Filomat 28:9 (2014), 1773–1781
DOI 10.2298/FIL1409773Z
Published by Faculty of Sciences and Mathematics,
University of Niš, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
Additive Property of Pseudo Drazin Inverse of
Elements in Banach Algebras
Huihui Zhua,b , Jianlong Chena
a Department
b CMAT-Centro
of Mathematics, Southeast University, Nanjing 210096, China
de Matemática, Universidade do Minho, Braga 4710-057, Portugal
Abstract. This article concerns the pseudo Drazin inverse of the sums (resp. differences) and the products
of elements in a Banach algebra A . Some equivalent conditions for the existence of the pseudo Drazin
inverse of a + b (resp. a − b) are characterized. Moreover, the representations for the pseudo Drazin inverse
are given. Some related known results are generalized.
1. Introduction
Throughout this paper, A is a complex Banach algebra with unity 1. The symbols J(A ), A # , A nil denote
the Jacobson radical, the sets of all group invertible, nilpotent elements of A , respectively. An element
a ∈ A is said to have a Drazin inverse [10] if there exists b ∈ A satisfying
ab = ba, bab = b, a − a2 b ∈ A nil .
The element b above is unique and is denoted by aD , and the nilpotency index of a − a2 b is called the Drazin
index of a, denoted by ind(a). If ind(a) = 1, then b is the group inverse of a and is denoted by a# . In 2012,
Wang and Chen [16] introduced the notion of pseudo Drazin inverse (abbr. p-Drazin inverse) in associative
rings and Banach algebras. An element a ∈ A is called p-Drazin invertible if there exists b ∈ A such that
ab = ba, bab = b, ak − ak+1 b ∈ J(A ) for some integer k > 1.
Any element b ∈ A satisfying the conditions above is called a p-Drazin inverse of a, denoted by a‡ . The set
of all p-Drazin invertible elements of A is denoted by A pD . By aΠ = 1 − aa‡ we mean the strongly spectral
idempotent of a.
Representations for the Drazin inverse of the sums and the products of two elements in certain algebras
have attracted wide interest. In general, it is a challenging task to characterize the Drazin inverse of a + b
or ab without additional hypothesis. Given a and b in a ring R with Drazin inverses aD and bD , respectively.
2010 Mathematics Subject Classification. 15A09, 15A27, 16H99
Keywords. pseudo Drazin inverse, strongly spectral idempotent, Jacobson radical.
Received: 25 July 2013; Accepted: 15 September 2014
Communicated by Dragana Cvetković-Ilić
Research supported by the National Natural Science Foundation of China (No. 11201063 and 11371089), the Specialized Research
Fund for the Doctoral Program of Higher Education (20120092110020), the Natural Science Foundation of Jiangsu Province (No.
BK20141327), the Foundation of Graduate Innovation Program of Jiangsu Province(CXLX13-072) and the Fundamental Research
Funds for the Central Universities (22420135011).
Email addresses: ahzhh08@sina.com (Huihui Zhu), jlchen@seu.edu.cn (Jianlong Chen)
H. Zhu, J. Chen / Filomat 28:9 (2014), 1773–1781
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If ab = ba = 0 then it follows that a + b is Drazin invertible with (a + b)D = aD + bD (see [10, Corollary 1]).
Wei and Deng [17] presented the expressions on the Drazin inverse of two commutative square complex
matrices. Later, Zhuang, Chen et al. [18] extended the results in [17] to the ring case and proved that a + b
is Drazin invertible if and only if 1 + aD b is Drazin invertible. Further, Deng [7] explored the Drazin inverse
in the ring B(X) of bounded operators of a Banach space X. Under the condition that PQ = λQP, they gave
explicit representations of the Drazin inverses (P − Q)D (resp. (P + Q)D ) and (PQ)D in terms of P, PD , Q and
QD . More results on (generalized) Drazin inverse can be found in [1-6, 8, 9, 11-15].
This article is motivated by Deng [7], Wei, Deng [17] and Zhuang et al. [18]. We investigate the
representations for p-Drazin inverse of the sums (resp. differences) and the products of two elements in a
Banach algebra. Moreover, some equivalent conditions for existence of p-Drazin inverse of the sums and
the differences of two elements in a Banach algebra are given.
2. p-Drazin Inverse Under the Condition ab = ba
In this section, we give some elementary properties on p-Drazin inverse.
Lemma 2.1. ([16, Proposition 5.2] ) Let a, b ∈ A . If ab = ba and a‡ , b‡ exist, then (ab)‡ = a‡ b‡ = b‡ a‡ . Moreover,
ab‡ = b‡ a.
In particular, if a ∈ A pD , then (a2 )‡ = (a‡ )2 by Lemma 2.1.
Lemma 2.2. Let a, b ∈ A . The following statements hold:
(1) If a ∈ J(A ), then ab, ba ∈ J(A ).
(2) If a, b ∈ J(A ), then (a + b)k ∈ J(A ) for integer k > 1.
In [10], some properties of Drazin inverse were presented. One may suspect that if the similar properties
can be inherited to p-Drazin inverse in a Banach algebra. The following result illustrates the possibility.
Theorem 2.3. Let a ∈ A pD . Then
(1) (an )‡ = (a‡ )n , n = 1, 2, · · · .
(2) (a‡ )‡ = a2 a‡ .
(3) ((a‡ )‡ )‡ = a‡ .
(4) a‡ (a‡ )‡ = aa‡ .
Proof. (1) It is obvious when n = 1.
Assume the result holds for n − 1, i.e., (an−1 )‡ = (a‡ )n−1 .
For n, by Lemma 2.1, we have (an )‡ = (aan−1 )‡ = a‡ (an−1 )‡ = a‡ (a‡ )n−1 = (a‡ )n .
Hence, an is p-Drazin invertible and (an )‡ = (a‡ )n .
(2) It is easy to check a‡ a2 a‡ = a2 a‡ a‡ and a2 a‡ a‡ a2 a‡ = a2 a‡ .
Since (a‡ )k −(a‡ )k+1 a2 a‡ = (a‡ )k −(a‡ )k+1 a = (a‡ )k −(a‡ )k = 0 ∈ J(A ) for some k > 1, it follows that (a‡ )‡ = a2 a‡ .
(3) By (2) and Lemma 2.1.
(4) According to (2).
Corollary 2.4. Let a ∈ A pD . Then (a‡ )‡ = a if and only if a ∈ A # .
Theorem 2.5. Let a, b ∈ A pD with ab = ba = 0. Then (a + b)‡ = a‡ + b‡ .
Proof. Since ab = ba = 0, it follows that ab‡ = ba‡ = 0 and a‡ b = b‡ a = 0.
Thus, we obtain
(i) (a‡ + b‡ )(a + b) = (a + b)(a‡ + b‡ ).
(ii) By a‡ aa‡ = a‡ , we have
(a‡ + b‡ )(a + b)(a‡ + b‡ ) =
=
(a‡ a + b‡ b)(a‡ + b‡ )
a‡ + b‡ .
H. Zhu, J. Chen / Filomat 28:9 (2014), 1773–1781
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(iii) According to ak − ak+1 a‡ ∈ J(A ) and bk − bk+1 b‡ ∈ J(A ) for some k > 1, we obtain
(a + b)k − (a + b)k+1 (a‡ + b‡ ) =
(ak + bk ) − (ak+1 + bk+1 )(a‡ + b‡ )
=
ak + bk − ak+1 a‡ − bk+1 b‡
=
ak − ak+1 a‡ + bk − bk+1 b‡
∈ J(A ).
Hence, (a + b)‡ = a‡ + b‡ .
Corollary 2.6. If a1 , a2 , · · · , an ∈ A pD such that ai a j = 0 (i, j = 1, · · · , n; i , j), then a1 + a2 + · · · + an is p-Drazin
‡
‡
‡
invertible and (a1 + a2 + · · · + an )‡ = a1 + a2 + · · · + an .
Proof. It is true for n = 2 by Theorem 2.5.
‡
‡
Assume that the result holds for n − 1. Then (a1 + · · · + an−1 )‡ = a1 + · · · + an−1 .
For n case, Theorem 2.5 guarantees that
(a1 + · · · + an−1 + an )‡
‡
= (a1 + · · · + an−1 )‡ + an
‡
‡
= a1 + · · · + an .
This completes the proof.
In [17], Wei and Deng presented the formula for the Drazin inverse of two square matrices that commute
with each other. We consider the result in [17] for p-Drazin inverse in a Banach algebra as follows.
Theorem 2.7. If a, b ∈ A pD and ab = ba, then a + b is p-Drazin invertible if and only if 1 + a‡ b is p-Drazin invertible.
In this case, we have
∞
X
(a + b)‡ = (1 + a‡ b)‡ a‡ + b‡
(−b‡ aaΠ )i aΠ ,
i=0
and
(1 + a‡ b)‡ = aΠ + a2 a‡ (a + b)‡ .
Proof. Suppose that a + b is p-Drazin invertible. We prove that 1 + a‡ b is p-Drazin invertible. Write
1 + a‡ b = a1 + b1 with a1 = aΠ and b1 = a‡ (a + b).
Note that a, b, a‡ and b‡ commute with each other. We obtain (b1 )‡ = (a‡ (a + b))‡ = a2 a‡ (a + b)‡ by Lemma
2.1 and Theorem 2.3(2).
Since a1 is idempotent, (a1 )‡ = a1 = aΠ . Observing that a1 b1 = b1 a1 = 0, it follows that (1 + a‡ b)‡ =
aΠ + a2 a‡ (a + b)‡ by Theorem 2.5.
∞
X
Conversely, let ξ = 1 + a‡ b ∈ A pD and x = ξ‡ a‡ + b‡ (−b‡ aaΠ )i aΠ .
i=0
We prove that x is the p-Drazin inverse of a + b by showing the following conditions hold: (i) x(a + b) =
(a + b)x, (ii) x(a + b)x = x, (iii) (a + b)k − (a + b)k+1 x ∈ J(A ).
(i) By Lemma 2.1, a + b commutes with x.
H. Zhu, J. Chen / Filomat 28:9 (2014), 1773–1781
∞
X
(ii) Note that
Π i
(−b aa ) = 1 +
‡
i=0
x(a + b) =
∞
X
Π i
(−b aa ) = 1 − b aa
‡
‡
Π
i=1
∞
X
1776
(−b‡ aaΠ )i . We have
i=0


∞
X


‡ Π i Π
ξ‡ a‡ + b‡
(−b aa ) a  (a + b)

i=0
=
ξ‡ a‡ (a + b) + b‡ aaΠ
∞
X
i=0
∞
X
=
ξ‡ a‡ (a + b) + b‡ aaΠ
=
ξ a (a + b) + bb a .
i=0
(−b‡ aaΠ )i + bb‡ aΠ
∞
X
(−b‡ aaΠ )i
i=0


∞
X


‡ Π i
‡ Π
‡
Π
i
‡
Π

(−b aa ) + bb a 1 − b aa
(−b aa ) 
i=0
‡ Π
‡ ‡
Hence,
x(a + b)x
=
h


∞
X
i 

‡
Π
i
Π
‡
‡
‡

ξ a (a + b) + bb a ξ a + b
(−b aa ) a 
‡ Π
‡ ‡
i=0
=
(ξ‡ )2 (a‡ )2 (a + b) + aΠ b‡
∞
X
(−b‡ aaΠ )i
i=0
=
(ξ‡ )2 a‡ ξ + aΠ b‡
∞
X
(−b‡ aaΠ )i
i=0
= ξ‡ a‡ + b‡
∞
X
(−b‡ aaΠ )i aΠ
i=0
= x.
(iii) We have (a + b)k − (a + b)k+1 x ∈ J(A ). Indeed,
a + b − (a + b)2 x
= a + b − (a + b)[ξ‡ a‡ (a + b) + bb‡ aΠ ]
= a + b − ξ‡ a‡ (a + b)2 − (a + b)bb‡ aΠ
= a + b − ξ‡ (a‡ (a + b))2 a − aaΠ bb‡ − b2 b‡ (1 − aa‡ )
= a + b − ξ‡ (ξ − aΠ )2 a − aaΠ bb‡ − b2 b‡ + aa‡ b2 b‡
=
a + b − ξ‡ ξ2 a + ξ‡ aaΠ − aaΠ bb‡ − b2 b‡ + aa‡ b(1 − bΠ )
=
b − b2 b‡ − aaΠ bb‡ + a + aa‡ b − aa‡ bbΠ + ξ‡ aaΠ − ξ‡ ξ2 a
=
b − b2 b‡ − aaΠ bb‡ + aξ − aa‡ bbΠ + ξ‡ aaΠ − ξ‡ ξ2 a
=
bbΠ + (ξ‡ − bb‡ )aaΠ − aa‡ bbΠ + aξξΠ
=
aΠ bbΠ + (ξ‡ − bb‡ )aaΠ + aξξΠ .
Since (aaΠ )k1 ∈ J(A ), (bbΠ )k2 ∈ J(A ) and (ξξΠ )k3 ∈ J(A ) for some positive integers k1 , k2 and k3 , take
suitable k > k1 + k2 + k3 , it follows that (a + b)k − (a + b)k+1 x = [a + b − (a + b)2 x]k ∈ J(A ) by Lemma 2.2(2).
The proof is completed.
3. p-Drazin Inverse Under the Condition ab = λba
In this section, we give some results on p-Drazin inverse under the condition that ab = λba.
Lemma 3.1. Let a, b ∈ A with ab = λba (λ , 0). Then
(1) abn = λn bn a, an b = λn ban .
−n(n−1)
n(n+1)
(2) (ab)n = λ 2 an bn = λ 2 bn an .
n(n−1)
−n(n+1)
(3) (ba)n = λ 2 bn an = λ 2 an bn .
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1777
Proof. By induction, it is easy to obtain the results.
Let A qnil = {a ∈ A : 1 + ax ∈ A −1 f or every x ∈ comm(a)}. Then we have (see [13, p. 138]) that a ∈ A qnil if
1
and only if kan k n → 0 (n → ∞) and that J(A ) ⊂ A qnil . It is well known that ck ∈ J(A ) implies that c ∈ A qnil
for k > 1. Indeed, for any x ∈ comm(c), 1 − (cx)k = (1 − cx)(1 + cx + · · · + (cx)k−1 ) ∈ A −1 . Then, 1 − cx ∈ A −1
implies that c ∈ A qnil . Hence, we have a(1 − aa‡ ) ∈ A qnil .
Lemma 3.2. Let a, b ∈ A pD with ab = λba (λ , 0). Then
(1) aa‡ b = baa‡ .
(2) bb‡ a = abb‡ .
1
Proof. (1) Since a(1 − aa‡ ) ∈ A qnil , it follows that k(a(1 − aa‡ ))n k n → 0 (n → ∞). Suppose p = aa‡ . We have
1
kpb − pbpk n
1
=
ka‡ ab(1 − aa‡ )k n
=
k(a‡ )n an b(1 − aa‡ )k n
=
k(a‡ )n λn ban (1 − aa‡ )k n
=
kλn (a‡ )n b(a(1 − aa‡ ))n k n
6
| λ | ka‡ kkbk n k(a(1 − aa‡ ))n k n .
1
1
1
1
1
1
Hence, kpb − pbpk n → 0 (n → ∞), it follows that pb = pbp.
Similarly, bp = pbp.
Thus, aa‡ b = baa‡ .
(2) The proof is similar to the proof of (1).
Authors [16] proved that (ab)‡ = b‡ a‡ under the condition ab = ba in a Banach algebra. We can obtain
some generalized results under weak commutative condition ab = λba.
Theorem 3.3. Let a, b ∈ A pD with ab = λba (λ , 0). Then
(1) a‡ b = λ−1 ba‡ .
(2) ab‡ = λ−1 b‡ a.
(3) (ab)‡ = b‡ a‡ = λ−1 a‡ b‡ .
Proof. (1) Since aa‡ b = baa‡ , we have
a‡ b =
a‡ aa‡ b = a‡ baa‡ = a‡ λ−1 aba‡
=
λ−1 a‡ aba‡ = λ−1 baa‡ a‡
=
λ−1 ba‡ .
(2) The proof is similar to (1).
(3) We first prove that b‡ a‡ = λ−1 a‡ b‡ . By Lemma 3.2, it follows that
b‡ a‡
=
b‡ (aa‡ )a‡ = (aa‡ )b‡ a‡
=
a‡ (ab‡ )a‡ = a‡ (λ−1 b‡ a)a‡
=
λ−1 a‡ b‡ (aa‡ ) = λ−1 a‡ (aa‡ )b‡
=
λ−1 a‡ b‡ .
We now prove that x = b‡ a‡ is the p-Drazin inverse of ab.
(i) By Lemma 3.2, we obtain
(ab)x
=
abb‡ a‡ = aa‡ b‡ b
=
b‡ aa‡ b = b‡ a‡ ab
=
x(ab).
H. Zhu, J. Chen / Filomat 28:9 (2014), 1773–1781
1778
(ii) x(ab)x = b‡ (a‡ a)bb‡ a‡ = b‡ bb‡ (a‡ a)a‡ = b‡ a‡ = x.
(iii) We first present an useful equality, i.e.,
bk+1 b‡ ak+1 a‡
= bk bb‡ ak aa‡ = bk ak aa‡ bb‡
= bk ak+1 a‡ bb‡ = bk ak+1 bb‡ a‡
= bk λk+1 bak+1 b‡ a‡
= λk+1 bk+1 ak+1 b‡ a‡ .
Hence, we have
(ab)k − (ab)k+1 b‡ a‡
λ
k(k+1)
2
bk a k − λ
=
λ
k(k+1)
2
(bk ak − λk+1 bk+1 ak+1 b‡ a‡ )
=
λ
k(k+1)
2
(bk ak − bk+1 b‡ ak+1 a‡ )
λ
k(k+1)
2
[−(bk − bk+1 b‡ )(ak − ak+1 a‡ ) + (bk − bk+1 b‡ )ak
=
=
(k+1)(k+2)
2
bk+1 ak+1 b‡ a‡
+bk (ak − ak+1 a‡ )]
∈ J(A )
for some k > 1.
Therefore, (ab)‡ = b‡ a‡ = λ−1 a‡ b‡ .
Corollary 3.4. Let a, b ∈ A pD with ab = λba (λ , 0). Then
−n(n+1)
n(n−1)
(1) (a‡ b)n = λ 2 (a‡ )n bn = λ 2 bn (a‡ )n .
n(n−1)
−n(n+1)
(2) (ab‡ )n = λ 2 an (b‡ )n = λ 2 (b‡ )n an .
Proof. By induction and Theorem 3.3.
Theorem 3.5. Let a, b ∈ A pD with ab = λba (λ , 0). Then a−b is p-Drazin invertible if and only if w = aa‡ (a−b)bb‡
is p-Drazin invertible. In this case, we have
(a − b)‡ = w‡ + a‡
∞
∞
X
X
(ba‡ )i bΠ − aΠ
(b‡ a)i b‡ ,
i=0
i=0
and w‡ = aa‡ (a − b)‡ bb‡ .
Proof. Assume that a − b is p-Drazin invertible. Since aa‡ is idempotent, aa‡ ∈ A pD . By Lemma 3.2, we have
aa‡ (a − b) = (a − b)aa‡ . Hence, aa‡ (a − b) ∈ A pD according to Lemma 2.1.
Again, Lemma 3.2 guarantees that aa‡ (a − b)bb‡ = bb‡ aa‡ (a − b). Thus, aa‡ (a − b)bb‡ ∈ A pD and w‡ =
‡
aa (a − b)‡ bb‡ according to Lemma 2.1.
Conversely, let
∞
∞
X
X
x = w‡ + a‡
(ba‡ )i bΠ − aΠ
(b‡ a)i b‡ .
i=0
i=0
We prove that x is the p-Drazin inverse of a − b, i.e., the following conditions hold: (i) x(a − b) = (a − b)x,
(ii) x(a − b)x = x, (iii) (a − b)k − (a − b)k+1 x ∈ J(A ).
(i) By Lemma 3.2, we have
(a − b)w =
(a − b)aa‡ (a − b)bb‡
=
aa‡ (a − b)(a − b)bb‡
=
aa‡ (a − b)bb‡ (a − b)
=
w(a − b).
H. Zhu, J. Chen / Filomat 28:9 (2014), 1773–1781
1779
Hence, (a − b)w‡ = w‡ (a − b). Moreover,
(a − b)a‡
∞
X
(ba‡ )i bΠ
= (aa‡ − ba‡ )
i=0
∞
X
(ba‡ )i bΠ = aa‡
i=0
∞
X
(ba‡ )i bΠ − ba‡
i=0
∞
X
(ba‡ )i bΠ
i=0
= aa‡ (1 + ba‡ + (ba‡ )2 + · · · )bΠ − ba‡ (1 + ba‡ + (ba‡ )2 + · · · )bΠ
= (aa‡ + ba‡ + (ba‡ )2 + · · · )bΠ − (ba‡ + (ba‡ )2 + · · · )bΠ
= aa‡ bΠ .


∞
 X

Similarly, (a − b)aΠ
(b‡ a)i b‡  = −bb‡ aΠ .
i=0
We have
(a − b)x
∞
X
‡ i Π
=
(a − b)(w + a
=
(a − b)w‡ + aa‡ bΠ + bb‡ aΠ .
‡
‡
(ba ) b − a
i=0
Π
∞
X
(b‡ a)i b‡ )
i=0
Note that Lemma 3.2. We get
a‡
∞
X
(ba‡ )i bΠ (a − b) = a‡
i=0
∞
X
(ba‡ )i (a − b)bΠ = a‡
i=0
Π
∞
X
(ba‡ )i abΠ − a‡
i=0
‡ 2
∞
X
(ba‡ )i bbΠ
i=0
Π
=
a (1 + ba + (ba ) + · · · )ab − a (1 + ba + (ba ) + · · · )bb
=
a‡ (abΠ + ba‡ abΠ + (ba‡ )2 abΠ + · · · ) − a‡ (bbΠ + ba‡ bbΠ + (ba‡ )2 bbΠ + · · · )
=
a‡ (abΠ + a‡ abbΠ + aa‡ ba‡ bbΠ + aa‡ (ba‡ )2 bbΠ + · · · )
‡
‡
‡ 2
‡
‡
−a‡ (bbΠ + ba‡ bbΠ + (ba‡ )2 bbΠ + · · · )
=
a‡ (abΠ + bbΠ + ba‡ bbΠ + (ba‡ )2 bbΠ + · · · )
−a‡ (bbΠ + ba‡ bbΠ + (ba‡ )2 bbΠ + · · · )
=
aa‡ bΠ .
Similarly, aΠ
∞
X
(b‡ a)i b‡ (a − b) = −bb‡ aΠ .
i=0
Therefore, we have


∞
∞
X
X


‡
‡
‡
i
Π
Π
‡
i
‡

x(a − b) = w + a
(ba ) b − a
(b a) b  (a − b)
i=0
i=0
‡ Π
‡ Π
=
(a − b)w + aa b + bb a
=
(a − b)x.
‡
(ii) aa‡ , bb‡ indeed commute with any elements of A from Lemma 2.1. Since aa‡ and bb‡ are idempotents,
w = aa‡ w‡ bb‡ and
‡
w‡ (a − b)w‡
= aa‡ w‡ bb‡ (a − b)aa‡ w‡ bb‡
= (w‡ )2 w
= w‡ .
Hence, we get w‡ aΠ = aΠ w‡ = 0 and w‡ bΠ = bΠ w‡ = 0. Also, w‡ (aa‡ bΠ + bb‡ aΠ ) = w‡ bΠ aa‡ + w‡ aΠ bb‡ = 0.
H. Zhu, J. Chen / Filomat 28:9 (2014), 1773–1781
1780
According to Lemma 3.2 and (a − b)w‡ = w‡ (a − b), we have
 ∞

 ∞

∞
∞
X
X
 X

 X

‡ i Π
Π
‡ i ‡
‡
‡
‡
i
Π
Π
‡
i
‡


a‡
(ba ) b − a
(b a) b  (a − b)w = a
(ba ) b − a
(b a) b  w‡ (a − b)

i=0
=
i=0
i=0
i=0
0
and
[a‡
∞
X
+a‡
∞
X
(ba‡ )i bΠ − aΠ
i=0
∞
X
(b‡ a)i b‡ ](aa‡ bΠ + bb‡ aΠ ) = a‡
i=0
(ba‡ )i bΠ bb‡ aΠ − aΠ
=
a‡
∞
X
(b‡ a)i b‡ aa‡ bΠ − aΠ
∞
X
i=0
(ba‡ )i bΠ − aΠ
i=0
∞
X
(ba‡ )i bΠ aa‡ bΠ
i=0
i=0
∞
X
∞
X
(b‡ a)i b‡ bb‡ aΠ
i=0
(b‡ a)i b‡ .
i=0
Therefore,
x(a − b)x = [w‡ + a‡
∞
X
(ba‡ )i bΠ − aΠ
∞
X
i=0
=
w‡ (a − b)w‡ + w‡ (aa‡ bΠ + bb‡ aΠ ) + [a
(b‡ a)i b‡ ][(a − b)w‡ + aa‡ bΠ + bb‡ aΠ ]
i=0
∞
X
‡
(ba‡ )i bΠ − aΠ
i=0
+[a‡
∞
X
(ba‡ )i bΠ − aΠ
i=0
=
w‡ + a‡
=
x.
(b‡ a)i b‡ ](a − b)w‡
i=0
(b‡ a)i b‡ ](aa‡ bΠ + bb‡ aΠ )
i=0
w‡ + [a‡
=
∞
X
∞
X
∞
X
(ba‡ )i bΠ − aΠ
∞
X
i=0
i=0
∞
X
∞
X
(ba‡ )i bΠ − aΠ
i=0
(b‡ a)i b‡ ](aa‡ bΠ + bb‡ aΠ )
(b‡ a)i b‡
i=0
(iii) Note that w‡ = aa‡ w‡ bb‡ . Then w‡ (a − b)2 = w‡ aa‡ (a − b)2 bb‡ = w‡ w2 and (a − b)(aa‡ bΠ + bb‡ aΠ ) =
aa‡ (a − b)bΠ + aΠ (a − b)bb‡ .
Writing wΠ = 1 − ww‡ , we obtain
(a − b)2 x
=
(a − b)(w‡ (a − b) + aa‡ bΠ + bb‡ aΠ )
=
w‡ (a − b)2 + (a − b)(aa‡ bΠ + bb‡ aΠ )
=
w‡ w2 + aa‡ (a − b)bΠ + aΠ (a − b)bb‡
=
w − wwΠ + aa‡ (a − b)bΠ + aΠ (a − b)bb‡
=
aa‡ (a − b)bb‡ + aa‡ (a − b)bΠ + aΠ (a − b)bb‡ − wwΠ
=
aa‡ (a − b) + aΠ (a − b)bb‡ − wwΠ .
Thus,
(a − b) − (a − b)2 x
= (a − b) − (aa‡ (a − b) + aΠ (a − b)bb‡ − wwΠ )
= (a − b) − aa‡ (a − b) − aΠ (a − b)bb‡ + wwΠ
= aΠ (a − b) − aΠ (a − b)bb‡ + wwΠ
= aΠ (a − b)bΠ + wwΠ
= aΠ abΠ − aΠ bbΠ + wwΠ .
H. Zhu, J. Chen / Filomat 28:9 (2014), 1773–1781
1781
There exist some integers k1 , k2 and k3 such that (aaΠ )k1 ∈ J(A ), (bbΠ )k2 ∈ J(A ) and (wwΠ )k3 ∈ J(A ). It
follows from Lemma 2.2 that (aΠ abΠ −aΠ bbΠ )k4 ∈ J(A ) for integer k4 > k1 +k2 . Finally, as wwΠ commutes with
aΠ abΠ −aΠ bbