Inequalities for M-tensors

Abstract

In this paper, we establish some important properties of M-tensors. We derive upper and lower bounds for the minimum eigenvalue of M-tensors, bounds for eigenvalues of M-tensors except the minimum eigenvalue are also presented; finally, we give the Ky Fan theorem for M-tensors.

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

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 [1–7].

If there are a complex number λ and a nonzero complex vector x that are solutions of the following homogeneous polynomial equations:

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

then λ is called the eigenvalue of and x the eigenvector of associated with λ, where $\mathcal{A}{x}^{m-1}$ and $x^{[m-1]}$ are vectors, whose i th component is

$$\begin{array}{c}\mathcal{A}{x}^{m-1}:={(\sum _{i_2,\dots ,i_m=1}^n{a}_{i{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_n})}_{1\le i\le n},\hfill \\ x^{[m-1]}:={\left(x_i^{m-1}\right)}_{1\le i\le n}.\hfill \end{array}$$

This definition was introduced by Qi and Lim [8, 9] where they supposed that is an order m dimension n symmetric tensor and m is even. First, we introduce some results of nonnegative tensors [10–12], which are generalized from nonnegative matrices.

Definition 1.1 The tensor is called reducible if there exists a nonempty proper index subset $\mathbb{J}\subset \{1,2,\dots ,n\}$ such that $a_{i_1,i_2,\dots ,i_m}=0$, $\mathrm{\forall }{i}_1\in \mathbb{J}$, $\mathrm{\forall }{i}_2,\dots ,i_m\notin \mathbb{J}$. If is not reducible, then we call to be irreducible.

Let $\rho (\mathcal{A})=max\{|\lambda |:\lambda \text{ is an eigenvalue of }\mathcal{A}\}$, where $|\lambda |$ denotes the modulus of λ. We call $\rho (\mathcal{A})$ the spectral radius of tensor .

Theorem 1.2 If is irreducible and nonnegative, then there exists a number $\rho (\mathcal{A})>0$ and a vector $x_0>0$ such that $\mathcal{A}{x}_0^{m-1}=\rho (\mathcal{A})x_0^{[m-1]}$. Moreover, if λ is an eigenvalue with a nonnegative eigenvector, then $\lambda =\rho (\mathcal{A})$. If λ is an eigenvalue of , then $|\lambda |\le \rho (\mathcal{A})$.

The authors in [13, 14] extended the notion of M-matrices to higher-order tensors and introduced the definition of an M-tensor.

Definition 1.3 Let be an m-order and n-dimensional tensor. is called an M-tensor if there exist a nonnegative tensor ℬ and a real number $c>\rho (\mathcal{B})$, where ℬ is the spectral radius of ℬ, such that

$$\mathcal{A}=c\mathcal{I}-\mathcal{B}.$$

Theorem 1.4 Let be an M-tensor and denote by $\tau (\mathcal{A})$ the minimal value of the real part of all eigenvalues of . Then $\tau (\mathcal{A})>0$ is an eigenvalue of with a nonnegative eigenvector. Moreover, there exist a nonnegative tensor ℬ and a real number $c>\rho (\mathcal{B})$ such that $\mathcal{A}=c\mathcal{I}-\mathcal{B}$. If is irreducible, then $\tau (\mathcal{A})$ is the unique eigenvalue with a positive eigenvector.

In this paper, let $N=\{1,2,\dots ,n\}$, we define the i th row sum of as $R_i(\mathcal{A})={\sum }_{i_2,\dots ,i_m=1}^n{a}_{i{i}_2\cdots i_m}$, and denote the largest and the smallest row sums of by

$$R_{max}(\mathcal{A})=\underset{i=1,\dots ,n}{max}{R}_i(\mathcal{A}),R_{min}(\mathcal{A})=\underset{i=1,\dots ,n}{min}{R}_i(\mathcal{A}).$$

Furthermore, a real tensor of order m dimension n is called the unit tensor, if its entries are ${\delta }_{i_1\cdots i_m}$ for $i_1,\dots ,i_m\in N$, where

$${\delta }_{i_1\cdots i_m}=\{\begin{array}{ll}1,& \text{if }{i}_1=\cdots =i_m,\\ 0,& \text{otherwise}.\end{array}$$

