1. No category

Generalized Drazin inverse of certain block matrices
in Banach algebras
Milica Z. Kolundˇzija, Dijana Mosi´c and Dragan S. Djordjevi´c∗
Abstract
Several representations of the generalized Drazin inverse of an antitriangular block matrix in Banach algebra are given in terms of the
generalized Banachiewicz–Schur form.
Key words and phrases: generalized Drazin inverse, Schur complement, block matrix.
2010 Mathematics subject classification: 46H05, 47A05, 15A09.
1
Introduction
The Drazin inverse plays an important role in Markov chains, singular differential and difference equations, iterative methods in numerical linear algebra, etc. Representations for the Drazin inverse of block matrices under
certain conditions where given in the literature [2, 3, 4, 10, 11, 12, 14, 15, 17].
Deng [7] investigated necessary and sufficient conditions for a partitioned
operator matrix on a Hilbert space to have the Drazin inverse with the generalized Banachiewicz–Schur form. In [8], a representation for the Drazin inverse of an anti-triangular block matrix under some conditions was obtained,
generalizing in different ways results from [6, 14]. Block anti-triangular matrices arise in numerous applications, ranging form constrained optimization
problems to solution of differential equations, etc. Deng [9] presented some
formulas for the
Drazin inverse of an anti-triangular operator
· generalized
¸
A B
matrix M =
, acting on a Banach space, with the assumption
C 0
that CAd B is invertible.
∗
The authors are supported by the Ministry of Education and Science, Republic of
Serbia, grant no. 174007.
1
In this paper, we study the generalized Drazin inverse of anti-triangular
matrices in a Banach space, getting as particular cases recent results from
[7, 8, 9].
Let A be a complex unital Banach algebra with unit 1. For a ∈ A, we
use σ(a), r(a) and ρ(a), respectively, to denote the spectrum, the spectral
radius and the resolvent set of a. The sets of all invertible, nilpotent and
quasinilpotent elements (σ(a) = {0}) of A will be denoted by A−1 , Anil and
Aqnil , respectively.
The generalized Drazin inverse of a ∈ A (or Koliha–Drazin inverse of a)
is the element b ∈ A which satisfies
bab = b,
ab = ba,
a − a2 b ∈ Aqnil .
If the generalized Drazin inverse of a exists, it is unique and denoted by
ad , and a is generalized Drazin invertible. The set of all generalized Drazin
invertible elements of A is denoted by Ad . The Drazin inverse is a special
case of the generalized Drazin inverse for which a − a2 b ∈ Anil insteed of
a − a2 b ∈ Aqnil . Obviously, if a is Drazin invertible, then it is generalized
Drazin invertible. The group inverse is the Drazin inverse for which the
conditon a − a2 b ∈ Anil is replaced with a = aba. We use a# to denote the
group inverse of a, and we use A# to denote the set of all group invertible
elements of A.
Recall that a ∈ A is generalized Drazin invertible if and only if there
exists an idempotent p = p2 ∈ A such that
ap = pa ∈ Aqnil ,
a + p ∈ A−1 .
Then p = 1 − aad is the spectral idempotent of a corresponding to the set
{0}, and it will be denoted by aπ . The generalized Drazin inverse ad double
commutes with a, that is, ax = xa implies ad x = xad .
Let p = p2 ∈ A be an idempotent. Then we can represent element a ∈ A
as
·
¸
a11 a12
a=
,
a21 a22
where a11 = pap, a12 = pa(1 − p), a21 = (1 − p)ap, a22 = (1 − p)a(1 − p).
We use the following lemmas.
Lemma 1.1. [5, Lemma 2.4] Let b, q ∈ Aqnil and let qb = 0. Then q + b ∈
Aqnil .
Lemma 1.2. Let b ∈ Ad and a ∈ Aqnil .
2
(i) [5, Corollary 3.4] If ab = 0, then a+b ∈ Ad and (a+b)d =
+∞
P
(bd )n+1 an .
n=0
(ii) If ba = 0, then a + b ∈ Ad and (a + b)d =
+∞
P
an (bd )n+1 .
n=0
Specializing [5, Corollary 3.4] (with multiplication reversed) to bounded
linear operators N. Castro Gonz´alez and J. J. Koliha [5] recovered [13, Theorem 2.2] which is a spacial case of Lemma 1.2(ii).
