JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 10.1155/2014/535716 535716 Research Article Inequalities for the Minimum Eigenvalue of Doubly Strictly Diagonally Dominant M-Matrices Xu Ming 1,2 Li Suhua 2 Li Chaoqian 2 Wu Shi-Liang 1 School of Mathematical Sciences Kaili University Kaili 556011 China kluniv.edu.cn 2 School of Mathematics and Statistics Yunnan University Kunming 650091 China ynu.edu.cn 2014 1482014 2014 28 06 2014 31 07 2014 17 8 2014 2014 Copyright © 2014 Ming Xu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Let A be a doubly strictly diagonally dominant M-matrix. Inequalities on upper and lower bounds for the entries of the inverse of A are given. And some new inequalities on the lower bound for the minimal eigenvalue of A and the corresponding eigenvector are presented to establish an upper bound for the L1-norm of the solution x(t) for the linear differential system dx/dt=-Ax(t), x(0)=x0>0.

1. Introduction

For a positive integer n, N denotes the set {1,2,,n}. For A=[aij]Rn×n, we write A0 (A>0) if all aij0 (aij>0), i,jN. If A0 (A>0), we say A is nonnegative (positive, resp.). Let Zn denote the class of all n×n real matrices all of whose off-diagonal entries are nonpositive. A matrix A is called an M-matrix  if AZn and the inverse of A, denoted by A-1, is nonnegative.

Let A be an M-matrix. Then there exist a positive eigenvalue of A, τ(A)=ρ(A-1)-1, and a corresponding eigenvector x=[x1,x2,,xn]T0, where ρ(A-1) is the Perron eigenvalue of the nonnegative matrix A-1, τ(A)=min{|λ|:λσ(A)}, and σ(A) denotes the spectrum of A. τ(A) is called the minimum eigenvalue of A [2, 3]. If, in addition, A is irreducible, then A-1>0 and τ(A) is simple and x>0, which is unique if we assume that the L1-norm of x equals 1; that is, x1=i=1n|xi|=1 . If D is the diagonal matrix of an M-matrix A and C=D-A, then the spectral radius of the Jacobi iterative matrix JA=D-1C of A is denoted by ρ(JA). For a set Ω, we denote by |Ω| the cardinality of Ω. Note that Ω= if and only if |Ω|=0.

For convenience, we employ the following notations throughout. Let A=[aij]Rn×n be nonsingular with aii0, for all iN, and A-1=[βij]. We denote, for any i,jN, (1)hi=1|aii|jin|aij|,si=aii+jinaijhj,ri=1|aii|j=i+1n|aij|,li=1|aii|j=i+1n|aji|,Ri(A)=j=1naij,R(A)=maxiNRi(A),r(A)=miniNRi(A),M=maxiN{j=1nβij},m=miniN{j=1nβij},Δ-={iN:hi>1},Δ0={iN:hi=1},Δ+={iN:hi<1}.

Definition 1 (see [<xref ref-type="bibr" rid="B8">4</xref>]).

A matrix A=[aij]Cn×n is called

(strictly) diagonally dominant, if hi1 (hi<1, resp.) for all iN, and A is called doubly (strictly) diagonally dominant if hihj1 (hihj<1, resp.) for all i,jN,ij;

weakly chained diagonally dominant, if hi1, J(A)={iN:hi<1} and for all iN/J(A), there exist indices i1,i2,,ik in N with airir+10, 0rk-1, where i0=i and ikJ(A).

Remark. (i) It is well known that a doubly strictly diagonally dominant matrix A is nonsingular and that |Δ-Δ0|1 . If |Δ-Δ0|=1, we denote by i0 the unique element throughout; that is, Δ-Δ0={i0}. Meanwhile, if A is doubly strictly diagonally dominant and Δ-Δ0=, then A is strictly diagonally dominant.

(ii) It is clear that a strictly diagonally dominant matrix is doubly strictly diagonally dominant and also weakly chained diagonally dominant. Also clearly, for a doubly strictly diagonally dominant matrix A, if Δ-=, then A is weakly chained diagonally dominant; otherwise, A is not weakly chained diagonally dominant.

