Some new bounds for the minimum eigenvalue of the Hadamard product of an M-matrix and an inverse M-matrix

Abstract

Let A and B be nonsingular M-matrices. Several new bounds on the minimum eigenvalue for the Hadamard product of B and the inverse matrix of A are given. These bounds can improve considerably some previous results.

MSC:15A42, 15B34.

1 Introduction

Let ${\mathbb{C}}^{n\times n}$ (${\mathbb{R}}^{n\times n}$) denote the set of all $n\times n$ complex (real) matrices, $A=(a_{ij})\in {\mathbb{C}}^{n\times n}$, $N=\{1,2,\dots ,n\}$. We write $A\ge 0$ if $a_{ij}\ge 0$ for any $i,j\in N$. If $A\ge 0$, A is called a nonnegative matrix. The spectral radius of A is denoted by $\rho (A)$.

We denote by $Z_n$ the class of all $n\times n$ real matrices, whose off-diagonal entries are nonpositive. A matrix $A=(a_{ij})\in Z_n$ is called a nonsingular M-matrix if there exist a nonnegative matrix B and a nonnegative real number s such that $A=sI-B$ with $s>\rho (B)$, where I is the identity matrix. $M_n$ will be used to denote the set of all $n\times n$ nonsingular M-matrices. Let us denote $\tau (A)=min\{Re(\lambda ):\lambda \in \sigma (A)\}$, where $\sigma (A)$ denotes the spectrum of A.

The Hadamard product of two matrices $A=(a_{ij})\in {\mathbb{C}}^{n\times n}$ and $B=(b_{ij})\in {\mathbb{C}}^{n\times n}$ is the matrix $A\circ B=(a_{ij}{b}_{ij})\in {\mathbb{C}}^{n\times n}$. If $A,B\in M_n$, then $B\circ A^{-1}$ is also an M-matrix (see [1]).

Let $A=(a_{ij})$ be an $n\times n$ matrix with all diagonal entries being nonzero throughout. For $i,j,k\in N$, $i\ne j$, denote

$$\begin{array}{c}{R}_i=\sum _{j\ne i}|a_{ij}|,d_i=\frac{R_i}{|a_{ii}|};\hfill \\ r_{ji}=\frac{|a_{ji}|}{|a_{jj}|-{\sum }_{k\ne j,i}|a_{jk}|},r_i=\underset{j\ne i}{max}\{r_{ji}\};\hfill \\ m_{ji}=\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|r_i}{|a_{jj}|},m_i=\underset{j\ne i}{max}\{m_{ij}\};\hfill \\ u_{ji}=\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}}{|a_{jj}|},u_i=\underset{j\ne i}{max}\{u_{ij}\}.\hfill \end{array}$$

In 2013, Zhou et al. [2] obtained the following result: If $A=(a_{ij})\in M_n$ is a strictly row diagonally dominant matrix, $B=(b_{ij})\in M_n$ and $A^{-1}=({\alpha }_{ij})$, then

$$\tau (B\circ A^{-1})\ge \underset{i\in N}{min}\left\{\frac{b_{ii}-m_i{\sum }_{j\ne i}|b_{ji}|}{a_{ii}}\right\}.$$

(1)

In 2013, Cheng et al. [3] presented the following result: If $A=(a_{ij})\in M_n$ and $A^{-1}=({\alpha }_{ij})$ is a doubly stochastic matrix, then

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

(2)

In this paper, we present some new lower bounds of $\tau (B\circ A^{-1})$ and $\tau (A\circ A^{-1})$, which improve (1) and (2).

2 Main results

In this section, we present our main results. Firstly, we give some lemmas.