Lemma 1.3. Let A be a complex unital Banach algebra with unit 1, and let
p be an idempotent of A. If x ∈ pAp, then σpAp (x) ∪ {0} = σA (x), where
σA (x) denotes the spectrum of x in the algebra A, and σpAp (x) denotes the
spectrum of x in the algebra pAp.
·
¸
a b
Let x =
∈ A relative to the idempotent p ∈ A. It is well
c d
known that if a ∈ (pAp)−1 and the Schur complement s = d − ca−1 b ∈
((1 − p)A(1 − p))−1 , then the inverse of x has Banachiewicz–Schur form
· −1
¸
a + a−1 bs−1 ca−1 −a−1 bs−1
−1
x =
.
−s−1 ca−1
s−1
We investigate equivalent conditions under which xd has the generalized
Banachiewicz–Schur form in Banach algebra. Also, we obtain several representations
·
¸for the generalized Drazin inverse of an anti-triangular matrix
a b
x=
under different conditions. Thus, we extend some results from
c 0
[7, 8, 9] to more general settings.
2
Results
In the following·lemma,¸ we present necessary and sufficient conditions for
a b
an element x =
of Banach algebra to have the generalized Drazin
c d
inverse with the generalized Banachiewicz–Schur form. We recover recent
result concerning the Drazin inverse of Hilbert space operators (see [7, Corollary 3]).
·
¸
a b
Lemma 2.1. Let x =
∈ A relative to the idempotent p ∈ A,
c d
a ∈ (pAp)# , and let s = d − ca# b ∈ ((1 − p)A(1 − p))# be the generalized
Schur complement of a in x. Then the following statements are equivalent:
3
(i) x ∈ Ad and
·
¸
a# + a# bs# ca# −a# bs#
x =
;
(1)
−s# ca#
s#
·
¸
0 bsπ
π
#
#
π
π
#
#
π
and z =
∈ Aqnil ;
(ii) a bs = a bs , s ca = s ca
caπ 0
·
¸
0 aπ b
π
π
π
π
(iii) a b = bs , s c = ca
and z =
∈ Aqnil .
sπ c 0
d
Proof. (i) ⇔ (ii): If the right side of (1) is denoted by y, then we obtain
·
¸
aa# − aπ bs# ca# aπ bs#
xy =
,
sπ ca#
ss#
· #
¸
a a − a# bs# caπ a# bsπ
yx =
.
s# caπ
s# s
So, xy = yx if and only if aπ bs# = a# bsπ and sπ ca# = s# caπ , because
these equalities imply (aπ bs# )ca# = a# b(sπ ca# ) = a# bs# caπ . Further,
we can verify that yxy = y. Using s = d − ca# b, aπ bs# = a# bsπ and
sπ ca# = s# caπ , we have
·
¸
−bs# caπ bsπ
2
x−x y =
.
caπ
0
From a# bsπ = aπ bs# = (p − aa# )bs# = bs# − aa# bs# , we obtain bs# =
a# bsπ + aa# bs# which gives caπ bs# = 0 = bs# caπ bs# and
·
¸ ·
¸
p −bs#
p bs#
2
x−x y =
z
.
0 1−p
0 1−p
·
¸·
¸
p bs#
p −bs#
2
Since r(x − x y) = r(
z) = r(z), we deduce that
0 1−p
0 1−p
x − x2 y ∈ Aqnil is equivalent to z ∈ Aqnil .
(ii) ⇔ (iii): We prove that aπ bs# = a# bsπ is equivalent to aπ b = bsπ .
Indeed, multiplying aπ bs# = a# bsπ from the right side by s and from the left
side by a, respectively, we obtain aπ bs# s = 0 and aa# bsπ = 0. Therefore,
bs# s = aa# bs# s = aa# b and
aπ b = b − aa# b = b − bs# s = bsπ .
On the other hand, if aπ b = bsπ , then (aπ b)s# = bsπ s# = 0 and a# (bsπ ) =
a# aπ b = 0, i.e. aπ bs# = a# bsπ .
In the same manner, we can verify that sπ ca# = s# caπ is equivalent to
sπ c = caπ . Hence, the equivalence (ii) ⇔ (iii) holds.
