Research ArticleRepresentations for the Generalized Drazin Inverse ofthe Sum in a Banach Algebra and Its Application for SomeOperator Matrices

Xiaoji Liu and Xiaolan Qin

Faculty of Science Guangxi University for Nationalities Nanning 530006 China

Correspondence should be addressed to Xiaoji Liu xiaojiliu72126com

Received 26 September 2014 Accepted 8 January 2015

Academic Editor Predrag S Stanimirovic

Copyright copy 2015 X Liu and X Qin This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We investigate additive properties of the generalized Drazin inverse in a Banach algebra A We find explicit expressions for thegeneralized Drazin inverse of the sum 119886+119887 under new conditions on 119886 119887 isin A As an application we give some new representationsfor the generalized Drazin inverse of an operator matrix

1 Introduction

LetA be a complex Banach algebra with unite 1 We use 120590(119886)to denote the spectrum of 119886 isin A The sets of all nilpotent andquasinilpotent elements (120590(119886) = 0) ofA will be denoted byAnil andAqnil respectively

The generalized Drazin inverse of 119886 isin A (introduced byKoliha in [1]) is the element 119887 isin A which satisfies

119909119886119909 = 119909 119886119909 = 119909119886 119886 minus 1198862119909 isin A

qnil (1)

If there exists the generalized Drazin inverse then thegeneralized Drazin inverse of 119886 is unique and is denoted by119886119889 The set of all generalized Drazin invertible elements ofA

is denoted byA119889 For interesting properties of the generalizedDrazin inverse see [2ndash6] For a complete treatment of thegeneralized Drazin inverse see [7 Chapter 2]

If 119901 = 1199012 isin A is an idempotent we denote 119901 = 1 minus 119901 Wecan represent element 119886 isin A as

119886 = [

1198861111988612

1198862111988622

]

119901

(2)

where 11988611= 119901119886119901 119886

12= 119901119886119901 119886

21= 119901119886119901 and 119886

22= 119901119886119901

Let 119886 isin A119889 and 119886120587 = 1 minus 119886119886119889 be the spectral idempotentof 119886 corresponding to 0 It is well known that 119886 isin A can berepresented in the following matrix form ([7 Chapter 2])

119886 = [

11988610

0 1198862

]

119901

(3)

relative to 119901 = 119886119886119889 where 119886

1is invertible in the algebra

119901A119901 119886119889 is its inverse in 119901A119901 and 1198862is quasinilpotent in the

algebra 119901A119901 Thus the generalized Drazin inverse of 119886 canbe expressed as

119886119889= [119886119889

10

0 0

]

119901

(4)

Obviously if 119886 isin Aqnil then 119886 is generalizedDrazin invertibleand 119886119889 = 0

In this paper we first give the formulas of (119886 + 119887)119889 underthe conditions 119886119887 = 119887119886119887120587 and 119886119887 = 119886120587119887119886119887120587 respectivelyThenwe will apply these formulas to provide some representationsfor the generalizedDrazin inverse of the operatormatrix119872 =

[119860 119861

119862 119863] under some conditions

Hindawi Publishing Corporatione Scientific World JournalVolume 2015 Article ID 156934 8 pageshttpdxdoiorg1011552015156934

2 The Scientific World Journal

2 Main Results

First we start the following result which is proved in [8] formatrices extended in [9] for a bounded linear operator andin [10] for arbitrary elements in a Banach algebra

Lemma 1 (see [10Theorem 23]) Let 119909 119910 isin A and 119901 isin A bean idempotent Assume that 119909 and 119910 are represented as

119909 = [

119886 0

119888 119887]

119901

119910 = [

119887 119888

0 119886]

119901

(5)

(i) If 119886 isin (119901A119901)119889 and 119887 isin (119901A119901)119889 then 119909 and 119910 are gen-eralized Drazin invertible and

119909119889= [

1198861198890

119906 119887119889]

119901

119910119889= [

119887119889119906

0 119886119889]

119901

(6)

where

119906 =

infin

sum

119899=0

(119887119889)

119899+2

119888119886119899119886120587+

infin

sum

119899=0

119887120587119887119899119888 (119886119889)

119899+2

minus 119887119889119888119886119889 (7)

(ii) If 119909 isin A119889 and 119886 isin (119901A119901)119889 then 119887 isin (119901A119901)119889 and 119909119889

and 119910119889 are given by (6) and (7)

Lemma 2 (see [11 Lemma 21]) Let 119886 119887 isin A119902119899119894119897 If 119886119887 = 119887119886or 119886119887 = 0 then 119886 + 119887 isin A119902119899119894119897

The following result is a generalization of [10 Corollary34]

Theorem 3 If 119886 isin A119902119899119894119897 119887 isin A119889 and 119886119887 = 119887119886119887120587 then 119886 + 119887 isinA119889 and

(119886 + 119887)119889= 119887119889+

infin

sum

119899=0

(119887119889)

119899+2

119886 (119886 + 119887)119899 (8)

Proof First suppose that 119887 isin Aqnil Therefore 119887120587 = 1 andfrom 119886119887 = 119887119886119887120587 we obtain 119886119887 = 119887119886 Using Lemma 2 119886 + 119887 isinAqnil and (8) holds

Now we assume 119887 is not quasinilpotent using matrixrepresentations of 119886 and 119887 relative to 119901 = 119887119887119889 We have

119887 = [

11988710

0 1198872

]

119901

119887119889= [119887119889

10

0 0

]

119901

(9)

where 1198871isin (119901A119901)

minus1 1198872isin (119901A119901)

qnilLet us represent

119886 = [

11988611198862

11988631198864

]

119901

(10)

From 119886119887 = 119887119886119887120587 and

119886119887 = [

1198861119887111988621198872

1198863119887111988641198872

]

119901

119887119886119887120587= [

0 11988711198862

0 11988721198864

]

119901

(11)

we obtain 11988611198871= 0 and 119886

31198871= 0 Since 119887

1is invertible we

have 1198861= 0 and 119886

3= 0

Hence we have

119886 + 119887 = [

1198871

1198862

0 1198864+ 1198872

]

119901

(12)

The condition 119886119887 = 119887119886119887120587 implies that 11988641198872= 11988721198864 Hence

using Lemma 2 we get 1198864+1198872isin Aqnil By Lemma 1 we obtain

that 119886 + 119887 isin A119889 and

(119886 + 119887)119889= [119887119889

1119906

0 0

]

119901

(13)

where

119906 =

infin

sum

119899=0

(119887119889

1)

119899+2

1198862(1198864+ 1198872)119899

(14)

Now from (14) using the matrix representation of 119887119889 119886 and119886 + 119887 we easily obtain formula (8) of the theorem

Thenext result is a generalization of [12Theorem22] and[10 Example 45]

Theorem 4 Let 119886 119887 isin A119889 If 119886119887 = 119886120587119887119886119887120587 then 119886 + 119887 isin A119889

and

(119886 + 119887)119889= 119887120587119886119889+ 119887119889119886120587+

infin

sum

119899=0

(119887119889)

119899+2

119886 (119886 + 119887)119899119886120587

+ 119887120587

infin

sum

119899=0

(119886 + 119887)119899119887 (119886119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119887119889)

119896+1

119886 (119886 + 119887)119899+119896119887 (119886119889)

119899+2

minus

infin

sum

119899=0

(119887119889)

119899+2

119886 (119886 + 119887)119899119887119886119889

(15)

Proof If 119886 is quasinilpotent we can apply Theorem 3 andwe obtain (15) for this particular case Now we assume that119886 is neither invertible nor quasinilpotent and consider thefollowingmatrix representations of 119886 119886119889 and 119887 relative to the119901 = 119886119886

119889

119886 = [

11988610

0 1198862

]

119901

119886119889= [119886119889

10

0 0

]

119901

119887 = [

11988711198872

11988731198874

]

119901

(16)

The condition 119886119887 = 119886120587119887119886119887120587 implies that 11988611198871= 0 and

11988611198872= 0 Since 119886

1is invertible we have 119887

1= 0 and 119887

2= 0

Thus 119887 can be represented as

119887 = [

0 0

11988731198874

]

119901

(17)

Therefore 1198874isin (119901A119901)

119889 and from Lemma 1 we have

119887119889= [

0 0

(119887119889

4)

2

1198873119887119889

4

]

119901

119887120587= [

119901 0

minus119887119889

41198873119887120587

4

]

119901

(18)

The Scientific World Journal 3

From 119886119887 = 119886120587119887119886119887120587 and

119886119887 = [

0 0

1198862119887311988621198874

]

119901

119886120587119887119886119887120587= [

0 0

11988731198861minus 11988741198862119887119889

4119887311988741198862119887120587

4

]

119901

(19)

we obtained 11988621198874= 11988741198862119887120587

4 FromTheorem 3 we get 119886

2+ 1198874isin

A119889 and

(1198862+ 1198874)119889

= 119887119889

4+

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

(20)

Further applying Lemma 1 to 119886 + 119887 we get

(119886 + 119887)119889= [

119886119889

10

119906 (1198862+ 1198874)119889] (21)

where

119906 =

infin

sum

119899=0

[(1198862+ 1198874)119889

]

119899+2

1198873119886119899

1119886120587

1

+

infin

sum

119899=0

(1198862+ 1198874)120587

(1198862+ 1198874)119899

1198873(119886119889

1)

119899+2

minus (1198862+ 1198874)119889

1198873119886119889

1

(22)

Observe that since 1198861isin (119901A119901)

minus1 then 1198861205871= 0

Hence the expression of 119906 reduces to

119906 =

infin

sum

119899=0

(1198862+ 1198874)120587

(1198862+ 1198874)119899

1198873(119886119889

1)

119899+2

minus (1198862+ 1198874)119889

1198873119886119889

1

(23)

From 11988621198874= 11988741198862119887120587

4we get = 119886

21198874(119887119889

4)2= 11988741198862119887120587

4(119887119889

4)2= 0

Hence from formula (20) and 1198862119887119889

4= 0 we have

(1198862+ 1198874)120587

= 119901 minus (1198862+ 1198874)

sdot (119887119889

4+

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

)

= 119901 minus 1198874(119887119889

4+

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

)

= 119887120587

4minus

infin

sum

119899=0

(119887119889

4)

119899+1

1198862(1198862+ 1198874)119899

(24)

Then substituting (20) and (24) in (22) we get

119906 =

infin

sum

119899=0

119887120587

4(1198862+ 1198874)119899

1198873(119886119889

1)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119887119889

4)

119896+1

1198862(1198862+ 1198874)119899+119896

1198873(119886119889

1)

119899+2

minus 119887119889

41198873119886119889

1minus

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

1198873119886119889

1

(25)

Now replacing 119906 by the above expression and consideringmatrix representations of 119886 and 119887 after direct computationswe obtain the formula (15) for (119886 + 119887)119889

3 Applications

In this section we give some formulas for the generalizedDrazin inverse of a 2 times 2 operator matrix under someconditions

Finding an explicit representation for the generalizedDrazin inverse of an operator matrix 119872 = [

119860 119861

119862 119863] in

terms of 119860 119861 119862 119863 and related generalized Drazin inversehas been studied by several authors [9 13ndash15] Djordjevicand Stanimirovic [9] generalize the well-known result in[8 16] concerning the Drazin inverse of block 2 times 2 uppertriangular matrices to the generalized Drazin inverse for ablock triangular operator matrix Further they consider thecase where 119861119862 = 0 119861119863 = 0 and119863119862 = 0

This section is devoted to the generalized Drazin inverseof 2 times 2 operator matrix

119872 = [

119860 119861

119862 119863] (26)

where 119860 isin B(119883) and 119863 isin B(119884) are generalized Drazininvertible

Next we will state some auxiliary lemmas

Lemma 5 (see [2 3]) Let 119860 and 119863 be generalized Drazininvertible and let 119872 be matrix of form (26) If 119861119862 = 0 and119861119863 = 0 then

119872119889= [119860119889(119860119889)

2

119861

1198830119863119889+ 1198831119861

] (27)

where

119883119899=

infin

sum

119894=0

(119863119889)

119894+119899+2

119862119860119894119860120587

+ 119863120587

infin

sum

119894=0

119863119894119862 (119860119889)

119894+119899+2

minus

119899

sum

119894=0

(119863119889)

119894+1

119862 (119860119889)

119899minus119894+1

119899 ge 0

(28)

Lemma 6 (see [17 Lemma 31]) If 119872 is matrix of form(26) such that 119860 is generalized Drazin invertible 119863 isquasinilpotent and 119861119863119899119862 = 0 for any nonnegative integer 119899then119872 is generalized Drazin invertible and

119872119889= [119860119889

Γ

Δ Δ119860Γ

] (29)

where

Γ =

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899 Δ =

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

(30)

4 The Scientific World Journal

Lemma 7 Let 119860 isin C119899times119899 Then (119860119860120587)119889 = 0 (1198602119860119889)119889 = 119860119889(1198602119860119889)120587= 119860120587 and Ind(119860119860120587) = Ind(119860) and Ind(1198602119860119889) = 1

Proof The Jordan canonical form of 119883 permits us to write119860 = 119878(119862 oplus 119873)119878

minus1 where 119878 and 119862 are nonsingular and 119873 isnilpotent with index Ind(119860) Thus 119860

119889= 119878(119862

minus1oplus 0)119878minus1 Now

it is evident that1198602119860119889 = 119878(119862oplus0)119878minus1 and119860119860120587 = 119878(0oplus119873)119878minus1which lead to the affirmations of this lemma

In [9 Theorem 53] authors gave an explicit representa-tion for119872119889 under conditions 119861119862 = 0 119863119862 = 0 and 119861119863 = 0Here we replace the last two conditions by the two weakerconditions119863119862 = 119863120587119862119860119860120587 and 119861119863 = 119860119860120587119861

Theorem 8 Let119860 and119863 be generalized Drazin invertible andlet119872 be matrix of form (26) If 119860119860120587119861 = 119861119863 119863119862 = 119863120587119862119860119860120587and 119861119862 = 0 Then

119872119889=

[

[

[

[

119860119889

(119860119889)

2

119861 +

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

119862 (119860119889)

2

119863119889+ 119862 (119860

119889)

3

119861 +

infin

sum

119899=1

119899

sum

119894=1

119863119894minus1119862119860119899minus119894119861 (119863119889)

119899+2

]

]

]

]

(31)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

1198601198601205870

0 119863] 119876 = [

1198602119860119889119861

119862 0

]

119875119889= [

0 0

0 119863119889] 119875

120587= [

119868 0

0 119863120587]

(32)

Since119863119862 = 119863120587119862119860119860120587 and 119860119860120587119861 = 119861119863 we have

119863119889119862 = (119863

119889)

2

119863119862 = (119863119889)

2

119863120587119862119860119860120587= 0

119863119862119860119889= 119863120587119862119860119860120587119860119889= 0

119860119889119861119863 = 119860

119889119860119860120587119861 = 0

(33)

From 119861119862 = 0 and applying Lemma 5 to 119876 we obtain

(119876119889)

119894

= [(119860119889)

119894

(119860119889)

119894+1

119861

119883119894minus1

119883119894119861

]

119876120587= [

119860120587

minus119860119889119861

minus119862119860119889119868 minus 119862 (119860

119889)

2

119861

]

(34)

where 119883119899is defined in (28) From 119863119889119862 = 0 and 119863119862119860119889 = 0

we get119883119899= 119862(119860

119889)119899+2

Since 119860119860120587119861 = 119861119863 and 119863119862 = 119863120587119862119860119860120587 we obtain 119875119876 =119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(35)

From 119860119889119861119863 = 0 we have

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889 119876

119889119875 = 0 (36)

Hence from (35) we obtain

(119875 + 119876)119889= 119875119889+ 119876119889+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(37)

Since 119860119889119861119863 = 0 we have

119876120587119876(119875119889)

2

= 119861 (119863119889)

2

(38)

The conditions 119861119862 = 0 and 119861119863 = 119860119860120587119861 imply that

119861119863119899119862 = 0 From 119861119863119899119862 = 0 and 119860119889119861119863 = 0 we get

119876120587

infin

sum

119899=1

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=1

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899

sum

119894=1

119863119894minus1119862119860119899minus119894119861 (119863119889)

119899+2

]

]

]

]

]

(39)

From (36) (38) and (39) it follows (31)The proof is finished

Since

[

119860 119861

119862 119863] = [

0 119868119899

1198681198980] [

119863 119862

119861 119860][

0 119868119898

1198681198990] (40)

we can obtain the following result applying Theorem 8 to[119863 119862

119861 119860]

