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# Some new inequalities for the Hadamard product of M-matrices

## Abstract

If A and B are $n×n$ nonsingular M-matrices, a new lower bound for the minimum eigenvalue $τ(B∘ A − 1 )$ for the Hadamard product of B and $A − 1$ is derived. As a consequence, a new lower bound for the minimum eigenvalue $τ(A∘ A − 1 )$ for the Hadamard product of A and its inverse $A − 1$ is given. Theoretical results and an example demonstrate that the new bounds are better than some existing ones.

MSC:15A06, 15A18, 15A48.

## 1 Introduction

For convenience, for any positive integer n, let $N={1,2,…,n}$ throughout. The set of all $n×n$ real matrices is denoted by $R n × n$ and $C n × n$ denotes the set of all $n×n$ complex matrices.

A matrix $A=( a i j )∈ R n × n$ is called a nonnegative matrix if $a i j ≥0$. The spectral radius of A is denoted by $ρ(A)$. If A is a nonnegative matrix, the Perron-Frobenius theorem guarantees that $ρ(A)$ is an eigenvalue of A.

$Z n$ denotes the class of all $n×n$ real matrices all of whose off-diagonal entries are nonpositive. An $n×n$ matrix A is called an M-matrix if there exists an $n×n$ nonnegative matrix B and a nonnegative real number λ such that $A=λI−B$ and $λ≥ρ(B)$, I is the identity matrix; if $λ>ρ(B)$, we call A a nonsingular M-matrix; if $λ=ρ(B)$, we call A a singular M-matrix. Denote by $M n$ the set of nonsingular M-matrices.

Let $A∈ Z n$, and let $τ(A)=min{Re(λ):λ∈σ(A)}$. Basic for our purpose are the following simple facts (see Problems 16, 19 and 28 in Section 2.5 of ):

1. (1)

$τ(A)∈σ(A)$; $τ(A)$ is called the minimum eigenvalue of A.

2. (2)

If $A,B∈ M n$, and $A≥B$, then $τ(A)≥τ(B)$.

3. (3)

If $A∈ M n$, then $ρ( A − 1 )$ is the Perron eigenvalue of the nonnegative matrix $A − 1$, and $τ(A)= 1 ρ ( A − 1 )$ is a positive real eigenvalue of A.

For two matrices $A=( a i j )$ and $B=( b i j )$, the Hadamard product of A and B is the matrix $A∘B=( a i j b i j )$. If A and B are two nonsingular M-matrices, then $B∘ A − 1$ is also a nonsingular M-matrix .

Let $A,B∈ M n$ and $A − 1 =( β i j )$, in [, Theorem 5.7.31] the following classical result is given:

$τ ( B ∘ A − 1 ) ≥τ(B) min 1 ≤ i ≤ n β i i .$
(1.1)

Huang [, Theorem 9] improved this result and obtained the following result:

$τ ( B ∘ A − 1 ) ≥ 1 − ρ ( J A ) ρ ( J B ) 1 + ρ 2 ( J A ) min 1 ≤ i ≤ n b i i a i i ,$
(1.2)

where $ρ( J A )$, $ρ( J B )$ are the spectral radii of $J A$ and $J B$.

The lower bound (1.1) is simple, but not accurate enough. The lower bound (1.2) is difficult to evaluate.

Recently, Li [, Theorem 2.1] improved these two results and gave a new lower bound for $τ(B∘ A − 1 )$, that is,

$τ ( B ∘ A − 1 ) ≥ min i { b i i − s i ∑ k ≠ i | b k i | a i i } ,$
(1.3)

where $r l i = | a l i | | a l l | − ∑ k ≠ l , i | a l k |$, $l≠i$; $r i = max l ≠ i { r l i }$, $i∈N$; $s j i = | a j i | + ∑ k ≠ j , i | a j k | r k a j j$, $j≠i$, $j∈N$; $s i = max j ≠ i { s i j }$, $i∈N$.

For an M-matrix A, Fiedler et al. showed in  that $τ(A∘ A − 1 )≤1$. Subsequently, Fiedler and Markham [, Theorem 3] gave a lower bound on $τ(A∘ A − 1 )$,

$τ ( A ∘ A − 1 ) ≥ 1 n ,$
(1.4)

and proposed the following conjecture:

$τ ( A ∘ A − 1 ) ≥ 2 n .$
(1.5)

Yong  and Song  have independently proved this conjecture.

