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

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.

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 B0 and s>0 such that

A=s I n Bands>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)=sρ(B),

if A=s I n B, where s>ρ(B), B0.

For any two n×n matrices A=( a i j ) and B=( b i j ), the Hadamard product of A and B is AB=( 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

PA 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 |,iN.

If A 1 =( b i j ), then

| b j i | k j | a j k | | a j j | | b i i |,ji,jN.

Lemma 2.2 Let A R n × n be a strictly diagonally dominant M-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 ,ji,iN.

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

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 iN. 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 ,ji,jN,

i.e.,

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

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 ,ji,jN.

This proof is completed. □

Lemma 2.3 Let 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 ,iN.

Proof Let B= A 1 . Since A is an M-matrix, then B0. By AB=BA= 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 ,iN.

Hence

1 a i i b i i ,iN,

or equivalently,

1 a i i b i i ,iN.

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 ,iN,

i.e.,

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

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 of A lie 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 Ae=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 3.1 If 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 ,iN.

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

Theorem 3.1 Let 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 |+1and a i i >1,iN.

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 } ,jN.

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 3.2 If 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 ji, 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 ,ji,jN.
(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 ,ji,jN,
(2)

and

u i s i ,iN.

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 ,iN.

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 3.3 If 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 ji, 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 ,ji,jN.
(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 ,ji,jN,
(4)

and

u i m i ,iN.

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 ,iN.

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 ,iN.

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 ,iN.

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 3.4 If 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. 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

  1. 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.

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Acknowledgements

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

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Correspondence to Guanghui Cheng.

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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.

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Cheng, G., Tan, Q. & Wang, Z. Some inequalities for the minimum eigenvalue of the Hadamard product of an M-matrix and its inverse. J Inequal Appl 2013, 65 (2013). https://doi.org/10.1186/1029-242X-2013-65

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Keywords

  • Hadamard product
  • M-matrix
  • inverse M-matrix
  • strictly diagonally dominant matrix
  • eigenvalue