And we define $\sigma (\mathcal{A})$ as the set of all the eigenvalues of and

$$r_i(\mathcal{A})=\sum _{{\delta }_{i{i}_2\cdots i_m}=0}|a_{i{i}_2\cdots i_m}|,r_i^j(\mathcal{A})=\underset{{\delta }_{j{i}_2\cdots i_m}=0}{\sum _{{\delta }_{i{i}_2\cdots i_m}=0,}}|a_{i{i}_2\cdots i_m}|=r_i(\mathcal{A})-|a_{ij\cdots j}|.$$

In this paper, we continue this research on the eigenvalue problems for tensors. In Section 2, some bounds for the minimum eigenvalue of M-tensors are obtained, and proved to be tighter than those in Theorem 1.1 in [15]. In Section 3, some bounds for eigenvalues of M-tensors except the minimum eigenvalue are given. Moreover, the Ky Fan theorem for M-tensors is presented in Section 4.

2 Bounds for the minimum eigenvalue of M-tensors

Theorem 2.1 Let be an irreducible M-tensor. Then

$$\tau (\mathcal{A})\le min\{a_{i\cdots i}\},$$

(1)

$$R_{min}(\mathcal{A})\le \tau (\mathcal{A})\le R_{max}(\mathcal{A}).$$

(2)

Proof Let $x>0$ be an eigenvector of corresponding to $\tau (\mathcal{A})$, i.e., $\mathcal{A}{x}^{m-1}=\tau (\mathcal{A})x^{[m-1]}$. For each $i\in N$, we can get

$$(a_{i\cdots i}-\tau (\mathcal{A}))x_i^{m-1}=-\sum _{{\delta }_{i{i}_2\cdots i_m}=0}{a}_{i{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_m}\ge 0,$$

then

$$\tau (\mathcal{A})\le min\{a_{i\cdots i}\}.$$

Assume that $x_s$ is the smallest component of x,

$$(a_{s\cdots s}-\tau (\mathcal{A}))x_s^{m-1}=-\sum _{{\delta }_{s{i}_2\cdots i_m}=0}{a}_{s{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_m}\ge 0.$$

That is,

$$\tau (\mathcal{A})\le \sum _{{\delta }_{s{i}_2\cdots i_m}=0}{a}_{s{i}_2\cdots i_m}+a_{s\cdots s},$$

so

$$\tau (\mathcal{A})\le R_{max}(\mathcal{A}).$$

Similarly, if we assume that $x_t=\{max{x}_i,i\in N\}$, then we can get

$$\tau (\mathcal{A})\ge \sum _{{\delta }_{t{i}_2\cdots i_m}=0}{a}_{t{i}_2\cdots i_m}+a_{t\cdots t}\ge R_{min}(\mathcal{A}).$$

Thus, we complete the proof. □

Theorem 2.2 Let be an irreducible M-tensor. Then

$$\begin{array}{r}\underset{i,j\in N,j\ne i}{min}\frac{1}{2}\{a_{i\cdots i}+a_{j\cdots j}-r_i^j(\mathcal{A})-{\mathrm{\Delta }}_{i,j}^{\frac{1}{2}}(\mathcal{A})\}\\ \le \tau (\mathcal{A})\le \underset{i,j\in N,j\ne i}{max}\frac{1}{2}\{a_{i\cdots i}+a_{j\cdots j}-r_i^j(\mathcal{A})-{\mathrm{\Delta }}_{i,j}^{\frac{1}{2}}(\mathcal{A})\},\end{array}$$

(3)

where

$${\mathrm{\Delta }}_{i,j}(\mathcal{A})={(a_{i\cdots i}-a_{j\cdots j}+r_i^j(\mathcal{A}))}^2-4a_{ij\cdots j}{r}_j(\mathcal{A}).$$

Proof Because $\tau (\mathcal{A})$ is an eigenvalue of , from Theorem 2.1 in [15], there are $i,j\in N$, $j\ne i$, such that

$$(|\tau (\mathcal{A})-a_{i\cdots i}|-r_i^j(\mathcal{A}))|\tau (\mathcal{A})-a_{j\cdots j}|\le |a_{ij\cdots j}|r_j(\mathcal{A}).$$

From Theorem 2.1, we can get

$$(a_{i\cdots i}-\tau (\mathcal{A})-r_i^j(\mathcal{A}))(a_{j\cdots j}-\tau (\mathcal{A}))\le -a_{ij\cdots j}{r}_j(\mathcal{A}),$$

equivalently,

$$\tau {(\mathcal{A})}^2-(a_{i\cdots i}+a_{j\cdots j}-r_i^j(\mathcal{A}))\tau (\mathcal{A})+a_{j\cdots j}(a_{i\cdots i}-r_i^j(\mathcal{A}))+a_{ij\cdots j}{r}_j(\mathcal{A})\le 0.$$