Theorem 9 Let119860 and119863 be generalized Drazin invertible andlet119872 be matrix of form (26) If119863119863120587119862 = 119862119860 119860119861 = 119860120587119861119863119863120587and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+ 119861 (119863

119889)

3

119862 +

infin

sum

119899=1

119899

sum

119894=1

119860119899minus119894119861119863119894minus1119862(119860119889)

119899+2

119861 (119863119889)

2

(119863119889)

2

119862 +

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

119863119889

]

]

]

]

]

(41)Theorem 10 Let 119860 119863 and 119861119862 be generalized Drazin invert-ible and let119872 be matrix of form (26) If 119860119861 = 119860120587119861119863 119863119862 =119863120587119862119860 and 119861119862 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

119863119889+

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(42)

The Scientific World Journal 5

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

119860 0

0 119863] 119876 = [

0 119861

119862 0] (43)

119875119889= [

1198601198890

0 119863119889] 119875

120587= [

1198601205870

0 119863120587] (44)

Since

1198762= [

119861119862 0

0 119862119861] 119876

3= [

0 119861119862119861

119862119861119862 0] (45)

from 119861119862 = 0 it is easy to get 1198763 = 0 Since 119876 is nilpotent wehave 119876119889 = 0 Applying Theorem 4 to the particular case weget

(119875 + 119876)119889= 119875119889+

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(46)

The conditions 119860119861 = 119860120587119861119863 and 119861119862 = 0 imply that

119861119863119899119862 = 0 for 119899 ge 0 so we get

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(47)

From (44) and (47) it follows (42)The proof is finished

Theorem 11 Let119860 and119863 be generalizedDrazin invertible andlet119872 be matrix of form (26) If 119860119860120587119861 = 1198611198632119863119889 1198632119863119889119862 =119863120587119862119860119860120587 and 119861119863119899119862 = 0 for any nonnegative integer 119899 Then

119872119889=

[

[

[

[

[

119860119889

Γ +

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

Δ 119863119889+ Δ119860Γ +

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(48)

where Γ and Δ are defined in (30)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

119860119860120587

0

0 1198632119863119889] 119876 = [

1198602119860119889

119861

119862 119863119863120587]

119875119889= [

0 0

0 119863119889] 119875

120587= [

119868 0

0 119863120587]

(49)

From 119860119860120587119861 = 1198611198632119863119889 and1198632119863119889119862 = 119863120587119862119860119860120587 we have

119863119889119862 = (119863

119889)

3

1198632119862 = (119863

119889)

2

1198632119863119889119862

= (119863119889)

2

119863120587119862119860119860120587= 0

(50)

119860119889119861119863119889= 1198601198891198611198632(119863119889)

3

= 1198601198891198611198632119863119889(119863119889)

2

= 119860119889119860119860120587119861 (119863119889)

2

= 0

(51)

so we get119863120587119862 = 119862Note that 119863119863120587 is quasinilpotent 119863120587119862 = 119862 and

119861(119863119863120587)119899119862 = 119861119863

119899119863120587119862 = 119861119863

119899119862 = 0 for any nonnegative

integer 119899 we can apply Lemma 6 to 119876 with 119863 replaced by119863119863120587 we have

119876119889= [119860119889

Γ1015840

Δ1015840Δ1015840119860Γ1015840] (52)

where

Γ1015840=

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899119863120587 Δ

1015840=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

(53)

Observe that (50) and (51) yield

Γ =

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899 Δ =

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

(54)

so we get

119876119889= [119860119889

Γ

Δ Δ119860Γ

] (55)

The condition 119861119863119899119862 = 0 implies that

119861Δ = 119861

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

= 0 (56)

Hence we have

119876119876119889= [

119860119860119889+ 119861Δ 119860

2119860119889Γ + 119861Δ119860Γ

119862119860119889+ 119863119863

120587Δ 119862Γ + 119863119863

120587Δ119860Γ

]

= [

119860119860119889

119860Γ

119862119860119889+ 119863Δ 119862Γ + 119863Δ119860Γ

]

119876120587= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

(57)

6 The Scientific World Journal

From 119860119860120587119861 = 119861119863

2119863119889 and 1198632119863119889119862 = 119863

120587119862119860119860120587 we

obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(58)

where

Γ1198632119863119889=

infin

sum

119899=0

(119860119889)

119899+2

1198611198631198991198631205871198632119863119889= 0

Δ119860119860120587=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

119860119860120587= 0

(59)

so we get

119876119889119875 = [

119860119889

Γ

Δ Δ119860Γ

] [

119860119860120587

0

0 1198632119863119889]

= [

119860119889119860119860120587

Γ1198632119863119889

Δ119860119860120587Δ119860Γ119863

2119863119889] = [

0 0

0 0]

(60)

Hence from (58) and (60) we obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(61)

By direct computation we verify that

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889 (62)

From 119861119863119899119862 = 0 we haveinfin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(63)

Observe that (51) and 119861119863119899119862 = 0 yield

119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0 119860120587

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

minus 119860Γ

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

0 (minus119862119860119889minus 119863Δ)

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

+ (119868 minus 119862Γ minus 119863Δ119860Γ)

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

(64)

From (62) and (64) it follows (48) The proof is finished

The Scientific World Journal 7

Theorem 12 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119860120587119861119863120587 = 119861119863 119861119862 = 0119862119860119889= 0 and 119862119861119863120587 = 0 Then

119872119889=[

[

[

119860119889

(119860119889)

2

119861

1198830119863119889+

infin

sum

119899=0

119883119899+1119861119863119899

]

]

]

(65)

where

119883119899=

infin

sum

119894=0

(119863119889)

119894+119899+2

119862119860119894 119899 ge 0 (66)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [1198602119860119889119861

0 0

] 119876 = [

1198601198601205870

119862 119863]

119875119889= [119860119889(119860119889)

2

119861

0 0

] 119875120587= [119860120587minus119860119889119861

0 119868

]

(67)

Applying Lemma 7 we have (119860119860120587)119889 = 0 so we get

(119876119889)

119899

= [

0 0

119883119899minus1

(119863119889)

119899] 119876120587= [

119868 0

minus1198631198830119863120587] (68)

where119883119899is defined in (28)

Since 119860119860120587119861119863120587 = 119861119863 119861119862 = 0 119862119861119863120587 = 0 and 1198621198602119860119889 =0 we obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(69)

From 119862119860119889= 0 we have 119876119875119889 = 0 Hence from (69) we

obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

(70)

where119883119899119860119889= 0 we get

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889

(119876119889)

2

119875119875120587= [

0 0

1198831(119863119889)

2] [1198602119860119889119861

0 0

] [119860120587minus119860119889119861

0 119868

]

= [

0 0

0 1198831119861]

(71)

The conditions 119860119860120587119861119863120587 = 119861119863 and 119861119862 = 0 imply that119861119863119894119862 = 0 So we get

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

=[

[

[

0 0

0

infin

sum

119899=1

119883119899+1119861119863119899

]

]

]

119899 ge 1

(72)

From (71) and (72) it follows (65)The proof is finished

Using (40) andTheorem 12 we have the following result

Theorem 13 If 119862119860 = 119863120587119863119862119860120587 119861119863119889 = 0 119862119861 = 0 and

119861119862119860120587= 0 Then

119872119889=[

[

[

119860119889+

infin

sum

119899=0

119883119899+21198621198601198991198831

(119863119889)

2

119862 119863119889

]

]

]

(73)

where

119883119899=

infin

sum

119894=0

(119860119889)

119894+119899+1

119861119863119894 119899 ge 1 (74)

Using the case ofTheorem 3 we get the following results

Theorem 14 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119863119862119860120587 = 119862119860 119860119861119863 = 0119861119862 = 0 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+

infin

sum

119899=0

119861 (119863119889)

119899+3

119862119860119899(119860119889)

2

119861 + 119861 (119863119889)

2

infin

sum

119899=0

(119863119889)

119899+2

119862119860119899

119863119889

]

]

]

]

]

(75)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

0 0

119862 0] 119876 = [

119860 119861

0 119863] (76)

From 119860119861119863 = 0 we have

(119876119889)

119894

=[

[

(119860119889)

119894

119883119894

0 (119863119889)

119894]

]

119876120587= [119860120587minus1198601198831minus 119861119863119889

0 119863120587 ]

(77)

where

119883119899= (119860119889)

119899+2

119861 + 119861 (119863119889)

119899+2

119899 ge 0 (78)

Note that 119875 is quasinilpotent since 119863119862119860120587 = 119862119860 119860119861119863 =0 119861119862 = 0 and 119862119861 = 0 we obtain 119875119876 = 119876119875119876

120587 ApplyingTheorem 3 we get

(119875 + 119876)119889= 119876119889+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899 (79)

8 The Scientific World Journal

From 119861119862 = 0 we have

(119876119889)

2

119875 =[

[

(119860119889)

2

1198832

0 (119863119889)

2]

]

[

0 0

119862 0]

= [

1198832119862 0

(119863119889)

2

119862 0

] =[

[

119861 (119863119889)

3

0

(119863119889)

2

119862 0

]

]

(80)

The conditions 119863119862119860120587 = 119862119860 and 119862119861 = 0 imply that119862119860119894119861 = 0 From 119860119861119863 = 0 119862119860119894119861 = 0 and 119861119862 = 0 we getinfin

sum

119899=1

(119876119889)

119899+2

119875 (119875 + 119876)119899

=

[

[

[

[

[

infin

sum

119899=1

119883119899+2119862119860119899

infin

sum

119899=1

119883119899+2119862119860119899minus1119861

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899minus1119861

]

]

]

]

]

=

[

[

[

[

[

infin

sum

119899=1

119861 (119863119889)

119899+3

1198621198601198990

infin

sum

119899=1

(119863119889)

119899+2

1198621198601198990

]

]

]

]

]

(81)

From (77) (80) and (81) it follows (75)

Using (40) andTheorem 14 we have the following result

Theorem 15 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119861119863120587 = 119861119863 119863119862119860 = 0119861119862 = 119860119861119862119860

119889 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899

(119863119889)

2

119862 + 119862 (119860119889)

2

119863119889+

infin

sum

119899=0

119862 (119860119889)

119899+3

119861119863119899

]

]

]

]

]

(82)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J J Koliha ldquoA generalized Drazin inverserdquo Glasgow Mathemat-ical Journal vol 38 no 3 pp 367ndash381 1991

[2] C Deng D S Cvetkovic-Ilic and Y Wei ldquoSome results on thegeneralized Drazin inverse of operator matricesrdquo Linear andMultilinear Algebra vol 58 no 3-4 pp 503ndash521 2010

[3] E Dopazo and M F Martınez-Serrano ldquoFurther results on therepresentation of the Drazin inverse of a 2 times 2 block matrixrdquoLinear Algebra and its Applications vol 432 no 8 pp 1896ndash1904 2010

[4] J J Koliha and V Rakocevic ldquoHolomorphic and meromorphicproperties of the g-Drazin inverserdquoDemonstratio Mathematicavol 38 no 3 pp 657ndash666 2005

[5] N Castro Gonzalez ldquoAdditive perturbation results for theDrazin inverserdquo Linear Algebra and Its Applications vol 397 pp279ndash297 2005

[6] X Liu L Xu and Y Yu ldquoThe representations of the Drazininverse of differences of twomatricesrdquoAppliedMathematics andComputation vol 216 no 12 pp 3652ndash3661 2010

[7] D S Djordjevic and V Rakocevic Lectures on GeneralizedInverses University of Nis 2008

[8] J Meyer Jr and N J Rose ldquoThe index and the Drazininverse of block triangular matricesrdquo SIAM Journal on AppliedMathematics vol 33 no 1 pp 1ndash7 1977

[9] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[10] N Castro-Gonzalez and J J Koliha ldquoNew additive resultsfor the 119892-Drazin inverserdquo Proceedings of the Royal Society ofEdinburgh Section A Mathematics vol 134 no 6 pp 1085ndash1097 2004

[11] D S Cvetkovic-Ilic D S Djordjevic and Y Wei ldquoAdditiveresults for the generalized Drazin inverse in a Banach algebrardquoLinear Algebra and Its Applications vol 418 no 1 pp 53ndash612006

[12] D S Djordjevic and YWei ldquoAdditive results for the generalizedDrazin inverserdquo Journal of the Australian Mathematical Societyvol 73 no 1 pp 115ndash125 2002

[13] C Deng and Y Wei ldquoA note on the Drazin inverse of an anti-triangular matrixrdquo Linear Algebra and its Applications vol 431no 10 pp 1910ndash1922 2009

[14] N Castro-Gonzalez E Dopazo and M F Martınez-SerranoldquoOn the Drazin inverse of the sum of two operators andits application to operator matricesrdquo Journal of MathematicalAnalysis and Applications vol 350 no 1 pp 207ndash215 2009

[15] R E Hartwig G Wang and Y Wei ldquoSome additive results onDrazin inverserdquo Linear Algebra and its Applications vol 322 no1ndash3 pp 207ndash217 2001

[16] R E Hartwig and J Shoaf ldquoGroup inverses and Drazin inversesof bidiagonal and triangular Toeplitzmatricesrdquo vol 24 no 1 pp10ndash34 1977

[17] R E Hartwig and J M Shoaf ldquoGroup inverse and Drazininverse of bidiagonal and triangular Toeplitz matricesrdquo Journalof the AustralianMathematical Society A vol 24 pp 10ndash34 1977

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Differential EquationsInternational Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 The Scientific World Journal

2 Main Results

First we start the following result which is proved in [8] formatrices extended in [9] for a bounded linear operator andin [10] for arbitrary elements in a Banach algebra

Lemma 1 (see [10Theorem 23]) Let 119909 119910 isin A and 119901 isin A bean idempotent Assume that 119909 and 119910 are represented as

119909 = [

119886 0

119888 119887]

119901

119910 = [

119887 119888

0 119886]

119901

(5)

(i) If 119886 isin (119901A119901)119889 and 119887 isin (119901A119901)119889 then 119909 and 119910 are gen-eralized Drazin invertible and

119909119889= [

1198861198890

119906 119887119889]

119901

119910119889= [

119887119889119906

0 119886119889]

119901

(6)

where

119906 =

infin

sum

119899=0

(119887119889)

119899+2

119888119886119899119886120587+

infin

sum

119899=0

119887120587119887119899119888 (119886119889)

119899+2

minus 119887119889119888119886119889 (7)

(ii) If 119909 isin A119889 and 119886 isin (119901A119901)119889 then 119887 isin (119901A119901)119889 and 119909119889

and 119910119889 are given by (6) and (7)

Lemma 2 (see [11 Lemma 21]) Let 119886 119887 isin A119902119899119894119897 If 119886119887 = 119887119886or 119886119887 = 0 then 119886 + 119887 isin A119902119899119894119897

The following result is a generalization of [10 Corollary34]

Theorem 3 If 119886 isin A119902119899119894119897 119887 isin A119889 and 119886119887 = 119887119886119887120587 then 119886 + 119887 isinA119889 and

(119886 + 119887)119889= 119887119889+

infin

sum

119899=0

(119887119889)

119899+2

119886 (119886 + 119887)119899 (8)

Proof First suppose that 119887 isin Aqnil Therefore 119887120587 = 1 andfrom 119886119887 = 119887119886119887120587 we obtain 119886119887 = 119887119886 Using Lemma 2 119886 + 119887 isinAqnil and (8) holds

Now we assume 119887 is not quasinilpotent using matrixrepresentations of 119886 and 119887 relative to 119901 = 119887119887119889 We have

119887 = [

11988710

0 1198872

]

119901

119887119889= [119887119889

10

0 0

]

119901

(9)

where 1198871isin (119901A119901)

minus1 1198872isin (119901A119901)

qnilLet us represent

119886 = [

11988611198862

11988631198864

]

119901

(10)

From 119886119887 = 119887119886119887120587 and

119886119887 = [

1198861119887111988621198872

1198863119887111988641198872

]

119901

119887119886119887120587= [

0 11988711198862

0 11988721198864

]

119901

(11)

we obtain 11988611198871= 0 and 119886

31198871= 0 Since 119887

1is invertible we

have 1198861= 0 and 119886

3= 0

Hence we have

119886 + 119887 = [

1198871

1198862

0 1198864+ 1198872

]

119901

(12)

The condition 119886119887 = 119887119886119887120587 implies that 11988641198872= 11988721198864 Hence

using Lemma 2 we get 1198864+1198872isin Aqnil By Lemma 1 we obtain

that 119886 + 119887 isin A119889 and

(119886 + 119887)119889= [119887119889

1119906

0 0

]

119901

(13)

where

119906 =

infin

sum

119899=0

(119887119889

1)

