The New Estimates of Diagonally Dominant Degree for the Perron Complement of Three Known Subclasses of H-matrices

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The New Estimates of Diagonally Dominant

Degree for the Perron Complement of Three

Known Subclasses of

H

-matrices

Jingjing Cui, Guohua Peng, Quan Lu and Zhong Xu

Abstract—The theory of Perron complement is very impor-tant in many fields such as control theory and computational mathematics. In this paper, some new estimates of diagonally dominant degree for the Perron complement of matrices are derived by using the entries and spectral radius of the original matrix, which are better than the corresponding ones obtained by Wang and Liu(J. Inequal. Appl. 2015:9, 2015). Finally, a numerical example is provided to confirm the theoretical results obtained in this paper.

Index Terms—Perron complement, diagonally dominant de-gree,H-matrix, nonnegative irreducible matrix, spectral radius

I. INTRODUCTION

F

OR a positive integern,N denotes the set{1,2,· · ·, n}, andRn×n(Cn×n) denotes the set of all real (complex)

matrices throughout. LetA= (aij)∈Cn×n(n≥2). Denote

|A|= (|aij|)and

Ri(A) =

X

j6=i

|aij|, Si(A) =

X

j6=i

|aji|, i∈N,

Nr(A) ={i:|aii|> Ri(A), i∈N}, Nc(A) ={i:|aii|> Si(A), i∈N}.

The comparison matrix of A, denoted by µ(A) = (µij)n×n, is defined to be

µij =

|aij| if i=j, −|aij|, if i6=j.

A matrixAis called a nonsingularM-matrix if there exist a nonnegative matrixB and a real numbers > ρ(B)such that A =sI−B, where ρ(B) is the spectral radius ofB. It is well known thatAis a nonsingular H-matrix if and only if µ(A)is a nonsingularM-matrix. We denote byHn andMn the sets of alln×n H-matrix andM-matrix, respectively.

Recall thatAis a (row) diagonally dominant matrix (Dn) if for alli= 1,2,· · ·, n,

|aii| ≥Ri(A). (1)

Manuscript received December 1, 2015; revised February 2, 2016. This work was supported by the National Natural Science Foundations of China (No.10802068). Paper titles should be written in uppercase and lowercase letters, not all uppercase. Avoid writing long formulas with subscripts in the title; short formulas that identify the elements are fine.

Jingjing Cui is with the Department of Applied Mathematics, Northwest-ern Polytechnical University, Xi’an, Shaanxi, 710072, PR China. e-mail: [email protected].

Guohua Peng (e-mail: [email protected]) , Quan Lu (e-mail: [email protected]) and Zhengge Huang (e-mail: [email protected])are with the Department of Applied Mathematics, Northwestern Polytechnical University,Xi’an, Shaanxi, 710072, PR China.

Ais a doubly diagonally dominant matrix (DDn) if for all i, j∈N, i6=j,

|aii||ajj| ≥Ri(A)Rj(A). (2)

A is a γ-diagonally dominant matrix (Dγ

n) if there exists γ∈[0,1]such that

|aii| ≥γRi(A) + (1−γ)Ci(A), ∀i∈N. (3)

Ais a productγ-diagonally dominant matrix (P Dγ

n) if there existsγ∈[0,1]such that

|aii| ≥[Ri(A)]γ[Ci(A)]1−γ, ∀i∈N. (4)

If all inequalities in (1)-(4) hold, then A is said to be a strictly (row) diagonally dominant matrix (SDn), a strictly doubly diagonally dominant matrix (SDDn), a strictly γ-diagonally dominant matrix (SDγ

n) and a strictly productγ-diagonally dominant matrix (SP Dγ

n), respectively. As in [1] and [2], for 1 ≤ i ≤ n and γ ∈ [0,1], we call |aii| −Ri(A), |aii| −γRi(A)−(1−γ)Ci(A) and |aii| −[Ri(A)]γ[Ci(A)]1−γ thei-th (row) dominant degree, γ-dominant degree and product γ-dominant degree of A, respectively.

ForA∈Cn×n, nonempty index setsα, β⊆N, we denote

by|α|the cardinality ofαandα0=N−αthe complement αof in N . LetA(α, β) denote the sub-matrix of A lying in the rows indexed by α and the columns indexed by β. A(α, α)is abbreviated toA(α). IfA(α)is nonsingular, then the Schur complement ofA(α)inAis given by

A/α=A/A(α) =A(α0)−A(α0, α)[A(α)]−1A(α, α0).

Meyer [3] introduced, for an n ×n non-negative and irreducible matrix A, the notion of the Perron complement. Again, let ∅ 6= α , and α0 = N −α, then the Perron complement of A with respect to A(α), which is denoted byP(A/A(α))or simplyP(A/α), is defined as

A(α0) +A(α0, α)[ρ(A)I−A(α)]−1A(α, α0),

where ρ(·) denotes the spectral radius of a matrix. Recall that asA is irreducible,ρ(A)> ρ(A(α)), so the expression of the above definition is well defined.

The Perron complement has been proved to be a useful tool in many fields such as control theory, statistics and computational mathematics, such as solving the constrained optimization problems [4], realized range-based threshold estimation for Jump-diffusion models [5] and so forth. Meyer [3], [6] has derived several interesting and useful properties of P(A/A(α)), obtained that the Perron complement of a nonnegative irreducible matrix is nonnegative irreducible,

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and first used the closure property of a nonnegative irre-ducible matrix to construct a divided and conquer algorithm to compute the Perron vector for a Markov chain. Moreover, many works have been done on it (see [7], [8], [9]). So far as we know, if a given matrix has a sharper diagonally dominant degree, then the designed iterative algorithms have faster convergent rate than the ordinary ones [10]. At the same time, if a given matrix has a sharper diagonally dominant degree, then we may discuss more properties about generalized nonlinear diagonal dominance in [11]. Motivated by the useful applications, we will study the diagonal dominant degree for the perron complement of several cases based on the nonnegative and irreducible nature. In this paper, we exhibit some new estimates of diagonally dominant, γ -diagonally and productγ-diagonally dominant degree for the Perron complement of matrices. These bounds improve the related results.

The remainder of this paper is organized as follows. In Section II, we present several new estimates for the diagonally dominant degree of the Perron complement of matrices. In Section III, we propose some estimates for the γ-diagonally and product γ-diagonally degree for the Perron complement of matrices. In Section IV is devoted to a numerical experiment to show the advantage of our derived results. Finally, we demonstrate our conclusions in Section V.

II. THE DIAGONALLY DOMINANT DEGREE FORPERRON COMPLEMENT

In this section, we start with some lemmas which are utilized in the next proofs. Based on these lemmas and using the entries and spectral radius of the original matrix, the estimations of the diagonally dominant degree for the Perron complement are presented.

Lemma 2.1: [12] If A is a nonsingular H-matrix, then

[µ(A)]−1≥ |A−1|.

Lemma 2.2: [12]If Ais a SDn or Ais a SDDn. Then µ(A) is a nonsingular M-matrix, i.e., A is a nonsingular H-matrix.

Lemma 2.3: [13] Let A ∈ _Rn×n_{. If} _A _∈

Mn , then

detA >0.

Lemma 2.4: [14] Let A = (aij)n×n,∅ 6= α ∈ N and assume thatA(α)is nonsingular. Then

detA= detA(α) detA/α.

Lemma 2.5: Let A = (aij) ∈ Rn×n be nonnegative

irreducible with spectral radiusρ(A),α={i1, i2,· · ·, ik} ⊆ Nr(A)6=∅,α0={j1, j2,· · ·, jl},|α|< n, and denote

Bjt =



       

x −|ajti1| · · · −|ajtik|

− l

P

u=1

|ai1ju| ..

. µ[ρ(A)I−A(α)] −

l

P

u=1

|aikju|



        ,

then for ρ(A)≥2|aii|(i∈α) and anyjt∈ α0, Bjt is an M-matrix ofk+ 1anddetBjt >0, if

x > h k

X

w=1

|ajtiw|

Piw(A) ρ(A)− |aiwiw|

(5)

where

r= max 1≤w≤k

l

P

u=1

|aiwju|

ρ(A)− |aiwiw| −

k

P

u=1,u6=w |aiwiu|

. (6)

Piw(A) =r

k

X

u=1,u6=w

|aiwiu|+

l

X

u=1

|aiwju| (7)

h= max 1≤w≤k

l

P

u=1

|aiwju|

Piw(A)−

k

P

u=1,u6=w |aiwiu|

Piu(A)

ρ(A)−|aiuiu|

(8)

Proof Denote Bjt ≡B = (bpq). Inasmuch as α⊆Nr(A)

andρ(A)≥2|aii|(i∈α), we deduce that

k

X

u=1,u6=w

|aiwiu|>0(1≤w≤k),

for all1≤w≤k, it follows from Equation (6) that

r≥

l

P

u=1

|aiwju|

k

P

u=1,u6=w |aiwiu|

,

that is,

r(ρ(A)− |aiwiw|)≥

l

X

u=1

|aiwju|+r

k

X

u=1,u6=w

|aiwiu|=Piw(A).