Then, solving for $\tau (\mathcal{A})$,

$$\tau (\mathcal{A})\ge \frac{1}{2}\{a_{i\cdots i}+a_{j\cdots j}-r_i^j(\mathcal{A})-{\mathrm{\Delta }}_{i,j}^{\frac{1}{2}}(\mathcal{A})\}\ge \underset{i,j\in N,j\ne i}{min}\frac{1}{2}\{a_{i\cdots i}+a_{j\cdots j}-r_i^j(\mathcal{A})-{\mathrm{\Delta }}_{i,j}^{\frac{1}{2}}(\mathcal{A})\}.$$

Let $x>0$ be an eigenvector of corresponding to $\tau (\mathcal{A})$, i.e., $\mathcal{A}{x}^{m-1}=\tau (\mathcal{A})x^{[m-1]}$, $x_s$ is the smallest component of x. For each $s,t\in N$, $s\ne t$, we can get

$$(a_{t\cdots t}-\tau (\mathcal{A}))x_t^{m-1}=-\sum _{{\delta }_{t{i}_2\cdots i_m}=0}{a}_{t{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_m}\ge r_t(\mathcal{A})x_s^{m-1},$$

(4)

$$\begin{array}{c}(a_{s\cdots s}-\tau (\mathcal{A}))x_s^{m-1}=-\underset{{\delta }_{s{i}_2\cdots i_m}=0}{\sum _{{\delta }_{t{i}_2\cdots i_m}=0,}}{a}_{t{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_m}-a_{st\cdots t}{x}_t^{m-1}\ge r_t^s(\mathcal{A})x_s^{m-1}-a_{st\cdots t}{x}_t^{m-1},\hfill \\ (a_{s\cdots s}-\tau (\mathcal{A})-r_t^s(\mathcal{A}))x_s^{m-1}\ge -a_{st\cdots t}{x}_t^{m-1}.\hfill \end{array}$$

(5)

Multiplying equations (4) and (5), we get

$$(a_{t\cdots t}-\tau (\mathcal{A}))(a_{s\cdots s}-\tau (\mathcal{A})-r_t^s(\mathcal{A}))\ge -a_{st\cdots t}{r}_t(\mathcal{A}).$$

Then, solving for $\tau (\mathcal{A})$,

$$\tau (\mathcal{A})\le \frac{1}{2}\{a_{t\cdots t}+a_{s\cdots s}-r_t^s(\mathcal{A})-{\mathrm{\Delta }}_{t,s}^{\frac{1}{2}}(\mathcal{A})\}\le \underset{i,j\in N,j\ne i}{max}\frac{1}{2}\{a_{i\cdots i}+a_{j\cdots j}-r_i^j(\mathcal{A})-{\mathrm{\Delta }}_{i,j}^{\frac{1}{2}}(\mathcal{A})\}.$$

Thus, we complete the proof. □

We now show that the bounds in Theorem 2.2 are tight and sharper than those in Theorem 1.1 in [15] by the following example. Consider the M-tensor $\mathcal{A}=(a_{ijkl})$ of order 4 dimension 2 with entries defined as follows:

$$\begin{array}{c}{a}_{1111}=3,a_{1222}=-1,\hfill \\ a_{2111}=-2,a_{2222}=2,\hfill \end{array}$$

other $a_{ijkl}=0$. By Theorem 1.1 in [15], we have

$$-2\le \tau (\mathcal{A})\le 4.$$

By Theorem 2.1, we have

$$0\le \tau (\mathcal{A})\le 2.$$

By Theorem 2.2, we have

$$\frac{1}{2}(5-\sqrt{17})\le \tau (\mathcal{A})\le \frac{1}{2}(5-\sqrt{5}).$$

In fact, $\tau (\mathcal{A})=1$. Hence, the bounds in Theorem 2.2 are tight and sharper than those in Theorem 1.1 in [15].

3 Bounds for eigenvalues of M-tensors except the minimum eigenvalue

In this section, we introduce the stochastic M-tensor, which is a generalization of the nonnegative stochastic tensor.

