Power Function Method for Finding the Spectral Radius of Weakly Irreducible Nonnegative Tensors

Abstract

Since the eigenvalue problem of real supersymmetric tensors was proposed, there have been many results regarding the numerical algorithms for computing the spectral radius of nonnegative tensors, among which the NQZ method is the most studied. However, the NQZ method is only suitable for calculating the spectral radius of a special weakly primitive tensor, or a weakly irreducible primitive tensor that satisfies certain conditions. In this paper, by means of diagonal similarrity transformation of tensors, we construct a numerical algorithm for computing the spectral radius of nonnegative tensors with the aid of power functions. This algorithm is suitable for the calculation of the spectral radius of all weakly irreducible nonnegative tensors. Furthermore, it is efficient and can be widely applied.

1. Introduction

Let $\mathbb{R}$ be the real field. Consider an m-th order n dimensional tensor $\mathcal{A}$ which consists of $n^m$ entries in $\mathbb{R}$ as follows.

$$\mathcal{A}={\left(a_{i_1{i}_2\cdots i_m}\right)}_{i_1,i_2,\dots ,i_m=1}^n,\mathrm{where}{a}_{i_1{i}_2\cdots i_m}\in \mathbb{R}.$$

A tensor $\mathcal{A}$ is called nonnegative if $a_{i_1{i}_2\cdots i_m}\ge 0,$$i_j=1,2,\cdots ,n,j=1,2,\cdots ,m.$ Denote the set of all m-th order n-dimensional (nonnegative) tensors as ${\mathbb{R}}^{[m,n]}$ (${\mathbb{R}}_{+}^{[m,n]}$), and the set of all n-dimensional nonnegative (positive) vectors as ${\mathbb{R}}_{+}^n({\mathbb{R}}_{++}^n$).

If there exists a complex number $\lambda$ and a nonzero complex vector $\mathbf{x}={(x_1,x_2,\dots ,x_n)}^{\mathrm{T}}$ such that

$$\mathcal{A}{\mathbf{x}}^{m-1}=\lambda {\mathbf{x}}^{[m-1]},$$

(1)

then $\lambda$ is called an eigenvalue of $\mathcal{A}$ and $\mathbf{x}$ is termed as an eigenvetor of $\mathcal{A}$ associated with $\lambda ,$$\mathcal{A}{\mathbf{x}}^{m-1}$ and ${\mathbf{x}}^{[m-1]}$ are vectors whose i-th entries are

$${\left(\mathcal{A}{\mathbf{x}}^{m-1}\right)}_i=\sum _{i_2,\dots ,i_m=1}^n{a}_{i{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_m},i\in \langle n\rangle =\{1,2,\dots ,n\}$$

and ${\left({\mathbf{x}}^{[m-1]}\right)}_i=x_i^{m-1}$, respectively, (see [1]).

This definition was introduced by Qi [2] for cases where m is even and $\mathcal{A}$ is symmetric. Independently, this definition was also given by Lim [3], where $\mathbf{x}$ and $\lambda$ were restricted to be a real vector and a real number, respectively. Similar to nonnegative matrix theory, the spectral radius of a tensor $\mathcal{A}$ is defined as $\rho \left(\mathcal{A}\right)=\sup \left\{\right|$$\lambda$$|:\lambda \in \mathrm{spec}(\mathcal{A}\left)\right\},$ where $\mathrm{spec}\left(\mathcal{A}\right)$ is the set of eigenvalues of a tensor $\mathcal{A}$(see [4]).

In recent years, the eigenvalue problems of nonnegative tensors have arisen in a wide range of practical applications such as algebraic geometry [5], spectral hypergraph theory [6,7], higher-order Markov chain [8,9] and so on. In particular, the spectral theory of tensors remains an important research topic in this field.

A nonnegative tensor is an important generalization of a nonnegative matrix, and there have been many results regarding the upper- and lower-bound estimations and numerical calculation of its spectral radius [10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26]. Ng et al. [11] proposed the NQZ algorithm (see Algorithm 1) and proved that the algorithm is applicable to the calculation of spectral radius of primitive tensors.

Algorithm 1:NQZ algorithm [11].

Step 0. Choose ${\mathbf{x}}^{\left(0\right)}>0,{\mathbf{x}}^{\left(0\right)}\in {\mathbb{R}}^n.$ Let ${\mathbf{y}}^{\left(0\right)}=\mathcal{A}{\left({\mathbf{x}}^{\left(0\right)}\right)}^{m-1}$ and set $k:=0$. Step 1. Compute $${\mathbf{x}}^{(k+1)}=\frac{{\left({\mathbf{y}}^{\left(k\right)}\right)}^{\frac{1}{m-1}}}{\parallel {\left({\mathbf{y}}^{\left(k\right)}\right)}^{\frac{1}{m-1}}\parallel },$$ $${\mathbf{y}}^{(k+1)}=\mathcal{A}{\left({\mathbf{x}}^{(k+1)}\right)}^{m-1},$$ $${\underline{\lambda }}_{k+1}=\underset{x_i^{(k+1)}>0}{min}\frac{{\left({\mathbf{y}}^{(k+1)}\right)}_i}{{\left(x_i^{(k+1)}\right)}^{m-1}},$$ $${\overline{\lambda }}_{k+1}=\underset{x_i^{(k+1)}>0}{max}\frac{{\left({\mathbf{y}}^{(k+1)}\right)}_i}{{\left(x_i^{(k+1)}\right)}^{m-1}}.$$ Step 2. If ${\overline{\lambda }}_{k+1}={\underline{\lambda }}_{k+1}$, stop. Otherwise, replace k by $k+1$ and go to Step 1.

   Chang et al. [14] proved the convergence of the NQZ method for primitive tensors. Zhang and Qi  [21] gave the linear convergence rate of the NQZ method for essentially positive tensors. Hu et al. [16] provided the R-linear convergence rate of the NQZ method for weakly primitive tensors, and proposed an algorithm for finding the spectral radius of weakly irreducible nonnegative tensors. In order to hasten the convergence of the NQZ method, Yang et al. [19] proposed a method with parameters, and proved that the algorithm has an explicit linear convergence rate for indirectly positive tensors, and the algorithm coincides with the NQZ method assuming the parameters are properly selected.

In order to render the NQZ algorithm applicable to the calculation of spectral radius of more types of nonnegative tensors, many scholars have performed an abundance of research. For example, in the literature [23], the authors first transform the general nonnegative tensor into the weakly irreducible form of the tensor, and then use the NQZ method to calculate the spectral radius of the weakly irreducible block successively. Similar to the literature [23], the authors in [22] transform a general symmetric nonnegative tensor into a weakly irreducible form of the tensor and calculate its spectral radius. In [24], an algorithm for calculating the spectral radius of a class of irreducible nonnegative tensors and generalized weakly irreducible nonnegative tensors (Definition 4.2 of [24]) is proposed. In [25], an inverse iterative method with local quadratic convergence for calculating the spectral radius of weakly irreducible nonnegative tensors is given, in which an LU decomposition of the matrix is needed in each iteration.

The above methods, except for the research work [25], all stem from the ideas of the NQZ method. Different from the NQZ method, we will apply the diagonal similarity transformation of tensors to construct an algorithm for computing the spectral radius of a weakly irreducible tensor with the aid of a power function. We also prove that this algorithm is convergent for all weakly irreducible nonnegative tensors.

2. Preliminaries

In this section, we mainly introduce some related concepts and important properties of tensors and matrices.

