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 $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ is called a nonnegative matrix if $a_{ij}\ge 0$. A matrix $A\in {\mathbb{R}}^{n\times n}$ is called a nonsingular M-matrix [1] if there exist $B\ge 0$ and $s>0$ such that

$$A=s{I}_n-B\text{and}s>\rho (B),$$

where $\rho (B)$ is a spectral radius of the nonnegative matrix B, $I_n$ is the $n\times n$ identity matrix. Denote by ${\mathcal{M}}_n$ the set of all $n\times n$ nonsingular M-matrices. The matrices in ${\mathcal{M}}_n^{-1}:=\{A^{-1}:A\in {\mathcal{M}}_n\}$ are called inverse M-matrices. Let us denote

$$\tau (A)=min\{Re\lambda :\lambda \in \sigma (A)\},$$

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

$$\tau (A)=\frac{1}{\rho \left(A^{-1}\right)}$$

is a positive real eigenvalue of $A\in {\mathcal{M}}_n$ and the corresponding eigenvector is nonnegative. Indeed

$$\tau (A)=s-\rho (B),$$

if $A=s{I}_n-B$, where $s>\rho (B)$, $B\ge 0$.

For any two $n\times n$ matrices $A=(a_{ij})$ and $B=(b_{ij})$, the Hadamard product of A and B is $A\circ B=(a_{ij}{b}_{ij})$. If $A,B\in {\mathcal{M}}_n$, 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

$$PA{P}^T=\left[\begin{array}{cc}{A}_{1,1}& A_{1,2}\\ 0& A_{2,2}\end{array}\right],$$

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 $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ 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 $A\in {\mathcal{M}}_n$, for example, $\tau (A\circ A^{-1})\le 1$ has been proven by Fiedler et al. in [4]. Subsequently, $\tau (A\circ A^{-1})>\frac{1}{n}$ was given by Fiedler and Markham in [3], and they conjectured that $\tau (A\circ A^{-1})>\frac{2}{n}$. Song [5], Yong [6] and Chen [7] have independently proven this conjecture. In [8], Li et al. improved the conjecture $\tau (A\circ A^{-1})\ge \frac{2}{n}$ when $A^{-1}$ is a doubly stochastic matrix and gave the following result:

$$\tau (A\circ A^{-1})\ge \underset{i}{min}\left\{\frac{a_{ii}-s_i{R}_i}{1+{\sum }_{j\ne i}{s}_{ji}}\right\}.$$

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

$$\tau (A\circ A^{-1})\ge \underset{i}{min}\left\{\frac{a_{ii}-m_i{R}_i}{1+{\sum }_{j\ne i}{m}_{ji}}\right\}.$$

Furthermore, if $a_{11}=a_{22}=\cdots =a_{nn}$, they have obtained

$$\underset{i}{min}\left\{\frac{a_{ii}-m_i{R}_i}{1+{\sum }_{j\ne i}{m}_{ji}}\right\}\ge \underset{i}{min}\left\{\frac{a_{ii}-s_i{R}_i}{1+{\sum }_{j\ne i}{s}_{ji}}\right\},$$

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 $\tau (A\circ A^{-1})$. 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 $A\in {\mathbb{R}}^{n\times n}$ be a strictly diagonally dominant matrix by row, i.e.,

$$|a_{ii}|>\sum _{j\ne i}|a_{ij}|,\mathrm{\forall }i\in N.$$

If $A^{-1}=(b_{ij})$, then

$$|b_{ji}|\le \frac{{\sum }_{k\ne j}|a_{jk}|}{|a_{jj}|}|b_{ii}|,j\ne i,\mathrm{\forall }j\in N.$$

Lemma 2.2 Let $A\in {\mathbb{R}}^{n\times n}$ be a strictly diagonally dominant M-matrix by row. If $A^{-1}=(b_{ij})$, then

$$b_{ji}\le \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}}{a_{jj}}{b}_{ii}\le u_j{b}_{ii},j\ne i,\mathrm{\forall }i\in N.$$

Proof Firstly, we consider $A\in {\mathbb{R}}^{n\times n}$ is a strictly diagonally dominant M-matrix by row. For $i\in N$, let

$$r_i(\epsilon )=\underset{j\ne i}{max}\left\{\frac{|a_{ji}|+\epsilon }{a_{jj}-{\sum }_{k\ne j,i}|a_{jk}|}\right\}$$

and

$$m_{ji}(\epsilon )=\frac{r_i(\epsilon )({\sum }_{k\ne j,i}|a_{jk}|+\epsilon )+|a_{ji}|}{a_{jj}},j\ne i.$$

