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

Guanghui Cheng; Qin Tan; Zhuande Wang

doi:10.1186/1029-242X-2013-65

Research
Open Access

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

Guanghui Cheng¹Email author,
Qin Tan¹ and
Zhuande Wang¹

Journal of Inequalities and Applications20132013:65

https://doi.org/10.1186/1029-242X-2013-65

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

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}) \in R^{n \times n}

is called a nonnegative matrix if

a_{i j} \geq 0

. A matrix

A \in R^{n \times n}

is called a nonsingular M-matrix [1] if there exist

B \geq 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 \times n

identity matrix. Denote by

M_{n}

the set of all

n \times n

nonsingular M-matrices. The matrices in

M_{n}^{- 1} : = {A^{- 1} : A \in M_{n}}

are called inverse M-matrices. Let us denote

τ (A) = min {Re λ : λ \in σ (A)},

and

σ (A)

denotes the spectrum of A. It is known that [2]

τ (A) = \frac{1}{ρ (A^{- 1})}

is a positive real eigenvalue of

A \in M_{n}

and the corresponding eigenvector is nonnegative. Indeed

τ (A) = s - ρ (B),

if $A = s I_{n} - B$ , where $s > ρ (B)$ , $B \geq 0$ .

For any two $n \times n$ matrices $A = (a_{i j})$ and $B = (b_{i j})$ , the Hadamard product of A and B is $A \circ B = (a_{i j} b_{i j})$ . If $A, B \in M_{n}$ , then $A \circ B^{- 1}$ 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} = [\begin{array}{c} A_{1, 1} & A_{1, 2} \\ 0 & A_{2, 2} \end{array}],

where $A_{1, 1}$ and $A_{2, 2}$ are square matrices.

For convenience, the set

{1, 2, \dots, n}

is denoted by N, where n (≥3) is any positive integer. Let

A = (a_{i j}) \in R^{n \times 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 \in M_{n}

, for example,

τ (A \circ A^{- 1}) \leq 1

has been proven by Fiedler et al. in [4]. Subsequently,

τ (A \circ A^{- 1}) > \frac{1}{n}

was given by Fiedler and Markham in [3], and they conjectured that

τ (A \circ A^{- 1}) > \frac{2}{n}

. Song [5], Yong [6] and Chen [7] have independently proven this conjecture. In [8], Li et al. improved the conjecture

τ (A \circ A^{- 1}) \geq \frac{2}{n}

when

A^{- 1}

is a doubly stochastic matrix and gave the following result:

τ (A \circ A^{- 1}) \geq min_{i} {\frac{a_{i i} - s_{i} R_{i}}{1 + \sum_{j \neq i} s_{j i}}} .

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

τ (A \circ A^{- 1}) \geq min_{i} {\frac{a_{i i} - m_{i} R_{i}}{1 + \sum_{j \neq i} m_{j i}}} .

Furthermore, if

a_{11} = a_{22} = \dots = a_{n n}

, they have obtained

min_{i} {\frac{a_{i i} - m_{i} R_{i}}{1 + \sum_{j \neq i} m_{j i}}} \geq min_{i} {\frac{a_{i i} - s_{i} R_{i}}{1 + \sum_{j \neq 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 $τ (A \circ A^{- 1})$ . 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 \in R^{n \times n}

be a strictly diagonally dominant matrix by row, i.e.,

| a_{i i} | > \sum_{j \neq i} | a_{i j} |, \forall i \in N .

If

A^{- 1} = (b_{i j})

, then

| b_{j i} | \leq \frac{\sum_{k \neq j} | a_{j k} |}{| a_{j j} |} | b_{i i} |, j \neq i, \forall j \in N .

Lemma 2.2 Let

A \in R^{n \times n}

be a strictly diagonally dominant M-matrix by row. If

A^{- 1} = (b_{i j})

, then

b_{j i} \leq \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | m_{k i}}{a_{j j}} b_{i i} \leq u_{j} b_{i i}, j \neq i, \forall i \in N .

Proof Firstly, we consider

A \in R^{n \times n}

is a strictly diagonally dominant M-matrix by row. For

i \in N

, let

r_{i} (ε) = max_{j \neq i} {\frac{| a_{j i} | + ε}{a_{j j} - \sum_{k \neq j, i} | a_{j k} |}}

and

m_{j i} (ε) = \frac{r_{i} (ε) (\sum_{k \neq j, i} | a_{j k} | + ε) + | a_{j i} |}{a_{j j}}, j \neq 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), \dots, m_{i - 1, i} (ε), 1, m_{i + 1, i} (ε), \dots, 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 \in N

. Furthermore, by Lemma 2.1, we can obtain the following result:

m_{j i}^{- 1} (ε) b_{j i} \leq \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | m_{k i} (ε)}{m_{j i} (ε) a_{j j}} b_{i i}, j \neq i, j \in N,

i.e.,

b_{j i} \leq \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | m_{k i} (ε)}{a_{j j}} b_{i i}, j \neq i, j \in N .

Let

ε ⟶ 0^{+}

to get

b_{j i} \leq \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | m_{k i}}{a_{j j}} b_{i i} \leq u_{j} b_{i i}, j \neq i, j \in N .

This proof is completed. □

Lemma 2.3 Let

A = (a_{i j}) \in M_{n}

be a strictly diagonally dominant matrix by row and

A^{- 1} = (b_{i j})

, then we have

\frac{1}{a_{i i}} \leq b_{i i} \leq \frac{1}{a_{i i} - \sum_{j \neq i} | a_{i j} | u_{j i}}, \forall i \in N .

Proof Let

B = A^{- 1}

. Since A is an M-matrix, then

B \geq 0

. By

A B = B A = I_{n}

, we have

1 = \sum_{j = 1}^{n} a_{i j} b_{j i} = a_{i i} b_{i i} - \sum_{j \neq i} | a_{i j} | b_{j i}, \forall i \in N .

Hence

1 \leq a_{i i} b_{i i}, \forall i \in N,

or equivalently,

\frac{1}{a_{i i}} \leq b_{i i}, \forall i \in N .

Furthermore, by Lemma 2.2, we get

1 = a_{i i} b_{i i} - \sum_{j \neq i} | a_{i j} | b_{j i} \geq (a_{i i} - \sum_{j \neq i} | a_{i j} | u_{j i}) b_{i i}, \forall i \in N,

i.e.,

b_{i i} \leq \frac{1}{a_{i i} - \sum_{j \neq i} | a_{i j} | u_{j i}}, \forall i \in N .

Thus the proof is completed. □

Lemma 2.4 [10]

Let

A \in C^{n \times n}

and

x_{1}, x_{2}, \dots, x_{n}

be positive real numbers. Then all the eigenvalues of A lie in the region

⋃_{1}^{n} {z \in C : | z - a_{i i} | \leq x_{i} \sum_{j \neq i} \frac{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 $e = {(1, 1, \dots, 1)}^{T}$ .

3 Main results

In this section, we give two new lower bounds for $τ (A \circ A^{- 1})$ which improve the ones in [8] and [9].

Lemma 3.1 If

A \in M_{n}

and

A^{- 1} = (b_{i j})

is a doubly stochastic matrix, then

b_{i i} \geq \frac{1}{1 + \sum_{j \neq i} u_{j i}}, \forall i \in N .

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 \in M_{n}

and

A^{- 1} = (b_{i j})

be a doubly stochastic matrix. Then

τ (A \circ A^{- 1}) \geq min_{i} {\frac{a_{i i} - u_{i} R_{i}}{1 + \sum_{j \neq i} u_{j i}}} .

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

a_{i i} = \sum_{j \neq i} | a_{i j} | + 1 = \sum_{j \neq i} | a_{j i} | + 1 and a_{i i} > 1, i \in N .

Denote

u_{j} = max_{i \neq j} {u_{j i}} = max {\frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | m_{k i}}{a_{j j}}}, j \in N .

Since A is an irreducible matrix, we know that

0 < u_{j} \leq 1

. So, by Lemma 2.4, there exists

i_{0} \in N

such that

| λ - a_{i_{0} i_{0}} b_{i_{0} i_{0}} | \leq u_{i_{0}} \sum_{j \neq i_{0}} \frac{1}{u_{j}} | a_{j i_{0}} b_{j i_{0}} |,

or equivalently,

\begin{array}{rcl} | λ | & \geq & a_{i_{0} i_{0}} b_{i_{0} i_{0}} - u_{i_{0}} \sum_{j \neq i_{0}} \frac{1}{u_{j}} | a_{j i_{0}} b_{j i_{0}} | \\ \geq & a_{i_{0} i_{0}} b_{i_{0} i_{0}} - u_{i_{0}} \sum_{j \neq i_{0}} \frac{1}{u_{j}} | a_{j i_{0}} | u_{j} b_{i_{0} i_{0}} (by Lemma 2.2) \\ \geq & (a_{i_{0} i_{0}} - u_{i_{0}} \sum_{j \neq 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}} \\ \geq & \frac{a_{i_{0} i_{0}} - u_{i_{0}} R_{i_{0}}}{1 + \sum_{j \neq i_{0}} u_{j i_{0}}} (by Lemma 3.1) \\ \geq & min_{i} {\frac{a_{i i} - u_{i} R_{i}}{1 + \sum_{j \neq i} u_{j i}}} . \end{array}