Lemma 1 [4]

Let $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$. If A is a strictly row diagonally dominant matrix, then $A^{-1}=({\alpha }_{ij})$ satisfies

$$|{\alpha }_{ji}|\le d_j|{\alpha }_{ii}|,j,i\in N,j\ne i.$$

Lemma 2 Let $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$. If A is a strictly row diagonally dominant M-matrix, then $A^{-1}=({\alpha }_{ij})$ satisfies

$${\alpha }_{ji}\le w_{ji}{\alpha }_{ii},j,i\in N,j\ne i,$$

where

$$w_{ji}=\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|m_{ki}{h}_i}{|a_{jj}|},h_i=\underset{j\ne i}{max}\left\{\frac{|a_{ji}|}{|a_{jj}|m_{ji}-{\sum }_{k\ne j,i}|a_{jk}|m_{ki}}\right\}.$$

Proof This proof is similar to the one of Lemma 2.2 in [3]. □

Lemma 3 If $A=(a_{ij})\in M_n$ and $A^{-1}=({\alpha }_{ij})$ is a doubly stochastic matrix, then

$${\alpha }_{ii}\ge \frac{1}{1+{\sum }_{j\ne i}{w}_{ji}},i\in N,$$

where $w_{ji}$ is defined as in Lemma  2.

Proof This proof is similar to the one of Lemma 3.1 in [3]. □

Lemma 4 [4]

If $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ is a strictly row diagonally dominant M-matrix, then, for $A^{-1}=({\alpha }_{ij})$,

$${\alpha }_{ii}\ge \frac{1}{a_{ii}},i\in N.$$

Lemma 5 [5]

If $A=(a_{ij})\in {\mathbb{C}}^{n\times n}$ and $x_1,x_2,\dots ,x_n$ are positive real numbers, then all the eigenvalues of A lie in the region

$$\bigcup _{i\ne j}\{z\in \mathbb{C}:|z-a_{ii}|\le x_i\sum _{k\ne i}\frac{1}{x_k}|a_{ki}|,i\in N\}.$$

Lemma 6 [6]

If $A=(a_{ij})\in {\mathbb{C}}^{n\times n}$ and $x_1,x_2,\dots ,x_n$ are positive real numbers, then all the eigenvalues of A lie in the region

$$\bigcup _{i\ne j}\{z\in \mathbb{C}:|z-a_{ii}||z-a_{jj}|\le (x_i\sum _{k\ne i}\frac{1}{x_k}|a_{ki}|)(x_j\sum _{k\ne j}\frac{1}{x_k}|a_{kj}|),i,j\in N\}.$$

Theorem 1 If $A=(a_{ij})$, $B=(b_{ij})\in M_n$ and $A^{-1}=({\alpha }_{ij})$, then

$$\tau (B\circ A^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{b_{ii}-w_i{\sum }_{j\ne i}|b_{ji}|}{a_{ii}}\right\},$$

(3)

where $w_i={max}_{j\ne i}\{w_{ij}\}$ and $w_{ij}$ is defined as in Lemma  2.

Proof It is evident that the result holds with equality for $n=1$.

We next assume that $n\ge 2$.

Since A is an M-matrix, there exists a positive diagonal matrix D such that $D^{-1}AD$ is a strictly row diagonally dominant M-matrix, and

$$\tau (B\circ A^{-1})=\tau \left(D^{-1}(B\circ A^{-1})D\right)=\tau (B\circ {\left(D^{-1}AD\right)}^{-1}).$$

Therefore, for convenience and without loss of generality, we assume that A is a strictly row diagonally dominant matrix.

(i) First, we assume that A and B are irreducible matrices. Then, for any $i\in N$, we have $0<w_i<1$. Since $\tau (B\circ A^{-1})$ is an eigenvalue of $B\circ A^{-1}$, then by Lemma 2 and Lemma 5, there exists an i such that

$$\begin{array}{rcl}|\tau (B\circ A^{-1})-b_{ii}{\alpha }_{ii}|& \le & w_i\sum _{j\ne i}\frac{1}{w_j}|b_{ji}{\alpha }_{ji}|\le w_i\sum _{j\ne i}\frac{1}{w_j}|b_{ji}|w_{ji}|{\alpha }_{ii}|\\ \le & w_i\sum _{j\ne i}\frac{1}{w_j}|b_{ji}|w_j|{\alpha }_{ii}|=w_i|{\alpha }_{ii}|\sum _{j\ne i}|b_{ji}|.\end{array}$$

