# Bound on the Minimum Eigenvalue ofH-Matrices Involving Hadamard Products

Authors: Kun Du; Guiding Gu; Guo Liu
Journal: Algebra
 (2013)

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## Abstract

We present a new lower bound on the minimum eigenvalue ofH-matrices involving Hadamard products τ(A1(α1)∘⋯∘Am(αm)), and we show that our lower bound is larger than the lower bound ∏k=1m[τ(Ak)]αk. Three examples verify our result.


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## Body

## 1. Introduction

In [1], it is shown by Theorem  5.7.15 that if Ak are n×n  H-matrices for all k∈{1,…,m}, and αk≥0 satisfy ∑k=1mαk≥1, then
(1)τ(A1(α1)∘⋯∘Am(αm))≥∏k=1m‍[τ(Ak)]αk,
where A(α) is defined as entrywise and any scalar definition of aα such that |aα|=|a|α is allowed. This theorem provided a beautiful result about the minimum eigenvalue of H-matrices involving Hadamard products, but sometimes this inequality could be very weak. For example,
(2)A=(100-21-3001-12001),B=(100-191000-81-171),C=(10001000100)
are H-matrices, and
(3)τ(A(2)∘B(2)∘C(2))=10000≫τ(A)2τ(B)2τ(C)2=1,
see the details in Section 3.A lot of works have been done on the minimum eigenvalue ofM-matrices and H-matrices involving Hadamard products, see the results in [1–6].In this paper, we present a new lower bound by including diagonal entries and prove that our bound is larger than the bound in (1).We now introduce some notations, see [1]. The Hadamard product of A=[aij]m×n and B=[bij]m×n is defined by A∘B≡[aijbij]m×n. We define |aα|=|a|α and 00=1. Let A=[aij]n×n and denote aij by A(i,j) for every i,j∈{1,…,n}. We denote A(r)=[aijr], where r≥0, and denote by Zn the class of all n×n real matrices all of whose off-diagonal entries are nonpositive. Let A∈Zn, then the minimum eigenvalue of A is defined by τ(A)=min{Re(λ):λ∈σ(A)}; τ(A) is a eigenvalue of A.For two real matricesA,B∈Zn, the Fan product of A and B, denoted by A⋆B=C=[cij], is defined by
(4)cij={-aijbij,ifi≠j,aiibii,ifi=j,
and then A⋆B∈Zn. The comparison matrix M(A)=[mij] of a given matrix A=[aij]∈ℂn×n is defined by
(5)mij={-|aij|,ifi≠j,|aii|,ifi=j.
A matrix A is an H-matrix if its comparison matrix M(A) is an M-matrix. For A∈ℂn×n, we define τ(A)≡τ(M(A)).Last, forA∈Zn, we introduce a new definition A[r]=[dij], where
(6)dij={-|aij|r,ifi≠j,|aii|r,ifi=j.