119899+2

1198862(1198864+ 1198872)119899

(14)

Now from (14) using the matrix representation of 119887119889 119886 and119886 + 119887 we easily obtain formula (8) of the theorem

Thenext result is a generalization of [12Theorem22] and[10 Example 45]

Theorem 4 Let 119886 119887 isin A119889 If 119886119887 = 119886120587119887119886119887120587 then 119886 + 119887 isin A119889

and

(119886 + 119887)119889= 119887120587119886119889+ 119887119889119886120587+

infin

sum

119899=0

(119887119889)

119899+2

119886 (119886 + 119887)119899119886120587

+ 119887120587

infin

sum

119899=0

(119886 + 119887)119899119887 (119886119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119887119889)

119896+1

119886 (119886 + 119887)119899+119896119887 (119886119889)

119899+2

minus

infin

sum

119899=0

(119887119889)

119899+2

119886 (119886 + 119887)119899119887119886119889

(15)

Proof If 119886 is quasinilpotent we can apply Theorem 3 andwe obtain (15) for this particular case Now we assume that119886 is neither invertible nor quasinilpotent and consider thefollowingmatrix representations of 119886 119886119889 and 119887 relative to the119901 = 119886119886

119889

119886 = [

11988610

0 1198862

]

119901

119886119889= [119886119889

10

0 0

]

119901

119887 = [

11988711198872

11988731198874

]

119901

(16)

The condition 119886119887 = 119886120587119887119886119887120587 implies that 11988611198871= 0 and

11988611198872= 0 Since 119886

1is invertible we have 119887

1= 0 and 119887

2= 0

Thus 119887 can be represented as

119887 = [

0 0

11988731198874

]

119901

(17)

Therefore 1198874isin (119901A119901)

119889 and from Lemma 1 we have

119887119889= [

0 0

(119887119889

4)

2

1198873119887119889

4

]

119901

119887120587= [

119901 0

minus119887119889

41198873119887120587

4

]

119901

(18)

The Scientific World Journal 3

From 119886119887 = 119886120587119887119886119887120587 and

119886119887 = [

0 0

1198862119887311988621198874

]

119901

119886120587119887119886119887120587= [

0 0

11988731198861minus 11988741198862119887119889

4119887311988741198862119887120587

4

]

119901

(19)

we obtained 11988621198874= 11988741198862119887120587

4 FromTheorem 3 we get 119886

2+ 1198874isin

A119889 and

(1198862+ 1198874)119889

= 119887119889

4+

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

(20)

Further applying Lemma 1 to 119886 + 119887 we get

(119886 + 119887)119889= [

119886119889

10

119906 (1198862+ 1198874)119889] (21)

where

119906 =

infin

sum

119899=0

[(1198862+ 1198874)119889

]

119899+2

1198873119886119899

1119886120587

1

+

infin

sum

119899=0

(1198862+ 1198874)120587

(1198862+ 1198874)119899

1198873(119886119889

1)

119899+2

minus (1198862+ 1198874)119889

1198873119886119889

1

(22)

Observe that since 1198861isin (119901A119901)

minus1 then 1198861205871= 0

Hence the expression of 119906 reduces to

119906 =

infin

sum

119899=0

(1198862+ 1198874)120587

(1198862+ 1198874)119899

1198873(119886119889

1)

119899+2

minus (1198862+ 1198874)119889

1198873119886119889

1

(23)

From 11988621198874= 11988741198862119887120587

4we get = 119886

21198874(119887119889

4)2= 11988741198862119887120587

4(119887119889

4)2= 0

Hence from formula (20) and 1198862119887119889

4= 0 we have

(1198862+ 1198874)120587

= 119901 minus (1198862+ 1198874)

sdot (119887119889

4+

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

)

= 119901 minus 1198874(119887119889

4+

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

)

= 119887120587

4minus

infin

sum

119899=0

(119887119889

4)

119899+1

1198862(1198862+ 1198874)119899

(24)

Then substituting (20) and (24) in (22) we get

119906 =

infin

sum

119899=0

119887120587

4(1198862+ 1198874)119899

1198873(119886119889

1)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119887119889

4)

119896+1

1198862(1198862+ 1198874)119899+119896

1198873(119886119889

1)

119899+2

minus 119887119889

41198873119886119889

1minus

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

1198873119886119889

1

(25)

Now replacing 119906 by the above expression and consideringmatrix representations of 119886 and 119887 after direct computationswe obtain the formula (15) for (119886 + 119887)119889

3 Applications

In this section we give some formulas for the generalizedDrazin inverse of a 2 times 2 operator matrix under someconditions

Finding an explicit representation for the generalizedDrazin inverse of an operator matrix 119872 = [

119860 119861

119862 119863] in

terms of 119860 119861 119862 119863 and related generalized Drazin inversehas been studied by several authors [9 13ndash15] Djordjevicand Stanimirovic [9] generalize the well-known result in[8 16] concerning the Drazin inverse of block 2 times 2 uppertriangular matrices to the generalized Drazin inverse for ablock triangular operator matrix Further they consider thecase where 119861119862 = 0 119861119863 = 0 and119863119862 = 0

This section is devoted to the generalized Drazin inverseof 2 times 2 operator matrix

119872 = [

119860 119861

119862 119863] (26)

where 119860 isin B(119883) and 119863 isin B(119884) are generalized Drazininvertible

Next we will state some auxiliary lemmas

Lemma 5 (see [2 3]) Let 119860 and 119863 be generalized Drazininvertible and let 119872 be matrix of form (26) If 119861119862 = 0 and119861119863 = 0 then

119872119889= [119860119889(119860119889)

2

119861

1198830119863119889+ 1198831119861

] (27)

where

119883119899=

infin

sum

119894=0

(119863119889)

119894+119899+2

119862119860119894119860120587

+ 119863120587

infin

sum

119894=0

119863119894119862 (119860119889)

119894+119899+2

minus

119899

sum

119894=0

(119863119889)

119894+1

119862 (119860119889)

119899minus119894+1

119899 ge 0

(28)

Lemma 6 (see [17 Lemma 31]) If 119872 is matrix of form(26) such that 119860 is generalized Drazin invertible 119863 isquasinilpotent and 119861119863119899119862 = 0 for any nonnegative integer 119899then119872 is generalized Drazin invertible and

119872119889= [119860119889

Γ

Δ Δ119860Γ

] (29)

where

Γ =

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899 Δ =

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

(30)

4 The Scientific World Journal

Lemma 7 Let 119860 isin C119899times119899 Then (119860119860120587)119889 = 0 (1198602119860119889)119889 = 119860119889(1198602119860119889)120587= 119860120587 and Ind(119860119860120587) = Ind(119860) and Ind(1198602119860119889) = 1

Proof The Jordan canonical form of 119883 permits us to write119860 = 119878(119862 oplus 119873)119878

minus1 where 119878 and 119862 are nonsingular and 119873 isnilpotent with index Ind(119860) Thus 119860

119889= 119878(119862

minus1oplus 0)119878minus1 Now

it is evident that1198602119860119889 = 119878(119862oplus0)119878minus1 and119860119860120587 = 119878(0oplus119873)119878minus1which lead to the affirmations of this lemma

In [9 Theorem 53] authors gave an explicit representa-tion for119872119889 under conditions 119861119862 = 0 119863119862 = 0 and 119861119863 = 0Here we replace the last two conditions by the two weakerconditions119863119862 = 119863120587119862119860119860120587 and 119861119863 = 119860119860120587119861

Theorem 8 Let119860 and119863 be generalized Drazin invertible andlet119872 be matrix of form (26) If 119860119860120587119861 = 119861119863 119863119862 = 119863120587119862119860119860120587and 119861119862 = 0 Then

119872119889=

[

[

[

[

119860119889

(119860119889)

2

119861 +

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

119862 (119860119889)

2

119863119889+ 119862 (119860

119889)

3

119861 +

infin

sum

119899=1

119899

sum

119894=1

119863119894minus1119862119860119899minus119894119861 (119863119889)

119899+2

]

]

]

]

(31)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

1198601198601205870

0 119863] 119876 = [

1198602119860119889119861

119862 0

]

119875119889= [

0 0

0 119863119889] 119875

120587= [

119868 0

0 119863120587]

(32)

Since119863119862 = 119863120587119862119860119860120587 and 119860119860120587119861 = 119861119863 we have

119863119889119862 = (119863

119889)

2

119863119862 = (119863119889)

2

119863120587119862119860119860120587= 0

119863119862119860119889= 119863120587119862119860119860120587119860119889= 0

119860119889119861119863 = 119860

119889119860119860120587119861 = 0

(33)

From 119861119862 = 0 and applying Lemma 5 to 119876 we obtain

(119876119889)

119894

= [(119860119889)

119894

(119860119889)

119894+1

119861

119883119894minus1

119883119894119861

]

119876120587= [

119860120587

minus119860119889119861

minus119862119860119889119868 minus 119862 (119860

119889)

2

119861

]

(34)

where 119883119899is defined in (28) From 119863119889119862 = 0 and 119863119862119860119889 = 0

we get119883119899= 119862(119860

119889)119899+2

Since 119860119860120587119861 = 119861119863 and 119863119862 = 119863120587119862119860119860120587 we obtain 119875119876 =119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(35)

From 119860119889119861119863 = 0 we have

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889 119876

119889119875 = 0 (36)

Hence from (35) we obtain

(119875 + 119876)119889= 119875119889+ 119876119889+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(37)

Since 119860119889119861119863 = 0 we have

119876120587119876(119875119889)

2

= 119861 (119863119889)

2

(38)

The conditions 119861119862 = 0 and 119861119863 = 119860119860120587119861 imply that

119861119863119899119862 = 0 From 119861119863119899119862 = 0 and 119860119889119861119863 = 0 we get

119876120587

infin

sum

119899=1

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=1

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899

sum

119894=1

119863119894minus1119862119860119899minus119894119861 (119863119889)

119899+2

]

]

]

]

]

(39)

From (36) (38) and (39) it follows (31)The proof is finished

Since

[

119860 119861

119862 119863] = [

0 119868119899

1198681198980] [

119863 119862

119861 119860][

0 119868119898

1198681198990] (40)

we can obtain the following result applying Theorem 8 to[119863 119862

119861 119860]

Theorem 9 Let119860 and119863 be generalized Drazin invertible andlet119872 be matrix of form (26) If119863119863120587119862 = 119862119860 119860119861 = 119860120587119861119863119863120587and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+ 119861 (119863

119889)

3

119862 +

infin

sum

119899=1

119899

sum

119894=1

119860119899minus119894119861119863119894minus1119862(119860119889)

119899+2

119861 (119863119889)

2

(119863119889)

2

119862 +

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

119863119889

]

]

]

]

]

(41)Theorem 10 Let 119860 119863 and 119861119862 be generalized Drazin invert-ible and let119872 be matrix of form (26) If 119860119861 = 119860120587119861119863 119863119862 =119863120587119862119860 and 119861119862 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

119863119889+

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(42)

The Scientific World Journal 5

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

119860 0

0 119863] 119876 = [

0 119861

119862 0] (43)

119875119889= [

1198601198890

0 119863119889] 119875

120587= [

1198601205870

0 119863120587] (44)

Since

1198762= [

119861119862 0

0 119862119861] 119876

3= [

0 119861119862119861

119862119861119862 0] (45)

from 119861119862 = 0 it is easy to get 1198763 = 0 Since 119876 is nilpotent wehave 119876119889 = 0 Applying Theorem 4 to the particular case weget

(119875 + 119876)119889= 119875119889+

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(46)

The conditions 119860119861 = 119860120587119861119863 and 119861119862 = 0 imply that

119861119863119899119862 = 0 for 119899 ge 0 so we get

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(47)

From (44) and (47) it follows (42)The proof is finished

Theorem 11 Let119860 and119863 be generalizedDrazin invertible andlet119872 be matrix of form (26) If 119860119860120587119861 = 1198611198632119863119889 1198632119863119889119862 =119863120587119862119860119860120587 and 119861119863119899119862 = 0 for any nonnegative integer 119899 Then

119872119889=

[

[

[

[

[

119860119889

Γ +

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

Δ 119863119889+ Δ119860Γ +

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(48)

where Γ and Δ are defined in (30)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

119860119860120587

0

0 1198632119863119889] 119876 = [

1198602119860119889

119861

119862 119863119863120587]

119875119889= [

0 0

0 119863119889] 119875

120587= [

119868 0

0 119863120587]

(49)

From 119860119860120587119861 = 1198611198632119863119889 and1198632119863119889119862 = 119863120587119862119860119860120587 we have

119863119889119862 = (119863

119889)

3

1198632119862 = (119863

119889)

2

1198632119863119889119862

= (119863119889)

2

119863120587119862119860119860120587= 0

(50)

119860119889119861119863119889= 1198601198891198611198632(119863119889)

3

= 1198601198891198611198632119863119889(119863119889)

2

= 119860119889119860119860120587119861 (119863119889)

2

= 0

(51)

so we get119863120587119862 = 119862Note that 119863119863120587 is quasinilpotent 119863120587119862 = 119862 and

119861(119863119863120587)119899119862 = 119861119863

119899119863120587119862 = 119861119863

119899119862 = 0 for any nonnegative

integer 119899 we can apply Lemma 6 to 119876 with 119863 replaced by119863119863120587 we have

119876119889= [119860119889

Γ1015840

Δ1015840Δ1015840119860Γ1015840] (52)

where

Γ1015840=

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899119863120587 Δ

1015840=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

(53)

Observe that (50) and (51) yield

Γ =

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899 Δ =

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

(54)

so we get

119876119889= [119860119889

Γ

Δ Δ119860Γ

] (55)

The condition 119861119863119899119862 = 0 implies that

119861Δ = 119861

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

= 0 (56)

Hence we have

119876119876119889= [

119860119860119889+ 119861Δ 119860

2119860119889Γ + 119861Δ119860Γ

119862119860119889+ 119863119863

120587Δ 119862Γ + 119863119863

120587Δ119860Γ

]

= [

119860119860119889

119860Γ

119862119860119889+ 119863Δ 119862Γ + 119863Δ119860Γ

]

119876120587= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

(57)

6 The Scientific World Journal

From 119860119860120587119861 = 119861119863

2119863119889 and 1198632119863119889119862 = 119863

120587119862119860119860120587 we

obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(58)

where

Γ1198632119863119889=

infin

sum

119899=0

(119860119889)

119899+2

1198611198631198991198631205871198632119863119889= 0

Δ119860119860120587=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

119860119860120587= 0

(59)

so we get

119876119889119875 = [

119860119889

Γ

Δ Δ119860Γ

] [

119860119860120587

0

0 1198632119863119889]

= [

119860119889119860119860120587

Γ1198632119863119889

Δ119860119860120587Δ119860Γ119863

2119863119889] = [

0 0

0 0]

(60)

Hence from (58) and (60) we obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(61)

By direct computation we verify that

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889 (62)

From 119861119863119899119862 = 0 we haveinfin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(63)

Observe that (51) and 119861119863119899119862 = 0 yield

119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0 119860120587

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

minus 119860Γ

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

0 (minus119862119860119889minus 119863Δ)

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

+ (119868 minus 119862Γ minus 119863Δ119860Γ)

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

(64)

From (62) and (64) it follows (48) The proof is finished

The Scientific World Journal 7

Theorem 12 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119860120587119861119863120587 = 119861119863 119861119862 = 0119862119860119889= 0 and 119862119861119863120587 = 0 Then

119872119889=[

[

[

119860119889

(119860119889)

2

119861

1198830119863119889+

infin

sum

119899=0

119883119899+1119861119863119899

]

]

]

(65)

where

119883119899=

infin

sum

119894=0

(119863119889)

119894+119899+2

119862119860119894 119899 ge 0 (66)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [1198602119860119889119861

0 0

] 119876 = [

1198601198601205870

119862 119863]

119875119889= [119860119889(119860119889)

2

119861

0 0

] 119875120587= [119860120587minus119860119889119861

0 119868

]

(67)

Applying Lemma 7 we have (119860119860120587)119889 = 0 so we get

(119876119889)

119899

= [

0 0

119883119899minus1

(119863119889)

119899] 119876120587= [

119868 0

minus1198631198830119863120587] (68)

where119883119899is defined in (28)

Since 119860119860120587119861119863120587 = 119861119863 119861119862 = 0 119862119861119863120587 = 0 and 1198621198602119860119889 =0 we obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(69)

From 119862119860119889= 0 we have 119876119875119889 = 0 Hence from (69) we

obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

(70)

where119883119899119860119889= 0 we get

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889

(119876119889)