Definition 3.1 An M-tensor of order m dimension n is called stochastic provided

$$R_i(\mathcal{A})=\sum _{i_2,\dots ,i_m=1}^n{a}_{i{i}_2\cdots i_m}\equiv 1,i=1,\dots ,n.$$

Obviously, when is a stochastic M-tensor, 1 is the minimum eigenvalue of and e is an eigenvector corresponding to 1, where e is an all-ones vector.

Theorem 3.2 Let be an order m dimension n irreducible M-tensor. Then there exists a diagonal matrix D with positive main diagonal entries such that

$$\tau (\mathcal{A})\cdot \mathcal{B}=\mathcal{A}\cdot D^{(1-m)}\cdot \stackrel{m-1}{\overbrace{D\cdot \dots \cdot D}},$$

where B is a stochastic irreducible M-tensor. Furthermore, B is unique, and the diagonal entries of D are exactly the components of the unique positive eigenvector corresponding to $\tau (\mathcal{A})$.

Proof Let x be the unique positive eigenvector corresponding to $\tau (\mathcal{A})$, i.e.,

$$\mathcal{A}{x}^{m-1}=\tau (\mathcal{A})x^{[m-1]}.$$

Let D be the diagonal matrix such that its diagonal entries are components of x, let us check the tensor $\mathcal{C}=\mathcal{A}\cdot D^{(1-m)}\cdot D\cdot \dots \cdot D$. It is clear that for $i=1,2,\dots ,n$,

$$\sum _{i_2,\dots ,i_m=1}^n{\mathcal{C}}_{i{i}_2\cdots i_m}={\left(\mathcal{C}{e}^{m-1}\right)}_i={(\mathcal{A}\cdot D^{(1-m)}\cdot \stackrel{m-1}{\overbrace{D\cdot \dots \cdot D}}{e}^{m-1})}_i=\tau (\mathcal{A}).$$

Hence $\mathcal{B}=\mathcal{C}/\tau (\mathcal{A})$ is the desired stochastic M-tensor. Since the positive eigenvector is unique, then B is unique, and the diagonal entries of D are exactly the components of the unique positive eigenvector corresponding to $\tau (\mathcal{A})$. □

Theorem 3.3 Let be an order m dimension n stochastic irreducible nonnegative tensor, $\omega =min{a}_{i\cdots i}$, $\lambda \in \sigma (\mathcal{A})$. Then

$$|\lambda -\omega |\le 1-\omega .$$

Proof Let λ be an eigenvalue of the stochastic irreducible nonnegative tensor , x is the eigenvector corresponding to λ, i.e.,

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

Assume that $0<|x_s|={max}_i|x_i|$, then we can get

$$(\lambda -a_{s\cdots s})x_s^{m-1}=\sum _{{\delta }_{s{i}_2\cdots i_m}=0}{a}_{s{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_m}.$$

Then

$$|\lambda -a_{s\cdots s}|\le \sum _{{\delta }_{s{i}_2\cdots i_m}=0}{a}_{s{i}_2\cdots i_m}=r_s(\mathcal{A})=1-a_{s\cdots s},$$

and therefore,

$$\begin{array}{rl}|\lambda -\omega |& \le |\lambda -a_{s\cdots s}+a_{s\cdots s}-\omega |\\ \le |\lambda -a_{s\cdots s}|+|a_{s\cdots s}-\omega |\\ \le (1-a_{s\cdots s})+(a_{s\cdots s}-\omega )\\ =1-\omega .\end{array}$$

(6)

Thus, we complete the proof. □

Theorem 3.4 Let be an order m dimension n irreducible M-tensor, $\mathrm{\Omega }=max{a}_{i\cdots i}$, $\lambda \in \sigma (\mathcal{A})$. Then

$$|\mathrm{\Omega }-\lambda |\le \mathrm{\Omega }-\tau (\mathcal{A}).$$

Proof From Theorem 3.2, we may evidently take $\tau (\mathcal{A})=1$, and after performing a similarity transformation with a positive diagonal matrix, we may assume that is stochastic. Then, for $\theta \in (0,1)$, the matrix $\mathcal{A}(\theta )=(1+\theta )\mathcal{I}-\theta \mathcal{A}$ is irreducible nonnegative stochastic, by Theorem 3.3, if $\lambda (\theta )\in \sigma (\mathcal{A}(\theta ))$, $\omega (\theta )=min{a}_{i\cdots i}(\theta )$, we can get

$$|\lambda (\theta )-\omega (\theta )|\le 1-\omega (\theta ).$$

That is,

$$|1+\theta -\theta \lambda -(1+\theta -\theta max{a}_{i\cdots i})|\le 1-(1+\theta -\theta max{a}_{i\cdots i}).$$

Then

$$|\mathrm{\Omega }-\lambda |\le \mathrm{\Omega }-1.$$

Transforming back to , we get

$$|\mathrm{\Omega }-\lambda |\le \mathrm{\Omega }-\tau (\mathcal{A}).$$

Thus, we complete the proof. □

4 Ky Fan theorem for M-tensors

In this section we give the Ky Fan theorem for M-tensors. Denote by ℤ the set of m-order and n-dimensional real tensors whose off-diagonal entries are nonpositive.