4
By Lemma 2.1, the following corollary recovers [1, Theorem 2]
·
¸
a b
Corollary 2.1. Let x =
∈ A relative to the idempotent p ∈ A,
c d
a ∈ (pAp)# , and let s = d − ca# b ∈ ((1 − p)A(1 − p))# be the generalized
Schur complement of a in x. Then x ∈ A# and
¸
· #
a + a# bs# ca# −a# bs#
#
x =
−s# ca#
s#
if and only if
aπ b = 0 = bsπ ,
sπ c = 0 = caπ .
Now, we extend the well known result concerning the Drazin inverse
of complex matrices to the generalized Drazin inverse of Banach algebra
elements, see [8, Theorem 3.5].
Theorem 2.1. Let
·
x=
a b
c 0
¸
∈A
(2)
relative to the idempotent p ∈ A, a ∈ (pAp)d and let s = −cad b ∈ ((1 −
p)A(1 − p))d . If
ssd caπ b = 0,
ssd caπ a = 0,
aad bsπ c = 0,
bsπ caπ = 0,
(3)
then x ∈ Ad and
Ã
x
d
!
¸n ·
¸
+∞ ·
X
aaπ aπ bsπ
0
aπ bssd
n+2
=
r+
r
sπ caπ
0
sπ caad
0
n=0
µ
·
¸¶
0
aad bsπ
×
1+r
,
ssd caπ
0
where
·
r=
ad + ad bsd cad −ad bsd
−sd cad
sd
¸
.
Proof. Applying aad + aπ = p and ssd + sπ = 1 − p, we have
· 2 d
¸ ·
¸
a a aad b
aaπ aπ b
x=
+
:= u + v.
sπ c 0
ssd c
0
The equalities ad aπ = 0 and (3) give uv = 0.
5
(4)
(5)
First, we show that u ∈ Ad . If we write
·
¸ ·
¸
a2 ad
aad bssd
0
aad bsπ
u=
+
:= u1 + u2 ,
ssd caad
0
ssd caπ
0
2 d
d
d
we can get u2 u1 = 0 and u22 = 0. Let
· Au1 ≡ a a¸ , Bu1 ≡ aa bss , Cu1 ≡
Au1 Bu1
ssd caad and Du1 ≡ 0. Then u1 =
and, by (a2 ad )# = ad ,
Cu1 Du1
2 d
Au1 ∈ (pAp)# . Also, from s = −cad b, Su1 ≡ Du1 − Cu1 A#
u1 Bu1 = s s ∈
#
((1 − p)A(1 − p))# and (s2 sd )# = sd . Consequently, Aπu1 Bu1 Su1 = 0 =
·
¸
0
Bu1 Suπ1
#
#
#
π
π
π
=0∈
Au1 Bu1 Su1 , Su1 Cu1 Au1 = 0 = Su1 Cu1 Au1 and
Cu1 Aπu1
0
Aqnil . By Lemma 2.1, notice that u1 ∈ Ad and
"
#
#
#
#
#
#
#
A
+
A
B
S
C
A
−A
B
S
u
u
u
u
u
u
u
u
u
1
1
1
1
1
1
1
1
1
ud1 =
= r.
−Su#1 Cu1 A#
Su#1
u1
Using Lemma 1.2(i), u ∈ Ad and ud = ud1 + (ud1 )2 u2 = r + r2 u2 .
To prove that v ∈ Aqnil , observe that
·
¸ ·
¸ ·
¸
aaπ aπ bsπ
0
0
0
aπ bssd
v =
+
+
0
0
sπ caπ 0
sπ caad
0
:= v1 + v2 + v3 .
·
¸
·
¸
m t
λp − m
−t
If z =
, then λ1 − z =
. Therefore
0 n
0
λ(1 − p) − n
λ ∈ ρpAp (m) ∩ ρ(1−p)A(1−p) (n) ⇒ λ ∈ ρ(z),
i.e.
σ(z) ⊆ σpAp (m) ∪ σ(1−p)A(1−p) (n).
Notice that, by aaπ ∈ (pAp)qnil , v1 ∈ Aqnil . It can be verified that
v1 v2 = 0 and v22 = 0, i.e. v2 ∈ Anil . Now, by Lemma 1.1, v1 + v2 ∈ Aqnil .
Using Lemma 1.1 again, from v32 = 0 and v3 (v1 + v2 ) = 0, we conclude that
v ∈ Aqnil .
Applying Lemma 1.2(ii), we deduce that x ∈ Ad and
!
Ã
!
Ã
+∞
+∞
X
X
v n+1 (ud )n+2 ud = 1 +
v n+1 (ud )n+2 r(1 + ru2 ).
xd = 1 +
n=0
n=0
6
Since u2 r = u2 ud1 = (u2·u1 )(ud1 )2 =¸0, then (ud )n+2 = ·(r + r2 u2 )n+2
¸ =
d
d
aa
0
aa
0
rn+2 (1 + ru2 ). From r =
r, we obtain vr = v
r=
0 ssd
0 ssd
·
¸
0
aπ bssd
. By v n+1 = (v1 + v2 )n v, we have v n+1 (ud )n+2 = (v1 +
sπ caad
0
·
¸
0
aπ bssd
n
v2 )
rn+1 (1 + ru2 ). Applying u2 r = 0 again, we get
sπ caad
0
(4).
From Theorem 2.1, we get the following consequence.
Corollary 2.2. Let x be defined as in (2), a ∈ (pAp)d and let r be defined
as in (5).
(1) If caπ = 0 and the generalized Schur complement s = −cad b is invertible, then x ∈ Ad and
¸n ·
¸
+∞ ·
X
aaπ 0
0 aπ b
d
x =r+
rn+2 .
0 0
0 0
n=0
(2) If caπ = 0, aπ b = 0 and the generalized Schur complement s = −cad b
is invertible, then x ∈ Ad and
· d
¸
a + ad bs−1 cad −ad bs−1
xd =
.
−s−1 cad
s−1
(3) If caπ b = 0, caπ a = 0 and the generalized Schur complement s = −cad b
is invertible, then x ∈ Ad and
Ã
!µ
¸
·
¸¶
+∞ ·
n aπ b
X
0
a
0 0
d
n+2
x = r+
r
1+r
.
0
0
caπ 0
n=0
In the following theorems, we assume that s = −cad b is the generalized
Drazin invertible, and we prove representations of the generalized Drazin
inverse of anti-triangular block matrices. Several results from [9] are extended.
Theorem 2.2. Let x be defined as in (2), a ∈ (pAp)d and let s = −cad b ∈
((1 − p)A(1 − p))d . If bcaπ = 0 and aad bsπ = 0, then x ∈ Ad and
¸n µ
·
¸ ¶
+∞ ·
X
aaπ aπ b
0 0
xd =
1
+
r rn+1 ,
(6)
caπ
0
sπ c 0
n=0
where r be defined as in (5).
7
Proof. We can write
· 2 d
¸ ·
¸
a a aad b
aaπ aπ b
x=
+
:= y + q.
caπ
0
caad
0
Now, we get yq = 0, by the assumption bcaπ = 0.
In order to prove that y ∈ Ad , note that
¸ ·
¸
·
0
aad bsπ
a2 ad
aad bssd
+
y =
sπ caad
0
ssd caad
0
·
¸
·
¸
2
d
d
d
a a
aa bss
0
0
=
+
:= y1 + y2 ,
sπ caad 0
ssd caad
0
y1 y2 = 0 and y22 = 0. Using Lemma 2.1, we have y1 ∈ Ad and y1d = r. By
Lemma 1.2(ii), y ∈ Ad and y d = y1d + y2 (y1d )2 = r + y2 r2 .
Further, we verify that q ∈ Aqnil . Let
·
¸ ·
¸
aaπ aπ b
0 0
q =
+
:= q1 + q2 .
0
0
caπ 0
Thus, we deduce that q1 ∈ Aqnil and q2 ∈ Anil , because aaπ ∈ (pAp)qnil and
q22 = 0. Since q1 q2 = 0, by Lemma 1.1, q ∈ Aqnil .
By Lemma 1.2(ii), x ∈ Ad and
xd =
+∞
X
q n (y d )n+1 =
n=0
·
The equality r =
aad 0
0 ssd
+∞
X
q n (1 + y2 r)rn+1 .
n=0
¸
·
r give y2 r =
0 0
sπ c 0
¸
r implying (6).