From the above inequality, for all1≤w≤k,

0≤ Piw(A) ρ(A)− |aiwiw|

≤r <1.

Together with Equation (7), for all1≤w≤k, we have l

P

u=1

|aiwju|

Piw(A)−

k

P

u=1,u6=w |aiwiu|

P_iu(A)

ρ(A)−|a_iuiu|

=

Piw(A)−r

k

P

u=1,u6=w |aiwiu|

Piw(A)−

k

P

u=1,u6=w |aiwiu|

Piu(A)

ρ(A)−|aiuiu|

≤1.

Combining Equation (8) and the above inequality results in

0≤h≤1.

Otherwise, for1≤w≤k,

h≥

l

P

u=1

|aiwju|

Piw(A)−

k

P

u=1,u6=w |aiwiu|

P_iu(A)

ρ(A)−|aiuiu| ,

we deduce that

hPiw(A)≥

l

X

u=1

|aiwju|+h

k

X

u=1,u6=w |aiwiu|

Piu(A) ρ(A)− |aiuiu|

. (9)

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Inequality (5) means that there exists ε >0such that

x > k

X

w=1

|ajtiw|(h

+ε). (10)

Choose a positive matrix D = diag(d1, d2,· · ·, dk+1) and Ct=BjtD= (csv), where

dv=

(

1, v= 1;

h Piv−1(A)

ρ(A)−|a_iv₋₁_iv₋₁|+ε, 2≤v≤k+ 1.

Ifs= 1, it follows from (10) that

|css| − k+1 X

w=1,w6=s |csw|

= |c11| −

k+1 X

w=2

|c1w|

= x− k

X

w=1

|ajtiw|[h

+ε]>0;

Ifs= 2,3,· · ·, k+ 1, it follows from (9) that

|css| − k+1

X

w=1,w6=s |csw|

=|bss|ds− |bs1| −

k+1

X

w=2,w6=s |bsw|dw

= ρ(A)− |ais−1is−1|

h Pis−1(A) ρ(A)− |ais−1is−1|

+ε

− l

X

u=1

|ais−1ju| −

k

X

w=1,w6=s−1

|ais−1iw|

h Piw(A) ρ(A)− |aiwiw|

+ε

=hPis−1(A) +



ρ(A)− |ais−1is−1| − k

X

w=1,w6=s−1

|ais−1iw|



ε

− l

X

u=1

|ais−1ju| −

k

X

w=1,w6=s−1

|ais−1iw|

hPiw(A) ρ(A)− |aiwiw|

≥ l

X

u=1

|ais−1ju|+h

k

X

u=1,u6=s−1

|ais−1iu|

Piu(A) ρ(A)− |aiuiu|

+



ρ(A)− |ais−1is−1| − k

X

w=1,w6=s−1

|ais−1iw|



ε

− l

X

u=1

|ais−1ju| −

k

X

w=1,w6=s−1

|ais−1iw|

=



ρ(A)− |ais−1is−1| − k

X

w=1,w6=s−1

|ais−1iw|



ε >0.

It follows thatCtis aSDk+1. Further,Bjt ∈Hk+1. Besides, µ(Bjt) = Bjt is an M-matrix of k+ 1. Therefore, from

Lemma 2.3, we obtaindetBjt>0.

Lemma 2.6: Let A = (aij) ∈ Rn×n be nonnegative

irreducible with spectral radiusρ(A),α={i1, i2,· · ·, ik} ⊆ Nr(A)6=∅,α0={j1, j2,· · ·, jl},|α|< n, P(A/α) = (a0ts), andρ(A)≥2|aii|(i∈α),

(i) Ifα⊆Nr(A), then for all1≤t≤l,

(ajti1,· · ·, ajtik)[ρ(A)I−A(α)]

−1   

ai1jt

.. . aikjt

   +

l

X

s=1,s6=t

ajtjs+ (ajti1,· · ·, ajtik)

×[ρ(A)I−A(α)]−1   

ai1js

.. . aikjs

  