Since A is strictly diagonally dominant, then $r_{ji}<1$ and $m_{ji}<1$. Therefore, there exists $\epsilon >0$ such that $0<r_i(\epsilon )<1$ and $0<m_{ji}(\epsilon )<1$. Let us define one positive diagonal matrix

$$M_i(\epsilon )=diag(m_{1i}(\epsilon ),\dots ,m_{i-1,i}(\epsilon ),1,m_{i+1,i}(\epsilon ),\dots ,m_{ni}(\epsilon )).$$

Similarly to the proofs of Theorem 2.1 and Theorem 2.4 in [8], we can prove that the matrix $A{M}_i(\epsilon )$ 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:

$$m_{ji}^{-1}(\epsilon )b_{ji}\le \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}(\epsilon )}{m_{ji}(\epsilon )a_{jj}}{b}_{ii},j\ne i,j\in N,$$

i.e.,

$$b_{ji}\le \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}(\epsilon )}{a_{jj}}{b}_{ii},j\ne i,j\in N.$$

Let $\epsilon ⟶0^{+}$ to get

$$b_{ji}\le \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}}{a_{jj}}{b}_{ii}\le u_j{b}_{ii},j\ne i,j\in N.$$

This proof is completed. □

Lemma 2.3 Let $A=(a_{ij})\in {\mathcal{M}}_n$ be a strictly diagonally dominant matrix by row and $A^{-1}=(b_{ij})$, then we have

$$\frac{1}{a_{ii}}\le b_{ii}\le \frac{1}{a_{ii}-{\sum }_{j\ne i}|a_{ij}|u_{ji}},\mathrm{\forall }i\in N.$$

Proof Let $B=A^{-1}$. Since A is an M-matrix, then $B\ge 0$. By $AB=BA=I_n$, we have

$$1=\sum _{j=1}^n{a}_{ij}{b}_{ji}=a_{ii}{b}_{ii}-\sum _{j\ne i}|a_{ij}|b_{ji},\mathrm{\forall }i\in N.$$

Hence

$$1\le a_{ii}{b}_{ii},\mathrm{\forall }i\in N,$$

or equivalently,

$$\frac{1}{a_{ii}}\le b_{ii},\mathrm{\forall }i\in N.$$

Furthermore, by Lemma 2.2, we get

$$1=a_{ii}{b}_{ii}-\sum _{j\ne i}|a_{ij}|b_{ji}\ge (a_{ii}-\sum _{j\ne i}|a_{ij}|u_{ji})b_{ii},\mathrm{\forall }i\in N,$$

i.e.,

$$b_{ii}\le \frac{1}{a_{ii}-{\sum }_{j\ne i}|a_{ij}|u_{ji}},\mathrm{\forall }i\in N.$$

Thus the proof is completed. □

Lemma 2.4 [10]

Let $A\in {\mathbb{C}}^{n\times n}$ and $x_1,x_2,\dots ,x_n$ be positive real numbers. Then all the eigenvalues of A lie in the region

$$\bigcup _1^n\{z\in C:|z-a_{ii}|\le x_i\sum _{j\ne i}\frac{1}{x_j}|a_{ji}|\}.$$

Lemma 2.5 [11]

If $A^{-1}$ is a doubly stochastic matrix, then $Ae=e$, $A^{T}e=e$, where $e={(1,1,\dots ,1)}^T$.

3 Main results

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

Lemma 3.1 If $A\in {\mathcal{M}}_n$ and $A^{-1}=(b_{ij})$ is a doubly stochastic matrix, then

$$b_{ii}\ge \frac{1}{1+{\sum }_{j\ne i}{u}_{ji}},\mathrm{\forall }i\in N.$$

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

Theorem 3.1 Let $A\in {\mathcal{M}}_n$ and $A^{-1}=(b_{ij})$ be a doubly stochastic matrix. Then

$$\tau (A\circ A^{-1})\ge \underset{i}{min}\left\{\frac{a_{ii}-u_i{R}_i}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}.$$

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

$$a_{ii}=\sum _{j\ne i}|a_{ij}|+1=\sum _{j\ne i}|a_{ji}|+1\text{and}{a}_{ii}>1,i\in N.$$

Denote

$$u_j=\underset{i\ne j}{max}\{u_{ji}\}=max\left\{\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}}{a_{jj}}\right\},j\in N.$$

Since A is an irreducible matrix, we know that $0<u_j\le 1$. So, by Lemma 2.4, there exists $i_0\in N$ such that

$$|\lambda -a_{i_0{i}_0}{b}_{i_0{i}_0}|\le u_{i_0}\sum _{j\ne i_0}\frac{1}{u_j}|a_{j{i}_0}{b}_{j{i}_0}|,$$

or equivalently,

$$\begin{array}{rcl}|\lambda |& \ge & a_{i_0{i}_0}{b}_{i_0{i}_0}-u_{i_0}\sum _{j\ne i_0}\frac{1}{u_j}|a_{j{i}_0}{b}_{j{i}_0}|\\ \ge & a_{i_0{i}_0}{b}_{i_0{i}_0}-u_{i_0}\sum _{j\ne i_0}\frac{1}{u_j}|a_{j{i}_0}|u_j{b}_{i_0{i}_0}\text{(by Lemma 2.2)}\\ \ge & (a_{i_0{i}_0}-u_{i_0}\sum _{j\ne i_0}|a_{j{i}_0}|)b_{i_0{i}_0}\\ =& (a_{i_0{i}_0}-u_{i_0}{R}_{i_0})b_{i_0{i}_0}\\ \ge & \frac{a_{i_0{i}_0}-u_{i_0}{R}_{i_0}}{1+{\sum }_{j\ne i_0}{u}_{j{i}_0}}\text{(by Lemma 3.1)}\\ \ge & \underset{i}{min}\left\{\frac{a_{ii}-u_i{R}_i}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}.\end{array}$$

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

$$A=\left[\begin{array}{cccc}{A}_{11}& A_{12}& \cdots & A_{1K}\\ 0& A_{22}& \cdots & A_{2K}\\ 0& 0& \cdots & \cdots \\ 0& 0& 0& A_{KK}\end{array}\right],$$

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

Theorem 3.2 If $A=(a_{ij})\in {\mathcal{M}}_n$ is a strictly diagonally dominant by row, then

$$\underset{i}{min}\left\{\frac{a_{ii}-u_i{R}_i}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}\ge \underset{i}{min}\left\{\frac{a_{ii}-s_i{R}_i}{1+{\sum }_{j\ne i}{s}_{ji}}\right\}.$$

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

$$\begin{array}{rcl}{d}_j-m_{ji}& =& \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|}{a_{jj}}-\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|r_i}{a_{jj}}\\ =& \frac{(1-r_i){\sum }_{k\ne j,i}|a_{jk}|}{a_{jj}}\\ \ge & 0,\end{array}$$

or equivalently,

$$d_j\ge m_{ji},j\ne i,\mathrm{\forall }j\in N.$$

(1)

So, we can obtain

$$u_{ji}=\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}}{a_{jj}}\le \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|d_k}{a_{jj}}=s_{ji},j\ne i,\mathrm{\forall }j\in N,$$

(2)

and

$$u_i\le s_i,\mathrm{\forall }i\in N.$$

Therefore, it is easy to obtain that

$$\frac{a_{ii}-u_i{R}_i}{1+{\sum }_{j\ne i}{u}_{ji}}\ge \frac{a_{ii}-s_i{R}_i}{1+{\sum }_{j\ne i}{s}_{ji}},\mathrm{\forall }i\in N.$$

Obviously, we have the desired result

$$\underset{i}{min}\left\{\frac{a_{ii}-u_i{R}_i}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}\ge \underset{i}{min}\left\{\frac{a_{ii}-s_i{R}_i}{1+{\sum }_{j\ne i}{s}_{ji}}\right\}.$$

This proof is completed. □

Theorem 3.3 If $A=(a_{ij})\in {\mathcal{M}}_n$ is strictly diagonally dominant by row, then

$$\underset{i}{min}\left\{\frac{a_{ii}-u_i{R}_i}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}\ge \underset{i}{min}\left\{\frac{a_{ii}-m_i{R}_i}{1+{\sum }_{j\ne i}{m}_{ji}}\right\}.$$