Definition 1

([10]). An m-th order n-dimensional tensor $\mathcal{A}$ is called reducible if there exists a nonempty proper index subset $I\subset \langle n\rangle$ such that

$$a_{i_1{i}_2\cdots i_m}=0,\forall i_1\in I,\forall i_2,\dots ,i_m\notin I.$$

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

Definition 2

([19]). Let $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}_{+}^{[m,n]}$.

(1) We call a nonnegative matrix $G\left(\mathcal{A}\right)$ the representation associated to nonnegative tensor $\mathcal{A}$, if the $(i,j)$-th element of $G\left(\mathcal{A}\right)$ is defined to be the summation of $a_{\{i{i}_2\cdots i_m\}}$ with indices $\{i_2\cdots i_m\}\ni j.$

(2) We call $\mathcal{A}$ weakly reducible if its representation $G\left(\mathcal{A}\right)$ is a reducible matrix, and weakly primitive if $G\left(\mathcal{A}\right)$ is a primitive matrix. If $\mathcal{A}$ is not weakly reducible, then it is called weakly irreducible.

Example 1.

Let $\mathcal{A}=\left(a_{i_1{i}_2{i}_3}\right)\in {\mathbb{R}}_{+}^{[3,3]},$ where $a_{123}=a_{221}=a_{223}=a_{312}=a_{332}=1,$ and zero elsewhere, then $G\left(\mathcal{A}\right)=\left(\begin{array}{ccc}0& 1& 1\\ 1& 2& 1\\ 1& 2& 1\end{array}\right).$ Clearly, $G\left(\mathcal{A}\right)$ is irreducible and primitive, then tensor $\mathcal{A}$ is weakly irreducible and weakly primitive. However, $a_{2i_2{i}_3}=0,$$i_2,i_3\in \{1,3\},$ therefore, $\mathcal{A}$ is reducible.

Theorem 1

([12]). Let $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}_{+}^{[m,n]}$, $\rho \left(\mathcal{A}\right)$ is the spectral radius of $\mathcal{A}$, then

$$\underset{i\in \langle n\rangle }{min}\sum _{i_2,\cdots ,i_m=1}^n{a}_{i{i}_2\cdots i_m}\le \rho \left(\mathcal{A}\right)\le \underset{i\in \langle n\rangle }{max}\sum _{i_2,\cdots ,i_m=1}^n{a}_{i{i}_2\cdots i_m}.$$

Theorem 2

([16]). For any nonnegative tensor $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}_{+}^{[m,n]}$, $\rho \left(\mathcal{A}\right)$ is an eigenvalue with a nonnegative eigenvector $\mathbf{x}\in {\mathbb{R}}_{+}^n$ corresponding to it.

Theorem 3

([26]). If $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}_{+}^{[m,n]}$ is weakly irreducible, then there exists a unique positive eigenvector of $\mathcal{A}$ associated with $\rho \left(\mathcal{A}\right)$.

Next, we state a special kind of similarity of tensors: the diagonal similarity. This similarity relation among tensors plays an important role in the study of the spectra of nonnegative irreducible tensors.

Definition 3

([1]). Let $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}^{[m,n]}$, $\mathcal{B}=\left(b_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}^{[m,n]}$ . The tensors $\mathcal{A}$ and $\mathcal{B}$ are said to be diagonal similar, if there exists some invertible diagonal matrix $D=\mathrm{diag}(d_1,d_2,\cdots ,d_n)$ of order n such that $\mathcal{B}=\mathcal{A}{\times }_1{D}^{-(m-1)}{\times }_{2}D{\times }_3\cdots {\times }_{m}D$  , where $b_{i_1{i}_2\cdots i_m}=a_{i_1{i}_2\cdots i_m}{d}_{i_1}^{-(m-1)}{d}_{i_2}\cdots d_{i_m}$.

Theorem 4

([1]). If the two m-th order n-dimensional tensors $\mathcal{A}$ and $\mathcal{B}$ are diagonal similar, then $\mathrm{spec}\left(\mathcal{A}\right)=\mathrm{spec}\left(\mathcal{B}\right)$.

Denote the set of $n\times n$ complex matrices as ${\mathbb{C}}^{n\times n}$. Let $A=\left(a_{ij}\right)\in {\mathbb{C}}^{n\times n}$. Denote the digragh of matrix A as $\Gamma \left(A\right)$(see [4]).

Definition 4

([27]). Let $A=\left(a_{ij}\right)\in {\mathbb{C}}^{n\times n}$. $\Gamma \left(A\right)$ is called strongly connected if for any pair of ordered nodes $(i,j)$ of $\Gamma \left(A\right)$, there is a directed path $i\to i_1\to i_2\to \cdots \to i_s\to j$ to connect them.

Theorem 5

([27]). If $A=\left(a_{ij}\right)\in {\mathbb{C}}^{n\times n},n\ge 2$, then A is irreducible if and only if the digraph $\Gamma \left(A\right)$ of A is strongly connected.

Next, we will describe the algorithm for computing the spectral radius of weakly irreducible nonnegative tensors, which expands the existing results.

3. Algorithm

Let $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}_{+}^{[m,n]}$, $0<\alpha <1.$ Denote ${\mathcal{A}}^{\left(0\right)}=\mathcal{A},{\mathcal{A}}^{\left(0\right)}=\left(a_{i_1{i}_2\cdots i_m}^{\left(0\right)}\right)$, ${\lambda }_i^{\left(0\right)}={\displaystyle \sum _{i_2,\dots ,i_m=1}^n}{a}_{i{i}_2\cdots i_m}^{\left(0\right)},$${\underline{\lambda }}^{\left(0\right)}=\underset{i\in \langle n\rangle }{min}{\lambda }_i^{\left(0\right)},$${\overline{\lambda }}^{\left(0\right)}=\underset{i\in \langle n\rangle }{max}{\lambda }_i^{\left(0\right)}$. By Theorem 1, we have ${\underline{\lambda }}^{\left(0\right)}\le \rho \left(\mathcal{A}\right)=\rho \left({\mathcal{A}}^{\left(0\right)}\right)\le {\overline{\lambda }}^{\left(0\right)}.$ For any ${\mathbf{x}}^{\left(0\right)}={(x_1^{\left(0\right)},x_2^{\left(0\right)},\cdots ,x_n^{\left(0\right)})}^{\mathrm{T}}\in {\mathbb{R}}_{++}^n,$ denote

$$D_0=\mathrm{diag}{\left(x_1^{\left(0\right)},x_2^{\left(0\right)},\cdots ,x_n^{\left(0\right)}\right)}^{\frac{\alpha }{m-1}}\in D_{+}^n,$$

then denote ${\mathcal{A}}^{\left(1\right)}={\mathcal{A}}^{\left(0\right)}{\times }_1{D}_0^{-(m-1)}{\times }_2{D}_0{\times }_3\cdots {\times }_m{D}_0,{\mathcal{A}}^{\left(1\right)}=\left(a_{i_1{i}_2\cdots i_m}^{\left(1\right)}\right)\in {\mathbb{R}}_{+}^{[m,n]},$${\lambda }_i^{\left(1\right)}={\displaystyle \sum _{i_2,\dots ,i_m=1}^n}{a}_{i{i}_2\cdots i_m}^{\left(1\right)},$${\underline{\lambda }}^{\left(1\right)}=\underset{i\in \langle n\rangle }{min}{\lambda }_i^{\left(1\right)},$${\overline{\lambda }}^{\left(1\right)}=\underset{i\in \langle n\rangle }{max}{\lambda }_i^{\left(1\right)}$. Use Theorem 1 and 4 to know that ${\underline{\lambda }}^{\left(1\right)}\le \rho \left(\mathcal{A}\right)=\rho \left({\mathcal{A}}^{\left(1\right)}\right)\le {\overline{\lambda }}^{\left(1\right)}.$ Denote

$$D_1={\left[\frac{\mathrm{diag}\left({\lambda }_1^{\left(1\right)},{\lambda }_2^{\left(1\right)},\cdots ,{\lambda }_n^{\left(1\right)}\right)}{\underset{i\in \langle n\rangle }{max}{\lambda }_i^{\left(1\right)}}\right]}^{\frac{\alpha }{m-1}}\in D_{+}^n,$$

then denote ${\mathcal{A}}^{\left(2\right)}={\mathcal{A}}^{\left(1\right)}{\times }_1{D}_1^{-(m-1)}{\times }_2{D}_1{\times }_3\cdots {\times }_m{D}_1,{\mathcal{A}}^{\left(2\right)}=\left(a_{i_1{i}_2\cdots i_m}^{\left(2\right)}\right)\in {\mathbb{R}}_{+}^{[m,n]},$${\mathbf{x}}^{\left(1\right)}=D_0{D}_1\mathbf{e}\in {\mathbb{R}}_{++}^n,$$\mathbf{e}={(1,1,\cdots ,1)}^{\mathrm{T}}\in {\mathbb{R}}_{++}^n,$ similarly, denote ${\mathcal{A}}^{\left(l\right)}={\mathcal{A}}^{(l-1)}{\times }_1{D}_{l-1}^{-(m-1)}{\times }_2{D}_{l-1}{\times }_3\cdots {\times }_m{D}_{l-1},{\mathcal{A}}^{\left(l\right)}=\left(a_{i_1{i}_2\cdots i_m}^{\left(l\right)}\right)\in {\mathbb{R}}_{+}^{[m,n]},$${\lambda }_i^{\left(l\right)}={\displaystyle \sum _{i_2,\dots ,i_m=1}^n}{a}_{i{i}_2\cdots i_m}^{\left(l\right)},$${\underline{\lambda }}^{\left(l\right)}=\underset{i\in \langle n\rangle }{min}{\lambda }_i^{\left(l\right)},$${\overline{\lambda }}^{\left(l\right)}=\underset{i\in \langle n\rangle }{max}{\lambda }_i^{\left(l\right)}$. By Theorem 1 and 4, we know ${\underline{\lambda }}^{\left(l\right)}\le \rho \left(\mathcal{A}\right)=\rho \left({\mathcal{A}}^{\left(l\right)}\right)\le {\overline{\lambda }}^{\left(l\right)}.$ Denote

$$D_l={\left[\frac{\mathrm{diag}\left({\lambda }_1^{\left(l\right)},{\lambda }_2^{\left(l\right)},\cdots ,{\lambda }_n^{\left(l\right)}\right)}{\underset{i\in \langle n\rangle }{max}{\lambda }_i^{\left(l\right)}}\right]}^{\frac{\alpha }{m-1}}\in D_{+}^n,$$

${\mathbf{x}}^{\left(l\right)}=D_0{D}_1\cdots D_l\mathbf{e}\in {\mathbb{R}}_{++}^n$$(l=0,1,2,\cdots ),$ we obtain a sequence of weakly irreducible nonnegative tensors with the same zero-element pattern $\left\{{\mathcal{A}}^{\left(l\right)}\right\}(l=0,1,2,\cdots )$ and the sequence of positive vectors $\left\{{\mathbf{x}}^{\left(l\right)}\right\}$$(l=0,1,2,\cdots )$.

From the construction process of the above nonnegative tensor sequences, we propose the following algorithm for finding the spectral radius of weakly irreducible nonnegative tensors. We call the algorithm the power function (PF) algorithm (see Algorithm 2).

Algorithm 2: PF algorithm.

Step 0. Given ${\mathbf{x}}^{\left(0\right)}={(x_1^{\left(0\right)},x_2^{\left(0\right)},\cdots ,x_n^{\left(0\right)})}^{\mathrm{T}}\in {\mathbb{R}}_{++}^n,$ $0<\alpha <1,\epsilon >0.$ Compute $${\lambda }_i^{\left(0\right)}=\frac{{\left(\mathcal{A}{\left({\mathbf{x}}^{\left(0\right)}\right)}^{m-1}\right)}_i}{{\left(x_i^{\left(0\right)}\right)}^{m-1}},$$ $${\mathbf{y}}^{\left(1\right)}={({\lambda }_1^{\left(0\right)},{\lambda }_2^{\left(0\right)},\cdots ,{\lambda }_n^{\left(0\right)})}^{\mathrm{T}},$$ $${\mathbf{x}}^{\left(1\right)}=\frac{{\mathbf{y}}^{\left(1\right)}}{\parallel {\mathbf{y}}^{\left(1\right)}{\parallel }_1}=:{(x_1^{\left(1\right)},x_2^{\left(1\right)},\cdots ,x_n^{\left(1\right)})}^{\mathrm{T}}.$$ Set $k=1$. Step 1. Compute $${\lambda }_i^{(k+1)}=\frac{{\left(\mathcal{A}{\left({\mathbf{x}}^{\left(k\right)}\right)}^{m-1}\right)}_i}{{\left(x_i^{\left(k\right)}\right)}^{m-1}},$$ $${\overline{\lambda }}^{(k+1)}=\underset{i\in \langle n\rangle }{max}\left\{{\lambda }_i^{(k+1)}\right\},{\underline{\lambda }}^{(k+1)}=\underset{i\in \langle n\rangle }{min}\left\{{\lambda }_i^{(k+1)}\right\},$$ $$y_i^{(k+1)}=x_i^{\left(k\right)}·{\left({\lambda }_i^{\left(k\right)}\right)}^{\frac{\alpha }{m-1}},i\in \langle n\rangle ,$$ $${\mathbf{x}}^{(k+1)}=\frac{{\mathbf{y}}^{(k+1)}}{\parallel {\mathbf{y}}^{(k+1)}{\parallel }_1}=:{(x_1^{(k+1)},x_2^{(k+1)},\cdots ,x_n^{(k+1)})}^{\mathrm{T}}.$$ Step 2. If ${\overline{\lambda }}^{(k+1)}-{\underline{\lambda }}^{(k+1)}<\epsilon$, then $\rho \left(\mathcal{A}\right)=\frac{{\overline{\lambda }}^{(k+1)}+{\underline{\lambda }}^{(k+1)}}{2}$, stop. Otherwise, set $k:=k+1$, go to Step 1.

Because $\forall {\mathbf{x}}^{\left(0\right)}={(x_1^{\left(0\right)},x_2^{\left(0\right)},\cdots ,x_n^{\left(0\right)})}^{\mathrm{T}}\in {\mathbb{R}}_{++}^n,$

$$D_{{\mathbf{x}}^{\left(0\right)}}=\mathrm{diag}(x_1^{\left(0\right)},x_2^{\left(0\right)},\cdots ,x_n^{\left(0\right)})$$

is a positive diagonal matrix. Let $\mathcal{B}=\mathcal{A}{\times }_1{D}_{{\mathbf{x}}^{\left(0\right)}}^{-(m-1)}{\times }_2{D}_{{\mathbf{x}}^{\left(0\right)}}{\times }_3\cdots {\times }_m{D}_{{\mathbf{x}}^{\left(0\right)}}.$ According to Theorem 7, Spec$\left(\mathcal{A}\right)=$ Spec$\left(\mathcal{B}\right).$ For convenience we can take ${\mathbf{x}}^{\left(0\right)}={({\lambda }_1^{\left(0\right)},{\lambda }_2^{\left(0\right)},\cdots ,{\lambda }_n^{\left(0\right)})}^{\mathrm{T}}\in {\mathbb{R}}_{++}^n$, where

$${\lambda }_i^{\left(0\right)}=\sum _{i_2,\cdots ,i_m=1}^n{a}_{i{i}_2\cdots i_m},i\in \langle n\rangle .$$

4. Convergence of the PF Algorithm

In this section, we prove that PF algorithm is convergent for weakly irreducible nonnegative tensors.