By Lemma 4, the above inequality and $0\le \tau (B\circ A^{-1})\le b_{ii}{\alpha }_{ii}$, for any $i\in N$, we obtain

$$\left|\tau (B\circ A^{-1})\right|\ge b_{ii}{\alpha }_{ii}-w_i|{\alpha }_{ii}|\sum _{j\ne i}|b_{ji}|\ge \frac{b_{ii}-w_i{\sum }_{j\ne i}|b_{ji}|}{a_{ii}}\ge \underset{1\le i\le n}{min}\left\{\frac{b_{ii}-w_i{\sum }_{j\ne i}|b_{ji}|}{a_{ii}}\right\}.$$

(ii) Now, assume that one of A and B is reducible. It is well known that a matrix in $Z_n$ is a nonsingular M-matrix if and only if all its leading principal minors are positive (see [7]). If we denote by $T=(t_{ij})$ the $n\times n$ permutation matrix with $t_{12}=t_{23}=\cdots =t_{n-1,n}=t_{n1}=1$, the remaining $t_{ij}$ zero, then both $A-ϵT$ and $B-ϵT$ are irreducible nonsingular M-matrices for any chosen positive real number ϵ sufficiently small such that all the leading principal minors of both $A-ϵT$ and $B-ϵT$ are positive. Now, we substitute $A-ϵT$ and $B-ϵT$ for A and B, respectively, in the previous case, and then letting $ϵ\to 0$, the result follows by continuity. □

From Lemma 3 and Theorem 1, we can easily obtain the following corollaries.

Corollary 1 If $A=(a_{ij}),B=(b_{ij})\in M_n$ and $A^{-1}=({\alpha }_{ij})$ is a doubly stochastic matrix, then

$$\tau (B\circ A^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{b_{ii}-w_i{\sum }_{j\ne i}|b_{ji}|}{1+{\sum }_{j\ne i}{w}_{ji}}\right\}.$$

Corollary 2 If $A=(a_{ij})\in M_n$ and $A^{-1}=({\alpha }_{ij})$ is a doubly stochastic matrix, then

$$\tau (A\circ A^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{a_{ii}-w_i{\sum }_{j\ne i}|a_{ji}|}{1+{\sum }_{j\ne i}{w}_{ji}}\right\}.$$

(4)

Remark 1 We next give a simple comparison between (3) and (1), (4) and (2), respectively. Since $m_{ji}{h}_i\le r_i$, $0\le h_i\le 1$, $j,i\in N$, $j\ne i$, then $w_{ji}\le m_{ji}$, $w_i\le m_i$ and $w_{ji}\le u_{ji}$, $w_i\le u_i$ for any $j,i\in N$, $j\ne i$. Therefore,

$$\begin{array}{c}\tau (B\circ A^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{b_{ii}-w_i{\sum }_{j\ne i}|b_{ji}|}{a_{ii}}\right\}\ge \underset{1\le i\le n}{min}\left\{\frac{b_{ii}-m_i{\sum }_{j\ne i}|b_{ji}|}{a_{ii}}\right\},\hfill \\ \tau (A\circ A^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{a_{ii}-w_i{\sum }_{j\ne i}|a_{ji}|}{1+{\sum }_{j\ne i}{w}_{ji}}\right\}\ge \underset{1\le i\le n}{min}\left\{\frac{a_{ii}-u_i{\sum }_{j\ne i}|a_{ji}|}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}.\hfill \end{array}$$

So, the bound in (3) is bigger than the bound in (1) and the bound in (4) is bigger than the bound in (2).