Replacing the hypothesis aad bsπ = 0 with sπ caad = 0 in Theorem 2.2,
we get the following theorem.
Theorem 2.3. Let x be defined as in (2), a ∈ (pAp)d and let s = −cad b ∈
((1 − p)A(1 − p))d . If bcaπ = 0 and sπ caad = 0, then x ∈ Ad and
¸n
µ
·
¸¶
+∞ ·
X
aaπ aπ b
0 bsπ
n+1
r
1+r
,
x =
caπ
0
0 0
d
n=0
where r is defined in the same way as in (5).
8
(7)
Proof. In the similar way as in the proof of Theorem 2.2, using
·
¸ ·
¸
a2 ad
aad bssd
0 aad bsπ
y=
+
:= y1 + y2
0
0
ssd caad
0
and y2 y1 = 0, we check this theorem.
If s = −cad b ∈ ((1 − p)A(1 − p))−1 and s0 = −s, then sπ = 0 and
(s0 )−1 = −s−1 . As a special case of Theorem 2.2 (or Theorem 2.3), we
obtain the following result which recovers [9, Theorem 3.1] for bounded
linear operators on a Banach space.
Corollary 2.3. Let x be defined as in (2), a ∈ (pAp)d and let s0 = cad b ∈
((1 − p)A(1 − p))−1 . If bcaπ = 0, then x ∈ Ad and
¸n
+∞ ·
X
aaπ aπ b
x =
tn+1 ,
caπ
0
d
n=0
·
where t =
ad − ad b(s0 )−1 cad ad b(s0 )−1
(s0 )−1 cad
−(s0 )−1
¸
.
Sufficient conditions under which the generalized Drazin inverse xd is
represented by (6) or (7) are investigated in the following result.
Theorem 2.4. Let x be defined as in (2), a ∈ (pAp)d and let s = −cad b ∈
((1 − p)A(1 − p))d . Suppose that aad bcaπ = 0 and caπ b = 0.
(1) If aad bsπ = 0 and (aaπ b = 0 or caaπ = 0), then x ∈ Ad and (6) is
satisfied.
(2) If sπ caad = 0 and (aaπ b = 0 or caaπ = 0), then x ∈ Ad and (7) is
satisfied.
Proof. This result can be proved similarly as Theorem 2.2 and Theorem 2.3,
applying q2 q1 = 0 when caaπ = 0, and the decomposition
·
¸ ·
¸
aaπ 0
0 aπ b
q=
+
caπ 0
0 0
when aaπ b = 0.
Remark. In the preceding theorem, if cad b ∈ ((1 − p)A(1 − p))−1 , then
we obtain as a particular case [9, Theorem 3.2] for Banach space operator.
The following result is well-known for complex matrices (see [16]).
9
·
¸
a b
∈ A relative to the idempotent p ∈ A,
Lemma 2.2. Let x =
c d
a ∈ (pAp)d and let w = aad + ad bcad be such that aw ∈ (pAp)d . If caπ = 0,
aπ b = 0 and the generalized Schur complement s = d − cad b is equal to 0,
then
·
¸·
¸·
¸
p 0
[(aω)d ]2 a 0
p ad b
xd =
.
(8)
cad 0
0
0
0 0
Proof. Denote by y the right side of (8). Then we obtain
·
¸
(a + bcad )[(aω)d ]2 a (a + bcad )[(aω)d ]2 b
xy =
,
(c + dcad )[(aω)d ]2 a (c + dcad )[(aω)d ]2 b
·
[(aω)d ]2 (a2 + bc)
[(aω)d ]2 (ab + bd)
d
d
2
2
ca [(aω) ] (a + bc) cad [(aω)d ]2 (ab + bd)
yx =
¸
.
By caπ = 0 and aπ b = 0, we can conclude that a + bcad commutes with aω.
Indeed,
(a + bcad )(aω) = (a2 + bcad a)(aad + ad bcad ) = (a2 + aad bc)ad (a + bcad )
= (aω)(a + bcad ).
Since a + bcad commutes with aω, it also commutes with (aω)d and we have
(a + bcad )[(aω)d ]2 a = [(aω)d ]2 (a + bcad )a = [(aω)d ]2 (a2 + bc).