Li [, Theorem 3.1] obtained the following result:

$τ ( A ∘ A − 1 ) ≥ min i { a i i − t i R i 1 + ∑ j ≠ i t j i } ,$
(1.6)

which only depends on the entries of $A=( a i j )$, where $R i = ∑ k ≠ i | a i k |$; $d i = R i | a i i |$, $i∈N$; $t j i = | a j i | + ∑ k ≠ j , i | a j k | d k | a j j |$, $j≠i$, $j∈N$; $t i = max j ≠ i { t i j }$, $i∈N$.

Li [, Theorem 3.2] improved the bound (1.6) and obtained the following result:

$τ ( A ∘ A − 1 ) ≥ min i { a i i − m i R i 1 + ∑ j ≠ i m j i } ,$
(1.7)

where $r l i = | a l i | | a l l | − ∑ k ≠ l , i | a l k |$, $l≠i$; $r i = max l ≠ i { r l i }$, $i∈N$; $m j i = | a j i | + ∑ k ≠ j , i | a j k | r i | a j j |$, $j≠i$, $j∈N$; $m i = max j ≠ i { m i j }$, $i∈N$.

Recently, Li [, Theorem 3.2] improved the bound (1.7) and gave a new lower bound for $τ(A∘ A − 1 )$, that is,

$τ ( A ∘ A − 1 ) ≥ min i { a i i − T i R i 1 + ∑ j ≠ i T j i } ,$
(1.8)

where $T j i =min{ m j i , s j i }$, $j≠i$; $T i = max j ≠ i { T i j }$, $i∈N$.

In the present paper, we present a new lower bound on $τ(B∘ A − 1 )$. As a consequence, we present a new lower bound on $τ(A∘ A − 1 )$. These bounds improve several existing results.

The following is the list of notations that we use throughout: For $i,j,k,l∈N$,

$R i = ∑ k ≠ i | a i k | , C i = ∑ k ≠ i | a k i | , d i = R i | a i i | , c ˆ i = C i | a i i | ; r l i = | a l i | | a l l | − ∑ k ≠ l , i | a l k | , l ≠ i ; r i = max l ≠ i { r l i } , i ∈ N ; c i l = | a i l | | a l l | − ∑ k ≠ l , i | a k l | , l ≠ i ; c i = max l ≠ i { c i l } , i ∈ N ; m j i = | a j i | + ∑ k ≠ j , i | a j k | r i | a j j | , j ≠ i ; m i = max j ≠ i { m i j } , i ∈ N ; s j i = | a j i | + ∑ k ≠ j , i | a j k | r k | a j j | , j ≠ i ; s i = max j ≠ i { s i j } , i ∈ N ; T j i = min { m j i , s j i } , j ≠ i ; T i = max j ≠ i { T i j } , i ∈ N .$

## 2 Some lemmas and the main results

In order to prove our results, we first give some lemmas.

Lemma 2.1 

If $A=( a i j )∈ R n × n$ is an M-matrix, then there exists a diagonal matrix D with positive diagonal entries such that $D − 1 AD$ is a strictly row diagonally dominant M-matrix.

Lemma 2.2 

Let $A,B=( a i j )∈ C n × n$ and suppose that $D∈ C n × n$ and $E∈ C n × n$ are diagonal matrices. Then

$D(A∘B)E=(DAE)∘B=(DA)∘(BE)=(AE)∘(DB)=A∘(DBE).$

Lemma 2.3 

If $A=( a i j )∈ R n × n$ is a strictly row diagonally dominant M-matrix, then $A − 1 =( β i j )$ satisfies

$β j i ≤ T j i β i i ,i,j∈N,i≠j.$

Lemma 2.4 

If $A − 1$ is a doubly stochastic matrix, then $Ae=e$, $A T e=e$, where $e= ( 1 , 1 , … , 1 ) T$.

Lemma 2.5 

Let $A=( a i j )∈ R n × n$ be a strictly row diagonally dominant M-matrix. Then, for $A − 1 =( β i j )$, we have

$β i i ≥ 1 a i i ,i∈N.$

Lemma 2.6 

If $A=( a i j )∈ R n × n$ is an M-matrix and $A − 1 =( β i j )$ is a doubly stochastic matrix, then

$β i i ≥ 1 1 + ∑ j ≠ i T j i ,i∈N.$

Lemma 2.7 

Let $A=( a i j )∈ C n × n$, and let $x 1 , x 2 ,…, x n$ be positive real numbers. Then all the eigenvalues of A lie in the region

$⋃ i , j = 1 i ≠ j n { z ∈ C : | z − a i i | | z − a j j | ≤ ( x i ∑ k ≠ i 1 x k | a k i | ) ( x j ∑ k ≠ j 1 x k | a k j | ) } .$

Theorem 2.1 Let $A,B=( b i j )∈ R n × n$ be two nonsingular M-matrices, and let $A − 1 =( β i j )$. Then

$τ ( B ∘ A − 1 ) ≥ min i ≠ j 1 2 { b i i β i i + b j j β j j − [ ( b i i β i i − b j j β j j ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) ( T j ∑ k ≠ j | b k j | β j j ) ] 1 2 } .$
(2.1)

Proof It is evident that (2.1) is an equality for $n=1$.

