Open Access Research

Some inequalities for the minimum eigenvalue of the Hadamard product of an M-matrix and its inverse

Guanghui Cheng*, Qin Tan and Zhuande Wang

Author Affiliations

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

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Journal of Inequalities and Applications 2013, 2013:65  doi:10.1186/1029-242X-2013-65


The electronic version of this article is the complete one and can be found online at: http://www.journalofinequalitiesandapplications.com/content/2013/1/65


Received:31 July 2012
Accepted:24 January 2013
Published:21 February 2013

© 2013 Cheng et al.; licensee Springer

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, some new inequalities for the minimum eigenvalue of the Hadamard product of an M-matrix and its inverse are given. These inequalities are sharper than the well-known results. A simple example is shown.

AMS Subject Classification: 15A18, 15A42.

Keywords:
Hadamard product; M-matrix; inverse M-matrix; strictly diagonally dominant matrix; eigenvalue

1 Introduction

A matrix A = ( a i j ) R n × n is called a nonnegative matrix if a i j 0 . A matrix A R n × n is called a nonsingular M-matrix [1] if there exist B 0 and s > 0 such that

A = s I n B and s > ρ ( B ) ,

where ρ ( B ) is a spectral radius of the nonnegative matrix B, I n is the n × n identity matrix. Denote by M n the set of all n × n nonsingular M-matrices. The matrices in M n 1 : = { A 1 : A M n } are called inverse M-matrices. Let us denote

τ ( A ) = min { Re λ : λ σ ( A ) } ,

and σ ( A ) denotes the spectrum of A. It is known that [2]

τ ( A ) = 1 ρ ( A 1 )

is a positive real eigenvalue of A M n and the corresponding eigenvector is nonnegative. Indeed

τ ( A ) = s ρ ( B ) ,

if A = s I n B , where s > ρ ( B ) , B 0 .

For any two n × n matrices A = ( a i j ) and B = ( b i j ) , the Hadamard product of A and B is A B = ( a i j b i j ) . If A , B M n , then A B 1 is also an M-matrix [3].

A matrix A is irreducible if there does not exist a permutation matrix P such that

P A P T = [ A 1 , 1 A 1 , 2 0 A 2 , 2 ] ,

where A 1 , 1 and A 2 , 2 are square matrices.

For convenience, the set { 1 , 2 , , n } is denoted by N, where n (≥3) is any positive integer. Let A = ( a i j ) R n × n be a strictly diagonally dominant by row, denote

Recently, some lower bounds for the minimum eigenvalue of the Hadamard product of an M-matrix and an inverse M-matrix have been proposed. Let A M n , for example, τ ( A A 1 ) 1 has been proven by Fiedler et al. in [4]. Subsequently, τ ( A A 1 ) > 1 n was given by Fiedler and Markham in [3], and they conjectured that τ ( A A 1 ) > 2 n . Song [5], Yong [6] and Chen [7] have independently proven this conjecture. In [8], Li et al. improved the conjecture τ ( A A 1 ) 2 n when A 1 is a doubly stochastic matrix and gave the following result:

τ ( A A 1 ) min i { a i i s i R i 1 + j i s j i } .

In [9], Li et al. gave the following result:

τ ( A A 1 ) min i { a i i m i R i 1 + j i m j i } .

Furthermore, if a 11 = a 22 = = a n n , they have obtained

min i { a i i m i R i 1 + j i m j i } min i { a i i s i R i 1 + j i s j i } ,

i.e., under this condition, the bound of [9] is better than the one of [8].

In this paper, our motives are to improve the lower bounds for the minimum eigenvalue τ ( A A 1 ) . The main ideas are based on the ones of [8] and [9].

2 Some preliminaries and notations

In this section, we give some notations and lemmas which mainly focus on some inequalities for the entries of the inverse M-matrix and the strictly diagonally dominant matrix.

Lemma 2.1[6]

Let A R n × n be a strictly diagonally dominant matrix by row, i.e.,

| a i i | > j i | a i j | , i N .

If A 1 = ( b i j ) , then

| b j i | k j | a j k | | a j j | | b i i | , j i , j N .

Lemma 2.2Let A R n × n be a strictly diagonally dominantM-matrix by row. If A 1 = ( b i j ) , then

b j i | a j i | + k j , i | a j k | m k i a j j b i i u j b i i , j i , i N .

Proof Firstly, we consider A R n × n is a strictly diagonally dominant M-matrix by row. For i N , let

r i ( ε ) = max j i { | a j i | + ε a j j k j , i | a j k | }

and

m j i ( ε ) = r i ( ε ) ( k j , i | a j k | + ε ) + | a j i | a j j , j i .