Estimating the bounds of the minimum eigenvalue τ(A) of an M-matrix A and its corresponding eigenvector is an interesting subject in matrix theory and has important applications in many practical problems; see [4, 68]. In particular, these bounds are used to estimate upper bounds of the L1-norm of the solution x(t) for the following system of ordinary differential equations: (2)dxdt=-Ax(t),x(0)=x0>0, where x(t), x0Rn, and ARn×n is a constant M-matrix. And it is proved in  that (3)x(t)1Qe-τ(A)tx01, where Q=maxi,jN(zi/zj) and z=[z1,z2,,zn]T is the positive eigenvector of AT corresponding to τ(A). When the order n of A is large, it is difficult to compute τ(A) and z. Hence it is necessary to estimate the bounds of τ(A) and z.

In , Shivakumar et al. obtained the following bounds of τ(A) when A is a weakly chained diagonally dominant M-matrix.

Theorem 2 (see [<xref ref-type="bibr" rid="B8">4</xref>, Theorem  4.1]).

Let A=[aij]Rn×n be a weakly chained diagonally dominant M-matrix and A-1=[βij]. Then (4)r(A)τ(A)R(A),τ(A)miniNaii,1Mτ(A)1m.

Recently, Tian and Huang  provided lower bounds of τ(A) by using the spectral radius of the Jacobi iterative matrix JA for a general M-matrix A.

Theorem 3 (see [<xref ref-type="bibr" rid="B9">9</xref>, Theorem  3.1]).

Let A=[aij]Rn×n be an M-matrix and A-1=[βij]. Then (5)τ(A)11+(n-1)ρ(JA)1maxiN{βii}.

Also in , a lower bound of τ(A), which depends only on the entries of A, has been presented when A is a strictly diagonally dominant M-matrix.

Theorem 4 (see [<xref ref-type="bibr" rid="B9">9</xref>, Corollary  3.4]).

Let A=[aij]Rn×n be a strictly diagonally dominant M-matrix. Then (6)τ(A)11+(n-1)maxiN{hi}miniN{si}.

As shown in , it is possible that r(A) equals zero or that 1/M is very small, and moreover, whenever A is not weakly chained diagonally dominant, Theorems 2 and 4 cannot be used to estimate the bounds of τ(A) effectively. On the other hand, it is difficult to estimate τ(A) by using Theorem 3 because of the difficulty of computing the diagonal elements of A-1 and ρ(JA) when n is very large.

In this paper, we continue to research the problems mentioned previously. For a doubly strictly diagonally dominant M-matrix A, we in Section 3 give some inequalities on the bounds of the entries of A-1. And in Section 4, some inequalities on bounds of τ(A) and the corresponding eigenvector are established. Lastly, an example, in which we estimate the L1-norm of the solution for the system (2) when A is a doubly strictly diagonally dominant M-matrix, is given in Section 5.

2. Preliminaries

In this section, we give a lemma which involves some results for a doubly strictly diagonally dominant M-matrix. First, some notations are listed: for a doubly strictly diagonally dominant matrix A=[aij]Rn×n and i,jN, (7)h^i={hi,if  (Δ-Δ0)=,hi,if  i(Δ-Δ0)={i0},1|aii|(|aii0|hi0+ji,i0|aij|),if  i(Δ-Δ0)={i0},s^i={si,if  (Δ-Δ0)=,aii+1h^ijiaijh^j,if  i(Δ-Δ0)={i0},aii+jiaijh^j,if  i(Δ-Δ0)={i0},r^i={ri,if  (Δ-Δ0)=,1ω|aii|j=i+1|aij|,if  (Δ-Δ0)={i0},i=i0,1|aii|(j=i+1,ji0|aij|+|aii0|hi0),if  (Δ-Δ0)={i0},i<i0,ri,if  (Δ-Δ0)={i0},i>i0, where (8)ω={1,if  (Δ-Δ0)=,minii01hi,if  (Δ-Δ0)={i0}. Note here that let 1/hi=+ if hi=0 (ii0).