## 2. Main Results

To prove the main theorem, we need several lemmas.Lemma 1.
Letxj=(xj(1),…,xj(n))T≥0 for all j∈{1,…,m}. If pj>0 and ∑j=1m(1/pj)=r≥1, then
(7)∑i=1n‍∏j=1m‍xj(i)≤∏j=1m‍{∑i=1n‍[xj(i)]pj}1/pj.Proof.
Whenr=1, we can prove this inequality by the Hölder inequality and by induction. When r>1, apparently, ∑j=1m(1/rpj)=1, similarly, we can prove this inequality.IfA∈Zn, then A=αI-B for some B≥0 and some α∈ℝ. Let σ(B)={ρ(B),λ2,…,λn}. Then σ(A)={α-ρ(B),α-λ2,…,α-λn}. If there is a λi such that α-ρ(B)=Re(α-ρ(B))>Re(α-λi)=α-Re(λi), then ρ(B)<Re(λi), but this is a contradiction. So, τ(A)=α-ρ(B). If x≥0 is the right Perron eigenvector of B, then Ax=(αI-B)x=αx-Bx=αx-ρ(B)x=(α-ρ(B))x=τ(A)x. If y≥0 is the left Perron eigenvector of B, then yTA=yT(αI-B)=αyT-yTB=αyT-ρ(B)yT=(α-ρ(B))yT=τ(A)yT. So, similar to the Perron-Frobenius theorem, we have the following: if A∈Zn became irreducible, then there exist positive vectors u and v such that Au=τ(A)u, and vTA=τ(A)vT, u and v being called right and left Perron eigenvectors of A, respectively.Lemma 2.
IfA∈Zn is irreducible, and Az≥kz for a nonnegative nonzero vector z, then k≤τ(A).Proof.
A
=
α
I
-
P, where P≥0 is irreducible and α∈ℝ. By the Perron-Frobenius theorem, P has a positive left Perron vector y, that is, yTP=ρ(P)yT. Note that Az-kz=(α-k)z-Pz≥0. Hence, yT(Az-kz)=(α-k)yTz-ρ(P)yTz=(α-k-ρ(P))yTz≥0. Since yTz>0, we have k≤α-ρ(P)=τ(A).Lemma 3.
LetAk∈Zn for all k∈{1,…,m}. If pk>0 and ∑k=1m(1/pk)≥1, then
(8)τ(A1⋆⋯⋆Am)≥min1≤i≤n{∏k=1m‍Ak(i,i)-∏k=1m‍[|Ak(i,i)|pk-τ(Ak[pk])]1/pk}.Proof.
It is quite evident that the conclusion holds with equality forn=1.
Forn≥2, we have two cases.
Case  1.
A
1
⋆
⋯
⋆
A
m
∈
Z
n is irreducible. Then Ak∈Zn are irreducible for all k∈{1,…,m}. Thus, Ak[pk]∈Zn are irreducible for all k∈{1,…,m}. Let uk(pk)=(uk(1)pk,…,uk(n)pk)T>0 be the right Perron eigenvectors of Ak[pk] for all k∈{1,…,m}. If uk(1)=uk=(uk(1),…,uk(n))T>0 for all k∈{1,…,m}, then for every i∈{1,…,n}, we have
(9)Ak[pk]uk(pk)=τ(Ak[pk])uk(pk),|Ak(i,i)|pkuk(i)pk-∑j≠i‍|Ak(i,j)|pkuk(j)pk=τ(Ak[pk])uk(i)pk,∑j≠i‍|Ak(i,j)|pkuk(j)pk=[|Ak(i,i)|pk-τ(Ak[pk])]uk(i)pk≥0.
Let C=A1⋆⋯⋆Am∈Zn. Let z=u1∘⋯∘um=(z(1),…,z(n))T>0. Then z(i)=∏k=1muk(i) for all i∈{1,…,n}. Then, for all i∈{1,…,n}, we have
(10)(Cz)i=∏k=1m‍Ak(i,i)z(i)-∑j≠i‍(∏k=1m‍|Ak(i,j)|)z(j)=∏k=1m‍Ak(i,i)z(i)-∑j≠i‍∏k=1m‍(|Ak(i,j)|uk(j))≥∏k=1m‍Ak(i,i)z(i)-∏k=1m‍{∑j≠i‍|Ak(i,j)|pkuk(j)pk}1/pk=∏k=1m‍Ak(i,i)z(i)-∏k=1m‍{[|Ak(i,i)|pk-τ(Ak[pk])]uk(i)pk}1/pk={∏k=1m‍Ak(i,i)-∏k=1m‍[|Ak(i,i)|pk-τ(Ak[pk])]1/pk}z(i).
By Lemma 1, the “≥” hold. By Lemma 2, we have
(11)τ(A1⋆⋯⋆Am)≥min1≤i≤n{∏k=1m‍[|Ak(i,i)|pk-τ(Ak[pk])]1/pk∏k=1m‍Ak(i,i)-∏k=1m‍[|Ak(i,i)|pk-τ(Ak[pk])]1/pk}.
Case  2.
A
1
⋆
⋯
⋆
A
m
∈
Z
n is reducible. We denote by T the n×n permutation matrix (tij) with
(12)t12=t23=⋯=tn-1,n=tn1=1,
the remaining tij zero, then both Ak-εT∈Zn are irreducible for any chosen positive real number ε. Now we substitute Ak-εT for Ak in the previous case, and then by letting ε→0, the result follows by continuity.Lemma 4 (see [7]).
Let for alli∈{1,…,n}, xi≥yi≥0 and pi>0 satisfying ∑i=1n(1/pi)≥1. Then
(13)∏i=1n‍xi-∏i=1n‍yi≥∏i=1n‍(xipi-yipi)1/pi.The following is our main theorem.Theorem 5.
Letαk≥0 and ∑k=1mαk≥1. If Ak is n×n  H-matrices for all k∈{1,…,m}, then
(14)τ(A1(α1)∘⋯∘Am(αm))≥min1≤i≤n{∏k=1m‍[|Ak(i,i)|-τ(Ak)]αk∏k=1m‍|Ak(i,i)|αk-∏k=1m‍[|Ak(i,i)|-τ(Ak)]αk}≥∏k=1m‍[τ(Ak)]αk.Proof.
Without loss of generality, we can assume thatαk>0 for all k∈{1,2,…,m}. Let pk=1/αk>0, then ∑k=1m(1/pk)=∑k=1mαk≥1. Now we have:
(15)τ(A1(α1)∘⋯∘Am(αm))=τ(M(A1(α1)∘⋯∘Am(αm)))=τ(M(A1(α1))⋆⋯⋆M(Am(αm)))≥min1≤i≤n{∏k=1m‍[|Ak(i,i)|-τ(M(Ak(αk))[pk])]1/pk∏k=1m‍|Ak(i,i)|αk-∏k=1m‍[|Ak(i,i)|-τ(M(Ak(αk))[pk])]1/pk}=min1≤i≤n{∏k=1m‍[|Ak(i,i)|-τ(Ak)]αk∏k=1m‍|Ak(i,i)|αk-∏k=1m‍[|Ak(i,i)|-τ(Ak)]αk}≥min1≤i≤n∏k=1m‍[|Ak(i,i)|-(|Ak(i,i)|-τ(Ak))]1/pk=∏k=1m‍[τ(Ak)]αk.
The first “≥” hold by Lemma 3. Because Ak is H-matrix, we have 0≤τ(Ak)≤|Ak(i,i)|, which means that |Ak(i,i)|≥|Ak(i,i)|-τ(Ak)≥0. So the second “≥” hold by |Ak(i,i)|αk≥[|Ak(i,i)|-τ(Ak)]αk≥0 and Lemma 4.


