can be computed. Here we cite a lemma for use below.

Lemma 2.1 

Suppose $\mathcal{A}$ is nonnegative such that $m\ge 3$ and x is its Perron vector x. Then
Define the i th row sum of $\mathcal{A}$ as
${R}_{i}\left(\mathcal{A}\right)=\sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{a}_{i{i}_{2}\cdots {i}_{m}},$
and denote the largest, the smallest, and the average row sums of $\mathcal{A}$ by
${R}_{max}\left(\mathcal{A}\right)=\underset{i=1,\dots ,n}{max}{R}_{i}\left(\mathcal{A}\right),\phantom{\rule{2em}{0ex}}{R}_{min}\left(\mathcal{A}\right)=\underset{i=1,\dots ,n}{min}{R}_{i}\left(\mathcal{A}\right).$
Let
$l=\underset{{i}_{1},\dots ,{i}_{m}}{min}{a}_{{i}_{1}\cdots {i}_{m}},\phantom{\rule{2em}{0ex}}L=\underset{{i}_{1},\dots ,{i}_{m}}{max}{a}_{{i}_{1}\cdots {i}_{m}}.$
Theorem 2.2 Suppose $\mathcal{A}$ is nonnegative with Perron vector x. Then
Proof Since, for any $i=1,2,\dots ,n$,
$\rho \left(\mathcal{A}\right){x}_{t}^{m-1}\le \rho \left(\mathcal{A}\right){x}_{i}^{m-1}=\sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{a}_{i{i}_{2}\cdots {i}_{m}}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}\le \sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{a}_{i{i}_{2}\cdots {i}_{m}}{x}_{s}^{m-1},$
we have
$\rho \left(\mathcal{A}\right){x}_{t}^{m-1}\le {R}_{min}\left(\mathcal{A}\right){x}_{s}^{m-1}.$
(4)
Similarly,
$\rho \left(\mathcal{A}\right){x}_{s}^{m-1}\ge {R}_{max}\left(\mathcal{A}\right){x}_{t}^{m-1}.$
(5)
It follows from (4) and (5) that
$\frac{{x}_{s}^{m-1}}{{x}_{t}^{m-1}}\ge \frac{{x}_{t}^{m-1}}{{x}_{s}^{m-1}}\frac{{R}_{max}\left(\mathcal{A}\right)}{{R}_{min}\left(\mathcal{A}\right)}$
and therefore
$\frac{{x}_{s}}{{x}_{t}}\ge {\left\{\frac{{R}_{max}\left(\mathcal{A}\right)}{{R}_{min}\left(\mathcal{A}\right)}\right\}}^{\frac{1}{2\left(m-1\right)}}.$
If we assume ${\parallel x\parallel }_{1}=1$, then, for any $i=1,2,\dots ,n$,

The result follows. □

we see that our new lower bound in (3) is sharper than that in (2).

If the tensor $\mathcal{A}$ is positive, we derive a new upper bound for ${x}_{s}/{x}_{t}$ in terms of the entries of $\mathcal{A}$ and the spectral radius $\rho \left(\mathcal{A}\right)$.