2

119875119875120587= [

0 0

1198831(119863119889)

2] [1198602119860119889119861

0 0

] [119860120587minus119860119889119861

0 119868

]

= [

0 0

0 1198831119861]

(71)

The conditions 119860119860120587119861119863120587 = 119861119863 and 119861119862 = 0 imply that119861119863119894119862 = 0 So we get

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

=[

[

[

0 0

0

infin

sum

119899=1

119883119899+1119861119863119899

]

]

]

119899 ge 1

(72)

From (71) and (72) it follows (65)The proof is finished

Using (40) andTheorem 12 we have the following result

Theorem 13 If 119862119860 = 119863120587119863119862119860120587 119861119863119889 = 0 119862119861 = 0 and

119861119862119860120587= 0 Then

119872119889=[

[

[

119860119889+

infin

sum

119899=0

119883119899+21198621198601198991198831

(119863119889)

2

119862 119863119889

]

]

]

(73)

where

119883119899=

infin

sum

119894=0

(119860119889)

119894+119899+1

119861119863119894 119899 ge 1 (74)

Using the case ofTheorem 3 we get the following results

Theorem 14 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119863119862119860120587 = 119862119860 119860119861119863 = 0119861119862 = 0 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+

infin

sum

119899=0

119861 (119863119889)

119899+3

119862119860119899(119860119889)

2

119861 + 119861 (119863119889)

2

infin

sum

119899=0

(119863119889)

119899+2

119862119860119899

119863119889

]

]

]

]

]

(75)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

0 0

119862 0] 119876 = [

119860 119861

0 119863] (76)

From 119860119861119863 = 0 we have

(119876119889)

119894

=[

[

(119860119889)

119894

119883119894

0 (119863119889)

119894]

]

119876120587= [119860120587minus1198601198831minus 119861119863119889

0 119863120587 ]

(77)

where

119883119899= (119860119889)

119899+2

119861 + 119861 (119863119889)

119899+2

119899 ge 0 (78)

Note that 119875 is quasinilpotent since 119863119862119860120587 = 119862119860 119860119861119863 =0 119861119862 = 0 and 119862119861 = 0 we obtain 119875119876 = 119876119875119876

120587 ApplyingTheorem 3 we get

(119875 + 119876)119889= 119876119889+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899 (79)

8 The Scientific World Journal

From 119861119862 = 0 we have

(119876119889)

2

119875 =[

[

(119860119889)

2

1198832

0 (119863119889)

2]

]

[

0 0

119862 0]

= [

1198832119862 0

(119863119889)

2

119862 0

] =[

[

119861 (119863119889)

3

0

(119863119889)

2

119862 0

]

]

(80)

The conditions 119863119862119860120587 = 119862119860 and 119862119861 = 0 imply that119862119860119894119861 = 0 From 119860119861119863 = 0 119862119860119894119861 = 0 and 119861119862 = 0 we getinfin

sum

119899=1

(119876119889)

119899+2

119875 (119875 + 119876)119899

=

[

[

[

[

[

infin

sum

119899=1

119883119899+2119862119860119899

infin

sum

119899=1

119883119899+2119862119860119899minus1119861

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899minus1119861

]

]

]

]

]

=

[

[

[

[

[

infin

sum

119899=1

119861 (119863119889)

119899+3

1198621198601198990

infin

sum

119899=1

(119863119889)

119899+2

1198621198601198990

]

]

]

]

]

(81)

From (77) (80) and (81) it follows (75)

Using (40) andTheorem 14 we have the following result

Theorem 15 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119861119863120587 = 119861119863 119863119862119860 = 0119861119862 = 119860119861119862119860

119889 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899

(119863119889)

2

119862 + 119862 (119860119889)

2

119863119889+

infin

sum

119899=0

119862 (119860119889)

119899+3

119861119863119899

]

]

]

]

]

(82)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J J Koliha ldquoA generalized Drazin inverserdquo Glasgow Mathemat-ical Journal vol 38 no 3 pp 367ndash381 1991

[2] C Deng D S Cvetkovic-Ilic and Y Wei ldquoSome results on thegeneralized Drazin inverse of operator matricesrdquo Linear andMultilinear Algebra vol 58 no 3-4 pp 503ndash521 2010

[3] E Dopazo and M F Martınez-Serrano ldquoFurther results on therepresentation of the Drazin inverse of a 2 times 2 block matrixrdquoLinear Algebra and its Applications vol 432 no 8 pp 1896ndash1904 2010

[4] J J Koliha and V Rakocevic ldquoHolomorphic and meromorphicproperties of the g-Drazin inverserdquoDemonstratio Mathematicavol 38 no 3 pp 657ndash666 2005

[5] N Castro Gonzalez ldquoAdditive perturbation results for theDrazin inverserdquo Linear Algebra and Its Applications vol 397 pp279ndash297 2005

[6] X Liu L Xu and Y Yu ldquoThe representations of the Drazininverse of differences of twomatricesrdquoAppliedMathematics andComputation vol 216 no 12 pp 3652ndash3661 2010

[7] D S Djordjevic and V Rakocevic Lectures on GeneralizedInverses University of Nis 2008

[8] J Meyer Jr and N J Rose ldquoThe index and the Drazininverse of block triangular matricesrdquo SIAM Journal on AppliedMathematics vol 33 no 1 pp 1ndash7 1977

[9] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[10] N Castro-Gonzalez and J J Koliha ldquoNew additive resultsfor the 119892-Drazin inverserdquo Proceedings of the Royal Society ofEdinburgh Section A Mathematics vol 134 no 6 pp 1085ndash1097 2004

[11] D S Cvetkovic-Ilic D S Djordjevic and Y Wei ldquoAdditiveresults for the generalized Drazin inverse in a Banach algebrardquoLinear Algebra and Its Applications vol 418 no 1 pp 53ndash612006

[12] D S Djordjevic and YWei ldquoAdditive results for the generalizedDrazin inverserdquo Journal of the Australian Mathematical Societyvol 73 no 1 pp 115ndash125 2002

[13] C Deng and Y Wei ldquoA note on the Drazin inverse of an anti-triangular matrixrdquo Linear Algebra and its Applications vol 431no 10 pp 1910ndash1922 2009

[14] N Castro-Gonzalez E Dopazo and M F Martınez-SerranoldquoOn the Drazin inverse of the sum of two operators andits application to operator matricesrdquo Journal of MathematicalAnalysis and Applications vol 350 no 1 pp 207ndash215 2009

[15] R E Hartwig G Wang and Y Wei ldquoSome additive results onDrazin inverserdquo Linear Algebra and its Applications vol 322 no1ndash3 pp 207ndash217 2001

[16] R E Hartwig and J Shoaf ldquoGroup inverses and Drazin inversesof bidiagonal and triangular Toeplitzmatricesrdquo vol 24 no 1 pp10ndash34 1977

[17] R E Hartwig and J M Shoaf ldquoGroup inverse and Drazininverse of bidiagonal and triangular Toeplitz matricesrdquo Journalof the AustralianMathematical Society A vol 24 pp 10ndash34 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

The Scientific World Journal 3

From 119886119887 = 119886120587119887119886119887120587 and

119886119887 = [

0 0

1198862119887311988621198874

]

119901

119886120587119887119886119887120587= [

0 0

11988731198861minus 11988741198862119887119889

4119887311988741198862119887120587

4

]

119901

(19)

we obtained 11988621198874= 11988741198862119887120587

4 FromTheorem 3 we get 119886

2+ 1198874isin

A119889 and

(1198862+ 1198874)119889

= 119887119889

4+

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

(20)

Further applying Lemma 1 to 119886 + 119887 we get

(119886 + 119887)119889= [

119886119889

10

119906 (1198862+ 1198874)119889] (21)

where

119906 =

infin

sum

119899=0

[(1198862+ 1198874)119889

]

119899+2

1198873119886119899

1119886120587

1

+

infin

sum

119899=0

(1198862+ 1198874)120587

(1198862+ 1198874)119899

1198873(119886119889

1)

119899+2

minus (1198862+ 1198874)119889

1198873119886119889

1

(22)

Observe that since 1198861isin (119901A119901)

minus1 then 1198861205871= 0

Hence the expression of 119906 reduces to

119906 =

infin

sum

119899=0

(1198862+ 1198874)120587

(1198862+ 1198874)119899

1198873(119886119889

1)

119899+2

minus (1198862+ 1198874)119889

1198873119886119889

1

(23)

From 11988621198874= 11988741198862119887120587

4we get = 119886

21198874(119887119889

4)2= 11988741198862119887120587

4(119887119889

4)2= 0

Hence from formula (20) and 1198862119887119889

4= 0 we have

(1198862+ 1198874)120587

= 119901 minus (1198862+ 1198874)

sdot (119887119889

4+

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

)

= 119901 minus 1198874(119887119889

4+

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

)

= 119887120587

4minus

infin

sum

119899=0

(119887119889

4)

119899+1

1198862(1198862+ 1198874)119899

(24)

Then substituting (20) and (24) in (22) we get

119906 =

infin

sum

119899=0

119887120587

4(1198862+ 1198874)119899

1198873(119886119889

1)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119887119889

4)

119896+1

1198862(1198862+ 1198874)119899+119896

1198873(119886119889

1)

119899+2

minus 119887119889

41198873119886119889

1minus

infin

sum

119899=0

(119887119889

4)

119899+2

1198862(1198862+ 1198874)119899

1198873119886119889

1

(25)

Now replacing 119906 by the above expression and consideringmatrix representations of 119886 and 119887 after direct computationswe obtain the formula (15) for (119886 + 119887)119889

3 Applications

In this section we give some formulas for the generalizedDrazin inverse of a 2 times 2 operator matrix under someconditions

Finding an explicit representation for the generalizedDrazin inverse of an operator matrix 119872 = [

119860 119861

119862 119863] in

terms of 119860 119861 119862 119863 and related generalized Drazin inversehas been studied by several authors [9 13ndash15] Djordjevicand Stanimirovic [9] generalize the well-known result in[8 16] concerning the Drazin inverse of block 2 times 2 uppertriangular matrices to the generalized Drazin inverse for ablock triangular operator matrix Further they consider thecase where 119861119862 = 0 119861119863 = 0 and119863119862 = 0

This section is devoted to the generalized Drazin inverseof 2 times 2 operator matrix

119872 = [

119860 119861

119862 119863] (26)

where 119860 isin B(119883) and 119863 isin B(119884) are generalized Drazininvertible

Next we will state some auxiliary lemmas

Lemma 5 (see [2 3]) Let 119860 and 119863 be generalized Drazininvertible and let 119872 be matrix of form (26) If 119861119862 = 0 and119861119863 = 0 then

119872119889= [119860119889(119860119889)

2

119861

1198830119863119889+ 1198831119861

] (27)

where

119883119899=

infin

sum

119894=0

(119863119889)

119894+119899+2

119862119860119894119860120587

+ 119863120587

infin

sum

119894=0

119863119894119862 (119860119889)

119894+119899+2

minus

119899

sum

119894=0

(119863119889)

119894+1

119862 (119860119889)

119899minus119894+1

119899 ge 0

(28)

Lemma 6 (see [17 Lemma 31]) If 119872 is matrix of form(26) such that 119860 is generalized Drazin invertible 119863 isquasinilpotent and 119861119863119899119862 = 0 for any nonnegative integer 119899then119872 is generalized Drazin invertible and

119872119889= [119860119889

Γ

Δ Δ119860Γ

] (29)

where

Γ =

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899 Δ =

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

(30)

4 The Scientific World Journal

Lemma 7 Let 119860 isin C119899times119899 Then (119860119860120587)119889 = 0 (1198602119860119889)119889 = 119860119889(1198602119860119889)120587= 119860120587 and Ind(119860119860120587) = Ind(119860) and Ind(1198602119860119889) = 1

Proof The Jordan canonical form of 119883 permits us to write119860 = 119878(119862 oplus 119873)119878

minus1 where 119878 and 119862 are nonsingular and 119873 isnilpotent with index Ind(119860) Thus 119860

119889= 119878(119862

minus1oplus 0)119878minus1 Now

it is evident that1198602119860119889 = 119878(119862oplus0)119878minus1 and119860119860120587 = 119878(0oplus119873)119878minus1which lead to the affirmations of this lemma

In [9 Theorem 53] authors gave an explicit representa-tion for119872119889 under conditions 119861119862 = 0 119863119862 = 0 and 119861119863 = 0Here we replace the last two conditions by the two weakerconditions119863119862 = 119863120587119862119860119860120587 and 119861119863 = 119860119860120587119861

Theorem 8 Let119860 and119863 be generalized Drazin invertible andlet119872 be matrix of form (26) If 119860119860120587119861 = 119861119863 119863119862 = 119863120587119862119860119860120587and 119861119862 = 0 Then

119872119889=

[

[

[

[

119860119889

(119860119889)

2

119861 +

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

119862 (119860119889)

2

119863119889+ 119862 (119860

119889)

3

119861 +

infin

sum

119899=1

119899

sum

119894=1

119863119894minus1119862119860119899minus119894119861 (119863119889)

119899+2

]

]

]

]

(31)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

1198601198601205870

0 119863] 119876 = [

1198602119860119889119861

119862 0

]

119875119889= [

0 0

0 119863119889] 119875

120587= [

119868 0

0 119863120587]

(32)

Since119863119862 = 119863120587119862119860119860120587 and 119860119860120587119861 = 119861119863 we have

119863119889119862 = (119863

119889)

2

119863119862 = (119863119889)

2

119863120587119862119860119860120587= 0

119863119862119860119889= 119863120587119862119860119860120587119860119889= 0

119860119889119861119863 = 119860

119889119860119860120587119861 = 0

(33)

From 119861119862 = 0 and applying Lemma 5 to 119876 we obtain

(119876119889)

119894

= [(119860119889)

119894

(119860119889)

119894+1

119861

119883119894minus1

119883119894119861

]

119876120587= [

119860120587

minus119860119889119861

minus119862119860119889119868 minus 119862 (119860

119889)

2

119861

]

(34)

where 119883119899is defined in (28) From 119863119889119862 = 0 and 119863119862119860119889 = 0

we get119883119899= 119862(119860

119889)119899+2

Since 119860119860120587119861 = 119861119863 and 119863119862 = 119863120587119862119860119860120587 we obtain 119875119876 =119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(35)

From 119860119889119861119863 = 0 we have

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889 119876

119889119875 = 0 (36)

Hence from (35) we obtain

(119875 + 119876)119889= 119875119889+ 119876119889+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(37)

Since 119860119889119861119863 = 0 we have

119876120587119876(119875119889)

2

= 119861 (119863119889)

2

(38)

The conditions 119861119862 = 0 and 119861119863 = 119860119860120587119861 imply that

119861119863119899119862 = 0 From 119861119863119899119862 = 0 and 119860119889119861119863 = 0 we get

119876120587

infin

sum

119899=1

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=1

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899

sum

119894=1

119863119894minus1119862119860119899minus119894119861 (119863119889)

119899+2

]

]

]

]

]

(39)

From (36) (38) and (39) it follows (31)The proof is finished

Since

[

119860 119861

119862 119863] = [

0 119868119899

1198681198980] [

119863 119862

119861 119860][

0 119868119898

1198681198990] (40)

we can obtain the following result applying Theorem 8 to[119863 119862

119861 119860]

Theorem 9 Let119860 and119863 be generalized Drazin invertible andlet119872 be matrix of form (26) If119863119863120587119862 = 119862119860 119860119861 = 119860120587119861119863119863120587and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+ 119861 (119863

119889)

3

119862 +

infin

sum

119899=1

119899

sum

119894=1

119860119899minus119894119861119863119894minus1119862(119860119889)

119899+2

119861 (119863119889)

2

(119863119889)

2

119862 +

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

119863119889

]

]

]

]

]

(41)Theorem 10 Let 119860 119863 and 119861119862 be generalized Drazin invert-ible and let119872 be matrix of form (26) If 119860119861 = 119860120587119861119863 119863119862 =119863120587119862119860 and 119861119862 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

119863119889+

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(42)

The Scientific World Journal 5

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

119860 0

0 119863] 119876 = [

0 119861

119862 0] (43)

119875119889= [

1198601198890

0 119863119889] 119875

120587= [

1198601205870

0 119863120587] (44)

Since

1198762= [

119861119862 0

0 119862119861] 119876

3= [

0 119861119862119861

119862119861119862 0] (45)

from 119861119862 = 0 it is easy to get 1198763 = 0 Since 119876 is nilpotent wehave 119876119889 = 0 Applying Theorem 4 to the particular case weget