Firstly, we show that the sequence $\left\{{\underline{\lambda }}^{\left(i\right)}\right\}(i=0,1,2,\cdots )$ is monotonically increasing and sequence $\left\{{\overline{\lambda }}^{\left(i\right)}\right\}(i=0,1,2,\cdots )$ is monotonically decreasing.

Lemma 1.

Let $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}_{+}^{[m,n]}$, $\rho \left(\mathcal{A}\right)$ be the spectral radius of $\mathcal{A}$. For tensor sequence ${\mathcal{A}}^{\left(l\right)}(l=0,1,2,\cdots ),$ we have

$${\underline{\lambda }}^{\left(0\right)}\le {\underline{\lambda }}^{\left(1\right)}\le \cdots \le {\underline{\lambda }}^{\left(l\right)}\le \rho \left(\mathcal{A}\right)\le {\overline{\lambda }}^{\left(l\right)}\le \cdots \le {\overline{\lambda }}^{\left(1\right)}\le {\overline{\lambda }}^{\left(0\right)}.$$

Proof. 

Since ${\mathcal{A}}^{(l+1)}={\mathcal{A}}^{\left(l\right)}{\times }_1{D}_l^{-(m-1)}{\times }_2{D}_l{\times }_3\cdots {\times }_m{D}_l=\left(a_{i_1{i}_2\cdots i_m}^{(l+1)}\right)\in {\mathbb{R}}_{+}^{[m,n]}(l=0,1,2,\cdots ),$$\forall i\in \langle n\rangle$, we have

$$\begin{array}{cc}\hfill {\lambda }_i^{(l+1)}& =\sum _{i_2,\cdots ,i_m=1}^n{a}_{i{i}_2\cdots i_m}^{(l+1)}=\sum _{i_2,\cdots ,i_m=1}^n{a}_{i{i}_2\cdots i_m}^{\left(l\right)}\frac{{\displaystyle \prod _{j=2}^m}{\left({\lambda }_{i_j}^{\left(l\right)}\right)}^{\frac{\alpha }{m-1}}}{{\left({\lambda }_i^{\left(l\right)}\right)}^{\alpha }}\hfill \\ \hfill & =\sum _{i_2,\cdots ,i_m=1}^n\frac{a_{i{i}_2\cdots i_m}^{\left(l\right)}}{{\lambda }_i^{\left(l\right)}}·\prod _{j=2}^m{\left({\lambda }_{i_j}^{\left(l\right)}\right)}^{\frac{\alpha }{m-1}}{\left({\lambda }_i^{\left(l\right)}\right)}^{1-\alpha },\hfill \end{array}$$

where ${\displaystyle \sum _{i_2,\cdots ,i_m=1}^n}\frac{a_{i{i}_2\cdots i_m}^{\left(l\right)}}{{\lambda }_i^{\left(l\right)}}=1.$ Therefore

$${\overline{\lambda }}^{(l+1)}=\underset{i\in \langle n\rangle }{max}{\lambda }_i^{(l+1)}\le \underset{i\in \langle n\rangle }{max}\left(\prod _{j=2}^m{\left({\lambda }_{i_j}^{\left(l\right)}\right)}^{\frac{\alpha }{m-1}}{\left({\lambda }_i^{\left(l\right)}\right)}^{1-\alpha }\right)\le {\overline{\lambda }}^{\left(l\right)},$$

i.e.,

$${\overline{\lambda }}^{(l+1)}\le {\overline{\lambda }}^{\left(l\right)}(l=0,1,2,\cdots ).$$

Similarly,

$${\underline{\lambda }}^{\left(l\right)}\le {\underline{\lambda }}^{(l+1)}(l=0,1,2,\cdots ).$$

By Theorem 1 and Lemma 4, we have

$${\underline{\lambda }}^{\left(l\right)}\le \rho \left({\mathcal{A}}^{(l+1)}\right)=\rho \left(\mathcal{A}\right)\le {\overline{\lambda }}^{\left(l\right)}(l=0,1,2,\cdots ).$$

Hence, we obtain

$${\underline{\lambda }}^{\left(0\right)}\le \cdots \le {\underline{\lambda }}^{\left(l\right)}\le \rho \left(\mathcal{A}\right)\le {\overline{\lambda }}^{\left(l\right)}\le \cdots \le {\overline{\lambda }}^{\left(0\right)}.$$

□

Before proving the convergence of PF algorithm, we first prove that the entries of the tensor sequence generated by PF algorithm have a lower bound.

Lemma 2.

Let $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}_{+}^{[m,n]}$ be weakly irreducible. For tensor sequence ${\mathcal{A}}^{\left(l\right)}=\left(a_{i_1{i}_2\cdots i_m}^{\left(l\right)}\right)(l=0,1,2,\cdots ),$ if $a_{i_1{i}_2\cdots i_m}>0,$ then $a_{i_1{i}_2\cdots i_m}^{\left(l\right)}>0(l=0,1,2,\cdots ).$ Furthermore, for any $a_{i_1{i}_2\cdots i_m}>0,$$a_{i_1{i}_2\cdots i_m}^{\left(l\right)}\ge \underline{a}{\left(\frac{\underline{a}}{\overline{\lambda }}\right)}^{\frac{{(m-1)}^{n-1}-1}{m-2}}$ holds true, where

$$\underline{a}=min\left\{a_{i_1{i}_2\cdots i_m}\right|a_{i_1{i}_2\cdots i_m}>0,i_1,i_2,\cdots ,i_m\in \langle n\rangle \},\overline{\lambda }={\overline{\lambda }}^{\left(0\right)}.$$

Proof. 

By PF algorithm, we know ${\mathbf{x}}^{\left(l\right)}={(x_1^{\left(l\right)},x_2^{\left(l\right)},\cdots ,x_n^{\left(l\right)})}^{\mathrm{T}}\in {\mathbb{R}}_{++}^n.$ Without loss of generality, assume

$$x_{t_n}^{\left(l\right)}\ge x_{t_{n-1}}^{\left(l\right)}\ge \cdots \ge x_{t_2}^{\left(l\right)}\ge x_{t_1}^{\left(l\right)}>0.$$

(2)

