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

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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

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