We next assume that $n≥2$.

If A is an M-matrix, then by Lemma 2.1 we know that there exists a diagonal matrix D with positive diagonal entries such that $D − 1 AD$ is a strictly row diagonally dominant M-matrix and satisfies

$τ ( B ∘ A − 1 ) =τ ( D − 1 ( B ∘ A − 1 ) D ) =τ ( B ∘ ( D − 1 A D ) − 1 ) .$

So, for convenience and without loss of generality, we assume that A is a strictly row diagonally dominant M-matrix. Therefore, $0< T i <1$, $i∈N$.

If $B∘ A − 1$ is irreducible, then B and A are irreducible. Let $τ(B∘ A − 1 )=λ$, so that $0<λ< b i i β i i$, $∀i∈N$. Thus, by Lemma 2.7, there is a pair $(i,j)$ of positive integers with $i≠j$ such that

$|λ− b i i β i i ||λ− b j j β j j |≤ ( T i ∑ k ≠ i 1 T k | b k i β k i | ) ( T j ∑ k ≠ j 1 T k | b k j β k j | ) .$

Observe that

$( T i ∑ k ≠ i 1 T k | b k i β k i | ) ( T j ∑ k ≠ j 1 T k | b k j β k j | ) ≤ ( T i ∑ k ≠ i 1 T k | b k i | T k i β i i ) ( T j ∑ k ≠ j 1 T k | b k j | T k j β j j ) ≤ ( T i ∑ k ≠ i | b k i | β i i ) ( T j ∑ k ≠ j | b k j | β j j ) .$

Thus, we have

$|λ− b i i β i i ||λ− b j j β j j |≤ ( T i ∑ k ≠ i | b k i | β i i ) ( T j ∑ k ≠ j | b k j | β j j ) .$

Then we have

$λ≥ 1 2 { b i i β i i + b j j β j j − [ ( b i i β i i − b j j β j j ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) ( T j ∑ k ≠ j | b k j | β j j ) ] 1 2 } .$

That is,

$τ ( B ∘ A − 1 ) ≥ 1 2 { b i i β i i + b j j β j j − [ ( b i i β i i − b j j β j j ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) ( T j ∑ k ≠ j | b k j | β j j ) ] 1 2 } ≥ min i ≠ j 1 2 { b i i β i i + b j j β j j − [ ( b i i β i i − b j j β j j ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) ( T j ∑ k ≠ j | b k j | β j j ) ] 1 2 } .$

Now, assume that $B∘ A − 1$ is reducible. It is known that a matrix in $Z n$ is a nonsingular M-matrix if and only if all its leading principal minors are positive (see condition (E17) of Theorem 6.2.3 of ). If we denote by $D=( d i j )$ the $n×n$ permutation matrix with $d 12 = d 23 =⋯= d n − 1 , n = d n 1 =1$, then both $A−tD$ and $B−tD$ are irreducible nonsingular M-matrices for any chosen positive real number t, sufficiently small such that all the leading principal minors of both $A−tD$ and $B−tD$ are positive. Now we substitute $A−tD$ and $B−tD$ for A and B, respectively in the previous case, and then letting $t⟶0$, the result follows by continuity. □

Theorem 2.2 Let $A,B=( b i j )∈ R n × n$ be two nonsingular M-matrices, and let $A − 1 =( β i j )$. Then

$min i ≠ j 1 2 { b i i β i i + b j j β j j − [ ( b i i β i i − b j j β j j ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) ( T j ∑ k ≠ j | b k j | β j j ) ] 1 2 } ≥ min 1 ≤ i ≤ n { b i i − s i ∑ k ≠ i | b k i | a i i } .$

Proof Since $T j i =min{ m j i , s j i }$, $j≠i$, $T i = max j ≠ i { T i j }$, so $T i ≤ s i$, $i∈N$. Without loss of generality, for $i≠j$, assume that

$b i i β i i − T i ∑ k ≠ i | b k i | β i i ≤ b j j β j j − T j ∑ k ≠ j | b k j | β j j .$
(2.2)