Theorem 4.1 Let $\mathcal{A},\mathcal{B}\in \mathbb{Z}$, assume that is an M-tensor and $\mathcal{B}\ge \mathcal{A}$. Then ℬ is an M-tensor, and

$$\tau (\mathcal{A})\le \tau (\mathcal{B}).$$

Proof If $x>0$, from assume that is an M-tensor and condition (D4) in [14], we know

$$\mathcal{A}{x}^{m-1}>0.$$

Because $\mathcal{B}\ge \mathcal{A}$, we can get

$$\mathcal{B}{x}^{m-1}\ge \mathcal{A}{x}^{m-1}>0,$$

then ℬ is an M-tensor.

Let $a={max}_{1\le i\le n}{\mathcal{B}}_{i\cdots i}$, from Theorem 3.1 and Corollary 3.2 in [13], assume that

$$\mathcal{B}=a\mathcal{I}-\mathcal{CB},\mathcal{A}=a\mathcal{I}-\mathcal{CA},$$

where $\mathcal{CA}$, $\mathcal{CB}$ are nonnegative tensors.

Because $\mathcal{A},\mathcal{B}\in \mathbb{Z}$ and $\mathcal{B}\ge \mathcal{A}$, then we can get

$$\mathcal{CA}\ge \mathcal{CB}.$$

From Lemma 3.5 in [12], we can get

$$\rho (\mathcal{CA})\ge \rho (\mathcal{CB}).$$

Therefore,

$$\tau (\mathcal{A})\le \tau (\mathcal{B}).$$

Thus, we complete the proof. □

Theorem 4.2 Let , ℬ be of order m dimension n, suppose that ℬ is an M-tensor and $|b_{i_1\cdots i_m}|\ge |a_{i_1\cdots i_m}|$ for all $i_1\ne \cdots \ne i_m$. Then, for any eigenvalue λ of , there exists $i\in 1,\dots ,n$ such that $|\lambda -a_{i\cdots i}|\le b_{i\cdots i}-\tau (\mathcal{B})$.

Proof We first suppose that ℬ is an M-tensor, $\tau (\mathcal{B})$ is an eigenvalue of ℬ with a positive corresponding eigenvector v. Denote

$$W=diag(v_1,\dots ,v_n),$$

where $v_i$ is the i th component of v. Let

$$\mathcal{C}=\mathcal{A}\cdot W^{1-m}\stackrel{[m-1]}{\overbrace{W\cdot \dots \cdot W}}$$

and let λ be an eigenvalue of with x, a corresponding eigenvector, i.e., $\mathcal{A}{x}^{m-1}=\lambda x^{[m-1]}$. Then, as in the proof of Theorem 4.1 in [12], we have

$$\mathcal{C}{\left(W^{-1}x\right)}^{m-1}=\lambda {\left(W^{-1}x\right)}^{m-1}.$$

By the definition of , we have $c_{i\cdots i}=a_{i\cdots i}$, $i=1,\dots ,n$. Applying the first conclusion of Theorem 6 of [8], we can get

$$\begin{array}{rl}|\lambda -c_{i\cdots i}|& \le \sum _{{\delta }_{i{i}_2\cdots i_m}=0}|c_{i{i}_2\cdots i_m}|\\ =v_i^{1-m}\sum |a_{i{i}_2\cdots i_m}|v_{i_2}\cdots v_{i_m}\\ \le v_i^{1-m}\sum |b_{i{i}_2\cdots i_m}|v_{i_2}\cdots v_{i_m}\\ =v_i^{1-m}(b_{i\cdots i}{v}^{m-1}-\sum _{i_1,\dots ,i_m=1}{b}_{i{i}_2\cdots i_m}{v}_{i_2}\cdots v_{i_m})\\ =b_{i\cdots i}-\tau (\mathcal{B}).\end{array}$$

(7)

Thus, we complete the proof. □