≤Rjt(A)−w

0

jt. (11)

(ii) Ifα⊆Nc(A), then for all1≤t≤l,

−1   

ai1jt

.. . aikjt

   +

l

X

s=1,s6=t

ajsjt+ (ajsi1,· · ·, ajsik)

×[ρ(A)I−A(α)]−1   

ai1jt

.. . aikjt

  

≤Sjt(A)−w

00

jt. (12)

Here we denote

w0jt =

k

X

w=1

|ajtiw|

ρ(A)− |aiwiw| −hPiw(A) ρ(A)− |aiwiw|

,

w00jt =

k

X

w=1

|aiwjt|

ρ(A)− |aiwiw| −ehQiw(A) ρ(A)− |aiwiw|

,

r= max 1≤w≤k

l

P

u=1

|aiwju|

k

P

u=1,u6=w |aiwiu|

,

e

r= max 1≤w≤k

l

P

u=1

|ajuiw|

k

P

u=1,u6=w |aiuiw|

,

Piw(A) =r

k

X

u=1,u6=w

|aiwiu|+

l

X

u=1

|aiwju|,

Qiw(A) =er

k

X

u=1,u6=w

|aiuiw|+

l

X

u=1

|ajuiw|,

h= max 1≤w≤k

l

P

u=1

|aiwju|

Piw(A)−

k

P

u=1,u6=w |aiwiu|

P_iu(A)

ρ(A)−|aiuiu| ,

e

h= max 1≤w≤k

l

P

u=1

|ajuiw|

Qiw(A)−

k

P

u=1,u6=w |aiuiw|

Q_iu(A)

ρ(A)−|aiuiu| .

Proof Inasmuch asα⊆Nr(A)=6 ∅, we haveA(α)∈SDk. Further, ρ(A)≥2aii(i∈α)yields ρ(A)I−A(α)∈SDk.

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By Lemma 2.1 and Lemma 2.2, it follows that

{µ[ρ(A)I−A(α)]}−1_{≥ |}_[_ρ₍_A₎_I₋_A₍_α_)]−1_|_.

Together with Lemma 2.4, ∀ε >0, we deduce that

a_,

−1   

ai1jt

.. . aikjt

   +

l

X

s=1,s6=t

×[ρ(A)I−A(α)]−1   

ai1js

.. . aikjs

  

≤ l

X

s=1,s6=t

|ajtjs|+

l

X

s=1

(ajti1,· · ·, ajtik)

×[ρ(A)I−A(α)]−1   

ai1js

.. . aikjs

  

≤ l

X

s=1,s6=t

|ajtjs|+

l

X

s=1

(|ajti1|,· · ·,|ajtik|)

×{µ[ρ(A)I−A(α)]}−1   

|ai1js|

.. . |aikjs|

  

=Rjt−

k

X

u=1

|ajtiu|+ (|ajti1|,· · ·,|ajtik|)

×{µ[ρ(A)I−A(α)]}−1       

l

P

s=1

|ai1js|

.. . l

P

s=1

|aikjs|       

=Rjt−w

0 jt+ε−

" _k X

u=1

|ajtiu| −w

0 jt+ε−

(|ajti1|,· · ·,|ajtik|){µ[ρ(A)I−A(α)]}

−1

×

      

l

P

s=1

|ai1js|

.. . l

P

s=1

|aikjs|              

=Rjt−w

0 jt+ε−

detBt

det{µ[ρ(A)I−A(α)]}.

where

Bt=



         

k

P

u=1

|ajtiu| −w 0

jt+ε−|ajti1| · · · −|ajtik|

− l

P

s=1

|ai1js| ..

. µ[ρ(A)I−A(α)] −

l

P

s=1

|aikjs|



          .

Furthermore,

k

X

u=1

|ajtiu| −w

0 jt+ε

=

k

X

u=1

|ajtiu| −

k

X

w=1

|ajtiw|

+ε

=

k

X

u=1

|ajtiu| −

k

X

w=1

|ajtiw|

1− hPiw(A) ρ(A)− |aiwiw|

+ε

=

k

X

w=1

|ajtiw|

+ε

> k

X

w=1

|ajtiw|

.

From Lemma 2.5, we havedetBt>0. Having in mind that

det{µ[ρ(A)I−A(α)]}>0, we obtain

a≤Rjt−w 0

jt+ε−

detBt

det{µ[ρ(A)I−A(α)]}≤Rjt−w 0

jt+ε.

