# Inequalities for the Minimum Eigenvalue of Doubly Strictly Diagonally DominantM-Matrices

Authors: Ming Xu; Suhua Li; Chaoqian Li
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

LetA 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.


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## Body

## 1. Introduction

For a positive integern, N denotes the set {1,2,…,n}. For A=[aij]∈Rn×n, we write A≥0 (A>0) if all aij≥0 (aij>0), i,j∈N. If A≥0 (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 [1] if A∈Zn and the inverse of A, denoted by A-1, is nonnegative.LetA 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]T≥0, where ρ(A-1) is the Perron eigenvalue of the nonnegative matrix A-1, τ(A)=min⁡⁡{|λ|:λ∈σ(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, ∥x∥1=∑i=1n|xi|=1 [3]. 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. LetA=[aij]∈Rn×n be nonsingular with aii≠0, for all i∈N, and A-1=[βij]. We denote, for any i,j∈N,
(1)hi=1|aii|∑j≠in‍|aij|,si=aii+∑j≠in‍aijhj,ri=1|aii|∑j=i+1n‍|aij|,li=1|aii|∑j=i+1n‍|aji|,Ri(A)=∑j=1n‍aij,R(A)=max⁡⁡i∈NRi(A),r(A)=min⁡⁡i∈NRi(A),M=max⁡⁡i∈N{∑j=1n‍βij},m=min⁡⁡i∈N{∑j=1n‍βij},Δ-={i∈N:hi>1},Δ0={i∈N:hi=1},Δ+={i∈N:hi<1}.Definition 1 (see [4]).
A matrixA=[aij]∈Cn×n is called(i)
(strictly) diagonally dominant, ifhi≤1 (hi<1, resp.) for all i∈N, and A is called doubly (strictly) diagonally dominant if hihj≤1 (hihj<1, resp.) for all i,j∈N,i≠j;(ii)
weakly chained diagonally dominant, ifhi≤1, J(A)={i∈N:hi<1}≠∅ and for all i∈N/J(A), there exist indices i1,i2,…,ik in N with airir+1≠0, 0≤r≤k-1, where i0=i and ik∈J(A).Remark. (i) It is well known that a doubly strictly diagonally dominant matrix A is nonsingular and that |Δ-⋃Δ0|≤1 [5]. 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 matrixA, if Δ-=∅, then A is weakly chained diagonally dominant; otherwise, A is not weakly chained diagonally dominant.Estimating the bounds of the minimum eigenvalueτ(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, 6–8]. 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), x0∈Rn, and A∈Rn×n is a constant M-matrix. And it is proved in [6] that
(3)∥x(t)∥1≤Qe-τ(A)t∥x0∥1,
where Q=max⁡⁡i,j∈N(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 [4], Shivakumar et al. obtained the following bounds of τ(A) when A is a weakly chained diagonally dominant M-matrix.Theorem 2 (see [4, Theorem  4.1]).
LetA=[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)≤min⁡⁡i∈Naii,1M≤τ(A)≤1m.Recently, Tian and Huang [9] 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 [9, Theorem  3.1]).
LetA=[aij]∈Rn×n be an M-matrix and A-1=[βij]. Then
(5)τ(A)≥11+(n-1)ρ(JA)1max⁡i∈N{βii}.Also in [9], 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 [9, Corollary  3.4]).