(119875 + 119876)119889= 119875119889+

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(46)

The conditions 119860119861 = 119860120587119861119863 and 119861119862 = 0 imply that

119861119863119899119862 = 0 for 119899 ge 0 so we get

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(47)

From (44) and (47) it follows (42)The proof is finished

Theorem 11 Let119860 and119863 be generalizedDrazin invertible andlet119872 be matrix of form (26) If 119860119860120587119861 = 1198611198632119863119889 1198632119863119889119862 =119863120587119862119860119860120587 and 119861119863119899119862 = 0 for any nonnegative integer 119899 Then

119872119889=

[

[

[

[

[

119860119889

Γ +

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

Δ 119863119889+ Δ119860Γ +

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(48)

where Γ and Δ are defined in (30)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

119860119860120587

0

0 1198632119863119889] 119876 = [

1198602119860119889

119861

119862 119863119863120587]

119875119889= [

0 0

0 119863119889] 119875

120587= [

119868 0

0 119863120587]

(49)

From 119860119860120587119861 = 1198611198632119863119889 and1198632119863119889119862 = 119863120587119862119860119860120587 we have

119863119889119862 = (119863

119889)

3

1198632119862 = (119863

119889)

2

1198632119863119889119862

= (119863119889)

2

119863120587119862119860119860120587= 0

(50)

119860119889119861119863119889= 1198601198891198611198632(119863119889)

3

= 1198601198891198611198632119863119889(119863119889)

2

= 119860119889119860119860120587119861 (119863119889)

2

= 0

(51)

so we get119863120587119862 = 119862Note that 119863119863120587 is quasinilpotent 119863120587119862 = 119862 and

119861(119863119863120587)119899119862 = 119861119863

119899119863120587119862 = 119861119863

119899119862 = 0 for any nonnegative

integer 119899 we can apply Lemma 6 to 119876 with 119863 replaced by119863119863120587 we have

119876119889= [119860119889

Γ1015840

Δ1015840Δ1015840119860Γ1015840] (52)

where

Γ1015840=

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899119863120587 Δ

1015840=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

(53)

Observe that (50) and (51) yield

Γ =

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899 Δ =

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

(54)

so we get

119876119889= [119860119889

Γ

Δ Δ119860Γ

] (55)

The condition 119861119863119899119862 = 0 implies that

119861Δ = 119861

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

= 0 (56)

Hence we have

119876119876119889= [

119860119860119889+ 119861Δ 119860

2119860119889Γ + 119861Δ119860Γ

119862119860119889+ 119863119863

120587Δ 119862Γ + 119863119863

120587Δ119860Γ

]

= [

119860119860119889

119860Γ

119862119860119889+ 119863Δ 119862Γ + 119863Δ119860Γ

]

119876120587= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

(57)

6 The Scientific World Journal

From 119860119860120587119861 = 119861119863

2119863119889 and 1198632119863119889119862 = 119863

120587119862119860119860120587 we

obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(58)

where

Γ1198632119863119889=

infin

sum

119899=0

(119860119889)

119899+2

1198611198631198991198631205871198632119863119889= 0

Δ119860119860120587=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

119860119860120587= 0

(59)

so we get

119876119889119875 = [

119860119889

Γ

Δ Δ119860Γ

] [

119860119860120587

0

0 1198632119863119889]

= [

119860119889119860119860120587

Γ1198632119863119889

Δ119860119860120587Δ119860Γ119863

2119863119889] = [

0 0

0 0]

(60)

Hence from (58) and (60) we obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(61)

By direct computation we verify that

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889 (62)

From 119861119863119899119862 = 0 we haveinfin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(63)

Observe that (51) and 119861119863119899119862 = 0 yield

119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0 119860120587

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

minus 119860Γ

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

0 (minus119862119860119889minus 119863Δ)

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

+ (119868 minus 119862Γ minus 119863Δ119860Γ)

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

(64)

From (62) and (64) it follows (48) The proof is finished

The Scientific World Journal 7

Theorem 12 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119860120587119861119863120587 = 119861119863 119861119862 = 0119862119860119889= 0 and 119862119861119863120587 = 0 Then

119872119889=[

[

[

119860119889

(119860119889)

2

119861

1198830119863119889+

infin

sum

119899=0

119883119899+1119861119863119899

]

]

]

(65)

where

119883119899=

infin

sum

119894=0

(119863119889)

119894+119899+2

119862119860119894 119899 ge 0 (66)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [1198602119860119889119861

0 0

] 119876 = [

1198601198601205870

119862 119863]

119875119889= [119860119889(119860119889)

2

119861

0 0

] 119875120587= [119860120587minus119860119889119861

0 119868

]

(67)

Applying Lemma 7 we have (119860119860120587)119889 = 0 so we get

(119876119889)

119899

= [

0 0

119883119899minus1

(119863119889)

119899] 119876120587= [

119868 0

minus1198631198830119863120587] (68)

where119883119899is defined in (28)

Since 119860119860120587119861119863120587 = 119861119863 119861119862 = 0 119862119861119863120587 = 0 and 1198621198602119860119889 =0 we obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(69)

From 119862119860119889= 0 we have 119876119875119889 = 0 Hence from (69) we

obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

(70)

where119883119899119860119889= 0 we get

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889

(119876119889)

2

119875119875120587= [

0 0

1198831(119863119889)

2] [1198602119860119889119861

0 0

] [119860120587minus119860119889119861

0 119868

]

= [

0 0

0 1198831119861]

(71)

The conditions 119860119860120587119861119863120587 = 119861119863 and 119861119862 = 0 imply that119861119863119894119862 = 0 So we get

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

=[

[

[

0 0

0

infin

sum

119899=1

119883119899+1119861119863119899

]

]

]

119899 ge 1

(72)

From (71) and (72) it follows (65)The proof is finished

Using (40) andTheorem 12 we have the following result

Theorem 13 If 119862119860 = 119863120587119863119862119860120587 119861119863119889 = 0 119862119861 = 0 and

119861119862119860120587= 0 Then

119872119889=[

[

[

119860119889+

infin

sum

119899=0

119883119899+21198621198601198991198831

(119863119889)

2

119862 119863119889

]

]

]

(73)

where

119883119899=

infin

sum

119894=0

(119860119889)

119894+119899+1

119861119863119894 119899 ge 1 (74)

Using the case ofTheorem 3 we get the following results

Theorem 14 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119863119862119860120587 = 119862119860 119860119861119863 = 0119861119862 = 0 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+

infin

sum

119899=0

119861 (119863119889)

119899+3

119862119860119899(119860119889)

2

119861 + 119861 (119863119889)

2

infin

sum

119899=0

(119863119889)

119899+2

119862119860119899

119863119889

]

]

]

]

]

(75)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

0 0

119862 0] 119876 = [

119860 119861

0 119863] (76)

From 119860119861119863 = 0 we have

(119876119889)

119894

=[

[

(119860119889)

119894

119883119894

0 (119863119889)

119894]

]

119876120587= [119860120587minus1198601198831minus 119861119863119889

0 119863120587 ]

(77)

where

119883119899= (119860119889)

119899+2

119861 + 119861 (119863119889)

119899+2

119899 ge 0 (78)

Note that 119875 is quasinilpotent since 119863119862119860120587 = 119862119860 119860119861119863 =0 119861119862 = 0 and 119862119861 = 0 we obtain 119875119876 = 119876119875119876

120587 ApplyingTheorem 3 we get

(119875 + 119876)119889= 119876119889+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899 (79)

8 The Scientific World Journal

From 119861119862 = 0 we have

(119876119889)

2

119875 =[

[

(119860119889)

2

1198832

0 (119863119889)

2]

]

[

0 0

119862 0]

= [

1198832119862 0

(119863119889)

2

119862 0

] =[

[

119861 (119863119889)

3

0

(119863119889)

2

119862 0

]

]

(80)

The conditions 119863119862119860120587 = 119862119860 and 119862119861 = 0 imply that119862119860119894119861 = 0 From 119860119861119863 = 0 119862119860119894119861 = 0 and 119861119862 = 0 we getinfin

sum

119899=1

(119876119889)

119899+2

119875 (119875 + 119876)119899

=

[

[

[

[

[

infin

sum

119899=1

119883119899+2119862119860119899

infin

sum

119899=1

119883119899+2119862119860119899minus1119861

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899minus1119861

]

]

]

]

]

=

[

[

[

[

[

infin

sum

119899=1

119861 (119863119889)

119899+3

1198621198601198990

infin

sum

119899=1

(119863119889)

119899+2

1198621198601198990

]

]

]

]

]

(81)

From (77) (80) and (81) it follows (75)

Using (40) andTheorem 14 we have the following result

Theorem 15 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119861119863120587 = 119861119863 119863119862119860 = 0119861119862 = 119860119861119862119860

119889 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899

(119863119889)

2

119862 + 119862 (119860119889)

2

119863119889+

infin

sum

119899=0

119862 (119860119889)

119899+3

119861119863119899

]

]

]

]

]

(82)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J J Koliha ldquoA generalized Drazin inverserdquo Glasgow Mathemat-ical Journal vol 38 no 3 pp 367ndash381 1991

[2] C Deng D S Cvetkovic-Ilic and Y Wei ldquoSome results on thegeneralized Drazin inverse of operator matricesrdquo Linear andMultilinear Algebra vol 58 no 3-4 pp 503ndash521 2010

[3] E Dopazo and M F Martınez-Serrano ldquoFurther results on therepresentation of the Drazin inverse of a 2 times 2 block matrixrdquoLinear Algebra and its Applications vol 432 no 8 pp 1896ndash1904 2010

[4] J J Koliha and V Rakocevic ldquoHolomorphic and meromorphicproperties of the g-Drazin inverserdquoDemonstratio Mathematicavol 38 no 3 pp 657ndash666 2005

[5] N Castro Gonzalez ldquoAdditive perturbation results for theDrazin inverserdquo Linear Algebra and Its Applications vol 397 pp279ndash297 2005

[6] X Liu L Xu and Y Yu ldquoThe representations of the Drazininverse of differences of twomatricesrdquoAppliedMathematics andComputation vol 216 no 12 pp 3652ndash3661 2010

[7] D S Djordjevic and V Rakocevic Lectures on GeneralizedInverses University of Nis 2008

[8] J Meyer Jr and N J Rose ldquoThe index and the Drazininverse of block triangular matricesrdquo SIAM Journal on AppliedMathematics vol 33 no 1 pp 1ndash7 1977

[9] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[10] N Castro-Gonzalez and J J Koliha ldquoNew additive resultsfor the 119892-Drazin inverserdquo Proceedings of the Royal Society ofEdinburgh Section A Mathematics vol 134 no 6 pp 1085ndash1097 2004

[11] D S Cvetkovic-Ilic D S Djordjevic and Y Wei ldquoAdditiveresults for the generalized Drazin inverse in a Banach algebrardquoLinear Algebra and Its Applications vol 418 no 1 pp 53ndash612006

[12] D S Djordjevic and YWei ldquoAdditive results for the generalizedDrazin inverserdquo Journal of the Australian Mathematical Societyvol 73 no 1 pp 115ndash125 2002

[13] C Deng and Y Wei ldquoA note on the Drazin inverse of an anti-triangular matrixrdquo Linear Algebra and its Applications vol 431no 10 pp 1910ndash1922 2009

[14] N Castro-Gonzalez E Dopazo and M F Martınez-SerranoldquoOn the Drazin inverse of the sum of two operators andits application to operator matricesrdquo Journal of MathematicalAnalysis and Applications vol 350 no 1 pp 207ndash215 2009

[15] R E Hartwig G Wang and Y Wei ldquoSome additive results onDrazin inverserdquo Linear Algebra and its Applications vol 322 no1ndash3 pp 207ndash217 2001

[16] R E Hartwig and J Shoaf ldquoGroup inverses and Drazin inversesof bidiagonal and triangular Toeplitzmatricesrdquo vol 24 no 1 pp10ndash34 1977

[17] R E Hartwig and J M Shoaf ldquoGroup inverse and Drazininverse of bidiagonal and triangular Toeplitz matricesrdquo Journalof the AustralianMathematical Society A vol 24 pp 10ndash34 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 The Scientific World Journal

Lemma 7 Let 119860 isin C119899times119899 Then (119860119860120587)119889 = 0 (1198602119860119889)119889 = 119860119889(1198602119860119889)120587= 119860120587 and Ind(119860119860120587) = Ind(119860) and Ind(1198602119860119889) = 1

Proof The Jordan canonical form of 119883 permits us to write119860 = 119878(119862 oplus 119873)119878

minus1 where 119878 and 119862 are nonsingular and 119873 isnilpotent with index Ind(119860) Thus 119860

119889= 119878(119862

minus1oplus 0)119878minus1 Now

it is evident that1198602119860119889 = 119878(119862oplus0)119878minus1 and119860119860120587 = 119878(0oplus119873)119878minus1which lead to the affirmations of this lemma

In [9 Theorem 53] authors gave an explicit representa-tion for119872119889 under conditions 119861119862 = 0 119863119862 = 0 and 119861119863 = 0Here we replace the last two conditions by the two weakerconditions119863119862 = 119863120587119862119860119860120587 and 119861119863 = 119860119860120587119861

Theorem 8 Let119860 and119863 be generalized Drazin invertible andlet119872 be matrix of form (26) If 119860119860120587119861 = 119861119863 119863119862 = 119863120587119862119860119860120587and 119861119862 = 0 Then

119872119889=

[

[

[

[

119860119889

(119860119889)

2

119861 +

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

119862 (119860119889)

2

119863119889+ 119862 (119860

119889)

3

119861 +

infin

sum

119899=1

119899

sum

119894=1

119863119894minus1119862119860119899minus119894119861 (119863119889)

119899+2

]

]

]

]

(31)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

1198601198601205870

0 119863] 119876 = [

1198602119860119889119861

119862 0

]

119875119889= [

0 0

0 119863119889] 119875

120587= [

119868 0

0 119863120587]

(32)

Since119863119862 = 119863120587119862119860119860120587 and 119860119860120587119861 = 119861119863 we have

119863119889119862 = (119863

119889)

2

119863119862 = (119863119889)

2

119863120587119862119860119860120587= 0

119863119862119860119889= 119863120587119862119860119860120587119860119889= 0

119860119889119861119863 = 119860

119889119860119860120587119861 = 0

(33)

From 119861119862 = 0 and applying Lemma 5 to 119876 we obtain

(119876119889)

119894

= [(119860119889)

119894

(119860119889)

119894+1

119861

119883119894minus1

119883119894119861

]

119876120587= [

119860120587

minus119860119889119861

minus119862119860119889119868 minus 119862 (119860

119889)

2

119861

]

(34)

where 119883119899is defined in (28) From 119863119889119862 = 0 and 119863119862119860119889 = 0

we get119883119899= 119862(119860

119889)119899+2

Since 119860119860120587119861 = 119861119863 and 119863119862 = 119863120587119862119860119860120587 we obtain 119875119876 =119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(35)

From 119860119889119861119863 = 0 we have

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889 119876

119889119875 = 0 (36)

Hence from (35) we obtain

(119875 + 119876)119889= 119875119889+ 119876119889+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(37)

Since 119860119889119861119863 = 0 we have

119876120587119876(119875119889)

2

= 119861 (119863119889)

2

(38)

The conditions 119861119862 = 0 and 119861119863 = 119860119860120587119861 imply that

119861119863119899119862 = 0 From 119861119863119899119862 = 0 and 119860119889119861119863 = 0 we get

119876120587

infin

sum

119899=1

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=1

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899

sum

119894=1

119863119894minus1119862119860119899minus119894119861 (119863119889)

119899+2

]

]

]

]

]

(39)

From (36) (38) and (39) it follows (31)The proof is finished

Since

[

119860 119861

119862 119863] = [

0 119868119899

1198681198980] [

119863 119862

119861 119860][

0 119868119898

1198681198990] (40)

we can obtain the following result applying Theorem 8 to[119863 119862

119861 119860]

Theorem 9 Let119860 and119863 be generalized Drazin invertible andlet119872 be matrix of form (26) If119863119863120587119862 = 119862119860 119860119861 = 119860120587119861119863119863120587and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+ 119861 (119863

119889)

3

119862 +

infin

sum

119899=1

119899

sum

119894=1

119860119899minus119894119861119863119894minus1119862(119860119889)

119899+2

119861 (119863119889)

2

(119863119889)

2

119862 +

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

119863119889

]

]

]

]

]

(41)Theorem 10 Let 119860 119863 and 119861119862 be generalized Drazin invert-ible and let119872 be matrix of form (26) If 119860119861 = 119860120587119861119863 119863119862 =119863120587119862119860 and 119861119862 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