Since A is strictly diagonally dominant, then r j i < 1 and m j i < 1 . Therefore, there exists ε > 0 such that 0 < r i ( ε ) < 1 and 0 < m j i ( ε ) < 1 . Let us define one positive diagonal matrix

M i ( ε ) = diag ( m 1 i ( ε ) , , m i 1 , i ( ε ) , 1 , m i + 1 , i ( ε ) , , m n i ( ε ) ) .

Similarly to the proofs of Theorem 2.1 and Theorem 2.4 in [8], we can prove that the matrix A M i ( ε ) is also a strictly diagonally dominant M-matrix by row for any i N . Furthermore, by Lemma 2.1, we can obtain the following result:

m j i 1 ( ε ) b j i | a j i | + k j , i | a j k | m k i ( ε ) m j i ( ε ) a j j b i i , j i , j N ,

i.e.,

b j i | a j i | + k j , i | a j k | m k i ( ε ) a j j b i i , j i , j N .

Let ε 0 + to get

b j i | a j i | + k j , i | a j k | m k i a j j b i i u j b i i , j i , j N .

This proof is completed. □

Lemma 2.3Let A = ( a i j ) M n be a strictly diagonally dominant matrix by row and A 1 = ( b i j ) , then we have

1 a i i b i i 1 a i i j i | a i j | u j i , i N .

Proof Let B = A 1 . Since A is an M-matrix, then B 0 . By A B = B A = I n , we have

1 = j = 1 n a i j b j i = a i i b i i j i | a i j | b j i , i N .

Hence

1 a i i b i i , i N ,

or equivalently,

1 a i i b i i , i N .

Furthermore, by Lemma 2.2, we get

1 = a i i b i i j i | a i j | b j i ( a i i j i | a i j | u j i ) b i i , i N ,

i.e.,

b i i 1 a i i j i | a i j | u j i , i N .

Thus the proof is completed. □

Lemma 2.4[10]

Let A C n × n and x 1 , x 2 , , x n be positive real numbers. Then all the eigenvalues ofAlie in the region

1 n { z C : | z a i i | x i j i 1 x j | a j i | } .

Lemma 2.5[11]

If A 1 is a doubly stochastic matrix, then A e = e , A T e = e , where e = ( 1 , 1 , , 1 ) T .

3 Main results

In this section, we give two new lower bounds for τ ( A A 1 ) which improve the ones in [8] and [9].

Lemma 3.1If A M n and A 1 = ( b i j ) is a doubly stochastic matrix, then

b i i 1 1 + j i u j i , i N .

Proof This proof is similar to the ones of Lemma 3.2 in [8] and Theorem 3.2 in [9]. □

Theorem 3.1Let A M n and A 1 = ( b i j ) be a doubly stochastic matrix. Then

τ ( A A 1 ) min i { a i i u i R i 1 + j i u j i } .

Proof Firstly, we assume that A is irreducible. By Lemma 2.5, we have

a i i = j i | a i j | + 1 = j i | a j i | + 1 and a i i > 1 , i N .

Denote

u j = max i j { u j i } = max { | a j i | + k j , i | a j k | m k i a j j } , j N .

Since A is an irreducible matrix, we know that 0 < u j 1 . So, by Lemma 2.4, there exists i 0 N such that

| λ a i 0 i 0 b i 0 i 0 | u i 0 j i 0 1 u j | a j i 0 b j i 0 | ,

or equivalently,

| λ | a i 0 i 0 b i 0 i 0 u i 0 j i 0 1 u j | a j i 0 b j i 0 | a i 0 i 0 b i 0 i 0 u i 0 j i 0 1 u j | a j i 0 | u j b i 0 i 0 (by Lemma 2.2) ( a i 0 i 0 u i 0 j i 0 | a j i 0 | ) b i 0 i 0 = ( a i 0 i 0 u i 0 R i 0 ) b i 0 i 0 a i 0 i 0 u i 0 R i 0 1 + j i 0 u j i 0 (by Lemma 3.1) min i { a i i u i R i 1 + j i u j i } .

Secondly, if A is reducible, without loss of generality, we may assume that A has the following block upper triangular form:

A = [ A 11 A 12 A 1 K 0 A 22 A 2 K 0 0 0 0 0 A K K ] ,

where A i i M n i is an irreducible diagonal block matrix, i = 1 , 2 , , K . Obviously, τ ( A A 1 ) = min i τ ( A i i A i i 1 ) . Thus the reducible case is converted into the irreducible case. This proof is completed. □

Theorem 3.2If A = ( a i j ) M n is a strictly diagonally dominant by row, then

min i { a i i u i R i 1 + j i u j i } min i { a i i s i R i 1 + j i s j i } .

