R E S E A R C H
Open Access
Some new inequalities for the minimum
eigenvalue of
M
-matrices
Feng Wang
*and De-shu Sun
*Correspondence:
College of Science, Guizhou Minzu University, Guiyang, Guizhou 550025, P.R. China
Abstract
Some new inequalities for the minimum eigenvalue ofM-matrices are obtained. These inequalities improve existing results, and the estimating formulas are easier to calculate since they only depend on the entries of matrices. Finally, some examples are also given to show that the bounds are better than some previous results.
MSC: 15A18; 15A42
Keywords: M-matrix; nonnegative matrix; Hadamard product; spectral radius; minimum eigenvalue
1 Introduction
LetCn×n(Rn×n) denote the set of alln×ncomplex (real) matrices,A= (aij)∈Cn×n,N= {, , . . . ,n}. We writeA≥ if allaij≥ (i,j∈N).Ais called nonnegative ifA≥. LetZn denote the class of alln×nreal matrices all of whose off-diagonal entries are nonpositive. A matrixAis called anM-matrix [] ifA∈Znand the inverse ofA, denoted byA–, is nonnegative.Mnwill be used to denote the set of alln×n M-matrices.
LetAbe anM-matrix. Then there exists a positive eigenvalue ofA,τ(A) =ρ(A–)–,
whereρ(A–) is the spectral radius of the nonnegative matrixA–,τ(A) =min{|λ|:λ∈ σ(A)},σ(A) denotes the spectrum ofA.τ(A) is called the minimum eigenvalue ofA[, ]. The Hadamard product of two matricesA= (aij)∈Rn×nandB= (bij)∈Rn×nis the ma-trixA◦B= (aijbij)∈Rn×n.
Ann×nmatrixAis said to be reducible if there exists a permutation matrixPsuch that
PTAP=
A
A A
,
whereA,Aare square matrices of order at least one. We callAirreducible if it is not
reducible. Note that any nonzero × matrix is irreducible.
Estimating the bounds for the minimum eigenvalueτ(A) of anM-matrixAis an interest-ing subject in matrix theory, and it has important applications in many practical problems [–]. Hence, it is necessary to estimate the bounds forτ(A).
In [], Shivakumaret al.obtained the following bound forτ(A): LetA= (aij)∈Rn×nbe a weakly chained diagonally dominantM-matrix. Then
r(A)≤τ(A)≤R(A), τ(A)≤min i∈N aii,
M≤τ(A)≤
m. ()
Subsequently, Tian and Huang [] provided a lower bound forτ(A) by using the spectral radius of the Jacobi iterative matrixJAofA: LetA= (aij)∈Rn×nbe anM-matrix andA–= (αij). Then
τ(A)≥ + (n– )ρ(JA)
maxi∈N{αii}
. ()
Recently, Liet al.[] improved () and gave the following result: LetB= (bij)∈Rn×nbe anM-matrix andB–= (βij). Then
τ(B)≥
maxi=j{βii+βjj+ [(βii–βjj)+ (n– )βiiβjjρ(JB)] }
. ()
In this paper, we continue to research the problems mentioned above. For anM-matrix B, we establish some new inequalities on the bounds forτ(B). Finally, some examples are given to illustrate our results.
For convenience, we employ the following notations throughout. LetA= (aij) be ann×n matrix. Fori,j,k∈N,i=j, denote
rji= | aji| |ajj|–k=j,i|ajk|
, ri=max j=i {rji}, mji=
|aji|+k=j,i|ajk|ri
|ajj| , hi=maxj=i
|
aji| |ajj|mji–
k=j,i|ajk|mki
,
uji=
|aji|+k=j,i|ajk|mkihi
|ajj| , ui=maxj=i {uij}. 2 Main results
In this section, we present our main results. Firstly, we give some notations and lemmas. LetA≥ andD=diag(aii). DenoteC=A–D,JA=D– C,D=diag(dii), where
dii=
, ifaii= , aii, ifaii= .
By the definition ofJA, we obtain
ρ(JAT) =ρ
D– CT =ρCD– =ρD– CD– D =ρ
D– C =ρ(JA).
Lemma [] Let A∈Cn×n,and let x,x, . . . ,xn be positive real numbers.Then all the eigenvalues of A lie in the region
i
z∈C:|z–aii| ≤xi
j=i xj
|aji|,i∈N
.
Lemma [] Let A∈Cn×n,and let x,x, . . . ,xnbe positive real numbers.Then all the eigenvalues of A lie in the region
j=i
z∈C:|z–aii||z–ajj| ≤
xi
k=i xk
|aki|
xj
l=j xl
|alj|
Lemma [] Let A,B∈Rn×n,and let X,Y∈Rn×nbe diagonal matrices.Then X(A◦B)Y= (XAY)◦B= (XA)◦(BY) = (AY)◦(XB) =A◦(XBY).
Lemma [] Let A= (aij)∈Mn.Then there exists a positive diagonal matrix X such that X–AX is a strictly diagonally dominant M-matrix
Lemma [] Let A= (aij)∈Rn×nbe a strictly diagonally dominant matrix and let A–=
(αij).Then for all i∈N,
αij≤ujiαjj, j∈N,j=i.
Theorem Let A= (aij)≥,B= (bij)∈Mn,and let B–= (βij).Then
ρA◦B– ≤max
≤i≤n
aii+uiρ(JA)dii βii
. ()
Proof It is evident that the result holds with equality forn= . We next assume thatn≥.
(i) First, we assume thatAandBare irreducible matrices. SinceBis anM-matrix, by Lemma , there exists a positive diagonal matrixX, such thatX–BXis a strictly row
di-agonally dominantM-matrix, and
ρA◦B– =ρX–A◦B– X =ρA◦X–BX – .
Hence, for convenience and without loss of generality, we assume thatBis a strictly diag-onally dominant matrix.
On the other hand, since A is irreducible and so isJAT. Then there exists a positive vectorx= (xi) such thatJATx=ρ(JAT)x=ρ(JA)x, thus, we obtainj=iajixj=ρ(JA)diixi. LetA= (a˜ij) =XAX–in whichXis the positive matrixX=diag(x,x, . . . ,xn). Then, we have
A= (a˜ij) =XAX–=
⎛ ⎜ ⎜ ⎜ ⎜ ⎝
a axx · · · anxxn
ax
x a · · · anx
xn ..
. ... . .. ... anxn
x
anxn
x · · · ann
⎞ ⎟ ⎟ ⎟ ⎟ ⎠.
From Lemma , we have
A◦B–=XAX– ◦B–=XA◦B– X–.
Thus, we obtainρ(A◦B–) =ρ(A◦B–). Letλ=ρ(A◦B–), so thatλ≥a
iiβii,∀i∈N. By Lemma , there existsi∈N, such that
|λ–aiiβii| ≤ui
t=i ut˜
atiβti≤ui
t=i ut˜
atiutiβii
≤ui
t=i ˜
atiβii=uiβii
t=i atixt
xi
Therefore,
λ≤aiiβii+uiρ(JA)diiβii=
aii+uiρ(JA)dii βii,
i.e.,
ρA◦B– ≤aii+uiρ(JA)dii βii ≤max
≤i≤n
aii+uiρ(JA)dii βii
.
(ii) Now, assume that one ofAandBis reducible. It is well known that a matrix inZnis a nonsingularM-matrix if and only if all its leading principal minors are positive (see []). If we denote byT= (tij) then×npermutation matrix witht=t=· · ·=tn–,n=tn= –,
the remainingtijzero, then bothA–εTandB+εTare irreducible matrices for any chosen positive real numberε, sufficiently small such that all the leading principal minors ofB+εT are positive. Now, we substituteA–εTandB+εTforAandB, respectively, in the previous case, and then lettingε→, the result follows by continuity.
Theorem Let B= (bij)∈Mnand B–= (βij).Then
τ(B)≥
max≤i≤n{( +ui(n– ))βii}. ()
Proof Let all entries ofAin () be . Thenaii= (∀i∈N),ρ(JA) =n– . Therefore, by (), we have
τ(B) =
ρ(B–)≥
max≤i≤n{( +ui(n– ))βii} .
The proof is completed.
Theorem Let A= (aij)≥,B= (bij)∈Mn,and let B–= (βij).Then
ρA◦B– ≤
maxi=j {aiiβii+ajjβjj+ ij}, () where ij= [(aiiβii–ajjβjj)+ uiujρ(JA)diidjjβiiβjj]
.
Proof It is evident that the result holds with equality forn= .
We next assume thatn≥. For convenience and without loss of generality, we assume thatBis a strictly row diagonally dominant matrix.
(i) First, we assume thatAandBare irreducible matrices. SinceAis irreducible and so is
JAT. Then there exists a positive vectory= (yi) such thatJATy=ρ(JAT)y=ρ(JA)y, thus, we obtain
k=i
akiyk=ρ(JA)diiyi,
k=j
LetA= (ˆaij) =YAY–in whichYis the positive matrixY=diag(y
,y, . . . ,yn). Then, we
have
A= (aˆij) =YAY–=
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
a ayy · · · anyyn
ay
y a · · · any yn .. . ... . .. ... anyn y anyn
y · · · ann
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ .
From Lemma , we get
A◦B–=YAY– ◦B–=YA◦B– Y–.
Thus, we obtainρ(A◦B–) =ρ(A◦B–). Letλ=ρ(A◦B–), so thatλ≥aiiβii(∀i∈N). By Lemma , there existi,j∈N,i=j, such that
|λ–aiiβii||λ–ajjβjj| ≤
ui
k=i uk
ˆ akiβki
uj
k=j uk
ˆ akjβkj
.
Note that
ui
k=i ukˆ
akiβki≤ui
k=i ukˆ
akiukiβii≤uiβii
k=i ˆ
aki=uiβiiρ(JA)dii,
uj
k=j uk
ˆ
akjβkj≤uj
k=j uk
ˆ
akjukjβjj≤ujβjj
k=j ˆ
akj=ujβjjρ(JA)djj.
Hence, we obtain
λ≤
(aiiβii+ajjβjj+ ij),
i.e.,
ρA◦B– ≤
(aiiβii+ajjβjj+ ij)≤
maxi=j {aiiβii+ajjβjj+ ij},
where ij= [(aiiβii–ajjβjj)+ uiujρ(JA)diidjjβiiβjj] .
(ii) Now, assume that one ofAandBis reducible. We substituteA–εTandB+εTfor AandB, respectively, in the previous case (as in the proof of Theorem ), and then letting
ε→, the result follows by continuity.
Theorem Let B= (bij)∈Mnand B–= (βij).Then
τ(B)≥
maxi=j{βii+βjj+ ij}
, ()
Proof Let all entries ofAin () be . Then
aii= (∀i∈N), ρ(JA) =n– , ij=
(βii–βjj)+ (n– )uiujβiiβjj
.
Therefore, by (), we have
τ(B) =
ρ(B–)≥
maxi=j{βii+βjj+ ij} .
The proof is completed.
Remark We next give a simple comparison between () and (), () and (), respectively. For convenience and without loss of generality, we assume that, fori,j∈N,i=j,
ajjβjj+ujdjjβjjρ(JA)≤aiiβii+uidiiβiiρ(JA),
i.e.,
ujdjjβjjρ(JA)≤aiiβii–ajjβjj+uidiiβiiρ(JA).
Hence,
ij=
(aiiβii–ajjβjj)+ uiujρ(JA)diidjjβiiβjj
≤(aiiβii–ajjβjj)+ uiρ(JA)diiβii
aiiβii–ajjβjj+uidiiβiiρ(JA)
=aiiβii–ajjβjj+ uidiiβiiρ(JA).
Further, we obtain
aiiβii+ajjβjj+ ij≤aiiβii+ uidiiβiiρ(JA),
i.e.,
ρA◦B– ≤
maxi=j {aiiβii+ajjβjj+ ij} ≤max≤i≤n
aii+uiρ(JA)dii βii
.
So, the bound in () is better than the bound in (). Similarly, we can prove that the bound in () is better than the bound in ().
3 Numerical examples
In this section, we present numerical examples to illustrate the advantages of our derived results.
Example Let
B=
⎛ ⎜ ⎝
. –. –. –. –. –. –. .
It is easy to see thatBis anM-matrix. By calculations with Matlab ., we have
τ(B)≥. (by ()), τ(B)≥. (by ()),
τ(B)≥. (by ()), τ(B)≥. (by ()),
τ(B)≥. (by ()),
respectively. In fact,τ(B) = .. It is obvious that the bound in () is the best result.
Example Let
B=
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
–. –. –. –. –. –. –. –. –. –. –. –. –. –. –. –. –. –. –. –.
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
.
It is easy to see thatBis anM-matrix. By calculations with Matlab ., we have
τ(B)≥. (by ()), τ(B)≥. (by ()),
τ(B)≥. (by ()), τ(B)≥. (by ()),
τ(B)≥. (by ()),
respectively. In fact,τ(B) = .. It is obvious that the bound in () is the best result.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The authors contributed equally to this work. All authors read and approved the final manuscript.
Acknowledgements
The authors are very indebted to the referees for their valuable comments and corrections, which improved the original manuscript of this paper. This work was supported by National Natural Science Foundations of China (11361074, 71161020), Applied Basic Research Programs of Science and Technology Department of Yunnan Province (2013FD002), and IRTSTYN, Research Foundation of Guizhou Minzu University (15XRY004).
Received: 21 April 2015 Accepted: 1 June 2015
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