Because tensor $\mathcal{A}$ is weakly irreducible, there exists a directed path $\gamma$ from $t_1$ to $t_n$ by Definition 2, and $\gamma$ is on the digraph $\Gamma \left(G\right(\mathcal{A}\left)\right)$ of matrix $G\left(\mathcal{A}\right)$, where $\gamma :t_1\to {\tilde{i}}_2\to {\tilde{i}}_3\to \cdots \to {\tilde{i}}_r\to t_n,2\le r\le n-1,$ such that $a_{t_1{i}_2\cdots {\tilde{i}}_2\cdots i_m}\ne 0,a_{{\tilde{i}}_2{i}_2\cdots {\tilde{i}}_3\cdots i_m}\ne 0,\cdots ,a_{{\tilde{i}}_r{i}_2\cdots t_n\cdots i_m}\ne 0.$ Therefore, combining Lemma 1, (2) and $l\in {\mathbb{Z}}^{+}\cup \left\{0\right\},$ we have

$$\begin{array}{cc}\hfill \overline{\lambda }{\left(x_{t_1}^{\left(l\right)}\right)}^{m-1}& \ge {\overline{\lambda }}^{(l+1)}{\left(x_{t_1}^{\left(l\right)}\right)}^{m-1}\ge {\overline{\lambda }}_{t_1}^{(l+1)}{\left(x_{t_1}^{\left(l\right)}\right)}^{m-1}\hfill \\ \hfill & =\sum _{i_2,\cdots ,i_m=1}^n{a}_{t_1{i}_2\cdots i_m}{x}_{i_2}^{\left(l\right)}\cdots x_{i_m}^{\left(l\right)}\hfill \\ \hfill & \ge a_{t_1{i}_2\cdots {\tilde{i}}_2\cdots i_m}{\left(x_{t_1}^{\left(l\right)}\right)}^{m-2}{x}_{{\tilde{i}}_2}^{\left(l\right)}\hfill \\ \hfill & \ge \underline{a}·{\left(x_{t_1}^{\left(l\right)}\right)}^{m-2}{x}_{{\tilde{i}}_2}^{\left(l\right)},\hfill \end{array}$$

i.e.,

$$\overline{\lambda }·x_{t_1}^{\left(l\right)}\ge \underline{a}·x_{{\tilde{i}}_2}^{\left(l\right)}.$$

(3)

Similarly, we get

$$\overline{\lambda }{\left(x_{{\tilde{i}}_2}^{\left(l\right)}\right)}^{m-1}\ge {\overline{\lambda }}^{(l+1)}{\left(x_{{\tilde{i}}_2}^{\left(l\right)}\right)}^{m-1}\ge a_{{\tilde{i}}_2{i}_2\cdots {\tilde{i}}_3\cdots i_m}{\left(x_{t_1}^{\left(l\right)}\right)}^{m-2}{x}_{{\tilde{i}}_3}^{\left(l\right)}\ge \underline{a}·{\left(x_{t_1}^{\left(l\right)}\right)}^{m-2}{x}_{{\tilde{i}}_3}^{\left(l\right)},$$

(4)

$$\begin{gathered} \overline{\lambda }{\left(x_{{\tilde{i}}_3}^{\left(l\right)}\right)}^{m-1}\ge {\overline{\lambda }}^{(l+1)}{\left(x_{{\tilde{i}}_3}^{\left(l\right)}\right)}^{m-1}\ge a_{{\tilde{i}}_3{i}_2\cdots {\tilde{i}}_4\cdots i_m}{\left(x_{t_1}^{\left(l\right)}\right)}^{m-2}{x}_{{\tilde{i}}_4}^{\left(l\right)}\ge \underline{a}·{\left(x_{t_1}^{\left(l\right)}\right)}^{m-2}{x}_{{\tilde{i}}_4}^{\left(l\right)}, \\ ⋮ \end{gathered}$$

(5)

$$\overline{\lambda }{\left(x_{{\tilde{i}}_r}^{\left(l\right)}\right)}^{m-1}\ge {\overline{\lambda }}^{(l+1)}{\left(x_{{\tilde{i}}_r}^{\left(l\right)}\right)}^{m-1}\ge a_{{\tilde{i}}_r{i}_2\cdots t_n\cdots i_m}{\left(x_{t_1}^{\left(l\right)}\right)}^{m-2}{x}_{t_n}^{\left(l\right)}\ge \underline{a}·{\left(x_{t_1}^{\left(l\right)}\right)}^{m-2}{x}_{t_n}^{\left(l\right)}.$$

(6)

So by (3) and (4), we obtain

$${\left(\overline{\lambda }·x_{t_1}^{\left(l\right)}\right)}^{m-1}\overline{\lambda }{\left(x_{{\tilde{i}}_2}^{\left(l\right)}\right)}^{m-1}\ge {\left(\underline{a}·x_{{\tilde{i}}_2}^{\left(l\right)}\right)}^{m-1}\underline{a}·{\left(x_{t_1}^{\left(l\right)}\right)}^{m-2}{x}_{{\tilde{i}}_3}^{\left(l\right)},$$

that is,

$${\overline{\lambda }}^{(m-1)+1}·x_{t_1}^{\left(l\right)}\ge {\underline{a}}^{(m-1)+1}·x_{{\tilde{i}}_3}^{\left(l\right)}.$$

(7)

By (7) and (5), we can get

$${\left({\overline{\lambda }}^{(m-1)+1}·x_{t_1}^{\left(l\right)}\right)}^{m-1}\overline{\lambda }{\left(x_{{\tilde{i}}_3}^{\left(l\right)}\right)}^{m-1}\ge {\left({\underline{a}}^{(m-1)+1}·x_{{\tilde{i}}_3}^{\left(l\right)}\right)}^{m-1}\underline{a}·{\left(x_{t_1}^{\left(l\right)}\right)}^{m-2}{x}_{{\tilde{i}}_4}^{\left(l\right)},$$

that is,

$${\overline{\lambda }}^{{(m-1)}^2+(m-1)+1}·x_{t_1}^{\left(l\right)}\ge {\underline{a}}^{{(m-1)}^2+(m-1)+1}·x_{{\tilde{i}}_4}^{\left(l\right)}.$$

Continue to proceed in turn, we obtain

$${\overline{\lambda }}^{{(m-1)}^{r-1}+{(m-1)}^{r-2}+\cdots +(m-1)+1}·x_{t_1}^{\left(l\right)}\ge {\underline{a}}^{{(m-1)}^{r-1}+{(m-1)}^{r-2}+\cdots +(m-1)+1}·x_{t_n}^{\left(l\right)},$$

i.e.,

$${\overline{\lambda }}^{\frac{{(m-1)}^r-1}{m-2}}·x_{t_1}^{\left(l\right)}\ge {\underline{a}}^{\frac{{(m-1)}^r-1}{m-2}}·x_{t_n}^{\left(l\right)}.$$

Thus

$$1\ge \frac{x_{t_1}^{\left(l\right)}}{x_{t_n}^{\left(l\right)}}\ge {\left(\frac{\underline{a}}{\overline{\lambda }}\right)}^{\frac{{(m-1)}^r-1}{m-2}}\ge {\left(\frac{\underline{a}}{\overline{\lambda }}\right)}^{\frac{{(m-1)}^{n-1}-1}{m-2}}=:a_0>0.$$

Therefore, for any $a_{i_1{i}_2\cdots i_m}\ne 0(>0),$ we have

$$a_{i_1{i}_2\cdots i_m}^{\left(l\right)}=a_{i_1{i}_2\cdots i_m}·\frac{{\displaystyle \prod _{j=2}^m}{x}_{i_j}^{\left(l\right)}}{{\left(x_{i_1}^{\left(l\right)}\right)}^{m-1}}\ge a_{i_1{i}_2\cdots i_m}·{\left(\frac{x_{t_1}^{\left(l\right)}}{x_{t_n}^{\left(l\right)}}\right)}^{m-1}\ge \underline{a}·{\left(a_0\right)}^{m-1}=:a>0.$$

From Lemma 1, we have

$$0<\frac{a}{\overline{\lambda }}\le 1.$$

□

Now we show our result on the convergence of the PF algorithm.

Theorem 6.

