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# 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

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

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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