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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 [17].

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

A x m 1 =λ x [ m 1 ] ,

then λ is called the eigenvalue of and x the eigenvector of associated with λ, where A x m 1 and x [ m 1 ] are vectors, whose i th component is

A x m 1 : = ( i 2 , , i m = 1 n a i i 2 i m x i 2 x i n ) 1 i n , x [ m 1 ] : = ( x i m 1 ) 1 i n .

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 [1012], which are generalized from nonnegative matrices.

Definition 1.1 The tensor is called reducible if there exists a nonempty proper index subset J{1,2,,n} such that a i 1 , i 2 , , i m =0, i 1 J, i 2 ,, i m J. If is not reducible, then we call to be irreducible.

Let ρ(A)=max{|λ|:λ is an eigenvalue of A}, where |λ| denotes the modulus of λ. We call ρ(A) the spectral radius of tensor .

Theorem 1.2 If is irreducible and nonnegative, then there exists a number ρ(A)>0 and a vector x 0 >0 such that A x 0 m 1 =ρ(A) x 0 [ m 1 ] . Moreover, if λ is an eigenvalue with a nonnegative eigenvector, then λ=ρ(A). If λ is an eigenvalue of , then |λ|ρ(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 and a real number c>ρ(B), where is the spectral radius of , such that

A=cIB.

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

In this paper, let N={1,2,,n}, we define the i th row sum of as R i (A)= i 2 , , i m = 1 n a i i 2 i m , and denote the largest and the smallest row sums of by

R max (A)= max i = 1 , , n R i (A), R min (A)= min i = 1 , , n R i (A).

Furthermore, a real tensor of order m dimension n is called the unit tensor, if its entries are δ i 1 i m for i 1 ,, i m N, where

δ i 1 i m = { 1 , if  i 1 = = i m , 0 , otherwise .

And we define σ(A) as the set of all the eigenvalues of and

r i (A)= δ i i 2 i m = 0 | a i i 2 i m |, r i j (A)= δ i i 2 i m = 0 , δ j i 2 i m = 0 | a i i 2 i m |= r i (A)| a i j 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

τ(A)min{ a i i },
(1)
R min (A)τ(A) R max (A).
(2)

Proof Let x>0 be an eigenvector of corresponding to τ(A), i.e., A x m 1 =τ(A) x [ m 1 ] . For each iN, we can get

( a i i τ ( A ) ) x i m 1 = δ i i 2 i m = 0 a i i 2 i m x i 2 x i m 0,

then

τ(A)min{ a i i }.

Assume that x s is the smallest component of x,

( a s s τ ( A ) ) x s m 1 = δ s i 2 i m = 0 a s i 2 i m x i 2 x i m 0.

That is,

τ(A) δ s i 2 i m = 0 a s i 2 i m + a s s ,

so

τ(A) R max (A).

Similarly, if we assume that x t ={max x i ,iN}, then we can get

τ(A) δ t i 2 i m = 0 a t i 2 i m + a t t R min (A).

Thus, we complete the proof. □

Theorem 2.2 Let be an irreducible M-tensor. Then

min i , j N , j i 1 2 { a i i + a j j r i j ( A ) Δ i , j 1 2 ( A ) } τ ( A ) max i , j N , j i 1 2 { a i i + a j j r i j ( A ) Δ i , j 1 2 ( A ) } ,
(3)

where

Δ i , j (A)= ( a i i a j j + r i j ( A ) ) 2 4 a i j j r j (A).

Proof Because τ(A) is an eigenvalue of , from Theorem 2.1 in [15], there are i,jN, ji, such that

( | τ ( A ) a i i | r i j ( A ) ) | τ ( A ) a j j | | a i j j | r j (A).

From Theorem 2.1, we can get

( a i i τ ( A ) r i j ( A ) ) ( a j j τ ( A ) ) a i j j r j (A),

equivalently,

τ ( A ) 2 ( a i i + a j j r i j ( A ) ) τ(A)+ a j j ( a i i r i j ( A ) ) + a i j j r j (A)0.

Then, solving for τ(A),

τ(A) 1 2 { a i i + a j j r i j ( A ) Δ i , j 1 2 ( A ) } min i , j N , j i 1 2 { a i i + a j j r i j ( A ) Δ i , j 1 2 ( A ) } .

Let x>0 be an eigenvector of corresponding to τ(A), i.e., A x m 1 =τ(A) x [ m 1 ] , x s is the smallest component of x. For each s,tN, st, we can get

( a t t τ ( A ) ) x t m 1 = δ t i 2 i m = 0 a t i 2 i m x i 2 x i m r t (A) x s m 1 ,
(4)
( a s s τ ( A ) ) x s m 1 = δ t i 2 i m = 0 , δ s i 2 i m = 0 a t i 2 i m x i 2 x i m a s t t x t m 1 r t s ( A ) x s m 1 a s t t x t m 1 , ( a s s τ ( A ) r t s ( A ) ) x s m 1 a s t t x t m 1 .
(5)

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

( a t t τ ( A ) ) ( a s s τ ( A ) r t s ( A ) ) a s t t r t (A).