If $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}_{+}^{[m,n]}$ is weakly irreducible, $\rho \left(\mathcal{A}\right)$ is the spectral radius of $\mathcal{A}$, then $\underset{l\to \infty }{lim}{\overline{\lambda }}^{\left(l\right)}=\underset{l\to \infty }{lim}{\underline{\lambda }}^{\left(l\right)}=\rho \left(\mathcal{A}\right)$ .

Proof. 

$\forall {\tilde{i}}_0\in \{t_s\in \langle n\rangle :{\lambda }_{t_s}^{\left(0\right)}=\underset{i\in \langle n\rangle }{min}{\lambda }_{i{i}_2\cdots i_m}^{\left(0\right)}\},$$\alpha \in (0,1),$ we have

$$\begin{array}{cc}\hfill {\lambda }_{{\tilde{i}}_0}^{\left(1\right)}& =\sum _{i_2,\cdots ,i_m=1}^n{a}_{{\tilde{i}}_0{i}_2\cdots i_m}^{\left(0\right)}\frac{{\displaystyle \prod _{j=2}^m}{\left({\lambda }_{i_j}^{\left(0\right)}\right)}^{\frac{\alpha }{m-1}}}{{\left({\lambda }_{{\tilde{i}}_0}^{\left(0\right)}\right)}^{\alpha }}\hfill \\ \hfill & =\sum _{i_2,\cdots ,i_m=1}^n{a}_{{\tilde{i}}_0{i}_2\cdots i_m}^{\left(0\right)}\frac{{\overline{\lambda }}^{\left(0\right)}-{\overline{\lambda }}^{\left(0\right)}+{\left({\lambda }_{{\tilde{i}}_0}^{\left(0\right)}\right)}^{1-\alpha }{\displaystyle \prod _{j=2}^m}{\left({\lambda }_{i_j}^{\left(0\right)}\right)}^{\frac{\alpha }{m-1}}}{{\lambda }_{{\tilde{i}}_0}^{\left(1\right)}}\hfill \\ \hfill & \le {\overline{\lambda }}^{\left(0\right)}-\sum _{i_2,\cdots ,i_m=1}^n{a}_{{\tilde{i}}_0{i}_2\cdots i_m}^{\left(0\right)}\frac{{\overline{\lambda }}^{\left(0\right)}-{\left({\lambda }_{{\tilde{i}}_0}^{\left(0\right)}\right)}^{1-\alpha }{\left({\overline{\lambda }}^{\left(0\right)}\right)}^{\alpha }}{{\lambda }_{{\tilde{i}}_0}^{\left(0\right)}}\hfill \\ \hfill & \le {\overline{\lambda }}^{\left(0\right)}-\frac{(1-\alpha )a}{{\overline{\lambda }}^{\left(0\right)}}\left({\overline{\lambda }}^{\left(0\right)}-{\underline{\lambda }}^{\left(0\right)}\right)\hfill \\ \hfill & =\overline{\lambda }-\frac{(1-\alpha )a}{\overline{\lambda }}\left(\overline{\lambda }-\underline{\lambda }\right).\hfill \end{array}$$

Similarly to the above formula, we can obtain

$${\lambda }_{{\tilde{i}}_0}^{(t+1)}\le \overline{\lambda }-{\left(\frac{(1-\alpha )a}{\overline{\lambda }}\right)}^t\left(\overline{\lambda }-\underline{\lambda }\right)(t=1,2,\cdots ,n-1).$$

(8)

Since $\mathcal{A}$ is irreducible, for any $i\notin \{t_s\in \langle n\rangle :{\lambda }_{t_s}^{\left(0\right)}=\underset{i\in \langle n\rangle }{min}{\lambda }_{i{i}_2\cdots i_m}^{\left(0\right)}\},$ there exists a directed path $i\to {\tilde{i}}_t\to \cdots \to {\tilde{i}}_2\to {\tilde{i}}_1\to {\tilde{i}}_0\in \{t_s\in \langle n\rangle :{\lambda }_{t_s}^{\left(0\right)}=\underset{i\in \langle n\rangle }{min}{\lambda }_{i{i}_2\cdots i_m}^{\left(0\right)}\},$ such that $a_{{\tilde{i}}_1\cdots {\tilde{i}}_0\cdots i_m}>0,a_{{\tilde{i}}_2\cdots {\tilde{i}}_1\cdots i_m}>0,\cdots ,a_{i\cdots {\tilde{i}}_t\cdots i_m}>0$ . By Lemma 2, for any $l\in {\mathbb{Z}}^{+}\cup \left\{0\right\}$, we have $a_{{\tilde{i}}_1\cdots {\tilde{i}}_0\cdots i_m}^{\left(l\right)}>a,a_{{\tilde{i}}_2\cdots {\tilde{i}}_1\cdots i_m}^{\left(l\right)}>a,\cdots ,a_{i\cdots {\tilde{i}}_t\cdots i_m}^{\left(l\right)}>a.$Hence, we obtain

$$\begin{array}{cc}\hfill {\lambda }_{{\tilde{i}}_1}^{\left(2\right)}& =\sum _{i_2,\cdots ,i_m=1}^n{a}_{{\tilde{i}}_1{i}_2\cdots i_m}^{\left(1\right)}\frac{{\displaystyle \prod _{j=2}^m}{\left({\lambda }_{i_j}^{\left(1\right)}\right)}^{\frac{\alpha }{m-1}}}{{\left({\lambda }_{{\tilde{i}}_1}^{\left(1\right)}\right)}^{\alpha }}\hfill \\ \hfill & \le {\overline{\lambda }}^{\left(0\right)}-a_{{\tilde{i}}_1\cdots {\tilde{i}}_0\cdots i_m}^{\left(1\right)}\frac{{\overline{\lambda }}^{\left(0\right)}-{\left({\lambda }_{{\tilde{i}}_1}^{\left(1\right)}\right)}^{1-\alpha }{\displaystyle \prod _{j=2,i_j\ne {\tilde{i}}_0}^m}{\left({\lambda }_{i_j}^{\left(1\right)}\right)}^{\frac{\alpha }{m-1}}{\left({\lambda }_{{\tilde{i}}_0}^{\left(1\right)}\right)}^{\frac{\alpha }{m-1}}}{{\lambda }_{{\tilde{i}}_1}^{\left(1\right)}}\hfill \\ \hfill & \le {\overline{\lambda }}^{\left(0\right)}-\frac{a}{{\overline{\lambda }}^{\left(0\right)}}\left({\overline{\lambda }}^{\left(0\right)}-{\left({\overline{\lambda }}^{\left(0\right)}\right)}^{1-\alpha }{\left({\overline{\lambda }}^{\left(0\right)}\right)}^{\frac{m-2}{m-1}\alpha }{\left({\lambda }_{{\tilde{i}}_0}^{\left(1\right)}\right)}^{\frac{\alpha }{m-1}}\right)\hfill \\ \hfill & \le {\overline{\lambda }}^{\left(0\right)}-\frac{a}{{\overline{\lambda }}^{\left(0\right)}}{\left({\overline{\lambda }}^{\left(0\right)}\right)}^{1-\frac{\alpha }{m-1}}\left({\left({\overline{\lambda }}^{\left(0\right)}\right)}^{\frac{\alpha }{m-1}}-{\left({\lambda }_{{\tilde{i}}_0}^{\left(1\right)}\right)}^{\frac{\alpha }{m-1}}\right)\hfill \\ \hfill & \le {\overline{\lambda }}^{\left(0\right)}-\frac{a}{{\overline{\lambda }}^{\left(0\right)}}\frac{\alpha }{m-1}\left({\overline{\lambda }}^{\left(0\right)}-{\lambda }_{{\tilde{i}}_0}^{\left(1\right)}\right)\hfill \\ \hfill & \le {\overline{\lambda }}^{\left(0\right)}-\left(\frac{a}{{\overline{\lambda }}^{\left(0\right)}}\right)\frac{\alpha (1-\alpha )}{m-1}\left({\overline{\lambda }}^{\left(0\right)}-{\underline{\lambda }}^{\left(0\right)}\right)\hfill \\ \hfill & =\overline{\lambda }-\left(\frac{a}{\overline{\lambda }}\right)\frac{\alpha (1-\alpha )}{m-1}\left(\overline{\lambda }-\underline{\lambda }\right).\hfill \end{array}$$