From s = 0, we get c + dcad = cad a + cad bcad = cad (a + bcad ). Thus,
(c + dcad )[(aω)d ]2 a = cad (a + bcad )[(aω)d ]2 a = cad [(aω)d ]2 (a2 + bc).
Also, ab + bd = ab + bcad b = (a + bcad )b and we obtain
(a + bcad )[(aω)d ]2 b = [(aω)d ]2 (ab + bd)
(c + dcad )[(aω)d ]2 b = cad [(aω)d ]2 (ab + bd).
So, we proved that
·
xy = yx =
[(aω)d ]2 (a + bcad )a
[(aω)d ]2 (a + bcad )b
d
d
2
d
ca [(aω) ] (a + bca )a cad [(aω)d ]2 (a + bcad )b
10
¸
.
Further, we can verify that yxy = y. Indeed, we have
·
¸
[(aω)d ]2 a
[(aω)d ]2 b
yxy =
cad [(aω)d ]2 a cad [(aω)d ]2 b
·
¸
[(aω)d ]2 (a + bcad )a
[(aω)d ]2 (a + bcad )b
×
cad [(aω)d ]2 (a + bcad )a cad [(aω)d ]2 (a + bcad )b
·
¸
[(aω)d ]4 (a + bcad )2 a
[(aω)d ]4 (a + bcad )2 b
=
.
cad [(aω)d ]4 (a + bcad )2 a cad [(aω)d ]4 (a + bcad )2 b
The equalities a+bcad = a−a2 ad +a2 ad +bcad = aaπ +aω and aπ ω = 0 = ωaπ
give (a + bcad )2 = a2 aπ + (aω)2 . Therefore,
(aω)d (a + bcad )2 = (aω)d (a2 aπ + (aω)2 )
= [(aω)d ]2 (aω)aπ a2 + (aω)d (aω)2 = (aω)d (aω)2
and [(aω)d ]4 (a + bcad )2 = [(aω)d ]4 (aω)2 = [(aω)d ]2 implying
·
¸
[(aω)d ]2 a
[(aω)d ]2 b
yxy =
= y.
cad [(aω)d ]2 a cad [(aω)d ]2 b
We obtain
·
¸ ·
¸·
¸
(aω)π a
(aω)π b
p 0
(aω)π a (aω)π b
x − x2 y =
=
.
0
0
cad (aω)π a cad (aω)π b
cad 0
Notice that, by a + bcad = aaπ + aω, (aω)π (a + bcad ) = aaπ + (aω)(aω)π .
Since aaπ , (aω)(aω)π ∈ (pAp)qnil and aaπ (aω)(aω)π = 0, by Lemma 1.1, we
have that aaπ + (aω)(aω)π ∈ (pAp)qnil and rpAp ((aω)π (a + bcad )) = 0. From
µ·
¸·
¸¶
p 0
(aω)π a (aω)π b
2
r(x − x y) = r
0
0
cad 0
µ·
¸¶
(aω)π (a + bcad ) 0
=r
= rpAp ((aω)π (a + bcad )) = 0,
0
0
we deduce that x − x2 y ∈ Aqnil and prove that xd = y.
In the following theorem, we extend [9, Theorem 3.3 and Theorem 3.4]
for Banach space operators to elements of a Banach algebra.
Theorem 2.5. Let x be defined as in (2), a ∈ (pAp)d and let k = a2 ad +
aad bcad ∈ (pAp)d . If cad b = 0 and if one of the following conditions holds:
11
(1) bcaπ = 0;
(2) aad bcaπ = 0, aaπ b = 0 and caπ b = 0;
(3) aad bcaπ = 0, caaπ = 0 and caπ b = 0;
then x ∈ Ad and
xd =
¸n ·
¸n+1
+∞ ·
X
aaπ aπ b
(k d )2 a
(k d )2 b
.
caπ
0
cad (k d )2 a cad (k d )2 b
(9)
n=0
Proof. To prove the part (1) suppose that x = y + q, where s and y are
defined as in the proof of Theorem 2.2. It follows that yq = 0 and q ∈ Aqnil .