119863119889+

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(42)

The Scientific World Journal 5

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

119860 0

0 119863] 119876 = [

0 119861

119862 0] (43)

119875119889= [

1198601198890

0 119863119889] 119875

120587= [

1198601205870

0 119863120587] (44)

Since

1198762= [

119861119862 0

0 119862119861] 119876

3= [

0 119861119862119861

119862119861119862 0] (45)

from 119861119862 = 0 it is easy to get 1198763 = 0 Since 119876 is nilpotent wehave 119876119889 = 0 Applying Theorem 4 to the particular case weget

(119875 + 119876)119889= 119875119889+

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(46)

The conditions 119860119861 = 119860120587119861119863 and 119861119862 = 0 imply that

119861119863119899119862 = 0 for 119899 ge 0 so we get

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(47)

From (44) and (47) it follows (42)The proof is finished

Theorem 11 Let119860 and119863 be generalizedDrazin invertible andlet119872 be matrix of form (26) If 119860119860120587119861 = 1198611198632119863119889 1198632119863119889119862 =119863120587119862119860119860120587 and 119861119863119899119862 = 0 for any nonnegative integer 119899 Then

119872119889=

[

[

[

[

[

119860119889

Γ +

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

Δ 119863119889+ Δ119860Γ +

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(48)

where Γ and Δ are defined in (30)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

119860119860120587

0

0 1198632119863119889] 119876 = [

1198602119860119889

119861

119862 119863119863120587]

119875119889= [

0 0

0 119863119889] 119875

120587= [

119868 0

0 119863120587]

(49)

From 119860119860120587119861 = 1198611198632119863119889 and1198632119863119889119862 = 119863120587119862119860119860120587 we have

119863119889119862 = (119863

119889)

3

1198632119862 = (119863

119889)

2

1198632119863119889119862

= (119863119889)

2

119863120587119862119860119860120587= 0

(50)

119860119889119861119863119889= 1198601198891198611198632(119863119889)

3

= 1198601198891198611198632119863119889(119863119889)

2

= 119860119889119860119860120587119861 (119863119889)

2

= 0

(51)

so we get119863120587119862 = 119862Note that 119863119863120587 is quasinilpotent 119863120587119862 = 119862 and

119861(119863119863120587)119899119862 = 119861119863

119899119863120587119862 = 119861119863

119899119862 = 0 for any nonnegative

integer 119899 we can apply Lemma 6 to 119876 with 119863 replaced by119863119863120587 we have

119876119889= [119860119889

Γ1015840

Δ1015840Δ1015840119860Γ1015840] (52)

where

Γ1015840=

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899119863120587 Δ

1015840=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

(53)

Observe that (50) and (51) yield

Γ =

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899 Δ =

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

(54)

so we get

119876119889= [119860119889

Γ

Δ Δ119860Γ

] (55)

The condition 119861119863119899119862 = 0 implies that

119861Δ = 119861

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

= 0 (56)

Hence we have

119876119876119889= [

119860119860119889+ 119861Δ 119860

2119860119889Γ + 119861Δ119860Γ

119862119860119889+ 119863119863

120587Δ 119862Γ + 119863119863

120587Δ119860Γ

]

= [

119860119860119889

119860Γ

119862119860119889+ 119863Δ 119862Γ + 119863Δ119860Γ

]

119876120587= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

(57)

6 The Scientific World Journal

From 119860119860120587119861 = 119861119863

2119863119889 and 1198632119863119889119862 = 119863

120587119862119860119860120587 we

obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(58)

where

Γ1198632119863119889=

infin

sum

119899=0

(119860119889)

119899+2

1198611198631198991198631205871198632119863119889= 0

Δ119860119860120587=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

119860119860120587= 0

(59)

so we get

119876119889119875 = [

119860119889

Γ

Δ Δ119860Γ

] [

119860119860120587

0

0 1198632119863119889]

= [

119860119889119860119860120587

Γ1198632119863119889

Δ119860119860120587Δ119860Γ119863

2119863119889] = [

0 0

0 0]

(60)

Hence from (58) and (60) we obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(61)

By direct computation we verify that

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889 (62)

From 119861119863119899119862 = 0 we haveinfin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(63)

Observe that (51) and 119861119863119899119862 = 0 yield

119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0 119860120587

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

minus 119860Γ

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

0 (minus119862119860119889minus 119863Δ)

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

+ (119868 minus 119862Γ minus 119863Δ119860Γ)

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

(64)

From (62) and (64) it follows (48) The proof is finished

The Scientific World Journal 7

Theorem 12 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119860120587119861119863120587 = 119861119863 119861119862 = 0119862119860119889= 0 and 119862119861119863120587 = 0 Then

119872119889=[

[

[

119860119889

(119860119889)

2

119861

1198830119863119889+

infin

sum

119899=0

119883119899+1119861119863119899

]

]

]

(65)

where

119883119899=

infin

sum

119894=0

(119863119889)

119894+119899+2

119862119860119894 119899 ge 0 (66)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [1198602119860119889119861

0 0

] 119876 = [

1198601198601205870

119862 119863]

119875119889= [119860119889(119860119889)

2

119861

0 0

] 119875120587= [119860120587minus119860119889119861

0 119868

]

(67)

Applying Lemma 7 we have (119860119860120587)119889 = 0 so we get

(119876119889)

119899

= [

0 0

119883119899minus1

(119863119889)

119899] 119876120587= [

119868 0

minus1198631198830119863120587] (68)

where119883119899is defined in (28)

Since 119860119860120587119861119863120587 = 119861119863 119861119862 = 0 119862119861119863120587 = 0 and 1198621198602119860119889 =0 we obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(69)

From 119862119860119889= 0 we have 119876119875119889 = 0 Hence from (69) we

obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

(70)

where119883119899119860119889= 0 we get

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889

(119876119889)

2

119875119875120587= [

0 0

1198831(119863119889)

2] [1198602119860119889119861

0 0

] [119860120587minus119860119889119861

0 119868

]

= [

0 0

0 1198831119861]

(71)

The conditions 119860119860120587119861119863120587 = 119861119863 and 119861119862 = 0 imply that119861119863119894119862 = 0 So we get

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

=[

[

[

0 0

0

infin

sum

119899=1

119883119899+1119861119863119899

]

]

]

119899 ge 1

(72)

From (71) and (72) it follows (65)The proof is finished

Using (40) andTheorem 12 we have the following result

Theorem 13 If 119862119860 = 119863120587119863119862119860120587 119861119863119889 = 0 119862119861 = 0 and

119861119862119860120587= 0 Then

119872119889=[

[

[

119860119889+

infin

sum

119899=0

119883119899+21198621198601198991198831

(119863119889)

2

119862 119863119889

]

]

]

(73)

where

119883119899=

infin

sum

119894=0

(119860119889)

119894+119899+1

119861119863119894 119899 ge 1 (74)

Using the case ofTheorem 3 we get the following results

Theorem 14 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119863119862119860120587 = 119862119860 119860119861119863 = 0119861119862 = 0 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+

infin

sum

119899=0

119861 (119863119889)

119899+3

119862119860119899(119860119889)

2

119861 + 119861 (119863119889)

2

infin

sum

119899=0

(119863119889)

119899+2

119862119860119899

119863119889

]

]

]

]

]

(75)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

0 0

119862 0] 119876 = [

119860 119861

0 119863] (76)

From 119860119861119863 = 0 we have

(119876119889)

119894

=[

[

(119860119889)

119894

119883119894

0 (119863119889)

119894]

]

119876120587= [119860120587minus1198601198831minus 119861119863119889

0 119863120587 ]

(77)

where

119883119899= (119860119889)

119899+2

119861 + 119861 (119863119889)

119899+2

119899 ge 0 (78)

Note that 119875 is quasinilpotent since 119863119862119860120587 = 119862119860 119860119861119863 =0 119861119862 = 0 and 119862119861 = 0 we obtain 119875119876 = 119876119875119876

120587 ApplyingTheorem 3 we get

(119875 + 119876)119889= 119876119889+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899 (79)

8 The Scientific World Journal

From 119861119862 = 0 we have

(119876119889)

2

119875 =[

[

(119860119889)

2

1198832

0 (119863119889)

2]

]

[

0 0

119862 0]

= [

1198832119862 0

(119863119889)

2

119862 0

] =[

[

119861 (119863119889)

3

0

(119863119889)

2

119862 0

]

]

(80)

The conditions 119863119862119860120587 = 119862119860 and 119862119861 = 0 imply that119862119860119894119861 = 0 From 119860119861119863 = 0 119862119860119894119861 = 0 and 119861119862 = 0 we getinfin

sum

119899=1

(119876119889)

119899+2

119875 (119875 + 119876)119899

=

[

[

[

[

[

infin

sum

119899=1

119883119899+2119862119860119899

infin

sum

119899=1

119883119899+2119862119860119899minus1119861

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899minus1119861

]

]

]

]

]

=

[

[

[

[

[

infin

sum

119899=1

119861 (119863119889)

119899+3

1198621198601198990

infin

sum

119899=1

(119863119889)

119899+2

1198621198601198990

]

]

]

]

]

(81)

From (77) (80) and (81) it follows (75)

Using (40) andTheorem 14 we have the following result

Theorem 15 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119861119863120587 = 119861119863 119863119862119860 = 0119861119862 = 119860119861119862119860

119889 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899

(119863119889)

2

119862 + 119862 (119860119889)

2

119863119889+

infin

sum

119899=0

119862 (119860119889)

119899+3

119861119863119899

]

]

]

]

]

(82)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J J Koliha ldquoA generalized Drazin inverserdquo Glasgow Mathemat-ical Journal vol 38 no 3 pp 367ndash381 1991

[2] C Deng D S Cvetkovic-Ilic and Y Wei ldquoSome results on thegeneralized Drazin inverse of operator matricesrdquo Linear andMultilinear Algebra vol 58 no 3-4 pp 503ndash521 2010

[3] E Dopazo and M F Martınez-Serrano ldquoFurther results on therepresentation of the Drazin inverse of a 2 times 2 block matrixrdquoLinear Algebra and its Applications vol 432 no 8 pp 1896ndash1904 2010

[4] J J Koliha and V Rakocevic ldquoHolomorphic and meromorphicproperties of the g-Drazin inverserdquoDemonstratio Mathematicavol 38 no 3 pp 657ndash666 2005

[5] N Castro Gonzalez ldquoAdditive perturbation results for theDrazin inverserdquo Linear Algebra and Its Applications vol 397 pp279ndash297 2005

[6] X Liu L Xu and Y Yu ldquoThe representations of the Drazininverse of differences of twomatricesrdquoAppliedMathematics andComputation vol 216 no 12 pp 3652ndash3661 2010

[7] D S Djordjevic and V Rakocevic Lectures on GeneralizedInverses University of Nis 2008

[8] J Meyer Jr and N J Rose ldquoThe index and the Drazininverse of block triangular matricesrdquo SIAM Journal on AppliedMathematics vol 33 no 1 pp 1ndash7 1977

[9] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[10] N Castro-Gonzalez and J J Koliha ldquoNew additive resultsfor the 119892-Drazin inverserdquo Proceedings of the Royal Society ofEdinburgh Section A Mathematics vol 134 no 6 pp 1085ndash1097 2004

[11] D S Cvetkovic-Ilic D S Djordjevic and Y Wei ldquoAdditiveresults for the generalized Drazin inverse in a Banach algebrardquoLinear Algebra and Its Applications vol 418 no 1 pp 53ndash612006

[12] D S Djordjevic and YWei ldquoAdditive results for the generalizedDrazin inverserdquo Journal of the Australian Mathematical Societyvol 73 no 1 pp 115ndash125 2002

[13] C Deng and Y Wei ldquoA note on the Drazin inverse of an anti-triangular matrixrdquo Linear Algebra and its Applications vol 431no 10 pp 1910ndash1922 2009

[14] N Castro-Gonzalez E Dopazo and M F Martınez-SerranoldquoOn the Drazin inverse of the sum of two operators andits application to operator matricesrdquo Journal of MathematicalAnalysis and Applications vol 350 no 1 pp 207ndash215 2009

[15] R E Hartwig G Wang and Y Wei ldquoSome additive results onDrazin inverserdquo Linear Algebra and its Applications vol 322 no1ndash3 pp 207ndash217 2001

[16] R E Hartwig and J Shoaf ldquoGroup inverses and Drazin inversesof bidiagonal and triangular Toeplitzmatricesrdquo vol 24 no 1 pp10ndash34 1977

[17] R E Hartwig and J M Shoaf ldquoGroup inverse and Drazininverse of bidiagonal and triangular Toeplitz matricesrdquo Journalof the AustralianMathematical Society A vol 24 pp 10ndash34 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

The Scientific World Journal 5

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

119860 0

0 119863] 119876 = [

0 119861

119862 0] (43)

119875119889= [

1198601198890

0 119863119889] 119875

120587= [

1198601205870

0 119863120587] (44)

Since

1198762= [

119861119862 0

0 119862119861] 119876

3= [

0 119861119862119861

119862119861119862 0] (45)

from 119861119862 = 0 it is easy to get 1198763 = 0 Since 119876 is nilpotent wehave 119876119889 = 0 Applying Theorem 4 to the particular case weget

(119875 + 119876)119889= 119875119889+

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(46)

The conditions 119860119861 = 119860120587119861119863 and 119861119862 = 0 imply that

119861119863119899119862 = 0 for 119899 ge 0 so we get

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(47)

From (44) and (47) it follows (42)The proof is finished

Theorem 11 Let119860 and119863 be generalizedDrazin invertible andlet119872 be matrix of form (26) If 119860119860120587119861 = 1198611198632119863119889 1198632119863119889119862 =119863120587119862119860119860120587 and 119861119863119899119862 = 0 for any nonnegative integer 119899 Then

119872119889=

[

[

[

[

[

119860119889

Γ +

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

Δ 119863119889+ Δ119860Γ +

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(48)

where Γ and Δ are defined in (30)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

119860119860120587

0

0 1198632119863119889] 119876 = [

1198602119860119889

119861

119862 119863119863120587]

119875119889= [

0 0

0 119863119889] 119875

120587= [

119868 0

0 119863120587]

(49)

From 119860119860120587119861 = 1198611198632119863119889 and1198632119863119889119862 = 119863120587119862119860119860120587 we have

119863119889119862 = (119863

119889)

3

1198632119862 = (119863

119889)

2

1198632119863119889119862

= (119863119889)

2

119863120587119862119860119860120587= 0

(50)

119860119889119861119863119889= 1198601198891198611198632(119863119889)

3

= 1198601198891198611198632119863119889(119863119889)

2

= 119860119889119860119860120587119861 (119863119889)

2

= 0

(51)

so we get119863120587119862 = 119862Note that 119863119863120587 is quasinilpotent 119863120587119862 = 119862 and

119861(119863119863120587)119899119862 = 119861119863

119899119863120587119862 = 119861119863

119899119862 = 0 for any nonnegative

integer 119899 we can apply Lemma 6 to 119876 with 119863 replaced by119863119863120587 we have

119876119889= [119860119889

Γ1015840

Δ1015840Δ1015840119860Γ1015840] (52)

where

Γ1015840=

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899119863120587 Δ

1015840=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

(53)

Observe that (50) and (51) yield

Γ =

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899 Δ =

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

(54)

so we get

119876119889= [119860119889

Γ

Δ Δ119860Γ

] (55)

The condition 119861119863119899119862 = 0 implies that

119861Δ = 119861

infin

sum

119899=0

119863119899119862 (119860119889)

119899+2

= 0 (56)

Hence we have

119876119876119889= [

119860119860119889+ 119861Δ 119860

2119860119889Γ + 119861Δ119860Γ

119862119860119889+ 119863119863

120587Δ 119862Γ + 119863119863

120587Δ119860Γ

]

= [

119860119860119889

119860Γ

119862119860119889+ 119863Δ 119862Γ + 119863Δ119860Γ

]

119876120587= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

(57)

6 The Scientific World Journal

From 119860119860120587119861 = 119861119863

2119863119889 and 1198632119863119889119862 = 119863

120587119862119860119860120587 we

obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(58)

where

Γ1198632119863119889=

infin

sum

119899=0

(119860119889)

119899+2

1198611198631198991198631205871198632119863119889= 0

Δ119860119860120587=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

119860119860120587= 0

(59)

so we get

119876119889119875 = [

119860119889

Γ

Δ Δ119860Γ

] [

119860119860120587

0

0 1198632119863119889]