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

$$\begin{array}{rcl}{r}_i-m_{ji}& =& r_i-\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|r_i}{a_{jj}}\\ =& \frac{r_i-|a_{ji}|-{\sum }_{k\ne j,i}|a_{jk}|}{a_{jj}}\\ =& \frac{r_i(a_{jj}-{\sum }_{k\ne j,i}|a_{jk}|)-|a_{ji}|}{a_{jj}}\\ =& \frac{a_{jj}-{\sum }_{k\ne j,i}|a_{jk}|}{a_{jj}}(r_i-\frac{|a_{ji}|}{a_{jj}-{\sum }_{k\ne j,i}|a_{jk}|})\\ \ge & 0,\end{array}$$

i.e.,

$$r_i\ge m_{ji},j\ne i,\mathrm{\forall }j\in N.$$

(3)

So, we can obtain

$$u_{ji}=\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}}{a_{jj}}\le \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|r_i}{a_{jj}}=m_{ji},j\ne i,\mathrm{\forall }j\in N,$$

(4)

and

$$u_i\le m_i,\mathrm{\forall }i\in N.$$

Therefore, it is easy to obtain that

$$\frac{a_{ii}-u_i{R}_i}{1+{\sum }_{j\ne i}{u}_{ji}}\ge \frac{a_{ii}-m_i{R}_i}{1+{\sum }_{j\ne i}{m}_{ji}},\mathrm{\forall }i\in N.$$

Obviously, we have the desired result

$$\underset{i}{min}\left\{\frac{a_{ii}-u_i{R}_i}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}\ge \underset{i}{min}\left\{\frac{a_{ii}-m_i{R}_i}{1+{\sum }_{j\ne i}{m}_{ji}}\right\}.$$

 □

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

$$b_{ji}\le \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}}{a_{jj}}{b}_{ii}\le \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|d_k}{a_{jj}}{b}_{ii},\mathrm{\forall }i\in N.$$

and

$$b_{ji}\le \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}}{a_{jj}}{b}_{ii}\le \frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|r_i}{a_{jj}}{b}_{ii},\mathrm{\forall }i\in N.$$

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 $A\in {\mathcal{M}}_n$ is strictly diagonally dominant by row, then

$$\tau (A\circ A^{-1})\ge \underset{i}{min}\{1-\frac{1}{a_{ii}}\sum _{j\ne i}|a_{ji}|u_{ji}\}.$$

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

$$1-\frac{1}{a_{ii}}\sum _{j\ne i}|a_{ji}|u_{ji}\ge 1-\frac{1}{a_{ii}}\sum _{j\ne i}|a_{ji}|s_{ji},$$

and

$$1-\frac{1}{a_{ii}}\sum _{j\ne i}|a_{ji}|u_{ji}\ge 1-\frac{1}{a_{ii}}\sum _{j\ne i}|a_{ji}|m_{ji}.$$

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:

$$A=\left[\begin{array}{cccc}4& -1& -1& -1\\ -2& 5& -1& -1\\ 0& -2& 4& -1\\ -1& -1& -1& 4\end{array}\right].$$

• 1. Estimate the upper bounds for entries of $A^{-1}=(b_{ij})$. Firstly, by Lemma 2.2(2) in [9], we have

  $$A^{-1}\le \left[\begin{array}{cccc}1& 0.5833& 0.5000& 0.5000\\ 0.6667& 1& 0.5000& 0.5000\\ 0.5000& 0.6667& 1& 0.5000\\ 0.5833& 0.5833& 0.5000& 1\end{array}\right]\circ \left[\begin{array}{cccc}{b}_{11}& b_{22}& b_{33}& b_{44}\\ b_{11}& b_{22}& b_{33}& b_{44}\\ b_{11}& b_{22}& b_{33}& b_{44}\\ b_{11}& b_{22}& b_{33}& b_{44}\end{array}\right].$$

By Lemma 2.2, we have

$$A^{-1}\le \left[\begin{array}{cccc}1& 0.5625& 0.5000& 0.5000\\ 0.6167& 1& 0.5000& 0.5000\\ 0.4792& 0.6458& 1& 0.5000\\ 0.5417& 0.5625& 0.5000& 1\end{array}\right]\circ \left[\begin{array}{cccc}{b}_{11}& b_{22}& b_{33}& b_{44}\\ b_{11}& b_{22}& b_{33}& b_{44}\\ b_{11}& b_{22}& b_{33}& b_{44}\\ b_{11}& b_{22}& b_{33}& b_{44}\end{array}\right].$$

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 $\tau (A\circ A^{-1})$.

By Theorem 3.2 in [9], we obtain

$$0.9755=\tau (A\circ A^{-1})\ge 0.8000.$$

By Theorem 3.1, we obtain

$$0.9755=\tau (A\circ A^{-1})\ge 0.8250.$$