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

A = [\begin{array}{c} A_{11} & A_{12} & \dots & A_{1 K} \\ 0 & A_{22} & \dots & A_{2 K} \\ 0 & 0 & \dots & \dots \\ 0 & 0 & 0 & A_{K K} \end{array}],

where $A_{i i} \in M_{n_{i}}$ is an irreducible diagonal block matrix, $i = 1, 2, \dots, K$ . Obviously, $τ (A \circ A^{- 1}) = {min}_{i} τ (A_{i i} \circ A_{i i}^{- 1})$ . Thus the reducible case is converted into the irreducible case. This proof is completed. □

Theorem 3.2 If

A = (a_{i j}) \in M_{n}

is a strictly diagonally dominant by row, then

min_{i} {\frac{a_{i i} - u_{i} R_{i}}{1 + \sum_{j \neq i} u_{j i}}} \geq min_{i} {\frac{a_{i i} - s_{i} R_{i}}{1 + \sum_{j \neq i} s_{j i}}} .

Proof Since A is strictly diagonally dominant by row, for any

j \neq i

, we have

\begin{array}{rcl} d_{j} - m_{j i} & = & \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} |}{a_{j j}} - \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | r_{i}}{a_{j j}} \\ = & \frac{(1 - r_{i}) \sum_{k \neq j, i} | a_{j k} |}{a_{j j}} \\ \geq & 0, \end{array}

or equivalently,

d_{j} \geq m_{j i}, j \neq i, \forall j \in N .

(1)

So, we can obtain

u_{j i} = \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | m_{k i}}{a_{j j}} \leq \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | d_{k}}{a_{j j}} = s_{j i}, j \neq i, \forall j \in N,

(2)

and

u_{i} \leq s_{i}, \forall i \in N .

Therefore, it is easy to obtain that

\frac{a_{i i} - u_{i} R_{i}}{1 + \sum_{j \neq i} u_{j i}} \geq \frac{a_{i i} - s_{i} R_{i}}{1 + \sum_{j \neq i} s_{j i}}, \forall i \in N .

Obviously, we have the desired result

min_{i} {\frac{a_{i i} - u_{i} R_{i}}{1 + \sum_{j \neq i} u_{j i}}} \geq min_{i} {\frac{a_{i i} - s_{i} R_{i}}{1 + \sum_{j \neq i} s_{j i}}} .

This proof is completed. □

Theorem 3.3 If

A = (a_{i j}) \in M_{n}

is strictly diagonally dominant by row, then

min_{i} {\frac{a_{i i} - u_{i} R_{i}}{1 + \sum_{j \neq i} u_{j i}}} \geq min_{i} {\frac{a_{i i} - m_{i} R_{i}}{1 + \sum_{j \neq i} m_{j i}}} .

Proof Since A is strictly diagonally dominant by row, for any

j \neq i

, we have

\begin{array}{rcl} r_{i} - m_{j i} & = & r_{i} - \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | r_{i}}{a_{j j}} \\ = & \frac{r_{i} - | a_{j i} | - \sum_{k \neq j, i} | a_{j k} |}{a_{j j}} \\ = & \frac{r_{i} (a_{j j} - \sum_{k \neq j, i} | a_{j k} |) - | a_{j i} |}{a_{j j}} \\ = & \frac{a_{j j} - \sum_{k \neq j, i} | a_{j k} |}{a_{j j}} (r_{i} - \frac{| a_{j i} |}{a_{j j} - \sum_{k \neq j, i} | a_{j k} |}) \\ \geq & 0, \end{array}

i.e.,

r_{i} \geq m_{j i}, j \neq i, \forall j \in N .

(3)

So, we can obtain

u_{j i} = \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | m_{k i}}{a_{j j}} \leq \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | r_{i}}{a_{j j}} = m_{j i}, j \neq i, \forall j \in N,

(4)

and

u_{i} \leq m_{i}, \forall i \in N .

Therefore, it is easy to obtain that

\frac{a_{i i} - u_{i} R_{i}}{1 + \sum_{j \neq i} u_{j i}} \geq \frac{a_{i i} - m_{i} R_{i}}{1 + \sum_{j \neq i} m_{j i}}, \forall i \in N .

Obviously, we have the desired result

min_{i} {\frac{a_{i i} - u_{i} R_{i}}{1 + \sum_{j \neq i} u_{j i}}} \geq min_{i} {\frac{a_{i i} - m_{i} R_{i}}{1 + \sum_{j \neq i} m_{j i}}} .