$$\begin{gathered} \begin{array}{cc}\hfill {\lambda }_{{\tilde{i}}_2}^{\left(3\right)}& \le {\overline{\lambda }}^{\left(0\right)}-a_{{\tilde{i}}_2\cdots {\tilde{i}}_1\cdots i_m}^{\left(2\right)}\frac{{\overline{\lambda }}^{\left(0\right)}-{\left({\lambda }_{{\tilde{i}}_2}^{\left(2\right)}\right)}^{1-\alpha }{\displaystyle \prod _{j=2,i_j\ne {\tilde{i}}_1}^m}{\left({\lambda }_{i_j}^{\left(2\right)}\right)}^{\frac{\alpha }{m-1}}{\left({\lambda }_{{\tilde{i}}_1}^{\left(2\right)}\right)}^{\frac{\alpha }{m-1}}}{{\lambda }_{{\tilde{i}}_2}^{\left(3\right)}}\hfill \\ \hfill & \le \overline{\lambda }-{\left(\frac{a}{\overline{\lambda }}\right)}^2{\left(\frac{\alpha (1-\alpha )}{m-1}\right)}^2\left(\overline{\lambda }-\underline{\lambda }\right). \\ ⋮\hfill \end{array} \end{gathered}$$

$$\begin{array}{cc}\hfill {\lambda }_{{\tilde{i}}_t}^{(t+1)}& \le {\overline{\lambda }}^{\left(0\right)}-a_{{\tilde{i}}_t{i}_2\cdots {\tilde{i}}_{t-1}\cdots i_m}^{\left(t\right)}\frac{{\overline{\lambda }}^{\left(0\right)}-{\left({\lambda }_{{\tilde{i}}_t}^{\left(t\right)}\right)}^{1-\alpha }{\displaystyle \prod _{j=2,i_j\ne {\tilde{i}}_{t-1}}^m}{\left({\lambda }_{i_j}^{\left(t\right)}\right)}^{\frac{\alpha }{m-1}}{\left({\lambda }_{{\tilde{i}}_{t-1}}^{\left(t\right)}\right)}^{\frac{\alpha }{m-1}}}{{\lambda }_{{\tilde{i}}_t}^{\left(t\right)}}\hfill \\ \hfill & \le \overline{\lambda }-{\left(\frac{a}{\overline{\lambda }}\right)}^t{\left(\frac{\alpha (1-\alpha )}{m-1}\right)}^t\left(\overline{\lambda }-\underline{\lambda }\right).\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\lambda }_i^{(t+2)}& \le {\overline{\lambda }}^{\left(0\right)}-a_{i{i}_2\cdots {\tilde{i}}_t\cdots i_m}^{(t+1)}\frac{{\overline{\lambda }}^{\left(0\right)}-{\left({\lambda }_i^{(t+1)}\right)}^{1-\alpha }{\displaystyle \prod _{j=2,i_j\ne {\tilde{i}}_t}^m}{\left({\lambda }_{i_j}^{(t+1)}\right)}^{\frac{\alpha }{m-1}}{\left({\lambda }_{{\tilde{i}}_t}^{(t+1)}\right)}^{\frac{\alpha }{m-1}}}{{\lambda }_i^{(t+1)}}\hfill \\ \hfill & \le \overline{\lambda }-{\left(\frac{a}{\overline{\lambda }}\right)}^{t+1}{\left(\frac{\alpha (1-\alpha )}{m-1}\right)}^{t+1}\left(\overline{\lambda }-\underline{\lambda }\right).\hfill \end{array}$$

If l is sufficiently large, we have ${\left(\frac{(1-\alpha )a}{\overline{\lambda }}\right)}^l>{\left(\frac{a}{\overline{\lambda }}\right)}^l{\left(\frac{\alpha (1-\alpha )}{m-1}\right)}^l.$ Together with (8), for any $l\in {\mathbb{Z}}^{+}\cup \left\{0\right\},$ we have

$$\begin{array}{c}\hfill {\lambda }_i^{(l+1)}\le \overline{\lambda }-{\left(\frac{a}{\overline{\lambda }}\right)}^l{\left(\frac{\alpha (1-\alpha )}{m-1}\right)}^l\left(\overline{\lambda }-\underline{\lambda }\right).\end{array}$$

Hence, we obtain

$$\begin{array}{cc}\hfill {\overline{\lambda }}^{\left(kn\right)}-{\underline{\lambda }}^{\left(kn\right)}& \le \underset{i\in \langle n\rangle }{max}{\lambda }_i^{\left(kn\right)}-{\underline{\lambda }}^{\left(\right(k-1\left)n\right)}\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \le \left(1-{\left(\frac{a}{\overline{\lambda }}\right)}^{n-1}{\left(\frac{\alpha (1-\alpha )}{m-1}\right)}^{n-1}\right)\left({\overline{\lambda }}^{\left(\right(k-1\left)n\right)}-{\underline{\lambda }}^{\left(\right(k-1\left)n\right)}\right)\hfill \\ \hfill & \le {\left(1-{\left(\frac{a}{\overline{\lambda }}\right)}^{n-1}{\left(\frac{\alpha (1-\alpha )}{m-1}\right)}^{n-1}\right)}^2\left({\overline{\lambda }}^{\left(\right(k-2\left)n\right)}-{\underline{\lambda }}^{\left(\right(k-2\left)n\right)}\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \le \cdots \hfill \\ \hfill & \le {\left(1-{\left(\frac{a}{\overline{\lambda }}\right)}^{n-1}{\left(\frac{\alpha (1-\alpha )}{m-1}\right)}^{n-1}\right)}^k\left(\overline{\lambda }-\underline{\lambda }\right).\hfill \end{array}$$

Thus, we have

$$\begin{array}{cc}\hfill \underset{k\to \infty }{lim}{\overline{\lambda }}^{\left(kn\right)}-\underset{k\to \infty }{lim}{\underline{\lambda }}^{\left(kn\right)}& =\underset{k\to \infty }{lim}\left({\overline{\lambda }}^{\left(kn\right)}-{\underline{\lambda }}^{\left(kn\right)}\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \le \underset{k\to \infty }{lim}{\left(1-{\left(\frac{a}{\overline{\lambda }}\right)}^{n-1}{\left(\frac{\alpha (1-\alpha )}{m-1}\right)}^{n-1}\right)}^k\left(\overline{\lambda }-\underline{\lambda }\right)\hfill \\ \hfill & =0.\hfill \end{array}$$

Hence, we obtain

$$\underset{l\to \infty }{lim}{\overline{\lambda }}^{\left(l\right)}=\underset{l\to \infty }{lim}{\underline{\lambda }}^{\left(l\right)}=\rho \left(\mathcal{A}\right).$$

□

Remark 1.