Theorem 2 If $A=(a_{ij}),B=(b_{ij})\in M_n$ and $A^{-1}=({\alpha }_{ij})$, then

$$\begin{array}{rl}\tau (B\circ A^{-1})\ge & \underset{i\ne j}{min}\frac{1}{2}\{{\alpha }_{ii}{b}_{ii}+{\alpha }_{jj}{b}_{jj}-[{({\alpha }_{ii}{b}_{ii}-{\alpha }_{jj}{b}_{jj})}^2\\ {+4(w_i\sum _{k\ne i}|b_{ki}|{\alpha }_{ii})(w_j\sum _{k\ne j}|b_{kj}|{\alpha }_{jj})]}^{\frac{1}{2}}\},\end{array}$$

where $w_i$ ($i\in N$) is defined as in Theorem  1.

Proof It is evident that the result holds with equality for $n=1$.

We next assume that $n\ge 2$. For convenience and without loss of generality, we assume that A is a strictly row diagonally dominant matrix.

(i)First, we assume that A and B are irreducible matrices. Let $R_j^{\sigma }={\sum }_{k\ne j}|a_{jk}|m_{ki}{h}_i$, $j,i\in N$, $j\ne i$. Then, for any $j,i\in N$, $j\ne i$, we have

$$R_j^{\sigma }=\sum _{k\ne j}|a_{jk}|m_{ki}{h}_i\le |a_{ji}|+\sum _{k\ne j,i}|a_{jk}|m_{ki}{h}_i\le R_j<a_{jj}.$$

Therefore, there exists a real number $z_{ji}$ ($0\le z_{ji}\le 1$) such that

$$|a_{ji}|+\sum _{k\ne j,i}|a_{jk}|m_{ki}{h}_i=z_{ji}{R}_j+(1-z_{ji})R_j^{\sigma },j,i\in N,j\ne i.$$

Hence,

$$w_{ji}=\frac{z_{ji}{R}_j+(1-z_{ji})R_j^{\sigma }}{a_{jj}},j\in N.$$

Let $z_j={max}_{i\ne j}{z}_{ji}$. Obviously, $0<z_j\le 1$ (if $z_j=0$, then A is reducible, which is a contradiction). Let

$$w_j=\underset{i\ne j}{max}\{w_{ji}\}=\frac{z_j{R}_j+(1-z_j)R_j^{\sigma }}{a_{jj}},j\in N.$$

Since A is irreducible, then $R_j>0$, $R_j^{\sigma }\ge 0$, and $0<w_j<1$. Let $\tau (B\circ A^{-1})=\lambda$. By Lemma 6, there exist $i_0,j_0\in N$, $i_0\ne j_0$ such that

$$|\lambda -{\alpha }_{i_0{i}_0}{b}_{i_0{i}_0}||\lambda -{\alpha }_{j_0{j}_0}{b}_{j_0{j}_0}|\le (w_{i_0}\sum _{k\ne i_0}\frac{1}{w_k}|{\alpha }_{k{i}_0}{b}_{k{i}_0}|)(w_{j_0}\sum _{k\ne j_0}\frac{1}{w_k}|{\alpha }_{k{j}_0}{b}_{k{j}_0}|).$$

And by Lemma 2, we have

$$\begin{array}{r}(w_{i_0}\sum _{k\ne i_0}\frac{1}{w_k}|{\alpha }_{k{i}_0}{b}_{k{i}_0}|)(w_{j_0}\sum _{k\ne j_0}\frac{1}{w_k}|{\alpha }_{k{j}_0}{b}_{k{j}_0}|)\\ \le (w_{i_0}\sum _{k\ne i_0}|b_{k{i}_0}|{\alpha }_{i_0{i}_0})(w_{j_0}\sum _{k\ne j_0}|b_{k{j}_0}|{\alpha }_{j_0{j}_0}).\end{array}$$

Therefore,

$$|\lambda -{\alpha }_{i_0{i}_0}{b}_{i_0{i}_0}||\lambda -{\alpha }_{j_0{j}_0}{b}_{j_0{j}_0}|\le (w_{i_0}\sum _{k\ne i_0}|b_{k{i}_0}|{\alpha }_{i_0{i}_0})(w_{j_0}\sum _{k\ne j_0}|b_{k{j}_0}|{\alpha }_{j_0{j}_0}).$$