Then, solving for τ(A),

τ(A) 1 2 { a t t + a s s r t s ( A ) Δ t , s 1 2 ( A ) } max i , j N , j i 1 2 { a i i + a j j r i j ( A ) Δ i , j 1 2 ( 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 A=( a i j k l ) of order 4 dimension 2 with entries defined as follows:

a 1111 = 3 , a 1222 = 1 , a 2111 = 2 , a 2222 = 2 ,

other a i j k l =0. By Theorem 1.1 in [15], we have

2τ(A)4.

By Theorem 2.1, we have

0τ(A)2.

By Theorem 2.2, we have

1 2 (5 17 )τ(A) 1 2 (5 5 ).

In fact, τ(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 (A)= i 2 , , i m = 1 n a i i 2 i m 1,i=1,,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

τ(A)B=A D ( 1 m ) D D m 1 ,

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 τ(A).

Proof Let x be the unique positive eigenvector corresponding to τ(A), i.e.,

A x m 1 =τ(A) x [ m 1 ] .

Let D be the diagonal matrix such that its diagonal entries are components of x, let us check the tensor C=A D ( 1 m ) DD. It is clear that for i=1,2,,n,

i 2 , , i m = 1 n C i i 2 i m = ( C e m 1 ) i = ( A D ( 1 m ) D D m 1 e m 1 ) i =τ(A).

Hence B=C/τ(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 τ(A). □

Theorem 3.3 Let be an order m dimension n stochastic irreducible nonnegative tensor, ω=min a i i , λσ(A). Then

|λω|1ω.

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

A x m 1 =λ x [ m 1 ] .

Assume that 0<| x s |= max i | x i |, then we can get

(λ a s s ) x s m 1 = δ s i 2 i m = 0 a s i 2 i m x i 2 x i m .

Then

|λ a s s | δ s i 2 i m = 0 a s i 2 i m = r s (A)=1 a s s ,

and therefore,

| λ ω | | λ a s s + a s s ω | | λ a s s | + | a s s ω | ( 1 a s s ) + ( a s s ω ) = 1 ω .
(6)

Thus, we complete the proof. □

Theorem 3.4 Let be an order m dimension n irreducible M-tensor, Ω=max a i i , λσ(A). Then

|Ωλ|Ωτ(A).

Proof From Theorem 3.2, we may evidently take τ(A)=1, and after performing a similarity transformation with a positive diagonal matrix, we may assume that is stochastic. Then, for θ(0,1), the matrix A(θ)=(1+θ)IθA is irreducible nonnegative stochastic, by Theorem 3.3, if λ(θ)σ(A(θ)), ω(θ)=min a i i (θ), we can get

| λ ( θ ) ω ( θ ) | 1ω(θ).

That is,

| 1 + θ θ λ ( 1 + θ θ max a i i ) | 1(1+θθmax a i i ).

Then

|Ωλ|Ω1.

Transforming back to , we get

|Ωλ|Ωτ(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 the set of m-order and n-dimensional real tensors whose off-diagonal entries are nonpositive.

Theorem 4.1 Let A,BZ, assume that is an M-tensor and BA. Then is an M-tensor, and

τ(A)τ(B).

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

A x m 1 >0.

Because BA, we can get

B x m 1 A x m 1 >0,

then is an M-tensor.

Let a= max 1 i n B i i , from Theorem 3.1 and Corollary 3.2 in [13], assume that

B=aICB,A=aICA,

where CA, CB are nonnegative tensors.

Because A,BZ and BA, then we can get

CACB.

From Lemma 3.5 in [12], we can get

ρ(CA)ρ(CB).

Therefore,

τ(A)τ(B).

Thus, we complete the proof. □

Theorem 4.2 Let , be of order m dimension n, suppose that is an M-tensor and | b i 1 i m || a i 1 i m | for all i 1 i m . Then, for any eigenvalue λ of , there exists i1,,n such that |λ a i i | b i i τ(B).

Proof We first suppose that is an M-tensor, τ(B) is an eigenvalue of with a positive corresponding eigenvector v. Denote

W=diag( v 1 ,, v n ),

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

C=A W 1 m W W [ m 1 ]

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

C ( W 1 x ) m 1 =λ ( W 1 x ) m 1 .

By the definition of , we have c i i = a i i , i=1,,n. Applying the first conclusion of Theorem 6 of [8], we can get

| λ c i i | δ i i 2 i m = 0 | c i i 2 i m | = v i 1 m | a i i 2 i m | v i 2 v i m v i 1 m | b i i 2 i m | v i 2 v i m = v i 1 m ( b i i v m 1 i 1 , , i m = 1 b i i 2 i m v i 2 v i m ) = b i i τ ( B ) .
(7)

Thus, we complete the proof. □

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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.

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He, J., Huang, TZ. Inequalities for M-tensors. J Inequal Appl 2014, 114 (2014). https://doi.org/10.1186/1029-242X-2014-114

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