Letε→0, thus we easily get

a≤Rjt−w

0 jt−

detBt

det{µ[ρ(A)I−A(α)]} ≤Rjt−w

0 jt,

which implies (11). Moreover, (12) can be proved with a similar method to the above techniques.

Remark 2.1: According to 0 ≤ r < 1,0 ≤ h ≤ 1, we have

Riw≥Piw(A)≥hPiw(A),1≤w≤k,

which implies that

k

X

w=1

|ajtiw|

Riw ρ(A)− |aiwiw|

≥ k

X

w=1

|ajtiw|

≥h k

X

w=1

|ajtiw|

,1≤t≤l.

It follows from the above inequalities that, for all1≤t≤l,

w_j0 t =

k

X

w=1

|ajtiw|

≥ k

X

w=1

|ajtiw|

ρ(A)− |aiwiw| −Piw(A) ρ(A)− |aiwiw|

≥ k

X

w=1

|ajtiw|

ρ(A)− |aiwiw| −Riw(A) ρ(A)− |aiwiw|

.

So it is obvious that the results of Lemma 2.5 and Lemma 2.6 are sharper than the ones of Proposition 1 and Proposition 2 in [15], respectively.

Theorem 2.1: Let A = (aij) ∈ Rn×n be nonnegative

irreducible with spectral radiusρ(A),α={i1, i2,· · ·, ik} ⊆ Nr(A)6=∅,α0={j1, j2,· · ·, jl},|α|< n, P(A/α) = (a0ts), then forρ(A)≥2|aii|(i∈α)and∀t: 1≤t≤l,

|a0_tt| −Rt(P(A/α)) ≥ |ajtjt| −Rjt(A) +w

0

jt ≥ |ajtjt| −Rjt(A), (13)

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and

|a0_tt|+Rt(P(A/α)) ≤ |ajtjt|+Rjt(A)−w

0

jt ≤ |ajtjt|+Rjt(A). (14) Proof Sinceα⊆Nr(A)6=∅, by the definition of the Perron complement and Lemma 2.6, ∀t: 1≤t≤l, we have

|a0tt| −Rt(P(A/α))

=|a0_tt| − l

X

s=1,s6=t |a0_ts|

=

ajtjt+ (ajti1,· · ·, ajtik)

×[ρ(A)I−A(α)]−1   

ai1jt

.. . aikjt

  

− l

X

s=1,s6=t

×[ρ(A)I−A(α)]−1   

ai1js

.. . aikjs

  

≥ |ajtjt| −

×[ρ(A)I−A(α)]−1   

ai1jt

.. . aikjt

  

− l

X

s=1,s6=t

×[ρ(A)I−A(α)]−1   

ai1js

.. . aikjs

  

≥ |ajtjt| −(Rjt−w

0 jt).

Thus we get (13). With the same manner applied in the proof of (13), we can prove the result (14) of this theorem.

Remark 2.2: According to Remark 2.1, we know that the results of Theorem 2.1 improve the ones of Theorem 1 in [15].

Corollary 2.1: Let A = (aij) ∈ Rn×n and take α =

{1,2,· · ·, n−1} ⊆Nr(A), then

0<|ann| −h n−1 X

i=1

|ani|

Pi(A) ρ(A)− |aii|

≤ |P(A/α)|

≤ |ann|+h n−1 X

i=1

|ani| Pi(A) ρ(A)− |aii|

.

Proof Notice thatα0contains only one elementjt=n. Thus, P(A/α)is nothing but a number, andRt(P(A/α)) = 0. By expression of w_j0_t, we have

w_j0_t =w_n0 =

n−1 X

i=1

|ani|ρ(A)− |aii| −hPi(A) ρ(A)− |aii|

=

n−1 X

i=1

|ani| −h n−1 X

i=1

|ani| Pi(A)

ρ(A)− |aii|. (15)

Substituting Equation (15) into Inequalities (13) and (14) results, we can obtain the result.

III. THEγ-DIAGONALLY AND PRODUCTγ-DIAGONALLY DOMINANT DEGREE FORPERRON COMPLEMENT

In this section, we obtain some estimates for the γ -diagonally and productγ-diagonally degree for Perron com-plement under some conditions.

Lemma 3.1: [1] Leta > b,c > b,b >0 and0≤r≤1. Then

arc1−r≥(a−b)r(c−b)1−r+b.

Theorem 3.1: If A = (aij) ∈ Rn×n is nonnegative

irreducible with spectral radiusρ(A),α={i1, i2,· · ·, ik} ⊆ Nr(A)TNc(A) 6= ∅, α0 = N − α = {j1, j2,· · ·, jl},

|α|< n, and P(A/α) = (a0

ts). Then for ρ(A)≥2|aii|(i∈ α),1≤t≤l and0≤γ≤1,

|a0_tt| −(Rt(P(A/α)))γ(St(P(A/α)))1−γ ≥ |ajtjt| −(Rjt(A)−w

0 jt)

γ₍_S

jt(A)−w

00 jt)

1−γ

≥ |ajtjt| −(Rjt(A))

γ

(Sjt(A)) 1−γ

and

|a0tt|+ (Rt(P(A/α)))γ(St(P(A/α)))1−γ ≤ |ajtjt|+ (Rjt(A)−w

0 jt)

γ₍_S

jt(A)−w

00 jt)

1−γ

≤ |ajtjt|+ (Rjt(A))

γ₍_S jt(A))

1−γ_.

Proof By the definition of the Perron complement

|a0_tt| −(Rt(P(A/α)))γ(St(P(A/α)))1−γ

=|a0_tt| −

 

l

X

s=1,s6=t |a0_ts|

 

γ 

l

X

s=1,s6=t |a0_st|

 

1−γ

=

ajtjt+ (ajti1,· · ·, ajtik)[ρ(A)I−A(α)]

−1   

ai1jt

.. . aikjt

  

−

" l

X

s=1,s6=t

× [ρ(A)I−A(α)]−1   

ai1js

.. . aikjs

     

γ

×

" l

X

s=1,s6=t

ajsjt + (ajsi1,· · ·, ajsik)

× [ρ(A)I−A(α)]−1   

ai1jt

.. . aikjt

     

1−γ

≥ |ajtjt| −

×[ρ(A)I−A(α)]−1   

ai1jt

.. . aikjt

  

− l

X

s=1,s6=t

"

|ajtjs|+

IAENG International Journal of Applied Mathematics, 46:3, IJAM_46_3_01

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× [ρ(A)I−A(α)]−1   

ai1js

.. . aikjs

     

  

γ

× l

X

s=1,s6=t

"

|ajsjt|+

(ajsi1,· · ·, ajsik)

× [ρ(A)I−A(α)]−1   

ai1jt

.. . aikjt

     

  

1−γ

.

Denote

f =

(ajti1,· · ·, ajtik)[ρ(A)I−(A(α))]

−1   

ai1jt

.. . aikjt

   .

From Lemma 2.6 and Lemma 3.1

|a0_tt| −(Rt(P(A/α)))γ(St(P(A/α)))1−γ ≥ |a0_tt| −f−

Rjt−w

0 jt−f

γ Sjt−w

00 jt−f

1−γ

≥ |a0_tt| −f−h Rjt−w

0 jt

γ

Sjt−w

00 jt

1−γ −fi

=|a0_tt| −

Rjt−w

0 jt

γ Sjt−w

00 jt

1−γ .

Thus we can get the first type of inequalities of Theorem 3.1. Similarly, we can immediately verify the other one.

Remark 3.1: According to 0 ≤ r < 1,0 ≤ h ≤ 1, we have

Riw ≥Piw(A)≥hPiw(A),1≤w≤k, Siw≥Qiw(A)≥hQe iw(A),1≤w≤k,

which imply that

k

X

w=1

|ajtiw|

Riw ρ(A)− |aiwiw|

≥ k

X

w=1

|ajtiw|

≥h k

X

w=1

|ajtiw|

,1≤t≤l,

k

X

w=1

|aiwjt|

Siw ρ(A)− |aiwiw|

≥ k

X

w=1

|aiwjt|

Qiw(A) ρ(A)− |aiwiw|

≥eh

k

X

w=1

|ajtiw|

Qiw(A) ρ(A)− |aiwiw|

,1≤t≤l.

It follows from the above inequalities that, for all1≤t≤l,

w0_j_t =

k

X

w=1

|ajtiw|

≥ k

X

w=1

|ajtiw|

ρ(A)− |aiwiw| −Piw(A) ρ(A)− |aiwiw|

≥ k

X

w=1

|ajtiw|

ρ(A)− |aiwiw| −Riw(A) ρ(A)− |aiwiw|

,

w00jt =

k

X

w=1

|aiwjt|

ρ(A)− |aiwiw| −ehQiw(A) ρ(A)− |aiwiw|

≥ k

X

w=1

|aiwjt|

ρ(A)− |aiwiw| −Qiw(A) ρ(A)− |aiwiw|

≥ k

X

w=1

|aiwjt|

ρ(A)− |aiwiw| −Siw(A) ρ(A)− |aiwiw|

.

From the above inequalities, we can conclude that the bounds of Theorem 3.1 are sharper than those of Theorem 3 in [15].

Using the same technique as the proof of Theorem 3.1, we can obtain the following Theorem.

Theorem 3.2: If A = (aij) ∈ Rn×n is nonnegative

irreducible with spectral radiusρ(A),α={i1, i2,· · ·, ik} ⊆ Nr(A)TNc(A) 6= ∅, α0 = N − α = {j1, j2,· · ·, jl}, |α|< n, and P(A/α) = (a0_ts). Then for ρ(A)≥2|aii|(i∈ α),1≤t≤l and0≤γ≤1,

|a0_tt| −γRt(P(A/α))−(1−γ)St(P(A/α)) ≥ |ajtjt| −γRjt(A)−(1−γ)Sjt(A) +γw

0

jt+ (1−γ)w

00 jt

≥ |ajtjt| −γRjt(A)−(1−γ)Sjt(A)

and

|a0_tt|+γRt(P(A/α)) + (1−γ)St(P(A/α)) ≤ |ajtjt|+γRjt(A) + (1−γ)Sjt(A)−γw

0

jt−(1−γ)w

00 jt

≤ |ajtjt|+γRjt(A) + (1−γ)Sjt(A).

Remark 3.2: From Remark 3.1, we can conclude that the bounds of Theorem 3.2 are sharper than those of Theorem 2 in [15].

IV. NUMERICAL EXAMPLE

In this section, we present a numerical example to illustrate the theory results in this paper and show the advantages of our derived results.

Example 4.1: Let

A=

   

3 1 1 0 1 3 0 1 2 1 4 1 1 2 1 4

   

, α={1,2} ⊆Nr(A).

By calculation with matlab 7.1, we haveρ(A) = 6.3028≥

2|aii|(i∈α). From Theorem 1 in [15], we obtain

w3= 2 X

u=1

|a3u|

ρ(A)− |auu| −Ru(A) ρ(A)− |auu|

= 1.1883,

w4= 2 X

u=1

|a4u|

ρ(A)− |auu| −Ru(A)

ρ(A)− |auu| = 1.1883,

and

|a0₁₁| −R1(P(A/α))≥ |a33| −R3(A) +w3= 1.1883,

|a022| −R2(P(A/α))≥ |a44| −R4(A) +w4= 1.1883.

By Theorem 2.1 in this paper, we have

w₃0 = 2 X

u=1

|a3u|

ρ(A)− |auu| −hPu(A)

ρ(A)− |auu| = 1.6972,

w04= 2 X

u=1

|a4u|

ρ(A)− |auu| −hPu(A)

ρ(A)− |auu| = 1.6972.

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Thus

|a011| −R1(P(A/α))≥ |a33| −R3(A) +w03= 1.6972,

|a0₂₂| −R2(P(A/α))≥ |a44| −R4(A) +w40 = 1.6972.

In fact, by calculation, we have

P(A/α) =

4.7676 1.5352 1.5352 4.7676

,

then,

|a0₁₁| −R1(P(A/α)) = 3.2324,

|a0₂₂| −R2(P(A/α)) = 3.2324.

Thus, by Theorem 2.1 in this paper, we can get a better bound for the diagonally dominant degree of the Perron complement of matrices than Theorem 1 in [15].

V. CONCLUSIONS

This paper studies the diagonally dominant degree for the Perron complement of three known subclasses of H -matrices and exhibits some new estimates of diagonally dominant, γ-diagonally and product γ-diagonally dominant degree for Perron complement of matrices. Furthermore, these estimations are more accurate than the existing ones in [15]. And numerical result given in Section IV also show that the derived results improve the related results.

In this paper, we do not give the error analysis. i.e., how accurately these bounds can be computed. At present, it is very difficult for the authors to do this. Next, we will continue to study this problem in the future.

ACKNOWLEDGMENTS

This research is supported by the National Natural Science Foundations of China (No.10802068).

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