## 3. Examples

In this section, we present three examples to illustrate our improved bound.Example 6.
We take the matricesA, B, and C in Section 1. It is easy to get τ(A)=τ(B)=τ(C)=1, so the low bound in (1) is τ(A)2τ(B)2τ(C)2=1; on the other side, we can get τ(A(2)∘B(2)∘C(2))=10000, which is greatly larger than the lower bound τ(A)=τ(B)=τ(C), that is,
(16)τ(A(2)∘B(2)∘C(2))=10000≫τ(A)2τ(B)2τ(C)2=1.
However, our lower bound in Theorem 5 is
(17)min1≤i≤3{[bii-τ(B)]2[cii-τ(C)]2aii2bii2cii2-[aii-τ(A)]2×[bii-τ(B)]2[cii-τ(C)]2}=min1≤i≤3{1002×12×12-(100-1)2×(1-1)2(1-1)2}=10000,
which is the exact value of τ(A(2)∘B(2)∘C(2)).Example 7.
Let(18)A=(11-5-8-214-7-1-39),B=(9-8-7-213-4-1-615).
It is easy to see that A and B are H-matrices and τ(A)=3.9152,τ(B)=4.1345. Thus, we have
(19)τ(A(2)∘B(2))=9799>τ(A)2τ(B)2=262.0319,
but our bound is
(20)min1≤i≤3{aii2bii2-[aii-τ(A)]2[bii-τ(B)]2}=8612.8,
which is close to the exact value of 9799.Example 8.
Take theH-matrices
(21)A=(15-12-4-314-2-4011),B=(27-14-6-420-8-10-523).
It is easy to see that
(22)τ(A(2)∘B(1))=2775.5>τ(A)2τ(B)1=244.0,
but our bound is
(23)min1≤i≤3{aii2bii1-[aii-τ(A)]2[bii-τ(B)]1}=2340.3.
Also, our bound is much closer to the exact value than the low bound in (1).



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