LetA=[aij]∈Rn×n be a strictly diagonally dominant M-matrix. Then
(6)τ(A)≥11+(n-1)max⁡i∈N{hi}min⁡⁡i∈N{si}.As shown in [9], 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 dominantM-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 dominantM-matrix. First, some notations are listed: for a doubly strictly diagonally dominant matrix A=[aij]∈Rn×n and i,j∈N,
(7)h^i={hi,if(Δ-⋃Δ0)=∅,hi,ifi∈(Δ-⋃Δ0)={i0},1|aii|(|aii0|hi0+∑j≠i,i0‍|aij|),ifi∉(Δ-⋃Δ0)={i0},s^i={si,if(Δ-⋃Δ0)=∅,aii+1h^i∑j≠i‍aijh^j,ifi∈(Δ-⋃Δ0)={i0},aii+∑j≠i‍aijh^j,ifi∉(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,j≠i0‍|aij|+|aii0|hi0),if(Δ-⋃Δ0)={i0},i<i0,ri,if(Δ-⋃Δ0)={i0},i>i0,
where
(8)ω={1,if(Δ-⋃Δ0)=∅,min⁡⁡i≠i01hi,if(Δ-⋃Δ0)={i0}.
Note here that let 1/hi=+∞ if hi=0 (i≠i0).Lemma 5.
LetA=[aij]∈Rn×n be a doubly strictly diagonally dominant M-matrix and (Δ-⋃Δ0)={i0}. And, for any ε∈(hi0,min⁡i≠i0(1/hi)), let X=diag⁡(x1,x2,…,xn), where xi0=ε and xi=1, i≠i0. Then AX is a strictly diagonally dominant M-matrix. Furthermore, h^i0≥1, h^i<1 for i≠i0 and s^i>0 for any i∈N.Proof.
SinceA is a doubly strictly diagonally dominant M-matrix and (Δ-⋃Δ0)={i0}, we have
(9)1≤hi0<min⁡i≠i01hi;
hence, from ε∈(hi0,min⁡i≠i0(1/hi)),
(10)1ai0i0ε∑j≠i0‍|ai0j|<1.
And, for any i≠i0, if ∑j≠i|aij|≠0,
(11)1<aiiε∑j≠i|aij|≤aii|aii0|ε+∑j≠i,i0|aij|,
and if ∑j≠i|aij|=0, inequality (11) is obvious.
From inequality (11), we have
(12)|aii0|ε+∑j≠i,i0|aij|aii<1,i≠i0.
Let AX=[a-ij]. Then
(13)a-ij={aijε,j=i0,i∈N,aij,j≠i0,i∈N.
From inequality (10), we have
(14)1a-i0i0∑j≠i0‍|a-i0j|=1ai0i0ε∑j≠i0‍|ai0j|<1.
And, for any i≠i0, from inequality (12), we have
(15)1a-ii∑j≠i‍|a-ij|=|aii0|ε+∑j≠i,i0|aij|aii<1.
From inequality (14) and inequality (15), AX is strictly diagonally dominant. Moreover, it is clear that AX∈Zn and (AX)-1=X-1A-1≥0, which implies that AX is an M-matrix.
Furthermore, from the definition ofh^i, we have that
(16)1≤hi0=h^i0
and for any i≠i0,
(17)h^i=|aii0|hi0+∑j≠i,i0|aij|aii≤|aii0|ε+∑j≠i,i0|aij|aii<1.
We now prove s^i>0 for any i∈N. Since A is doubly strictly diagonally dominant, we get that there is k∈N, k≠i0, such that ai0k≠0 (otherwise, a contradiction to the definition of doubly strictly diagonally dominant matrices). Hence
(18)0=ai0i0hi0-∑j≠i0‍|ai0j|<ai0i0hi0-∑j≠i0‍|ai0j|h^j,
and equivalently,
(19)s^i0=ai0i0-1h^i0∑j≠i0‍|ai0j|h^j>0.
And for any i≠i0,
(20)s^i=aii-∑j≠i0‍|aij|h^j=aii-(|aii0|h^i0+∑j≠i,i0‍|aij|h^j)≥aii-(|aii0|hi0+∑j≠i,i0‍|aij|)>0(byInequality(17)).
Hence, from inequality (19), inequality (20), and the fact that A is an M-matrix, we have that, for any i∈N,
(21)s^i>0.
The proof is completed.Lemma 6 (see [10, Page 719]).
LetA=[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{z∈C:|z-aii|≤xi∑j≠i‍1xj|aji|,i∈N}.