= [

119860119889119860119860120587

Γ1198632119863119889

Δ119860119860120587Δ119860Γ119863

2119863119889] = [

0 0

0 0]

(60)

Hence from (58) and (60) we obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(61)

By direct computation we verify that

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889 (62)

From 119861119863119899119862 = 0 we haveinfin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(63)

Observe that (51) and 119861119863119899119862 = 0 yield

119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0 119860120587

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

minus 119860Γ

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

0 (minus119862119860119889minus 119863Δ)

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

+ (119868 minus 119862Γ minus 119863Δ119860Γ)

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

(64)

From (62) and (64) it follows (48) The proof is finished

The Scientific World Journal 7

Theorem 12 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119860120587119861119863120587 = 119861119863 119861119862 = 0119862119860119889= 0 and 119862119861119863120587 = 0 Then

119872119889=[

[

[

119860119889

(119860119889)

2

119861

1198830119863119889+

infin

sum

119899=0

119883119899+1119861119863119899

]

]

]

(65)

where

119883119899=

infin

sum

119894=0

(119863119889)

119894+119899+2

119862119860119894 119899 ge 0 (66)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [1198602119860119889119861

0 0

] 119876 = [

1198601198601205870

119862 119863]

119875119889= [119860119889(119860119889)

2

119861

0 0

] 119875120587= [119860120587minus119860119889119861

0 119868

]

(67)

Applying Lemma 7 we have (119860119860120587)119889 = 0 so we get

(119876119889)

119899

= [

0 0

119883119899minus1

(119863119889)

119899] 119876120587= [

119868 0

minus1198631198830119863120587] (68)

where119883119899is defined in (28)

Since 119860119860120587119861119863120587 = 119861119863 119861119862 = 0 119862119861119863120587 = 0 and 1198621198602119860119889 =0 we obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(69)

From 119862119860119889= 0 we have 119876119875119889 = 0 Hence from (69) we

obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

(70)

where119883119899119860119889= 0 we get

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889

(119876119889)

2

119875119875120587= [

0 0

1198831(119863119889)

2] [1198602119860119889119861

0 0

] [119860120587minus119860119889119861

0 119868

]

= [

0 0

0 1198831119861]

(71)

The conditions 119860119860120587119861119863120587 = 119861119863 and 119861119862 = 0 imply that119861119863119894119862 = 0 So we get

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

=[

[

[

0 0

0

infin

sum

119899=1

119883119899+1119861119863119899

]

]

]

119899 ge 1

(72)

From (71) and (72) it follows (65)The proof is finished

Using (40) andTheorem 12 we have the following result

Theorem 13 If 119862119860 = 119863120587119863119862119860120587 119861119863119889 = 0 119862119861 = 0 and

119861119862119860120587= 0 Then

119872119889=[

[

[

119860119889+

infin

sum

119899=0

119883119899+21198621198601198991198831

(119863119889)

2

119862 119863119889

]

]

]

(73)

where

119883119899=

infin

sum

119894=0

(119860119889)

119894+119899+1

119861119863119894 119899 ge 1 (74)

Using the case ofTheorem 3 we get the following results

Theorem 14 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119863119862119860120587 = 119862119860 119860119861119863 = 0119861119862 = 0 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+

infin

sum

119899=0

119861 (119863119889)

119899+3

119862119860119899(119860119889)

2

119861 + 119861 (119863119889)

2

infin

sum

119899=0

(119863119889)

119899+2

119862119860119899

119863119889

]

]

]

]

]

(75)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

0 0

119862 0] 119876 = [

119860 119861

0 119863] (76)

From 119860119861119863 = 0 we have

(119876119889)

119894

=[

[

(119860119889)

119894

119883119894

0 (119863119889)

119894]

]

119876120587= [119860120587minus1198601198831minus 119861119863119889

0 119863120587 ]

(77)

where

119883119899= (119860119889)

119899+2

119861 + 119861 (119863119889)

119899+2

119899 ge 0 (78)

Note that 119875 is quasinilpotent since 119863119862119860120587 = 119862119860 119860119861119863 =0 119861119862 = 0 and 119862119861 = 0 we obtain 119875119876 = 119876119875119876

120587 ApplyingTheorem 3 we get

(119875 + 119876)119889= 119876119889+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899 (79)

8 The Scientific World Journal

From 119861119862 = 0 we have

(119876119889)

2

119875 =[

[

(119860119889)

2

1198832

0 (119863119889)

2]

]

[

0 0

119862 0]

= [

1198832119862 0

(119863119889)

2

119862 0

] =[

[

119861 (119863119889)

3

0

(119863119889)

2

119862 0

]

]

(80)

The conditions 119863119862119860120587 = 119862119860 and 119862119861 = 0 imply that119862119860119894119861 = 0 From 119860119861119863 = 0 119862119860119894119861 = 0 and 119861119862 = 0 we getinfin

sum

119899=1

(119876119889)

119899+2

119875 (119875 + 119876)119899

=

[

[

[

[

[

infin

sum

119899=1

119883119899+2119862119860119899

infin

sum

119899=1

119883119899+2119862119860119899minus1119861

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899minus1119861

]

]

]

]

]

=

[

[

[

[

[

infin

sum

119899=1

119861 (119863119889)

119899+3

1198621198601198990

infin

sum

119899=1

(119863119889)

119899+2

1198621198601198990

]

]

]

]

]

(81)

From (77) (80) and (81) it follows (75)

Using (40) andTheorem 14 we have the following result

Theorem 15 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119861119863120587 = 119861119863 119863119862119860 = 0119861119862 = 119860119861119862119860

119889 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899

(119863119889)

2

119862 + 119862 (119860119889)

2

119863119889+

infin

sum

119899=0

119862 (119860119889)

119899+3

119861119863119899

]

]

]

]

]

(82)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J J Koliha ldquoA generalized Drazin inverserdquo Glasgow Mathemat-ical Journal vol 38 no 3 pp 367ndash381 1991

[2] C Deng D S Cvetkovic-Ilic and Y Wei ldquoSome results on thegeneralized Drazin inverse of operator matricesrdquo Linear andMultilinear Algebra vol 58 no 3-4 pp 503ndash521 2010

[3] E Dopazo and M F Martınez-Serrano ldquoFurther results on therepresentation of the Drazin inverse of a 2 times 2 block matrixrdquoLinear Algebra and its Applications vol 432 no 8 pp 1896ndash1904 2010

[4] J J Koliha and V Rakocevic ldquoHolomorphic and meromorphicproperties of the g-Drazin inverserdquoDemonstratio Mathematicavol 38 no 3 pp 657ndash666 2005

[5] N Castro Gonzalez ldquoAdditive perturbation results for theDrazin inverserdquo Linear Algebra and Its Applications vol 397 pp279ndash297 2005

[6] X Liu L Xu and Y Yu ldquoThe representations of the Drazininverse of differences of twomatricesrdquoAppliedMathematics andComputation vol 216 no 12 pp 3652ndash3661 2010

[7] D S Djordjevic and V Rakocevic Lectures on GeneralizedInverses University of Nis 2008

[8] J Meyer Jr and N J Rose ldquoThe index and the Drazininverse of block triangular matricesrdquo SIAM Journal on AppliedMathematics vol 33 no 1 pp 1ndash7 1977

[9] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[10] N Castro-Gonzalez and J J Koliha ldquoNew additive resultsfor the 119892-Drazin inverserdquo Proceedings of the Royal Society ofEdinburgh Section A Mathematics vol 134 no 6 pp 1085ndash1097 2004

[11] D S Cvetkovic-Ilic D S Djordjevic and Y Wei ldquoAdditiveresults for the generalized Drazin inverse in a Banach algebrardquoLinear Algebra and Its Applications vol 418 no 1 pp 53ndash612006

[12] D S Djordjevic and YWei ldquoAdditive results for the generalizedDrazin inverserdquo Journal of the Australian Mathematical Societyvol 73 no 1 pp 115ndash125 2002

[13] C Deng and Y Wei ldquoA note on the Drazin inverse of an anti-triangular matrixrdquo Linear Algebra and its Applications vol 431no 10 pp 1910ndash1922 2009

[14] N Castro-Gonzalez E Dopazo and M F Martınez-SerranoldquoOn the Drazin inverse of the sum of two operators andits application to operator matricesrdquo Journal of MathematicalAnalysis and Applications vol 350 no 1 pp 207ndash215 2009

[15] R E Hartwig G Wang and Y Wei ldquoSome additive results onDrazin inverserdquo Linear Algebra and its Applications vol 322 no1ndash3 pp 207ndash217 2001

[16] R E Hartwig and J Shoaf ldquoGroup inverses and Drazin inversesof bidiagonal and triangular Toeplitzmatricesrdquo vol 24 no 1 pp10ndash34 1977

[17] R E Hartwig and J M Shoaf ldquoGroup inverse and Drazininverse of bidiagonal and triangular Toeplitz matricesrdquo Journalof the AustralianMathematical Society A vol 24 pp 10ndash34 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 The Scientific World Journal

From 119860119860120587119861 = 119861119863

2119863119889 and 1198632119863119889119862 = 119863

120587119862119860119860120587 we

obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(58)

where

Γ1198632119863119889=

infin

sum

119899=0

(119860119889)

119899+2

1198611198631198991198631205871198632119863119889= 0

Δ119860119860120587=

infin

sum

119899=0

119863119899119863120587119862 (119860119889)

119899+2

119860119860120587= 0

(59)

so we get

119876119889119875 = [

119860119889

Γ

Δ Δ119860Γ

] [

119860119860120587

0

0 1198632119863119889]

= [

119860119889119860119860120587

Γ1198632119863119889

Δ119860119860120587Δ119860Γ119863

2119863119889] = [

0 0

0 0]

(60)

Hence from (58) and (60) we obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

(61)

By direct computation we verify that

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889 (62)

From 119861119863119899119862 = 0 we haveinfin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

=

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

(63)

Observe that (51) and 119861119863119899119862 = 0 yield

119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

= [

119860120587

minus119860Γ

minus119862119860119889minus 119863Δ 119868 minus 119862Γ minus 119863Δ119860Γ

]

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0 119860120587

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

minus 119860Γ

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

0 (minus119862119860119889minus 119863Δ)

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

+ (119868 minus 119862Γ minus 119863Δ119860Γ)

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

=

[

[

[

[

[

[

0

infin

sum

119899=0

119860119899119861 (119863119889)

119899+2

0

infin

sum

119899=1

119899minus1

sum

119894=0

119863119894119862119860119899minus119894minus1

119861 (119863119889)

119899+2

]

]

]

]

]

]

(64)

From (62) and (64) it follows (48) The proof is finished

The Scientific World Journal 7

Theorem 12 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119860120587119861119863120587 = 119861119863 119861119862 = 0119862119860119889= 0 and 119862119861119863120587 = 0 Then

119872119889=[

[

[

119860119889

(119860119889)

2

119861

1198830119863119889+

infin

sum

119899=0

119883119899+1119861119863119899

]

]

]

(65)

where

119883119899=

infin

sum

119894=0

(119863119889)

119894+119899+2

119862119860119894 119899 ge 0 (66)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [1198602119860119889119861

0 0

] 119876 = [

1198601198601205870

119862 119863]

119875119889= [119860119889(119860119889)

2

119861

0 0

] 119875120587= [119860120587minus119860119889119861

0 119868

]

(67)

Applying Lemma 7 we have (119860119860120587)119889 = 0 so we get

(119876119889)

119899

= [

0 0

119883119899minus1

(119863119889)

119899] 119876120587= [

119868 0

minus1198631198830119863120587] (68)

where119883119899is defined in (28)

Since 119860119860120587119861119863120587 = 119861119863 119861119862 = 0 119862119861119863120587 = 0 and 1198621198602119860119889 =0 we obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(69)

From 119862119860119889= 0 we have 119876119875119889 = 0 Hence from (69) we

obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

(70)

where119883119899119860119889= 0 we get

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889

(119876119889)

2

119875119875120587= [

0 0

1198831(119863119889)

2] [1198602119860119889119861

0 0

] [119860120587minus119860119889119861

0 119868

]

= [

0 0

0 1198831119861]

(71)

The conditions 119860119860120587119861119863120587 = 119861119863 and 119861119862 = 0 imply that119861119863119894119862 = 0 So we get

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

=[

[

[

0 0

0

infin

sum

119899=1

119883119899+1119861119863119899

]

]

]

119899 ge 1

(72)

From (71) and (72) it follows (65)The proof is finished

Using (40) andTheorem 12 we have the following result

Theorem 13 If 119862119860 = 119863120587119863119862119860120587 119861119863119889 = 0 119862119861 = 0 and

119861119862119860120587= 0 Then

119872119889=[

[

[

119860119889+

infin

sum

119899=0

119883119899+21198621198601198991198831

(119863119889)

2

119862 119863119889

]

]

]

(73)

where

119883119899=

infin

sum

119894=0

(119860119889)

119894+119899+1

119861119863119894 119899 ge 1 (74)

Using the case ofTheorem 3 we get the following results

Theorem 14 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119863119862119860120587 = 119862119860 119860119861119863 = 0119861119862 = 0 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+

infin

sum

119899=0

119861 (119863119889)

119899+3

119862119860119899(119860119889)

2

119861 + 119861 (119863119889)

2

infin

sum

119899=0

(119863119889)

119899+2

119862119860119899

119863119889

]

]

]

]

]

(75)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

0 0

119862 0] 119876 = [

119860 119861

0 119863] (76)

From 119860119861119863 = 0 we have

(119876119889)

119894

=[

[

(119860119889)

119894

119883119894

0 (119863119889)

119894]

]

119876120587= [119860120587minus1198601198831minus 119861119863119889

0 119863120587 ]

(77)

where

119883119899= (119860119889)

119899+2

119861 + 119861 (119863119889)

119899+2

119899 ge 0 (78)

Note that 119875 is quasinilpotent since 119863119862119860120587 = 119862119860 119860119861119863 =0 119861119862 = 0 and 119862119861 = 0 we obtain 119875119876 = 119876119875119876

120587 ApplyingTheorem 3 we get

(119875 + 119876)119889= 119876119889+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899 (79)

8 The Scientific World Journal

From 119861119862 = 0 we have

(119876119889)

2

119875 =[

[

(119860119889)

2

1198832

0 (119863119889)

2]

]

[

0 0

119862 0]

= [

1198832119862 0

(119863119889)

2

119862 0

] =[

[

119861 (119863119889)

3

0

(119863119889)

2

119862 0

]

]

(80)

The conditions 119863119862119860120587 = 119862119860 and 119862119861 = 0 imply that119862119860119894119861 = 0 From 119860119861119863 = 0 119862119860119894119861 = 0 and 119861119862 = 0 we getinfin

sum

119899=1

(119876119889)

119899+2

119875 (119875 + 119876)119899

=

[

[

[

[

[

infin

sum

119899=1

119883119899+2119862119860119899

infin

sum

119899=1

119883119899+2119862119860119899minus1119861

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899minus1119861

]

]

]

]

]

=

[

[

[

[

[

infin

sum

119899=1

119861 (119863119889)

119899+3

1198621198601198990

infin

sum

119899=1

(119863119889)

119899+2

1198621198601198990

]

]

]

]

]

(81)

From (77) (80) and (81) it follows (75)

Using (40) andTheorem 14 we have the following result

Theorem 15 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119861119863120587 = 119861119863 119863119862119860 = 0119861119862 = 119860119861119862119860

119889 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899

(119863119889)

2

119862 + 119862 (119860119889)

2

119863119889+

infin

sum

119899=0

119862 (119860119889)

119899+3

119861119863119899

]

]

]

]

]

(82)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J J Koliha ldquoA generalized Drazin inverserdquo Glasgow Mathemat-ical Journal vol 38 no 3 pp 367ndash381 1991

[2] C Deng D S Cvetkovic-Ilic and Y Wei ldquoSome results on thegeneralized Drazin inverse of operator matricesrdquo Linear andMultilinear Algebra vol 58 no 3-4 pp 503ndash521 2010

[3] E Dopazo and M F Martınez-Serrano ldquoFurther results on therepresentation of the Drazin inverse of a 2 times 2 block matrixrdquoLinear Algebra and its Applications vol 432 no 8 pp 1896ndash1904 2010

[4] J J Koliha and V Rakocevic ldquoHolomorphic and meromorphicproperties of the g-Drazin inverserdquoDemonstratio Mathematicavol 38 no 3 pp 657ndash666 2005

[5] N Castro Gonzalez ldquoAdditive perturbation results for theDrazin inverserdquo Linear Algebra and Its Applications vol 397 pp279ndash297 2005