Thus, (2.2) is equivalent to

$T j ∑ k ≠ j | b k j | β j j ≤ T i ∑ k ≠ i | b k i | β i i + b j j β j j − b i i β i i .$
(2.3)

From (2.1) and (2.3), we have

$1 2 { b i i β i i + b j j β j j − [ ( b i i β i i − b j j β j j ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) ( T j ∑ k ≠ j | b k j | β j j ) ] 1 2 } ≥ 1 2 { b i i β i i + b j j β j j − [ ( b i i β i i − b j j β j j ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) ( T i ∑ k ≠ i | b k i | β i i + b j j β j j − b i i β i i ) ] 1 2 } = 1 2 { b i i β i i + b j j β j j − [ ( b i i β i i − b j j β j j ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) ( b j j β j j − b i i β i i ) ] 1 2 } = 1 2 { b i i β i i + b j j β j j − [ ( b j j β j j − b i i β i i + 2 T i ∑ k ≠ i | b k i | β i i ) 2 ] 1 2 } = 1 2 { b i i β i i + b j j β j j − ( b j j β j j − b i i β i i + 2 T i ∑ k ≠ i | b k i | β i i ) } = b i i β i i − T i ∑ k ≠ i | b k i | β i i = β i i ( b i i − T i ∑ k ≠ i | b k i | ) ≥ β i i ( b i i − s i ∑ k ≠ i | b k i | ) ≥ b i i − s i ∑ k ≠ i | b k i | a i i .$

Thus, we have

$τ ( B ∘ A − 1 ) ≥ min i ≠ j 1 2 { b i i β i i + b j j β j j − [ ( b i i β i i − b j j β j j ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) ( T j ∑ k ≠ j | b k j | β j j ) ] 1 2 } ≥ min 1 ≤ i ≤ n { b i i − s i ∑ k ≠ i | b k i | a i i } .$

This proof is completed. □

Remark 2.1 Theorem 2.2 shows that the result of Theorem 2.1 is better than the result of Theorem 2.1 in .

If $A=B$, according to Theorem 2.1, we can obtain the following corollary.

Corollary 2.1 Let $A=( a i j )∈ R n × n$ be an M-matrix, and let $A − 1 =( β i j )$ be a doubly stochastic matrix. Then

$τ ( A ∘ A − 1 ) ≥ min i ≠ j 1 2 { a i i β i i + a j j β j j − [ ( a i i β i i − a j j β j j ) 2 + 4 ( T i ∑ k ≠ i | a k i | β i i ) ( T j ∑ k ≠ j | a k j | β j j ) ] 1 2 } .$
(2.4)

Theorem 2.3 Let $A=( a i j )∈ R n × n$ be an M-matrix, and let $A − 1 =( β i j )$ be a doubly stochastic matrix. Then

$min i ≠ j 1 2 { a i i β i i + a j j β j j − [ ( a i i β i i − a j j β j j ) 2 + 4 ( T i ∑ k ≠ i | a k i | β i i ) ( T j ∑ k ≠ j | a k j | β j j ) ] 1 2 } ≥ min i { a i i − T i R i 1 + ∑ j ≠ i T j i } .$

Proof Since A is an irreducible M-matrix and $A − 1$ is a doubly stochastic matrix by Lemma 2.4, we have

$a i i = ∑ k ≠ i | a i k |+1= ∑ k ≠ i | a k i |+1,i∈N.$

Without loss of generality, for $i≠j$, assume that

$a i i β i i − T i ∑ k ≠ i | a k i | β i i ≤ a j j β j j − T j ∑ k ≠ j | a k j | β j j .$
(2.5)

Thus, (2.5) is equivalent to

$T j ∑ k ≠ j | a k j | β j j ≤ a j j β j j − a i i β i i + T i ∑ k ≠ i | a k i | β i i .$
(2.6)