Proof Since A is strictly diagonally dominant by row, for any j i , we have

d j m j i = | a j i | + k j , i | a j k | a j j | a j i | + k j , i | a j k | r i a j j = ( 1 r i ) k j , i | a j k | a j j 0 ,

or equivalently,

d j m j i , j i , j N . (1)

So, we can obtain

u j i = | a j i | + k j , i | a j k | m k i a j j | a j i | + k j , i | a j k | d k a j j = s j i , j i , j N , (2)

and

u i s i , i N .

Therefore, it is easy to obtain that

a i i u i R i 1 + j i u j i a i i s i R i 1 + j i s j i , i N .

Obviously, we have the desired result

min i { a i i u i R i 1 + j i u j i } min i { a i i s i R i 1 + j i s j i } .

This proof is completed. □

Theorem 3.3If A = ( a i j ) M n is strictly diagonally dominant by row, then

min i { a i i u i R i 1 + j i u j i } min i { a i i m i R i 1 + j i m j i } .

Proof Since A is strictly diagonally dominant by row, for any j i , we have

r i m j i = r i | a j i | + k j , i | a j k | r i a j j = r i | a j i | k j , i | a j k | a j j = r i ( a j j k j , i | a j k | ) | a j i | a j j = a j j k j , i | a j k | a j j ( r i | a j i | a j j k j , i | a j k | ) 0 ,

i.e.,

r i m j i , j i , j N . (3)

So, we can obtain

u j i = | a j i | + k j , i | a j k | m k i a j j | a j i | + k j , i | a j k | r i a j j = m j i , j i , j N , (4)

and

u i m i , i N .

Therefore, it is easy to obtain that

a i i u i R i 1 + j i u j i a i i m i R i 1 + j i m j i , i N .

Obviously, we have the desired result

min i { a i i u i R i 1 + j i u j i } min i { a i i m i R i 1 + j i m j i } .

 □

Remark 3.1 According to inequalities (1) and (3), it is easy to know that

b j i | a j i | + k j , i | a j k | m k i a j j b i i | a j i | + k j , i | a j k | d k a j j b i i , i N .

and

b j i | a j i | + k j , i | a j k | m k i a j j b i i | a j i | + k j , i | a j k | r i a j j b i i , i N .

That is to say, the result of Lemma 2.2 is sharper than the ones of Theorem 2.1 in [8] and Lemma 2.2 in [9]. Moreover, the results of Theorem 3.2 and Theorem 3.3 are sharper than the ones of Theorem 3.1 in [8] and Theorem 3.3 in [9], respectively.

Theorem 3.4If A M n is strictly diagonally dominant by row, then

τ ( A A 1 ) min i { 1 1 a i i j i | a j i | u j i } .

Proof This proof is similar to the one of Theorem 3.5 in [8]. □

Remark 3.2 According to inequalities (2) and (4), we get

1 1 a i i j i | a j i | u j i 1 1 a i i j i | a j i | s j i ,

and

1 1 a i i j i | a j i | u j i 1 1 a i i j i | a j i | m j i .

That is to say, the bound of Theorem 3.4 is sharper than the ones of Theorem 3.5 in [8] and Theorem 3.4 in [9], respectively.

Remark 3.3 Using the above similar ideas, we can obtain similar inequalities of the strictly diagonally M-matrix by column.

4 Example

For convenience, we consider the M-matrix A is the same as the matrix of [8]. Define the M-matrix A as follows:

A = [ 4 1 1 1 2 5 1 1 0 2 4 1 1 1 1 4 ] .

1. Estimate the upper bounds for entries of A 1 = ( b i j ) . Firstly, by Lemma 2.2(2) in [9], we have

A 1 [ 1 0.5833 0.5000 0.5000 0.6667 1 0.5000 0.5000 0.5000 0.6667 1 0.5000 0.5833 0.5833 0.5000 1 ] [ b 11 b 22 b 33 b 44 b 11 b 22 b 33 b 44 b 11 b 22 b 33 b 44 b 11 b 22 b 33 b 44 ] .

By Lemma 2.2, we have

A 1 [ 1 0.5625 0.5000 0.5000 0.6167 1 0.5000 0.5000 0.4792 0.6458 1 0.5000 0.5417 0.5625 0.5000 1 ] [ b 11 b 22 b 33 b 44 b 11 b 22 b 33 b 44 b 11 b 22 b 33 b 44 b 11 b 22 b 33 b 44 ] .

By Lemma 2.3 and Theorem 3.1 in [9], we get

By Lemma 2.3 and Lemma 3.1, we get

2. Lower bounds for τ ( A A 1 ) .

By Theorem 3.2 in [9], we obtain

0.9755 = τ ( A A 1 ) 0.8000 .

By Theorem 3.1, we obtain

0.9755 = τ ( A A 1 ) 0.8250 .

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.

Acknowledgements

This research is supported by National Natural Science Foundations of China (No. 11101069).

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