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Applied Mathematics and Computation 242 (2014) 196–201

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The representation of the Drazin inverse of anti-triangular operator matrices based on resolvent expansions Junjie Huang a, Yunfeng Shi b,⇑, Alatancang Chen a a b

School of Mathematical Sciences, Inner Mongolia University, Hohhot 010021, PR China School of Mathematical Sciences, Fudan University, Shanghai 200433, PR China

a r t i c l e

i n f o

a b s t r a c t

Keywords: Drazin inverse Anti-triangular operator matrix Resolvent expansion

This paper deals with the anti-triangular operator matrix M ¼



A C

B 0



with A2 ¼ A and

CAp B ¼ 0. Using the resolvent expansion technique, we obtain the explicit representation of the Drazin inverse of M, in terms of its entries and the Drazin inverses of the entries and their compositions. The result extends the main results in Bu et al. (2011) [2] and Castro-González and Dopazo (2005) [6] and the results on group inverses in Bu et al. (2008) [3] and Liu and Yang (2012) [14]. As an application, a new additive result is given of the Drazin inverse for two matrices P; Q 2 Cnn with P2 ¼ P and PQ 2 ¼ 0. Ó 2014 Elsevier Inc. All rights reserved.

1. Introduction Let X and Y be complex Banach spaces. The set BðX; YÞ consists of all bounded linear operators from X to Y and, if X ¼ Y, is abbreviated as BðXÞ. An operator T 2 BðXÞ is said to be Drazin invertible, if there exists some S 2 BðXÞ such that

ST ¼ TS;

STS ¼ S;

T kþ1 S ¼ T k ;

ð1:1Þ

where the nonnegative integer k is the index of T and is denoted by indðTÞ. In this case, we call S the Drazin inverse of T and write T D ¼ S. As we know, the solution (if exists) of (1.1) must be unique. The special case when indðTÞ ¼ 1 gives the group inverse denoted by T ] . Obviously, T is invertible if and only if indðTÞ ¼ 0. See, e.g., [1,16] for details. The Drazin inverse has been regarded as a convenient and effective tool in various applied mathematical areas such as singular differential or difference equations, Markov chains, cryptography and iterative methods (see [1,4]). Since Drazin introduced the concept of the Drazin inverse in [11], more and more authors have focused on this topic. See, e.g., [4,8–10,16]. In [4], Campbell and Meyer proposed the problem that how to establish the explicit representation of the Drazin inverse   A B in terms of its individual blocks and the Drazin inverses of the blocks and their comof the 2  2 block matrix M ¼ C D positions, where A and D are both square but need not be of the same size. In fact, it is very difficult to determine the explicit representation even for anti-triangular block matrices. In [6], the authors took almost ten pages to obtain the main result for   I I M¼ with I being the identity matrix, and they made further investigations later in [7]. Subsequently, Bu et al. E 0   A A with A2 ¼ A (see [2]). extended this result to the case M ¼ B 0 ⇑ Corresponding author at: School of Mathematical Sciences, Fudan University, Shanghai 200433, PR China. E-mail address: [email protected] (Y. Shi). http://dx.doi.org/10.1016/j.amc.2014.05.053 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.

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J. Huang et al. / Applied Mathematics and Computation 242 (2014) 196–201

In this paper, we study the representation of the Drazin inverse of anti-triangular operator matrices and its applications. Based on resolvent expansions, the explicit formula under the assumptions A2 ¼ A and CAp B ¼ 0 is obtained, which extends the relevant results in [2,3,6,14]. Using the representation, we also give a new additive result of the Drazin inverse. Finally, a numerical example is presented to illustrate the main result. It is shown that the resolvent expansion method is concise and effective in solving the representation problem of the Drazin inverse. Throughout this paper, we use the following notations and assumptions:Suppose that the complex Banach space X can be decomposed as X ¼ Y  Z, where Y and Z are the closed subspaces of X. Denote by P Y (resp. P Z ¼ I  P Y ) the projection on Y (resp. Z) along Z (resp. Y), where I is the identity operator. Let M 2 BðXÞ admit the block form:





A

B

C

0

 ;

where A 2 BðYÞ; B 2 BðZ; YÞ; C 2 BðY; ZÞ; 0 2 BðZÞ. For an operator T 2 BðXÞ, we write qðTÞ; rðTÞ; rðTÞ and Rðk; TÞ for the resolvent set, the spectrum, the spectral radius and the resolvent ðkI  TÞ1 of T, respectively; we also write T p ¼ I  TT D . Denote n! , and assume that Cðn; kÞ ¼ 0 whenever k > n. We define a sum to be zero if its lower by Cðn; kÞ the binomial coefficient k!ðnkÞ! limits is bigger than its upper limits. 2. Main result In this section, we first review a fundamental result, and then prove our main result, i.e., Theorem 2.3. Finally, as the special cases of Theorem 2.3, some earlier results are mentioned. Lemma 2.1 (see [5]). Let T 2 BðXÞ, then T is Drazin invertible if and only if 0 R rðTÞ n f0g and the point zero, provided 0 2 rðTÞ, is a pole of the resolvent Rðk; TÞ ¼ ðkI  TÞ1 , and in this case the following representation holds:

Rðk; TÞ ¼

indðTÞ X

kk T k1 T p 

k¼1

1 X kþ1 kk ðT D Þ ; k¼0

1

D

where 0 < jkj < ðrðT ÞÞ . Remark 2.2. From Lemma 2.1, we can obtain T D by determining the coefficient of k0 in the Laurent expansion of the resolvent Rðk; TÞ in some punctured neighborhood of zero, i.e.,

TD ¼ 

1 2p i

I

k1 Rðk; TÞdk;

c

where c ¼ fk : jkj ¼ eg and

e is sufficient small such that fk : jkj 6 eg \ rðTÞ ¼ f0g.

 A B and CAB be Drazin invertible and s ¼ indðCABÞ. If A2 ¼ A and CAp B ¼ 0, then the Drazin inverse   C 0 A B of the anti-triangular operator matrix M ¼ is given by C 0 Theorem 2.3. Let M ¼

MD ¼





E11 þ A E12 E21

E22

 ;

where 2

3

2

E11 ¼ ABU 20 þ BU 21 þ Ap BU 22  Ap BðððCABÞD Þ Cð2I  AÞ þ ððCABÞD Þ CAp Þ  ABððCABÞD CA þ ððCABÞD Þ CAp Þ; 2

E12 ¼ ABU 10 þ Ap BU 11 þ BðCABÞD þ Ap BððCABÞD Þ ; 2

E21 ¼ U 10 CA þ U 11 CAp þ ðCABÞD C þ ððCABÞD Þ CAp ; E22 ¼ U 00  ðCABÞD ; s X ð1Þjþkþm Cð2j  2 þ k; j þ mÞðCABÞj1 ðCABÞp : U km ¼ j¼1

Proof. Suppose that A – 0, since otherwise the result is trivial. We first assert DðkÞ1 ¼ ðk  1ÞRðkðk  1Þ; CABÞ for 0 < jkj < 1, 1 where DðkÞ ¼ kI  CRðk; AÞB. By A2 ¼ A, we have AD ¼ A; indðAÞ 6 1 and rðAÞ ¼ rðAD Þ ¼ limk!1 jjAk jjk ¼ 1. When 0 < jkj < 1, it p follows from Lemma 2.1 and CA B ¼ 0 that

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J. Huang et al. / Applied Mathematics and Computation 242 (2014) 196–201

DðkÞðk  1ÞRðkðk  1Þ; CABÞ ¼

kI þ

!  1 X kn CAB ðk  1ÞRðkðk  1Þ; CABÞ ¼ kI þ n¼0

 1 CAB ðk  1ÞRðkðk  1Þ; CABÞ ¼ I; 1k

similarly, ðk  1ÞRðkðk  1Þ; CABÞDðkÞ ¼ I. This proves the assertion. Next, we give the explicit representation of the resolvent Rðk; MÞ of M. To show this, let C ¼ fk : 0 < jkj < 1; 0 < jk2  kj < rððCABÞD Þ obtain

1

g. Evidently, there exists a punctured neighborhood U of zero such that U  C. For k 2 U, we

Rðk; AÞ þ Rðk; AÞBDðkÞ1 CRðk; AÞ Rðk; AÞBDðkÞ1

Rðk; MÞ ¼

DðkÞ1 CRðk; AÞ

!

DðkÞ1

by the Schur theorem. Then, we claim that M is Drazin invertible. Indeed, from Lemma 2.1 and the assumptions, it follows that

Rðk; MÞ ¼ ðRðk; AÞ þ Rðk; AÞBDðkÞ1 CRðk; AÞÞPY þ Rðk; AÞBDðkÞ1 PZ þ DðkÞ1 CRðk; AÞPY þ DðkÞ1 P Z :

ð2:1Þ

Note that PY and P Z are analytic (operator) functions, and the resolvents of the operators A and CAB are meromorphic functions (in some neighborhood of zero). As a result, the expression (2.1) is a sum of several meromorphic functions and, hence, Rðk; MÞ is meromorphic (with a unique pole at the point zero), i.e., M D exists. Finally, we compute MD . Let c  U be a Cauchy contour around the point zero. Using Lemma 2.1 and the representation of Rðk; MÞ yields

0

I

1 M ¼ 2pi D

k Rðk; MÞdk ¼ @ 1

c

 21pi

H

c ðk

1

Rðk; AÞ þ k1 E11 ðkÞÞdk  21pi

 21pi

H

ck

1

E21 ðkÞdk

 21pi

H H

ck

1

E12 ðkÞdk

ck

1

E22 ðkÞdk

1 A;

where

E11 ðkÞ ¼ Rðk; AÞBDðkÞ1 CRðk; AÞ; E21 ðkÞ ¼ DðkÞ1 CRðk; AÞ; Set Eij ¼  21pi

H

1

ck

E12 ðkÞ ¼ Rðk; AÞBDðkÞ1 ;

E22 ðkÞ ¼ DðkÞ1 :

Eij ðkÞdk. To show the representation of M D , we need determine the coefficients of k0 in the Laurent expan-

sions of Eij ðkÞ ði; j ¼ 1; 2Þ in some punctured neighborhood of zero. Define f ðkÞ ¼ ðk  1Þjþ1 for jkj < 1. Obviously, the complex P jþ1 n Cðj þ n  2; nÞ. From valued function f ðkÞ can be expanded as the power series f ðkÞ ¼ 1 n¼0 an k , where an ¼ ð1Þ

aj ¼ ð1Þjþ1 Cð2j  2; jÞ and DðkÞ1 ¼

s X

kj f ðkÞðCABÞj1 ðCABÞp 

j¼1

1 X jþ1 kj ðk  1Þjþ1 ððCABÞD Þ ; j¼0

it follows that

E22 ¼ 

1 2p i

I

s X k1 E22 ðkÞdk ¼  aj ðCABÞj1 ðCABÞp  ðCABÞD ¼ U 00  ðCABÞD :

c

j¼1

Analogously, we can define another two functions gðkÞ ¼ ðk  1Þj and hðkÞ ¼ ðk  1Þj1 ðjkj < 1Þ. To obtain E11 ; E12 ; E21 , it suffices to observe

E11 ðkÞ ¼ k2 Ap BDðkÞ1 CAp þ

k1 1 ðAp BDðkÞ1 CA þ ABDðkÞ1 CAp Þ þ ABDðkÞ1 CA; k1 ðk  1Þ2

E12 ðkÞ ¼ k1 Ap BDðkÞ1 þ

1 ABDðkÞ1 ; k1

E21 ðkÞ ¼ k1 DðkÞ1 CAp þ

1 DðkÞ1 CA: k1

The proof is completed.

h

The following corollaries immediately follow from Theorem 2.3. Their proofs are trivial and, hence, omitted.   A A Corollary 2.4 (see [2]). Let M ¼ with A; B 2 Cnn and A2 ¼ A. Then, B 0

MD ¼



M 11 M 21

 M 12 ; M 22

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J. Huang et al. / Applied Mathematics and Computation 242 (2014) 196–201

where indðMÞ ¼ s P 1, and

M 11 ¼ M 12  ABUð1Þ  ðABÞD ;

M 12 ¼ Uð0ÞA þ ðABÞD A; 2

M 21 ¼ M 22 þ B½Uð0Þ þ Uð1Þ þ ðBAÞD B þ B½ðABÞD  ; M 22 ¼ BUð1ÞA  ðBAÞD ; UðkÞ ¼

s1 X ð1Þj Cð2j  k þ 1; j þ 1ÞðABÞp ðABÞjk ;

k ¼ 0; 1:

j¼k

Corollary 2.5 (see [3]). Let M ¼



A B

A 0



with A; B 2 Cnn and A2 ¼ A. Then,

(i) M ] exists if and only if rankðBÞ ¼ rankðBABÞ; (ii) if M ] exists, then

A  ðABÞ] þ ðABÞ] A  ðABÞ] ABA A þ ðABÞ] A  ðABÞ] ABA

MD ¼

ðBAÞ] B þ ðBAÞ] ðABÞ] AB  ðBAÞ]

Corollary 2.6 (see [14]). Let M ¼

MD ¼



A þ M11 M 21



A C

B 0



!

ðBAÞ]

:

with A 2 Cnn ; A2 ¼ A; CAp B ¼ 0 and CB group invertible. Then,

 M12 ] ; ðCBÞ

where 2

2

2

M 11 ¼ ABðCBÞ] CA  2ABðCBÞp CA þ ABðCBÞp CAp  ABððCBÞ] Þ CAp  Ap BððCBÞ] Þ CAp þ Ap BðCBÞp CA  Ap BððCBÞ] Þ C 3

 Ap BððCBÞ] Þ CAp ; 2

M 12 ¼ ABðCBÞp þ BðCBÞ] þ Ap BððCBÞ] Þ ; 2

M 21 ¼ ðCBÞp CA þ ðCBÞ] C þ ððCBÞ] Þ CAp : 3. Application to additive results A topic closely related to the representation of the Drazin inverse of 2  2 operator matrices, is to express the Drazin inverse ðP þ Q ÞD of the sum of two operators P and Q in terms of P; Q ; PD ; Q D and ðPQ ÞD , i.e., the so-called additive result of the Drazin inverse. In 1958, Drazin [11] obtained the first additive result under the assumptions PQ ¼ QP ¼ 0. In 2001, Hartwig et al. [12] extended this result to the case of PQ ¼ 0. Many results are obtained in some special cases in the sequel (see, e.g., [15,13,17]). This problem, however, has not been solved until now in general case. In this section, we apply Theorem 2.3 to investigate the additive result of the Drazin inverse of matrices. We should note that our results can be easily extended to the operator case. 2

Lemma 3.1 (see [1]). Let A 2 Cnm and B 2 Cmn . Then, ðABÞD ¼ AððBAÞD Þ B. Lemma 3.2 (see [12]). Let P; Q 2 Cnn with PQ ¼ 0. Then,

ðP þ Q ÞD ¼

s1 t1 X X kþ1 kþ1 Q k Q p ðPD Þ þ ðQ D Þ Pk Pp ; k¼0

k¼0

where s ¼ indðPÞ and t ¼ indðQ Þ. Theorem 3.3. Let P; Q 2 Cnn with P 2 ¼ P and PQ 2 ¼ 0. Then, D

ðP þ Q Þ ¼ ð I  P where s ¼ ind I

8 " <X s1  P 0 0 QÞ : n¼0 0 Q n Q p I PQ 0

  P ; t ¼ indðQ Þ and I

PQ 0 PQ 0

D #nþ1 þ

D

 ¼

t1  X 0 n¼0

W 11 þ P W 21

0 nþ1 0 ðQ D Þ W 12 W 22





with

P I

PQ 0

p 

P I

PQ 0

9 n =2   P ; ; I

ð3:1Þ

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J. Huang et al. / Applied Mathematics and Computation 242 (2014) 196–201 2

W 11 ¼ PQ ðU 20 þ U 21  ðPQ ÞD P  ððPQÞD Þ ðI  PÞÞ;

W 12 ¼ PQðU 10 þ ðPQ ÞD Þ;

2

W 21 ¼ U 10 P þ U 11 ðI  PÞ þ ðPQÞD P þ ððPQÞD Þ ðI  PÞ; indðPQÞ X

U km ¼

W 22 ¼ U 00  ðPQ ÞD ;

ð1Þjþkþm Cð2j  2 þ k; mÞðPQÞj1 ðPQ Þp :

j¼1

Proof. According to Lemma 3.1,

 ðP þ QÞD ¼ ð I

Write F ¼



P I



P I

PQ 0

PQ

D



¼

Q



 D P ¼ ðI I

and G ¼



s1  X 0 n¼0

0 0

0 Q

"

P I

PQ Q

D #2   P : I

  . Since PQ 2 ¼ 0; FG ¼ 0 0

"

0



0 Q nQ p

P

PQ

I

D #nþ1

þ

0

t1  X 0 n¼0

ð3:2Þ

PQ 2 0



 ¼

0 0



0 D nþ1

P I

0 ðQ Þ

 0 . By Lemma 3.2, we have 0

  PQ p P 0

PQ

I

0

n :

This together with (3.2) demonstrates that (3.1) is valid. Furthermore, the fact P 2 ¼ P implies P p PQ ¼ 0, and hence   W 11 þ P W 12 by using Theorem 2.3. The proof is finished. h FD ¼ W 21 W 22 The statement below is a direct consequence of Theorem 3.3. Corollary 3.4. Let P; Q 2 Cnn with P 2 ¼ P; PQ 2 ¼ 0 and Q nilpotent. Then, D

ðP þ QÞ ¼ ð I

8 " s1  <X 0 0 P QÞ : n¼0 0 Q n I

where s and the calculation of



P I

PQ 0

D

9 D #nþ1 =2   P ; ; I 0

PQ

are the same as in Theorem 3.3.

4. Numerical example In this section, we conclude with an simple example to illustrate the previous main result.   A B Example 4.1. Let M ¼ with C D

1 1 0 1 0 C B B 0 1 0 0C C B A¼B C; B 2 0 2 0C A @

1 1 0 C B B0 1C C B B¼B C; B1 0C A @

0

0

3

0

0

0

0



1 2

1

1 2

0

 12 1  12 0

! ;



0 0 0 0

! :

1

p

2

One can easily show A ¼ A and CA B ¼ 0. Then, according to Theorem 2.3, we have

0

5 2

7

5 2

B B 1 3 1 B B B  7 11  7 B 2 2 MD ¼ B B B 4 11 4 B B 1 1 B 1 2 @ 2  12

1

0

7

9

1

C 4 C C C 0 11 15 C C C: C 0 11 14 C C C 0 1 1 C A 0

 12 0

3

1

1

Acknowledgment We are grateful to the referees for valuable comments on this paper. This work is supported by the National Natural Science Foundation of China (No. 11061019), the Natural Science Foundation of Inner Mongolia (No. 2013JQ01), and the Program for Young Talents of Science and Technology in Universities of Inner Mongolia (No. NJYT-12-B06).

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