□

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

b_{j i} \leq \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | m_{k i}}{a_{j j}} b_{i i} \leq \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | d_{k}}{a_{j j}} b_{i i}, \forall i \in N .

and

b_{j i} \leq \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | m_{k i}}{a_{j j}} b_{i i} \leq \frac{| a_{j i} | + \sum_{k \neq j, i} | a_{j k} | r_{i}}{a_{j j}} b_{i i}, \forall i \in 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 3.4 If

A \in M_{n}

is strictly diagonally dominant by row, then

τ (A \circ A^{- 1}) \geq min_{i} {1 - \frac{1}{a_{i i}} \sum_{j \neq 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 - \frac{1}{a_{i i}} \sum_{j \neq i} | a_{j i} | u_{j i} \geq 1 - \frac{1}{a_{i i}} \sum_{j \neq i} | a_{j i} | s_{j i},

and

1 - \frac{1}{a_{i i}} \sum_{j \neq i} | a_{j i} | u_{j i} \geq 1 - \frac{1}{a_{i i}} \sum_{j \neq 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 = [\begin{array}{c} 4 & - 1 & - 1 & - 1 \\ - 2 & 5 & - 1 & - 1 \\ 0 & - 2 & 4 & - 1 \\ - 1 & - 1 & - 1 & 4 \end{array}] .

1.

Estimate the upper bounds for entries of $A^{- 1} = (b_{i j})$ . Firstly, by Lemma 2.2(2) in [9], we have
$A^{- 1} \leq [\begin{array}{c} 1 & 0.5833 & 0.5000 & 0.5000 \\ 0.6667 & 1 & 0.5000 & 0.5000 \\ 0.5000 & 0.6667 & 1 & 0.5000 \\ 0.5833 & 0.5833 & 0.5000 & 1 \end{array}] \circ [\begin{array}{c} b_{11} & b_{22} & b_{33} & b_{44} \\ b_{11} & b_{22} & b_{33} & b_{44} \\ b_{11} & b_{22} & b_{33} & b_{44} \\ b_{11} & b_{22} & b_{33} & b_{44} \end{array}] .$

By Lemma 2.2, we have

A^{- 1} \leq [\begin{array}{c} 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 \end{array}] \circ [\begin{array}{c} 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} \end{array}] .

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 $τ (A \circ A^{- 1})$ .

By Theorem 3.2 in [9], we obtain

0.9755 = τ (A \circ A^{- 1}) \geq 0.8000 .

By Theorem 3.1, we obtain

0.9755 = τ (A \circ A^{- 1}) \geq 0.8250 .

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)

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

References

Berman A, Plemmons RJ Classics in Applied Mathematics 9. In Nonnegative Matrices in the Mathematical Sciences. SIAM, Philadelphia; 1994.View ArticleGoogle Scholar
Horn RA, Johnson CR: Topics in Matrix Analysis. Cambridge University Press, Cambridge; 1991.MATHView ArticleGoogle Scholar
Fiedler M, Markham TL: An inequality for the Hadamard product of an M -matrix and inverse M -matrix. Linear Algebra Appl. 1988, 101: 1–8.MATHMathSciNetView ArticleGoogle Scholar
Fiedler M, Johnson CR, Markham T, Neumann M: A trace inequality for M -matrices and the symmetrizability of a real matrix by a positive diagonal matrix. Linear Algebra Appl. 1985, 71: 81–94.MATHMathSciNetView ArticleGoogle Scholar
Song YZ: On an inequality for the Hadamard product of an M -matrix and its inverse. Linear Algebra Appl. 2000, 305: 99–105. 10.1016/S0024-3795(99)00224-4MATHMathSciNetView ArticleGoogle Scholar
Yong XR: Proof of a conjecture of Fiedler and Markham. Linear Algebra Appl. 2000, 320: 167–171. 10.1016/S0024-3795(00)00211-1MATHMathSciNetView ArticleGoogle Scholar
Chen SC: A lower bound for the minimum eigenvalue of the Hadamard product of matrix. Linear Algebra Appl. 2004, 378: 159–166.MATHMathSciNetView ArticleGoogle Scholar
Li HB, Huang TZ, Shen SQ, Li H: Lower bounds for the eigenvalue of Hadamard product of an M -matrix and its inverse. Linear Algebra Appl. 2007, 420: 235–247. 10.1016/j.laa.2006.07.008MATHMathSciNetView ArticleGoogle Scholar
Li YT, Chen FB, Wang DF: New lower bounds on eigenvalue of the Hadamard product of an M -matrix and its inverse. Linear Algebra Appl. 2009, 430: 1423–1431. 10.1016/j.laa.2008.11.002MATHMathSciNetView ArticleGoogle Scholar
Varga RS: Minimal Gerschgorin sets. Pac. J. Math. 1965, 15(2):719–729. 10.2140/pjm.1965.15.719MATHView ArticleGoogle Scholar
Sinkhorn R: A relationship between arbitrary positive matrices and doubly stochastic matrices. Ann. Math. Stat. 1964, 35: 876–879. 10.1214/aoms/1177703591MATHMathSciNetView ArticleGoogle Scholar

Copyright

This article is published under license to BioMed Central Ltd. 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.