Furthermore, we obtain

$$\begin{array}{rl}\lambda \ge & \frac{1}{2}\{{\alpha }_{i_0{i}_0}{b}_{i_0{i}_0}+{\alpha }_{j_0{j}_0}{b}_{j_0{j}_0}-[{({\alpha }_{i_0{i}_0}{b}_{i_0{i}_0}-{\alpha }_{j_0{j}_0}{b}_{j_0{j}_0})}^2\\ {+4(w_{i_0}\sum _{k\ne i_0}|b_{k{i}_0}|{\alpha }_{i_0{i}_0})(w_{j_0}\sum _{k\ne j_0}|b_{k{j}_0}|{\alpha }_{j_0{j}_0})]}^{\frac{1}{2}}\},\end{array}$$

that is,

$$\begin{array}{c}\tau (B\circ A^{-1})\hfill \\ \ge \frac{1}{2}\{{\alpha }_{i_0{i}_0}{b}_{i_0{i}_0}+{\alpha }_{j_0{j}_0}{b}_{j_0{j}_0}-[{({\alpha }_{i_0{i}_0}{b}_{i_0{i}_0}-{\alpha }_{j_0{j}_0}{b}_{j_0{j}_0})}^2\hfill \\ {+4(w_{i_0}\sum _{k\ne i_0}|b_{k{i}_0}|{\alpha }_{i_0{i}_0})(w_{j_0}\sum _{k\ne j_0}|b_{k{j}_0}|{\alpha }_{j_0{j}_0})]}^{\frac{1}{2}}\}\hfill \\ \ge \underset{i\ne j}{min}\frac{1}{2}\{{\alpha }_{ii}{b}_{ii}+{\alpha }_{jj}{b}_{jj}-{[{({\alpha }_{ii}{b}_{ii}-{\alpha }_{jj}{b}_{jj})}^2+4(w_i\sum _{k\ne i}|b_{ki}|{\alpha }_{ii})(w_j\sum _{k\ne j}|b_{kj}|{\alpha }_{jj})]}^{\frac{1}{2}}\}.\hfill \end{array}$$

(ii) Now, assume that one of A and B is reducible. We substitute $A-ϵT$ and $B-ϵT$ for A and B, respectively, in the previous case, and then letting $ϵ\to 0$, the result follows by continuity. □

Corollary 3 If $A=(a_{ij})\in M_n$ and $A^{-1}=({\alpha }_{ij})$, then

$$\begin{array}{rl}\tau (A\circ A^{-1})\ge & \underset{i\ne j}{min}\frac{1}{2}\{{\alpha }_{ii}{a}_{ii}+{\alpha }_{jj}{a}_{jj}-[{({\alpha }_{ii}{a}_{ii}-{\alpha }_{jj}{a}_{jj})}^2\\ {+4(w_i\sum _{k\ne i}|a_{ki}|{\alpha }_{ii})(w_j\sum _{k\ne j}|a_{kj}|{\alpha }_{jj})]}^{\frac{1}{2}}\}.\end{array}$$