From (2.4) and (2.6), we have

$1 2 { a i i β i i + a j j β j j − [ ( a i i β i i − a j j β j j ) 2 + 4 ( T i ∑ k ≠ i | a k i | β i i ) ( T j ∑ k ≠ j | a k j | β j j ) ] 1 2 } ≥ 1 2 { a i i β i i + a j j β j j − [ ( a i i β i i − a j j β j j ) 2 + 4 ( T i ∑ k ≠ i | a k i | β i i ) ( a j j β j j − a i i β i i + T i ∑ k ≠ i | a k i | β i i ) ] 1 2 } = 1 2 { a i i β i i + a j j β j j − [ ( a i i β i i − a j j β j j ) 2 + 4 ( T i ∑ k ≠ i | a k i | β i i ) 2 + 4 ( T i ∑ k ≠ i | a k i | β i i ) ( a j j β j j − a i i β i i ) ] 1 2 } = 1 2 { a i i β i i + a j j β j j − [ ( a j j β j j − a i i β i i + 2 T i ∑ k ≠ i | a k i | β i i ) 2 ] 1 2 } = 1 2 { a i i β i i + a j j β j j − ( a j j β j j − a i i β i i + 2 T i ∑ k ≠ i | a k i | β i i ) } = a i i β i i − T i ∑ k ≠ i | a k i | β i i = β i i ( a i i − T i ∑ k ≠ i | a k i | ) ≥ a i i − T i R i 1 + ∑ j ≠ i T j i .$

Thus, we have

$τ ( A ∘ A − 1 ) ≥ min i ≠ j 1 2 { a i i β i i + a j j β j j − [ ( a i i β i i − a j j β j j ) 2 + 4 ( T i ∑ k ≠ i | a k i | β i i ) ( T j ∑ k ≠ j | a k j | β j j ) ] 1 2 } ≥ min i { a i i − T i R i 1 + ∑ j ≠ i T j i } .$

This proof is completed. □

Remark 2.2 Theorem 2.3 shows that the result of Corollary 2.1 is better than the result of Theorem 3.2 in .

## 3 Example

For convenience, we consider that the M-matrices A and B are the same as the matrices of .

$A=[ 4 − 1 − 1 − 1 − 2 5 − 1 − 1 0 − 2 4 − 1 − 1 − 1 − 1 4 ],B=[ 1 − 0.5 0 0 − 0.5 1 − 0.5 0 0 − 0.5 1 − 0.5 0 0 − 0.5 1 ].$
1. (1)

We consider the lower bound for $τ(B∘ A − 1 )$.

If we apply (1.1), we have

$τ ( B ∘ A − 1 ) ≥τ(B) min 1 ≤ i ≤ n β i i =0.07.$

If we apply (1.2), we have

$τ ( B ∘ A − 1 ) ≥ 1 − ρ ( J A ) ρ ( J B ) 1 + ρ 2 ( J A ) min 1 ≤ i ≤ n b i i a i i =0.048.$

If we apply (1.3), we have

$τ ( B ∘ A − 1 ) ≥ min i { b i i − s i ∑ k ≠ i | b k i | a i i } =0.08.$

If we apply Theorem 2.1, we have

$τ ( B ∘ A − 1 ) ≥ min i ≠ j 1 2 { b i i β i i + b j j β j j − [ ( b i i β i i − b j j β j j ) 2 + 4 ( T i ∑ k ≠ i | b k i | β i i ) ( T j ∑ k ≠ j | b k j | β j j ) ] 1 2 } = 0.1753 .$

In fact, $τ(B∘ A − 1 )=0.2148$.

1. (2)

We consider the lower bound for $τ(A∘ A − 1 )$.

If we apply (1.5), we have

$τ ( A ∘ A − 1 ) ≥ 2 n = 1 2 =0.5.$

If we apply (1.6), we have

$τ ( A ∘ A − 1 ) ≥ min i { a i i − t i R i 1 + ∑ j ≠ i t j i } =0.6624.$

If we apply (1.7), we have

$τ ( A ∘ A − 1 ) ≥ min i { a i i − m i R i 1 + ∑ j ≠ i m j i } =0.7999.$

If we apply (1.8), we have

$τ ( A ∘ A − 1 ) ≥ min i { a i i − T i R i 1 + ∑ j ≠ i T j i } =0.85.$

If we apply Corollary 2.1, we have

$τ ( A ∘ A − 1 ) ≥ min i ≠ j 1 2 { a i i β i i + a j j β j j − [ ( a i i β i i − a j j β j j ) 2 + 4 ( T i ∑ k ≠ i | a k i | β i i ) ( T j ∑ k ≠ j | a k j | β j j ) ] 1 2 } = 0.9755 .$

In fact, $τ(A∘ A − 1 )=0.9755$.

Remark 3.1 The numerical example shows that the bounds of Theorem 2.1 and Corollary 2.1 are sharper than those of Theorem 2.1 in  and Theorem 3.2 in .

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## Acknowledgements

This work is supported by the Natural Science Foundation of China (No: 71161020).

## Author information

Correspondence to Fu-bin Chen.

### Competing interests

The author declares that he has no competing interests.

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