Theorem 2.3 If $\mathcal{A}$ is positive with Perron vector x, then
$\begin{array}{rl}\frac{l}{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l\left(m-1\right)n}& \le \frac{{x}_{t}^{m-1}}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}\\ \le \frac{L}{\rho \left(\mathcal{A}\right)-{R}_{max}\left(\mathcal{A}\right)+L\left(m-1\right)n},\end{array}$
(7)
$\begin{array}{rl}\frac{\rho \left(\mathcal{A}\right)-{R}_{max}\left(\mathcal{A}\right)+\left[\left(m-1\right)n-1\right]L}{\rho \left(\mathcal{A}\right)-{R}_{max}\left(\mathcal{A}\right)+L\left(m-1\right)n}\cdot \frac{1}{\left(m-1\right)n-1}& \le \frac{{x}_{s}^{m-1}}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}\\ \le \frac{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l}{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l\left(m-1\right)n},\end{array}$
(8)
$\begin{array}{r}{\left\{\frac{\rho \left(\mathcal{A}\right)-{R}_{max}\left(\mathcal{A}\right)+\left[\left(m-1\right)n-1\right]L}{L}\cdot \frac{1}{\left(m-1\right)n-1}\right\}}^{\frac{1}{m-1}}\\ \phantom{\rule{1em}{0ex}}\le \frac{{x}_{s}}{{x}_{t}}\le {\left\{\frac{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l}{l}\right\}}^{\frac{1}{m-1}}.\end{array}$
(9)
Proof First, we prove the right side of (9). Now we consider
$\begin{array}{rl}\rho \left(\mathcal{A}\right){x}_{t}^{m-1}-l\sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}& =\sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{a}_{t{i}_{2}\cdots {i}_{m}}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}-l\sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}\\ \ge {x}_{t}^{m-1}\left[{R}_{min}\left(\mathcal{A}\right)-l\left(m-1\right)n\right]\end{array}$
(10)
and
$\begin{array}{rl}\frac{{x}_{s}^{m-1}}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}& =1-\frac{\underset{\phantom{\mathrm{except}\phantom{\rule{0.25em}{0ex}}====}\mathrm{except}\phantom{\rule{0.25em}{0ex}}{i}_{2}=\cdots ={i}_{m}=s}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}\\ \le 1-\frac{\left[\left(m-1\right)n-1\right]}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}{x}_{t}^{m-1},\end{array}$
(11)
so we can get
$\begin{array}{r}\frac{{x}_{t}^{m-1}}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}\ge \frac{l}{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l\left(m-1\right)n},\\ \frac{{x}_{s}^{m-1}}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}\le \frac{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l}{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l\left(m-1\right)n},\end{array}$
then
$\frac{{x}_{s}}{{x}_{t}}\le {\left\{\frac{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l}{l}\right\}}^{\frac{1}{m-1}}.$
On the other hand,
$\begin{array}{rl}L\sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}-\rho \left(\mathcal{A}\right){x}_{t}^{m-1}& =L\sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}-\sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{a}_{t{i}_{2}\cdots {i}_{m}}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}\\ \ge {x}_{t}^{m-1}\left[L\left(m-1\right)n-{R}_{max}\left(\mathcal{A}\right)\right]\end{array}$
(12)
and
$\begin{array}{rl}\frac{{x}_{t}^{m-1}}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}& =1-\frac{\underset{\phantom{\mathrm{except}\phantom{\rule{0.25em}{0ex}}====}\mathrm{except}\phantom{\rule{0.25em}{0ex}}{i}_{2}=\cdots ={i}_{m}=t}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}\\ \le 1-\frac{\left[\left(m-1\right)n-1\right]}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}{x}_{s}^{m-1},\end{array}$
(13)
so we can get
$\begin{array}{r}\frac{{x}_{t}^{m-1}}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}\le \frac{L}{\rho \left(\mathcal{A}\right)-{R}_{max}\left(\mathcal{A}\right)+L\left(m-1\right)n},\\ \frac{{x}_{s}^{m-1}}{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}\ge \frac{\rho \left(\mathcal{A}\right)-{R}_{max}\left(\mathcal{A}\right)+\left[\left(m-1\right)n-1\right]L}{\rho \left(\mathcal{A}\right)-{R}_{max}\left(\mathcal{A}\right)+L\left(m-1\right)n}\cdot \frac{1}{\left(m-1\right)n-1},\end{array}$
then
$\frac{{x}_{s}}{{x}_{t}}\ge {\left\{\frac{\rho \left(\mathcal{A}\right)-{R}_{max}\left(\mathcal{A}\right)+\left[\left(m-1\right)n-1\right]L}{L}\cdot \frac{1}{\left(m-1\right)n-1}\right\}}^{\frac{1}{m-1}}.$

This completes the proof. □

In the following corollary, we get some bounds for ${x}_{t}/{\sum }_{j=1}^{n}{x}_{j}$ and ${x}_{s}/{\sum }_{j=1}^{n}{x}_{j}$ in terms of the entries of $\mathcal{A}$ and the spectral radius $\rho \left(\mathcal{A}\right)$.

Corollary 2.4 If $\mathcal{A}$ is positive with Perron vector x. Then
$\frac{{x}_{t}}{{\sum }_{{i}_{k}=1}^{n}{x}_{{i}_{k}}}\ge \frac{l}{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+nl},$
(14)
$\frac{{x}_{s}}{{\sum }_{{i}_{k}=1}^{n}{x}_{{i}_{k}}}\le \frac{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l}{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+nl},$
(15)
$\frac{{x}_{s}}{{x}_{t}}\le \frac{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l}{l}.$
(16)
Proof From
$\rho \left(\mathcal{A}\right){x}_{t}^{m-1}=\sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{a}_{t{i}_{2}\cdots {i}_{m}}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}\ge \left(\sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{a}_{t{i}_{2}\cdots {i}_{m}}{x}_{{i}_{k}}\right){x}_{t}^{m-2},$
we have
$\begin{array}{rl}\rho \left(\mathcal{A}\right){x}_{t}-l\sum _{{i}_{k}=1}^{n}{x}_{{i}_{k}}& \ge \sum _{{i}_{2},\dots ,{i}_{m}=1}^{n}{a}_{t{i}_{2}\cdots {i}_{m}}{x}_{{i}_{k}}-l\sum _{{i}_{k}=1}^{n}{x}_{{i}_{k}}\\ \ge {x}_{t}\left[{R}_{min}\left(\mathcal{A}\right)-nl\right].\end{array}$
(17)

Similar to the proof of Theorem 2.3, we can get the results. □

Remark Similar to the proof of Theorem 2.3, we can get the lower bound of $\frac{{x}_{s}}{{x}_{t}}$.

## 3 Application to the perturbation bound

Suppose $\mathcal{A}\ge 0$ and is another tensor satisfying $\mathcal{A}\le \mathcal{B}$. In this section we are interested in the bound of $\rho \left(\mathcal{B}\right)-\rho \left(\mathcal{A}\right)$.

Lemma 3.1 

$0\le \mathcal{A}\le \mathcal{B}$, then $\rho \left(\mathcal{A}\right)\le \rho \left(\mathcal{B}\right)$.

We let $\mathbb{P}=\left\{\left({y}_{1},{y}_{2},\dots ,{y}_{n}\right)|{y}_{i}\ge 0\right\}$ be the positive cone, and let the interior of be denoted by $int\left(\mathbb{P}\right)=\left\{\left({y}_{1},{y}_{2},\dots ,{y}_{n}\right)|{y}_{i}>0\right\}$.

Theorem 3.2 Let $\mathcal{A}$ be weakly irreducible. For a nonzero $x\in \mathbb{P}$, we define
$\mu \left(x\right)=\underset{{x}_{i}>0}{min}\frac{{\left(\mathcal{A}{x}^{m-1}\right)}_{i}}{{x}_{i}^{m-1}},\phantom{\rule{2em}{0ex}}\nu \left(x\right)=\underset{{x}_{i}>0}{max}\frac{{\left(\mathcal{A}{x}^{m-1}\right)}_{i}}{{x}_{i}^{m-1}}.$
Then
$\mu \left(x\right)\le \rho \left(\mathcal{A}\right)\le \nu \left(x\right).$
(18)
Proof Since
$\begin{array}{r}\underset{{x}_{i}>0}{min}\frac{{\left(\mathcal{A}{x}^{m-1}\right)}_{i}}{{x}_{i}^{m-1}}\le \frac{{\left(\mathcal{A}{x}^{m-1}\right)}_{i}}{{x}_{i}^{m-1}}\le \underset{{x}_{i}>0}{max}\frac{{\left(\mathcal{A}{x}^{m-1}\right)}_{i}}{{x}_{i}^{m-1}},\phantom{\rule{1em}{0ex}}{x}_{i}\ne 0,\\ \mu \left(x\right){x}^{m-1}\le \mathcal{A}{x}^{m-1}\le \nu \left(x\right){x}^{m-1}.\end{array}$
By the result of Lemma 5.4 in , we have
$\mu \left(x\right)\le \rho \left(\mathcal{A}\right)\le \nu \left(x\right).$

□

Theorem 3.3 Let $\mathcal{A}$, be irreducible and positive such that $\omega ={min}_{{i}_{1},\dots ,{i}_{m}}\left({b}_{{i}_{1},\dots ,{i}_{m}}-{a}_{{i}_{1},\dots ,{i}_{m}}\right)\ge 0$ and $\mu ={max}_{{i}_{1},\dots ,{i}_{m}}\left({b}_{{i}_{1},\dots ,{i}_{m}}-{a}_{{i}_{1},\dots ,{i}_{m}}\right)$. Then
$\frac{\omega \left[\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l\left(m-1\right)n\right]}{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l}\le \rho \left(\mathcal{B}\right)-\rho \left(\mathcal{A}\right)\le \frac{\mu \left[\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l\left(m-1\right)n\right]}{l}.$
Proof Let x be the Perron vector of $\mathcal{A}$. Define i as follows:
$\frac{{\left(\mathcal{B}{x}^{m-1}\right)}_{i}}{{x}_{i}^{m-1}}=\underset{k}{max}\frac{{\left(\mathcal{B}{x}^{m-1}\right)}_{k}}{{x}_{k}^{m-1}}.$
Then, by Theorem 3.2, we have
$\underset{k}{max}\frac{{\left(\mathcal{B}{x}^{m-1}\right)}_{k}}{{x}_{k}^{m-1}}\ge \rho \left(\mathcal{B}\right).$
From the simple equality
$\left(\mathcal{B}-\mathcal{A}\right){x}^{m-1}+\rho \left(\mathcal{A}\right){x}^{\left[m-1\right]}=\mathcal{B}{x}^{m-1},$
and by considering the i th coordinate,
$\frac{{\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}\left({b}_{i{i}_{2}\cdots {i}_{m}}-{a}_{i{i}_{2}\cdots {i}_{m}}\right){x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}{{x}_{i}^{m-1}}+\rho \left(\mathcal{A}\right)=\frac{{\left(\mathcal{B}{x}^{m-1}\right)}_{i}}{{x}_{i}^{m-1}}\ge \rho \left(\mathcal{B}\right).$
Since $\mu ={max}_{{i}_{1},\dots ,{i}_{m}}\left({b}_{{i}_{1},\dots ,{i}_{m}}-{a}_{{i}_{1},\dots ,{i}_{m}}\right)$ and ${x}_{t}={min}_{1\le i\le n}{x}_{i}$,
$\rho \left(\mathcal{B}\right)-\rho \left(\mathcal{A}\right)\le \frac{\mu {\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}{{x}_{t}^{m-1}}\le \frac{\mu \left[\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l\left(m-1\right)n\right]}{l}.$
Similarly, we can get
$\rho \left(\mathcal{B}\right)-\rho \left(\mathcal{A}\right)\ge \frac{\omega {\sum }_{{i}_{2},\dots ,{i}_{m}=1}^{n}{x}_{{i}_{2}}\cdots {x}_{{i}_{m}}}{{x}_{s}^{m-1}}\ge \frac{\omega \left[\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l\left(m-1\right)n\right]}{\rho \left(\mathcal{A}\right)-{R}_{min}\left(\mathcal{A}\right)+l}.$

□

Remark If we let x be the Perron vector of , then by a similar method, we can get the following bound:
$\frac{\omega \left[\rho \left(\mathcal{B}\right)-{R}_{min}\left(\mathcal{B}\right)+l\left(m-1\right)n\right]}{\rho \left(\mathcal{B}\right)-{R}_{min}\left(\mathcal{B}\right)+l}\le \rho \left(\mathcal{B}\right)-\rho \left(\mathcal{A}\right)\le \frac{\mu \left[\rho \left(\mathcal{B}\right)-{R}_{min}\left(\mathcal{B}\right)+l\left(m-1\right)n\right]}{l}.$

## 4 Conclusion

We have obtained a new and sharper bound of ${x}_{s}/{x}_{t}$, lower bounds for ${x}_{t}^{m-1}/\sum {x}_{{i}_{2}}\cdots {x}_{{i}_{m}}$, and upper bounds for ${x}_{s}^{m-1}/\sum {x}_{{i}_{2}}\cdots {x}_{{i}_{m}}$, where $x=\left({x}_{i}\right)>0$ is the Perron vector of a positive tensor $\mathcal{A}$ with ${x}_{s}={max}_{1\le i\le n}{x}_{i}$ and ${x}_{t}={min}_{1\le i\le n}{x}_{i}$. In addition we have given bounds of $\rho \left(\mathcal{B}\right)-\rho \left(\mathcal{A}\right)$ when $0<\mathcal{A}\le \mathcal{B}$ by these bounds.

## Declarations

### Acknowledgements

This research is supported by NSFC (61170311), Chinese Universities Specialized Research Fund for the Doctoral Program (20110185110020), Sichuan Province Sci. & Tech. Research Project (12ZC1802). The first author is supported by the Fundamental Research Funds for Central Universities.

## Authors’ Affiliations

(1)
School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan, 611731, P.R. China

## References 