Lemma 5.

Let A=[aij]Rn×n be a doubly strictly diagonally dominant M-matrix and (Δ-Δ0)={i0}. And, for any ε(hi0,minii0(1/hi)), let X=diag(x1,x2,,xn), where xi0=ε and xi=1, ii0. Then AX is a strictly diagonally dominant M-matrix. Furthermore, h^i01, h^i<1 for ii0 and s^i>0 for any iN.

Proof.

Since A is a doubly strictly diagonally dominant M-matrix and (Δ-Δ0)={i0}, we have (9)1hi0<minii01hi; hence, from ε(hi0,minii0(1/hi)), (10)1ai0i0εji0|ai0j|<1. And, for any ii0, if ji|aij|0, (11)1<aiiεji|aij|aii|aii0|ε+ji,i0|aij|, and if ji|aij|=0, inequality (11) is obvious.

From inequality (11), we have (12)|aii0|ε+ji,i0|aij|aii<1,ii0. Let AX=[a-ij]. Then (13)a-ij={aijε,j=i0,iN,aij,ji0,iN. From inequality (10), we have (14)1a-i0i0ji0|a-i0j|=1ai0i0εji0|ai0j|<1. And, for any ii0, from inequality (12), we have (15)1a-iiji|a-ij|=|aii0|ε+ji,i0|aij|aii<1. From inequality (14) and inequality (15), AX is strictly diagonally dominant. Moreover, it is clear that AXZn and (AX)-1=X-1A-10, which implies that AX is an M-matrix.

Furthermore, from the definition of h^i, we have that (16)1hi0=h^i0 and for any ii0, (17)h^i=|aii0|hi0+ji,i0|aij|aii|aii0|ε+ji,i0|aij|aii<1. We now prove s^i>0 for any iN. Since A is doubly strictly diagonally dominant, we get that there is kN, ki0, such that ai0k0 (otherwise, a contradiction to the definition of doubly strictly diagonally dominant matrices). Hence (18)0=ai0i0hi0-ji0|ai0j|<ai0i0hi0-ji0|ai0j|h^j, and equivalently, (19)s^i0=ai0i0-1h^i0ji0|ai0j|h^j>0. And for any ii0, (20)s^i=aii-ji0|aij|h^j=aii-(|aii0|h^i0+ji,i0|aij|h^j)aii-(|aii0|hi0+ji,i0|aij|)>0  (by  Inequality  (17)). Hence, from inequality (19), inequality (20), and the fact that A is an M-matrix, we have that, for any iN, (21)s^i>0. The proof is completed.

Lemma 6 (see [<xref ref-type="bibr" rid="B10">10</xref>, Page 719]).

Let A=[aij] be an n×n complex matrix and let x1,x2,,xn be positive real numbers. Then all the eigenvalues of A lie in the (22)i{zC:|z-aii|xiji1xj|aji|,iN}.

3. Bounds for the Entries of the Inverse of a Doubly Strictly Diagonally Dominant <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M221"><mml:mrow><mml:mi>M</mml:mi></mml:mrow></mml:math></inline-formula>-Matrix

In this section, upper and lower bounds for the entries of A-1 are given when A is a doubly strictly diagonally dominant M-matrix.

Lemma 7 (see [<xref ref-type="bibr" rid="B6">11</xref>, Lemma  2.2]).

Let A=[aij]Rn×n be a strictly diagonally dominant M-matrix and let A-1=[βij]. Then, for all iN, (23)βijki|aik|aiiβjj,jN,ji.

Next, we present a similar result for a doubly strictly diagonally dominant M-matrix.

Theorem 8.

Let A=[aij]Rn×n be a doubly strictly diagonally dominant M-matrix and let A-1=[βij]. Then, for all iN, (24)βijh^iβjj,jN,ji.

Proof.

If (Δ0Δ-)=, then A is strictly diagonally dominant and the conclusion follows from Lemma 7. We next suppose that (Δ0Δ-)={i0}. From Lemma 5, we get that AX is a strictly diagonally dominant M-matrix for any ε(hi0,minii0(1/hi)), where X=diag(x1,x2,,xn), xi0=ε, and xi=1, ii0. Let AX=[a-ij] and X-1A-1=[β-ij]. Then (25)a-ij={aijε,j=i0,iN,aij,ji0,iN,β-ij={βijε,i=i0,jN,βij,ii0,jN. If i=i0, from Lemma 7, we have that (26)βijε=β-ijki|a-ik|a-iiβ-jj=ki|aik|aiiεβjj,ji; that is, (27)βi0jki0|ai0k|ai0i0βjj=hi0βjj=h^i0βjj,ji0. If ii0 and j=i0, from Lemma 7, then (28)β-ij=βijki|a-ik|a-iiβ-jj=|aii0|ε+ki,i0|aik|aiiβjjε, that is, (29)βii0|aii0|ε+ki,i0|aik|aiiεβi0i0; moreover, by ε>hi01, we have (30)βii0|aii0|ε+ki,i0|aik|aiiβi0i0. And if ii0 and ji0, from Lemma 7, then (31)β-ij=βijki|a-ik|a-iiβ-jj=|aii0|ε+ki,i0|aik|aiiβjj; that is, (32)βij|aij|ε+ki,i0|aik|aiiβjj,ji. Hence, from inequality (30) and inequality (32) and letting εhi0, we have that, for any ii0, (33)βij|aii0|hi0+ki,i0|aik|aiiβjj=h^iβjj,ji. The conclusion follows from inequality (27) and inequality (33).

We next establish the upper and lower bounds for the diagonal entries of the inverse of a doubly strictly diagonally dominant M-matrix.

Theorem 9.

Let A=[aij]Rn×n be a doubly strictly diagonally dominant M-matrix and let A-1=[βij]. Then, for all iN, (34)1aiiβii1s^i.

Proof.

If (Δ0Δ-)=, then the conclusion follows from Lemma  2.2 of . We next suppose that (Δ0Δ-)={i0}. Since A is a doubly strictly diagonally dominant M-matrix, A-10 and aij0, i,jN, ij. By AA-1=I, we have that, for all iN, (35)1=aiiβii+jiaijβji, which implies (36)βii1aii. Moreover, from equality (35) and Theorem 8, we have that, for any ii0, (37)1aiiβii+jiaijh^jβii=(aii+jiaijh^j)βii=s^iβii. And similar to the proof of Theorem 8, AX=[a-ij] is a strictly diagonally dominant M-matrix, where X is given in Lemma 5. Let (AX)-1=[β-ij]. Then, from AX(AX)-1=I, we have that (38)1=a-i0i0β-i0i0+ji0a-i0jβ-ji0a-i0i0β-i0i0+ji0a-i0jkj|a-jk|a-jjβ-i0i0(by  Lemma  8)  =(a-i0i0+ji0a-i0jkj|a-jk|a-jj)β-i0i0=(ai0i0ε+ji0ai0j|aji0|ε+kj,i0|ajk|ajj)βi0i0ε=(ai0i0+1εji0ai0j|aji0|ε+kj,i0|ajk|ajj)βi0i0(ai0i0+1hi0ji0ai0j|aji0|hi0+kj,i0|ajk|ajj)×βi0i0(byε>hi0)=(ai0i0+1h^i0ji0ai0jh^j)βi0i0=s^i0βi0i0. Hence, from inequality (37), inequality (38), and Lemma 5, we obtain that for any iN(39)βii1s^i. The conclusion follows from inequality (36) and inequality (39).

Next a lower bound of the entries of the inverse of a doubly strictly diagonally dominant M-matrix will be established. Firstly, a lemma is given.

Lemma 10 (see [<xref ref-type="bibr" rid="B8">4</xref>, Theorem  3.5]).

Let A=[aij]Rn×n be a weakly chained diagonally dominant M-matrix and let A-1=[βij]. Then (40)minj,kβjk1anni=1n-1min{li,ri}.

Theorem 11.

Let A=[aij]Rn×n be a doubly strictly diagonally dominant M-matrix and let A-1=[βij]. Then (41)minj,kβjk1ωanni=1n-1min{li,r^i}, where (42)ω={1,if  (Δ-Δ0)=,minii01hi,if  (Δ-Δ0)={i0}.

Proof.

If (Δ0Δ-)=, then A is a strictly diagonally dominant M-matrix, also a weakly chained diagonally dominant M-matrix. The conclusion is evident from Lemma 10. We next suppose that (Δ0Δ-)={i0}. Similar to the proof of Theorem 8, AX is a strictly diagonally dominant M-matrix, where X is given in Lemma 5. Let AX=[a-ij] and (AX)-1=[β-ij]. By Lemma 10, we have that (43)minj,kβ-jk1a-nni=1n-1min{1a-iik=i+1n|a-ki|,1a-iik=i+1n|a-ik|}. Moreover, note that minj,kβjkminj,kβ-jk and 1/a-nn1/εann>1/ωann; we have (44)minj,kβjk1ωanni=1n-1min{1a-iik=i+1n|a-ki|,1a-iik=i+1n|a-ik|}. And also note that, for any iN, (45)1a-iik=i+1n|a-ki|=1aiik=i+1n|aki|=li. Hence, we need only prove that (1/a-ii)k=i+1n|a-ik|r^i for any iN. In fact, if i<i0, then (46)1a-iik=i+1n|a-ik|=1aii(k=i+1,ki0n|aik|+|aii0|ε)1aii(k=i+1,ki0n|aik|+|aii0|hi0)=r^i. If i=i0, then (47)1a-iik=i+1n|a-ik|=1εaiik=i+1n|aik|1ωaiik=i+1n|aik|=r^i. If i>i0, then (48)1a-iik=i+1n|a-ik|=1aiik=i+1n|aik|=ri=r^i. Hence, for any iN, (49)1a-iik=i+1n|a-ik|r^i. The conclusion follows from inequalities (44), (45), and (49).

4. Bounds for the Minimum Eigenvalue of a Doubly Strictly Diagonally Dominant <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M329"><mml:mrow><mml:mi>M</mml:mi></mml:mrow></mml:math></inline-formula>-Matrix

In this section, we give some lower bounds for τ(A) which depend only on the entries of A when A is a doubly strictly diagonally dominant M-matrix. First, for A-1=[βij], we give an upper bound of A-11, where A-11=maxiN{j=1n|βji|}.

Theorem 12.

Let A=[aij]Rn×n be a doubly strictly diagonally dominant M-matrix. Then (50)A-11maxiN{1s^i(1+jih^j)}.

Proof.

Let A-1=[βij]. Then (51)A-11=maxiN{jβji}maxiN{βii+jih^jβii}(by  Theorem  9)=maxiN{(1+jih^j)βii}maxiN{1s^i(1+jih^j)}(by  Theorem  10). The proof is completed.

Theorem 13.

Let A=[aij]Rn×n be a doubly strictly diagonally dominant M-matrix. Then (52)τ(A)minin{s^i1+(n-1)h^i}.

Proof.

If A is irreducible, then A-1>0; meanwhile, from the irreducibility of A and the definition of h^i, we have h^i>0 for any iN. We next consider the spectral radius ρ(A-1) of A-1. From Lemma 6, we have that there is k0N such that (53)|ρ(A-1)-βk0k0|h^k0kk0βkk0h^k, which, from ρ(A-1)>βk0k0 , leads to (54)ρ(A-1)βk0k0+h^k0kk0βkk0h^kβk0k0+h^k0kk0βk0k0(by  Theorem  9)=(1+(n-1)h^k0)βk0k01+(n-1)h^k0s^k0(by  Theorem  10)maxiN{1+(n-1)his^i}. Hence, (55)τ(A)=1ρ(A-1)minin{s^i1+(n-1)h^i}.

If A is reducible, then we can obtain a doubly strictly diagonally dominant M-matrix A(ϵ) such that A(ϵ) is irreducible by replacing some nondiagonal zero entries of A with sufficiently small negative real number -ϵ. Now replace A with A(ϵ) in the previous case. Let ϵ approach 0; the conclusion follows by the continuity of τ(A) about the entries of A.

From Theorems 12 and 13, we have the following result.

Theorem 14.

Let A=[aij]Rn×n be a doubly strictly diagonally dominant M-matrix. Then (56)τ(A)max{H-,H~}, where (57)H-=mini{s^i1+(n-1)h^i},H~=mini{s^i1+jih^j}.

Proof.

By Theorem 12 and the fact that ρ(A-1)A-11, we have that (58)τ(A)=1ρ(A-1)1A-11minin{s^i1+jih^j}=H~. Hence, from Theorem 13, τ(A)max{H-,H~}.

We now give upper and lower bounds for the components of the eigenvector z corresponding to the minimum eigenvalue τ(A) for an irreducible doubly strictly diagonally dominant M-matrix.

Theorem 15.

Let A=[aij]Rn×n be an irreducible doubly strictly diagonally dominant M-matrix and let A-1=[βij]. And let z=[z1,z2,,zn]T be the positive eigenvector of A corresponding to τ(A) with z1=1. Then, for all iN, (59)τ(A)minj,kβjkziτ(A)maxj,kβjk. Furthermore, (60)maxi,jzizjmaxi,j,kh^iβkkβjk.

Proof.

It is clear that A-1 exists and A-1>0. From Az=τ(A)z and z>0, we have A-1z=ρ(A-1)z=τ(A)-1z and z>0; hence, (61)zi=τ(A)k=1nβikzkτ(A)maxj,kβjkk=1nzk=τ(A)maxj,kβjk, where k=1nzk=1. The lower bound for zi is proved similarly. Furthermore, by Theorem  3.1 of , (62)maxi,jzizjmaxi,j,kβikβjk. By Theorem 8, βikh^iβkk. Hence, (63)maxi,jzizjmaxi,j,kh^iβkkβjk. The proof is completed.

5. Example

Consider the following system: (64)dxdt=-Ax(t),x(0)=x0, where (65)A=[1-0.2-0.2-0.2-0.6-0.21-0.2-0.2-0.2-0.2-0.21-0.2-0.2-0.2-0.2-0.21-0.2-0.2-0.2-0.2-0.21]. It is easy to verify that A is an irreducible doubly strictly diagonally dominant M-matrix and that Δ-={1}. Hence A is not a weakly chained diagonally dominant M-matrix. We now establish the upper bound for the L1-norm of the solution x(t). Let A-1=[βij]. By Theorems 8 and 9, we have (66)maxj,kβjk7.5000. By Theorem 11, we have (67)minj,kβjk0.0307. By Theorem 14, we have (68)τ(A)0.0276. Hence, by inequality (3) and Theorem 15, we have (69)Q=maxi,jzizjmaxi,j,kh^iβkkβjk244.1406. Hence, (70)x(t)1244.1406e-0.0276tx01. Note here that we cannot estimate the lower bound of τ(A) by using Theorem 2 (Theorem  4.1 of ) and Theorem 4 (Corollary  3.4 of ) because A is not a strictly diagonally dominant M-matrix and not a weakly chained diagonally dominant M-matrix.

Conflict of Interests

The authors declare that they have no conflict of interests.

Authors’ Contribution

Ming Xu, Suhua Li, and Chaoqian Li contributed equally to this work. All authors read and approved the final paper.

Acknowledgments

The authors are grateful to the referees for their useful and constructive suggestions. The first author is supported by Science Foundation of Guizhou Province (20132260, LKK201331, and LKK201424). The third author is supported by National Natural Science Foundations of China (11326242, 11361074), Natural Science Foundations of Yunnan Province (2013FD002), and IRTSTYN.

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