[6] X Liu L Xu and Y Yu ldquoThe representations of the Drazininverse of differences of twomatricesrdquoAppliedMathematics andComputation vol 216 no 12 pp 3652ndash3661 2010

[7] D S Djordjevic and V Rakocevic Lectures on GeneralizedInverses University of Nis 2008

[8] J Meyer Jr and N J Rose ldquoThe index and the Drazininverse of block triangular matricesrdquo SIAM Journal on AppliedMathematics vol 33 no 1 pp 1ndash7 1977

[9] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[10] N Castro-Gonzalez and J J Koliha ldquoNew additive resultsfor the 119892-Drazin inverserdquo Proceedings of the Royal Society ofEdinburgh Section A Mathematics vol 134 no 6 pp 1085ndash1097 2004

[11] D S Cvetkovic-Ilic D S Djordjevic and Y Wei ldquoAdditiveresults for the generalized Drazin inverse in a Banach algebrardquoLinear Algebra and Its Applications vol 418 no 1 pp 53ndash612006

[12] D S Djordjevic and YWei ldquoAdditive results for the generalizedDrazin inverserdquo Journal of the Australian Mathematical Societyvol 73 no 1 pp 115ndash125 2002

[13] C Deng and Y Wei ldquoA note on the Drazin inverse of an anti-triangular matrixrdquo Linear Algebra and its Applications vol 431no 10 pp 1910ndash1922 2009

[14] N Castro-Gonzalez E Dopazo and M F Martınez-SerranoldquoOn the Drazin inverse of the sum of two operators andits application to operator matricesrdquo Journal of MathematicalAnalysis and Applications vol 350 no 1 pp 207ndash215 2009

[15] R E Hartwig G Wang and Y Wei ldquoSome additive results onDrazin inverserdquo Linear Algebra and its Applications vol 322 no1ndash3 pp 207ndash217 2001

[16] R E Hartwig and J Shoaf ldquoGroup inverses and Drazin inversesof bidiagonal and triangular Toeplitzmatricesrdquo vol 24 no 1 pp10ndash34 1977

[17] R E Hartwig and J M Shoaf ldquoGroup inverse and Drazininverse of bidiagonal and triangular Toeplitz matricesrdquo Journalof the AustralianMathematical Society A vol 24 pp 10ndash34 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

The Scientific World Journal 7

Theorem 12 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119860120587119861119863120587 = 119861119863 119861119862 = 0119862119860119889= 0 and 119862119861119863120587 = 0 Then

119872119889=[

[

[

119860119889

(119860119889)

2

119861

1198830119863119889+

infin

sum

119899=0

119883119899+1119861119863119899

]

]

]

(65)

where

119883119899=

infin

sum

119894=0

(119863119889)

119894+119899+2

119862119860119894 119899 ge 0 (66)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [1198602119860119889119861

0 0

] 119876 = [

1198601198601205870

119862 119863]

119875119889= [119860119889(119860119889)

2

119861

0 0

] 119875120587= [119860120587minus119860119889119861

0 119868

]

(67)

Applying Lemma 7 we have (119860119860120587)119889 = 0 so we get

(119876119889)

119899

= [

0 0

119883119899minus1

(119863119889)

119899] 119876120587= [

119868 0

minus1198631198830119863120587] (68)

where119883119899is defined in (28)

Since 119860119860120587119861119863120587 = 119861119863 119861119862 = 0 119862119861119863120587 = 0 and 1198621198602119860119889 =0 we obtain 119875119876 = 119875120587119876119875119876120587 ApplyingTheorem 4 we get

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587

+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

+ 119876120587

infin

sum

119899=0

(119875 + 119876)119899119876(119875119889)

119899+2

minus

infin

sum

119899=0

infin

sum

119896=0

(119876119889)

119896+1

119875 (119875 + 119876)119899+119896119876(119875119889)

119899+2

minus

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119876119875119889

(69)

From 119862119860119889= 0 we have 119876119875119889 = 0 Hence from (69) we

obtain

(119875 + 119876)119889= 119876120587119875119889+ 119876119889119875120587+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

(70)

where119883119899119860119889= 0 we get

119876120587119875119889= 119875119889 119876

119889119875120587= 119876119889

(119876119889)

2

119875119875120587= [

0 0

1198831(119863119889)

2] [1198602119860119889119861

0 0

] [119860120587minus119860119889119861

0 119868

]

= [

0 0

0 1198831119861]

(71)

The conditions 119860119860120587119861119863120587 = 119861119863 and 119861119862 = 0 imply that119861119863119894119862 = 0 So we get

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899119875120587

=[

[

[

0 0

0

infin

sum

119899=1

119883119899+1119861119863119899

]

]

]

119899 ge 1

(72)

From (71) and (72) it follows (65)The proof is finished

Using (40) andTheorem 12 we have the following result

Theorem 13 If 119862119860 = 119863120587119863119862119860120587 119861119863119889 = 0 119862119861 = 0 and

119861119862119860120587= 0 Then

119872119889=[

[

[

119860119889+

infin

sum

119899=0

119883119899+21198621198601198991198831

(119863119889)

2

119862 119863119889

]

]

]

(73)

where

119883119899=

infin

sum

119894=0

(119860119889)

119894+119899+1

119861119863119894 119899 ge 1 (74)

Using the case ofTheorem 3 we get the following results

Theorem 14 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119863119862119860120587 = 119862119860 119860119861119863 = 0119861119862 = 0 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889+

infin

sum

119899=0

119861 (119863119889)

119899+3

119862119860119899(119860119889)

2

119861 + 119861 (119863119889)

2

infin

sum

119899=0

(119863119889)

119899+2

119862119860119899

119863119889

]

]

]

]

]

(75)

Proof We can split matrix119872 as119872 = 119875 + 119876 where

119875 = [

0 0

119862 0] 119876 = [

119860 119861

0 119863] (76)

From 119860119861119863 = 0 we have

(119876119889)

119894

=[

[

(119860119889)

119894

119883119894

0 (119863119889)

119894]

]

119876120587= [119860120587minus1198601198831minus 119861119863119889

0 119863120587 ]

(77)

where

119883119899= (119860119889)

119899+2

119861 + 119861 (119863119889)

119899+2

119899 ge 0 (78)

Note that 119875 is quasinilpotent since 119863119862119860120587 = 119862119860 119860119861119863 =0 119861119862 = 0 and 119862119861 = 0 we obtain 119875119876 = 119876119875119876

120587 ApplyingTheorem 3 we get

(119875 + 119876)119889= 119876119889+

infin

sum

119899=0

(119876119889)

119899+2

119875 (119875 + 119876)119899 (79)

8 The Scientific World Journal

From 119861119862 = 0 we have

(119876119889)

2

119875 =[

[

(119860119889)

2

1198832

0 (119863119889)

2]

]

[

0 0

119862 0]

= [

1198832119862 0

(119863119889)

2

119862 0

] =[

[

119861 (119863119889)

3

0

(119863119889)

2

119862 0

]

]

(80)

The conditions 119863119862119860120587 = 119862119860 and 119862119861 = 0 imply that119862119860119894119861 = 0 From 119860119861119863 = 0 119862119860119894119861 = 0 and 119861119862 = 0 we getinfin

sum

119899=1

(119876119889)

119899+2

119875 (119875 + 119876)119899

=

[

[

[

[

[

infin

sum

119899=1

119883119899+2119862119860119899

infin

sum

119899=1

119883119899+2119862119860119899minus1119861

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899minus1119861

]

]

]

]

]

=

[

[

[

[

[

infin

sum

119899=1

119861 (119863119889)

119899+3

1198621198601198990

infin

sum

119899=1

(119863119889)

119899+2

1198621198601198990

]

]

]

]

]

(81)

From (77) (80) and (81) it follows (75)

Using (40) andTheorem 14 we have the following result

Theorem 15 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119861119863120587 = 119861119863 119863119862119860 = 0119861119862 = 119860119861119862119860

119889 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899

(119863119889)

2

119862 + 119862 (119860119889)

2

119863119889+

infin

sum

119899=0

119862 (119860119889)

119899+3

119861119863119899

]

]

]

]

]

(82)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J J Koliha ldquoA generalized Drazin inverserdquo Glasgow Mathemat-ical Journal vol 38 no 3 pp 367ndash381 1991

[2] C Deng D S Cvetkovic-Ilic and Y Wei ldquoSome results on thegeneralized Drazin inverse of operator matricesrdquo Linear andMultilinear Algebra vol 58 no 3-4 pp 503ndash521 2010

[3] E Dopazo and M F Martınez-Serrano ldquoFurther results on therepresentation of the Drazin inverse of a 2 times 2 block matrixrdquoLinear Algebra and its Applications vol 432 no 8 pp 1896ndash1904 2010

[4] J J Koliha and V Rakocevic ldquoHolomorphic and meromorphicproperties of the g-Drazin inverserdquoDemonstratio Mathematicavol 38 no 3 pp 657ndash666 2005

[5] N Castro Gonzalez ldquoAdditive perturbation results for theDrazin inverserdquo Linear Algebra and Its Applications vol 397 pp279ndash297 2005

[6] X Liu L Xu and Y Yu ldquoThe representations of the Drazininverse of differences of twomatricesrdquoAppliedMathematics andComputation vol 216 no 12 pp 3652ndash3661 2010

[7] D S Djordjevic and V Rakocevic Lectures on GeneralizedInverses University of Nis 2008

[8] J Meyer Jr and N J Rose ldquoThe index and the Drazininverse of block triangular matricesrdquo SIAM Journal on AppliedMathematics vol 33 no 1 pp 1ndash7 1977

[9] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[10] N Castro-Gonzalez and J J Koliha ldquoNew additive resultsfor the 119892-Drazin inverserdquo Proceedings of the Royal Society ofEdinburgh Section A Mathematics vol 134 no 6 pp 1085ndash1097 2004

[11] D S Cvetkovic-Ilic D S Djordjevic and Y Wei ldquoAdditiveresults for the generalized Drazin inverse in a Banach algebrardquoLinear Algebra and Its Applications vol 418 no 1 pp 53ndash612006

[12] D S Djordjevic and YWei ldquoAdditive results for the generalizedDrazin inverserdquo Journal of the Australian Mathematical Societyvol 73 no 1 pp 115ndash125 2002

[13] C Deng and Y Wei ldquoA note on the Drazin inverse of an anti-triangular matrixrdquo Linear Algebra and its Applications vol 431no 10 pp 1910ndash1922 2009

[14] N Castro-Gonzalez E Dopazo and M F Martınez-SerranoldquoOn the Drazin inverse of the sum of two operators andits application to operator matricesrdquo Journal of MathematicalAnalysis and Applications vol 350 no 1 pp 207ndash215 2009

[15] R E Hartwig G Wang and Y Wei ldquoSome additive results onDrazin inverserdquo Linear Algebra and its Applications vol 322 no1ndash3 pp 207ndash217 2001

[16] R E Hartwig and J Shoaf ldquoGroup inverses and Drazin inversesof bidiagonal and triangular Toeplitzmatricesrdquo vol 24 no 1 pp10ndash34 1977

[17] R E Hartwig and J M Shoaf ldquoGroup inverse and Drazininverse of bidiagonal and triangular Toeplitz matricesrdquo Journalof the AustralianMathematical Society A vol 24 pp 10ndash34 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 The Scientific World Journal

From 119861119862 = 0 we have

(119876119889)

2

119875 =[

[

(119860119889)

2

1198832

0 (119863119889)

2]

]

[

0 0

119862 0]

= [

1198832119862 0

(119863119889)

2

119862 0

] =[

[

119861 (119863119889)

3

0

(119863119889)

2

119862 0

]

]

(80)

The conditions 119863119862119860120587 = 119862119860 and 119862119861 = 0 imply that119862119860119894119861 = 0 From 119860119861119863 = 0 119862119860119894119861 = 0 and 119861119862 = 0 we getinfin

sum

119899=1

(119876119889)

119899+2

119875 (119875 + 119876)119899

=

[

[

[

[

[

infin

sum

119899=1

119883119899+2119862119860119899

infin

sum

119899=1

119883119899+2119862119860119899minus1119861

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899

infin

sum

119899=1

(119863119889)

119899+2

119862119860119899minus1119861

]

]

]

]

]

=

[

[

[

[

[

infin

sum

119899=1

119861 (119863119889)

119899+3

1198621198601198990

infin

sum

119899=1

(119863119889)

119899+2

1198621198601198990

]

]

]

]

]

(81)

From (77) (80) and (81) it follows (75)

Using (40) andTheorem 14 we have the following result

Theorem 15 Let 119860 and 119863 be generalized Drazin invertibleand let119872 be matrix of form (26) If 119860119861119863120587 = 119861119863 119863119862119860 = 0119861119862 = 119860119861119862119860

119889 and 119862119861 = 0 Then

119872119889=

[

[

[

[

[

119860119889

infin

sum

119899=0

(119860119889)

119899+2

119861119863119899

(119863119889)

2

119862 + 119862 (119860119889)

2

119863119889+

infin

sum

119899=0

119862 (119860119889)

119899+3

119861119863119899

]

]

]

]

]

(82)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J J Koliha ldquoA generalized Drazin inverserdquo Glasgow Mathemat-ical Journal vol 38 no 3 pp 367ndash381 1991

[2] C Deng D S Cvetkovic-Ilic and Y Wei ldquoSome results on thegeneralized Drazin inverse of operator matricesrdquo Linear andMultilinear Algebra vol 58 no 3-4 pp 503ndash521 2010

[3] E Dopazo and M F Martınez-Serrano ldquoFurther results on therepresentation of the Drazin inverse of a 2 times 2 block matrixrdquoLinear Algebra and its Applications vol 432 no 8 pp 1896ndash1904 2010

[4] J J Koliha and V Rakocevic ldquoHolomorphic and meromorphicproperties of the g-Drazin inverserdquoDemonstratio Mathematicavol 38 no 3 pp 657ndash666 2005

[5] N Castro Gonzalez ldquoAdditive perturbation results for theDrazin inverserdquo Linear Algebra and Its Applications vol 397 pp279ndash297 2005

[6] X Liu L Xu and Y Yu ldquoThe representations of the Drazininverse of differences of twomatricesrdquoAppliedMathematics andComputation vol 216 no 12 pp 3652ndash3661 2010

[7] D S Djordjevic and V Rakocevic Lectures on GeneralizedInverses University of Nis 2008

[8] J Meyer Jr and N J Rose ldquoThe index and the Drazininverse of block triangular matricesrdquo SIAM Journal on AppliedMathematics vol 33 no 1 pp 1ndash7 1977

[9] D S Djordjevic and P S Stanimirovic ldquoOn the generalizedDrazin inverse and generalized resolventrdquo Czechoslovak Math-ematical Journal vol 51 no 3 pp 617ndash634 2001

[10] N Castro-Gonzalez and J J Koliha ldquoNew additive resultsfor the 119892-Drazin inverserdquo Proceedings of the Royal Society ofEdinburgh Section A Mathematics vol 134 no 6 pp 1085ndash1097 2004

[11] D S Cvetkovic-Ilic D S Djordjevic and Y Wei ldquoAdditiveresults for the generalized Drazin inverse in a Banach algebrardquoLinear Algebra and Its Applications vol 418 no 1 pp 53ndash612006

[12] D S Djordjevic and YWei ldquoAdditive results for the generalizedDrazin inverserdquo Journal of the Australian Mathematical Societyvol 73 no 1 pp 115ndash125 2002

[13] C Deng and Y Wei ldquoA note on the Drazin inverse of an anti-triangular matrixrdquo Linear Algebra and its Applications vol 431no 10 pp 1910ndash1922 2009

[14] N Castro-Gonzalez E Dopazo and M F Martınez-SerranoldquoOn the Drazin inverse of the sum of two operators andits application to operator matricesrdquo Journal of MathematicalAnalysis and Applications vol 350 no 1 pp 207ndash215 2009

[15] R E Hartwig G Wang and Y Wei ldquoSome additive results onDrazin inverserdquo Linear Algebra and its Applications vol 322 no1ndash3 pp 207ndash217 2001

[16] R E Hartwig and J Shoaf ldquoGroup inverses and Drazin inversesof bidiagonal and triangular Toeplitzmatricesrdquo vol 24 no 1 pp10ndash34 1977

[17] R E Hartwig and J M Shoaf ldquoGroup inverse and Drazininverse of bidiagonal and triangular Toeplitz matricesrdquo Journalof the AustralianMathematical Society A vol 24 pp 10ndash34 1977

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

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Stochastic AnalysisInternational Journal of