## 3. Bounds for the Entries of the Inverse of a Doubly Strictly Diagonally DominantM-Matrix

In this section, upper and lower bounds for the entries ofA-1 are given when A is a doubly strictly diagonally dominant M-matrix.Lemma 7 (see [11, Lemma  2.2]).
LetA=[aij]∈Rn×n be a strictly diagonally dominant M-matrix and let A-1=[βij]. Then, for all i∈N,
(23)βij≤∑k≠i|aik|aiiβjj,j∈N,j≠i.Next, we present a similar result for a doubly strictly diagonally dominantM-matrix.Theorem 8.
LetA=[aij]∈Rn×n be a doubly strictly diagonally dominant M-matrix and let A-1=[βij]. Then, for all i∈N,
(24)βij≤h^iβjj,j∈N,j≠i.Proof.
If(Δ0⋃Δ-)=∅, then A is strictly diagonally dominant and the conclusion follows from Lemma 7. We next suppose that (Δ0⋃Δ-)={i0}. From Lemma 5, we get that AX is a strictly diagonally dominant M-matrix for any ε∈(hi0,min⁡i≠i0(1/hi)), where X=diag⁡(x1,x2,…,xn), xi0=ε, and xi=1, i≠i0. Let AX=[a-ij] and X-1A-1=[β-ij]. Then
(25)a-ij={aijε,j=i0,i∈N,aij,j≠i0,i∈N,β-ij={βijε,i=i0,j∈N,βij,i≠i0,j∈N.
If i=i0, from Lemma 7, we have that
(26)βijε=β-ij≤∑k≠i|a-ik|a-iiβ-jj=∑k≠i|aik|aiiεβjj,j≠i;
that is,
(27)βi0j≤∑k≠i0|ai0k|ai0i0βjj=hi0βjj=h^i0βjj,j≠i0.
If i≠i0 and j=i0, from Lemma 7, then
(28)β-ij=βij≤∑k≠i|a-ik|a-iiβ-jj=|aii0|ε+∑k≠i,i0|aik|aiiβjjε,
that is,
(29)βii0≤|aii0|ε+∑k≠i,i0|aik|aiiεβi0i0;
moreover, by ε>hi0≥1, we have
(30)βii0≤|aii0|ε+∑k≠i,i0|aik|aiiβi0i0.
And if i≠i0 and j≠i0, from Lemma 7, then
(31)β-ij=βij≤∑k≠i|a-ik|a-iiβ-jj=|aii0|ε+∑k≠i,i0|aik|aiiβjj;
that is,
(32)βij≤|aij|ε+∑k≠i,i0|aik|aiiβjj,j≠i.
Hence, from inequality (30) and inequality (32) and letting ε→hi0, we have that, for any i≠i0,
(33)βij≤|aii0|hi0+∑k≠i,i0|aik|aiiβjj=h^iβjj,j≠i.
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 dominantM-matrix.Theorem 9.
LetA=[aij]∈Rn×n be a doubly strictly diagonally dominant M-matrix and let A-1=[βij]. Then, for all i∈N,
(34)1aii≤βii≤1s^i.Proof.
If(Δ0⋃Δ-)=∅, then the conclusion follows from Lemma  2.2 of [9]. We next suppose that (Δ0⋃Δ-)={i0}. Since A is a doubly strictly diagonally dominant M-matrix, A-1≥0 and aij≤0, i,j∈N, i≠j. By AA-1=I, we have that, for all i∈N,
(35)1=aiiβii+∑j≠i‍aijβji,
which implies
(36)βii≥1aii.
Moreover, from equality (35) and Theorem 8, we have that, for any i≠i0,
(37)1≥aiiβii+∑j≠i‍aijh^jβii=(aii+∑j≠i‍aijh^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+∑j≠i0‍a-i0jβ-ji0≥a-i0i0β-i0i0+∑j≠i0‍a-i0j∑k≠j|a-jk|a-jjβ-i0i0(byLemma8)=(a-i0i0+∑j≠i0‍a-i0j∑k≠j|a-jk|a-jj)β-i0i0=(ai0i0ε+∑j≠i0‍ai0j|aji0|ε+∑k≠j,i0|ajk|ajj)βi0i0ε=(ai0i0+1ε∑j≠i0‍ai0j|aji0|ε+∑k≠j,i0|ajk|ajj)βi0i0≥(ai0i0+1hi0∑j≠i0‍ai0j|aji0|hi0+∑k≠j,i0|ajk|ajj)×βi0i0(byε>hi0)=(ai0i0+1h^i0∑j≠i0‍ai0jh^j)βi0i0=s^i0βi0i0.
Hence, from inequality (37), inequality (38), and Lemma 5, we obtain that for any i∈N(39)βii≤1s^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 dominantM-matrix will be established. Firstly, a lemma is given.Lemma 10 (see [4, Theorem  3.5]).
LetA=[aij]∈Rn×n be a weakly chained diagonally dominant M-matrix and let A-1=[βij]. Then
(40)min⁡⁡j,kβjk≥1ann∏i=1n-1‍min⁡⁡{li,ri}.Theorem 11.
LetA=[aij]∈Rn×n be a doubly strictly diagonally dominant M-matrix and let A-1=[βij]. Then
(41)min⁡⁡j,kβjk≥1ωann∏i=1n-1‍min⁡⁡{li,r^i},
where
(42)ω={1,if(Δ-⋃Δ0)=∅,min⁡⁡i≠i01hi,if(Δ-⋃Δ0)={i0}.Proof.
If(Δ0⋃Δ-)=∅, 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 (Δ0⋃Δ-)={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)min⁡⁡j,kβ-jk≥1a-nn∏i=1n-1‍min⁡⁡{1a-ii∑k=i+1n‍|a-ki|,1a-ii∑k=i+1n‍|a-ik|}.
Moreover, note that min⁡j,kβjk≥min⁡j,kβ-jk and 1/a-nn≥1/εann>1/ωann; we have
(44)min⁡⁡j,kβjk≥1ωann∏i=1n-1‍min⁡⁡{1a-ii∑k=i+1n‍|a-ki|,1a-ii∑k=i+1n‍|a-ik|}.
And also note that, for any i∈N,
(45)1a-ii∑k=i+1n‍|a-ki|=1aii∑k=i+1n‍|aki|=li.
Hence, we need only prove that (1/a-ii)∑k=i+1n|a-ik|≥r^i for any i∈N. In fact, if i<i0, then
(46)1a-ii∑k=i+1n‍|a-ik|=1aii(∑k=i+1,k≠i0n‍|aik|+|aii0|ε)≥1aii(∑k=i+1,k≠i0n‍|aik|+|aii0|hi0)=r^i.
If i=i0, then
(47)1a-ii∑k=i+1n‍|a-ik|=1εaii∑k=i+1n‍|aik|≥1ωaii∑k=i+1n‍|aik|=r^i.
If i>i0, then
(48)1a-ii∑k=i+1n‍|a-ik|=1aii∑k=i+1n‍|aik|=ri=r^i.
Hence, for any i∈N,
(49)1a-ii∑k=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 DominantM-Matrix