By Theorem 4, we know that $\mathrm{spec}\left(\mathcal{A}\right)=\mathrm{spec}\left({\mathcal{A}}^{\left(0\right)}\right)=\mathrm{spec}\left({\mathcal{A}}^{\left(1\right)}\right).$ Therefore, we know from the proof of Theorem 6 that PF algorithm is convergent for any weakly irreducible nonnegative tensors and any initial value ${\mathbf{x}}^{\left(0\right)}\in {\mathbb{R}}_{++}^n$.

5. Numerical Examples

In this section, we provide some numerical results and comparison for convergence rate of calculating the spectral radius of weakly irreducible nonnegative between the PF algorithm and NQZ method [11]. All numerical experiments are conducted using Matlab R2016b on a PC with 4GB memory Intel CPU 15-4210.

The numbers m and n in $(m,n)$ denote the order and the dimension of $\mathcal{A}$, respectively. $\rho \left(\mathcal{A}\right)$ denotes the spectral radius of $\mathcal{A}$. ${\mathrm{iter}}_{\mathrm{PF}}$ and ${\mathrm{iter}}_{\mathrm{NQZ}}$ denote the iteration of the PF algorithm and NQZ method, ${\mathrm{CPU}}_{\mathrm{PF}}\left(s\right)$ and ${\mathrm{CPU}}_{\mathrm{NQZ}}\left(s\right)$ denote the time costing by the PF algorithm and NQZ method, respectively.

Example 2.

$\mathcal{A}=\left(a_{ijk}\right)\in {\mathbb{R}}_{+}^{[3,3]}$, where $a_{122}=2,a_{233}=3,a_{311}=4,$ and zero elsewhere.

We know from Example 2, $G\left(\mathcal{A}\right)=\left(\begin{array}{ccc}0& 4& 0\\ 0& 0& 6\\ 8& 0& 0\end{array}\right).$ Clearly, $G\left(\mathcal{A}\right)$ is irreducible but not primitive, then tensor $\mathcal{A}$ is weakly irreducible but not weakly primitive. However, the condition of convergence of PF algorithm is satisfied, therefore, we can use the PF algorithm to calculate the spectral radius and corresponding eigenvector of the tensor. Take $\alpha =\frac{1}{2}$, if $\epsilon ={10}^{-5}$, we obtain that the spectral radius of tensor $\mathcal{A}$ is $\rho \left(\mathcal{A}\right)=2.88450,$ and the corresponding eigenvector is $\mathbf{x}={(0.75161,0.90263,0.88509)}^{\mathrm{T}}$.

Remark 2.

For the NQZ algorithm, the authors only prove the convergence of special weak primitive tensor $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}_{+}^{[m,n]}(a_{ii\cdots i}>0,i\in \langle n\rangle or\mathcal{B}=\mathcal{A}+\alpha \mathcal{I},\alpha >0)$, so the NQZ algorithm cannot be directly used to calculate the spectral radius of weak irreducible non-primitive nonnegative tensor.

Example 3.

$\mathcal{A}=\left(a_{ijk}\right)\in {\mathbb{R}}_{+}^{[3,3]}$, where $a_{123}=1,a_{221}=a_{223}=2,a_{312}=a_{332}=4,$ and zero elsewhere.

We know by Example 3, $G\left(\mathcal{A}\right)=\left(\begin{array}{ccc}0& 1& 1\\ 2& 4& 2\\ 4& 8& 4\end{array}\right).$ Clearly, $G\left(\mathcal{A}\right)$ is irreducible and primitive, then tensor $\mathcal{A}$ is weakly irreducible and weakly primitive. Therefore, the PF algorithm and NQZ method can be used to calculate the spectral radius and corresponding eigenvector of the tensor. Take $\alpha =\frac{1}{2}$. The comparison results are presented in

Table 1. The comparison of the PF algorithm and NQZ method in Example 3.

$\mathit{\epsilon }$	${\mathbf{iter}}_{\mathbf{PF}}$	${\mathbf{CPU}}_{\mathbf{PF}}\left(\mathit{s}\right)$	${\mathbf{iter}}_{\mathbf{NQZ}}$	${\mathbf{CPU}}_{\mathbf{NQZ}}\left(\mathit{s}\right)$	$\mathit{\rho }\left(\mathcal{A}\right)$
${10}^{-5}$	19	0.006570	46	0.032884	4.01536
${10}^{-6}$	22	0.008410	55	0.033562	4.015358
${10}^{-7}$	25	0.013257	64	0.038537	4.0153578
${10}^{-8}$	29	0.015513	73	0.041672	4.01535784
${10}^{-9}$	32	0.018249	80	0.047824	4.015357837
${10}^{-10}$	35	0.021307	88	0.048362	4.0153578367

, where $\epsilon$ denotes error.

Example 4.

Consider a random tensor $\mathcal{A}\in {\mathbb{R}}_{+}^{[m,n]}$, whose all entries are random values drawn from the standard uniform distribution on (0, 1). Obviously, this is a nonnegative irreducible tensor. The comparison results are presented in

Table 2. The comparison of PF algorithm and NQZ method in Example 4.

$\mathit{\epsilon }$	$(\mathit{m},\mathit{n})$	${\mathbf{iter}}_{\mathbf{PF}}$	${\mathbf{CPU}}_{\mathbf{PF}}\left(\mathit{s}\right)$	${\mathbf{iter}}_{\mathbf{NQZ}}$	${\mathbf{CPU}}_{\mathbf{NQZ}}\left(\mathit{s}\right)$	$\mathit{\rho }\left(\mathcal{A}\right)$
${10}^{-12}$	(3,10)	8	0.02551	10	0.032884	49.5135
${10}^{-10}$	(3,50)	6	0.03456	8	0.073562	$1.2365\times {10}^3$
${10}^{-10}$	(3,60)	5	0.10735	6	0.163854	$1.7931\times {10}^3$
${10}^{-11}$	(3,70)	6	0.15338	8	0.191672	$2.3564\times {10}^3$
${10}^{-11}$	(3,80)	5	0.23721	5	0.277824	$3.2452\times {10}^3$
${10}^{-11}$	(3,90)	4	0.34425	5	0.388362	$4.3780\times {10}^3$
${10}^{-10}$	(3,100)	5	0.47075	7	0.538537	$5.2145\times {10}^3$

, where ε denotes error.

Table 1 and Table 2 show a comparison between the PF algorithm and the NQZ method given in [11], with the same error, the number of iterations and calculation time are significantly reduced, which further verifies that our proposed algorithm is more efficient.

6. Conclusions

For general numerical algorithms in calculating the spectral radius of nonnegative tensors $\mathcal{A}=\left(a_{i_1{i}_2\cdots i_m}\right)\in {\mathbb{R}}_{+}^{[m,n]}$, the research results regarding the NQZ algorithm are the richest. At present, however, it is only proved that the NQZ algorithm is applicable to the calculation of spectral radius of special weak primitive tensors, that is, $a_{ii\cdots i}>0,i=1,2,\cdots ,n$ or transformed $\mathcal{A}+\alpha \mathcal{I},\alpha >0$, where $\mathcal{A}$ is a weakly irreducible weak primitive tensor. In this paper, a numerical algorithm for calculating the spectral radius of nonnegative tensors, with the aid of power functions, is constructed by using the diagonal similarity transformation of tensors. The algorithm is applicable to the calculation of spectral radius of all weakly irreducible nonnegative tensors. Moreover, the algorithm not only bears a wider scope of application, but also has a higher computational efficiency compared with the NQZ method.