Example 1 Let

$$\begin{array}{c}A=\left(\begin{array}{cccccccccc}39& -16& -2& -3& -2& -5& -2& -3& -5& 0\\ -26& 44& -2& -4& -2& -1& 0& -2& -3& -3\\ -1& -9& 29& -3& -4& 0& -5& -4& -1& -1\\ -2& -3& -10& 36& -12& 0& -5& -1& -2& 0\\ 0& -3& -1& -9& 44& -16& -3& -4& -4& -3\\ -3& -4& -3& -4& -12& 48& -18& -1& 0& -2\\ -2& -1& -4& -3& -4& -16& 45& -9& -4& -1\\ -1& -2& -2& -2& -3& -1& -5& 38& -20& -1\\ -2& -1& 0& -3& -4& -5& -2& -10& 47& -19\\ -1& -4& -4& -4& 0& -3& -4& -3& -7& 31\end{array}\right),\hfill \\ B=\left(\begin{array}{cccccccccc}90& -3& -2& -7& -4& -7& -6& -3& -9& -3\\ -4& 100& -5& -4& -8& -7& -1& -9& -8& -8\\ -5& -9& 62& -4& -7& -9& -9& -1& -4& -8\\ -8& -8& -10& 99& 0& -6& -8& -9& -3& -6\\ -3& -8& -10& -6& 62& -3& -6& -7& -5& -1\\ -2& -3& -5& -10& -6& 55& -5& -1& -3& -10\\ -8& -5& -8& -8& -3& -3& 52& -6& -1& -4\\ -4& -5& -8& -4& -1& -1& -6& 57& -7& -7\\ -2& -1& -6& -10& -2& -6& -5& -9& 86& -5\\ -5& -7& -3& -9& -5& -7& -9& -5& -9& 72\end{array}\right).\hfill \end{array}$$

It is easily proved that A and B are nonsingular M-matrices and A is a doubly stochastic matrix.

(i) If we apply Theorem 4.8 of [2], we have

$$\tau (B\circ A^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{b_{ii}-m_i{\sum }_{j\ne i}|b_{ji}|}{a_{ii}}\right\}=0.0027.$$

If we apply Theorem 2.4 of [8], we have

$$\tau (B\circ A^{-1})\ge (1-\rho (J_A)\rho (J_B))\underset{1\le i\le n}{min}\frac{a_{ii}}{b_{ii}}=0.3485.$$

But, if we apply Theorem 1, we have

$$\tau (B\circ A^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{b_{ii}-w_i{\sum }_{j\ne i}|b_{ji}|}{a_{ii}}\right\}=0.0435.$$

If we apply Corollary 1, we have

$$\tau (B\circ A^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{b_{ii}-w_i{\sum }_{j\ne i}|b_{ji}|}{1+{\sum }_{j\ne i}{w}_{ji}}\right\}=0.2172.$$

If we apply Theorem 2, we have

$$\begin{array}{rl}\tau (B\circ A^{-1})\ge & \underset{i\ne j}{min}\frac{1}{2}\{{\alpha }_{ii}{b}_{ii}+{\alpha }_{jj}{b}_{jj}-[{({\alpha }_{ii}{b}_{ii}-{\alpha }_{jj}{b}_{jj})}^2\\ {+4(w_i\sum _{k\ne i}|b_{ki}|{\alpha }_{ii})(w_j\sum _{k\ne j}|b_{kj}|{\alpha }_{jj})]}^{\frac{1}{2}}\}\\ =& 0.7212.\end{array}$$

• (ii) If we apply Theorem 3.2 of [3], we get

  $$\tau (A\circ A^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{a_{ii}-u_i{\sum }_{j\ne i}|a_{ji}|}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}=0.3269.$$

But, if we apply Corollary 2, we get

$$\tau (A\circ A^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{a_{ii}-w_i{\sum }_{j\ne i}|a_{ji}|}{1+{\sum }_{j\ne i}{w}_{ji}}\right\}=0.3605.$$

If we apply Corollary 3, we get

$$\begin{array}{rl}\tau (A\circ A^{-1})\ge & \underset{i\ne j}{min}\frac{1}{2}\{{\alpha }_{ii}{a}_{ii}+{\alpha }_{jj}{a}_{jj}-[{({\alpha }_{ii}{a}_{ii}-{\alpha }_{jj}{a}_{jj})}^2\\ {+4(w_i\sum _{k\ne i}|b_{ki}|{\alpha }_{ii})(w_j\sum _{k\ne j}|b_{kj}|{\alpha }_{jj})]}^{\frac{1}{2}}\}\\ =& 0.4072.\end{array}$$