Applying Lemma 2.2, we conclude that y ∈ Ad and
·
¸· d 2 2 d
¸·
¸
p 0
(k ) a a 0
p ad b
d
y =
.
cad 0
0
0
0 0
Since kaad = k, then k d aad = k d and
·
¸
(k d )2 a
(k d )2 b
d
y =
.
cad (k d )2 a cad (k d )2 b
Using Lemma 1.2(ii), we conclude that x ∈ Ad and xd =
+∞
P
q n (y d )n+1 .
n=0
Thus, (9) holds.
The parts (2) and (3) can be checked in the similar manner as in the
part (1) and in the proof of Theorem 2.4.
kd
If c = 0 or b = 0 in Theorem 2.5, we have k = a2 ad ∈ (pAp)d and
= ad . As a consequence of Theorem 2.5, we obtain the following result.
Corollary 2.4. Let x be defined as in (2) and let a ∈ (pAp)d .
(1) If c = 0, then x ∈ Ad and
·
ad (ad )2 b
0
0
d
x =
¸
.
(2) If b = 0, then x ∈ Ad and
·
d
x =
ad
0
d
2
c(a ) 0
12
¸
.
References
[1] J. Ben´ıtez, N. Thome, The generalized Schur complement in group
inverses and (k + 1)-potent matrices Linear and Multilinear Algebra
54 (2006) 405-413.
[2] S. L. Campbell, The Drazin inverse of an infinite matrix, SIAM J.
Appl. Math. 31 (1976) 492-503.
[3] S. L. Campbell, The Drazin inverse of an operator, in Recent Applications of Generalized Inverses, Ed. by S.L. Campbell, Pitman Publishing Company, 1982.
[4] N. Castro-Gonzalez, E. Dopazo, J. Robles, Formulas for the Drazin
inverse of special block matrices, Appl. Math. Comput. 174 (2006)
252-270.
[5] N. Castro Gonz´alez, J. J. Koliha, New additive results for the g-Drazin
inverse, Proc. Roy. Soc. Edinburgh Sect. A 134 (2004) 1085-1097.
[6] C. G. Cao, Representations of the group inverse of some 2 × 2 block
matrices, International Mathematical Forum 1 (2006) 1511–1517.
[7] C. Deng, A note on the Drazin inverses with Banachiewicz-Schur
forms, Appl. Math. Comput. 213 (2009) 230-234.
[8] C. Deng, A comment on some recent results concerning the Drazin
inverse of an anti-triangular block matrix, Filomat 26:2 (2012) 135–
145.
[9] C. Y. Deng, Generalized Drazin inverses of anti-triangular block matrices J. Math. Anal. Appl. 368 (2010) 1-8.
[10] C. Deng, Y.Wei, A note on the Drazin inverse of an anti-triangular
matrix, Linear Algebra Appl. 431 (2009) 1910-1922.
[11] C. Deng, Y. Wei, Representations for the Drazin inverse of 2×2 blockoperator matrix with singular Schur complement, Linear Algebra and
its Applications 435 (2011) 2766-2783
[12] D. S. Djordjevi´c, P. S. Stanimirovi´c, On the generalized Drazin inverse
and generalized resolvent, Czechoslovak Math. J. 51 (2001) 617634.
[13] D. S. Djordjevi´c, Y. Wei, Additive results for the generalized Drazin
inverse, J. Austral. Math. Soc. 73 (2002) 115125.
13
[14] R. E. Hartwig, X. Li, Y. Wei, Representations for the Drazin inverse
of 2 × 2 block matrix, SIAM J. Matrix Anal. Appl. 27 (2006) 757-771.
[15] X. Li, Y. Wei, A note on the representations for the Drazin inverse of
2 × 2 block matrix, Linear Algebra Appl. 423 (2007) 332-338.
[16] J. Miao, Results of the Drazin inverse of block matrices, J. Shanghai
Normal Univ. 18 (1989) 25-31.
[17] Y. Wei, Expressions for the Drazin inverse of 2×2 block matrix, Linear
and Multilinear Algebra 45 (1998) 131-146.
Address:
Faculty of Sciences and Mathematics, University of Niˇs, P.O. Box 224,
18000 Niˇs, Serbia
E-mail
M. Z. Kolundˇzija: [email protected]
D. Mosi´c: [email protected]
D. S. Djordjevi´c: [email protected]
14