In this section, we give some lower bounds forτ(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-1∥1, where ∥A-1∥1=max⁡i∈N{∑j=1n|βji|}.Theorem 12.
LetA=[aij]∈Rn×n be a doubly strictly diagonally dominant M-matrix. Then
(50)∥A-1∥1≤max⁡i∈N{1s^i(1+∑j≠i‍h^j)}.Proof.
LetA-1=[βij]. Then
(51)∥A-1∥1=max⁡i∈N{∑j‍βji}≤max⁡i∈N{βii+∑j≠i‍h^jβii}(byTheorem9)=max⁡i∈N{(1+∑j≠i‍h^j)βii}≤max⁡i∈N{1s^i(1+∑j≠i‍h^j)}(byTheorem10).
The proof is completed.Theorem 13.
LetA=[aij]∈Rn×n be a doubly strictly diagonally dominant M-matrix. Then
(52)τ(A)≥min⁡⁡i∈n{s^i1+(n-1)h^i}.Proof.
IfA 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 i∈N. We next consider the spectral radius ρ(A-1) of A-1. From Lemma 6, we have that there is k0∈N such that
(53)|ρ(A-1)-βk0k0|≤h^k0∑k≠k0‍βkk0h^k,
which, from ρ(A-1)>βk0k0 [12], leads to
(54)ρ(A-1)≤βk0k0+h^k0∑k≠k0‍βkk0h^k≤βk0k0+h^k0∑k≠k0‍βk0k0(byTheorem9)=(1+(n-1)h^k0)βk0k0≤1+(n-1)h^k0s^k0(byTheorem10)≤max⁡⁡i∈N{1+(n-1)his^i}.
Hence,
(55)τ(A)=1ρ(A-1)≥min⁡i∈n{s^i1+(n-1)h^i}.
IfA 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 Theorems12 and 13, we have the following result.Theorem 14.
LetA=[aij]∈Rn×n be a doubly strictly diagonally dominant M-matrix. Then
(56)τ(A)≥max⁡⁡{H-,H~},
where
(57)H-=min⁡i{s^i1+(n-1)h^i},H~=min⁡i{s^i1+∑j≠ih^j}.Proof.
By Theorem12 and the fact that ρ(A-1)≤∥A-1∥1, we have that
(58)τ(A)=1ρ(A-1)≥1∥A-1∥1≥min⁡⁡i∈n{s^i1+∑j≠ih^j}=H~.
Hence, from Theorem 13, τ(A)≥max⁡⁡{H-,H~}.We now give upper and lower bounds for the components of the eigenvectorz corresponding to the minimum eigenvalue τ(A) for an irreducible doubly strictly diagonally dominant M-matrix.Theorem 15.
LetA=[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 ∥z∥1=1. Then, for all i∈N,
(59)τ(A)min⁡⁡j,kβjk≤zi≤τ(A)max⁡⁡j,kβjk.
Furthermore,
(60)max⁡⁡i,jzizj≤max⁡⁡i,j,kh^iβkkβjk.Proof.
It is clear thatA-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=1n‍βikzk≤τ(A)max⁡⁡j,kβjk∑k=1n‍zk=τ(A)max⁡⁡j,kβjk,
where ∑k=1nzk=1. The lower bound for zi is proved similarly. Furthermore, by Theorem  3.1 of [12],
(62)max⁡i,jzizj≤max⁡i,j,kβikβjk.
By Theorem 8, βik≤h^iβkk. Hence,
(63)max⁡i,jzizj≤max⁡i,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)max⁡⁡j,kβjk≤7.5000.
By Theorem 11, we have
(67)min⁡⁡j,kβjk≥0.0307.
By Theorem 14, we have
(68)τ(A)≥0.0276.
Hence, by inequality (3) and Theorem 15, we have
(69)Q=max⁡i,jzizj≤max⁡i,j,kh^iβkkβjk≈244.1406.
Hence,
(70)∥x(t)∥1≤244.1406e-0.0276t∥x0∥1.
Note here that we cannot estimate the lower bound of τ(A) by using Theorem 2 (Theorem  4.1 of [4]) and Theorem 4 (Corollary  3.4 of [9]) because A is not a strictly diagonally dominant M-matrix and not a weakly chained diagonally dominant M-matrix.



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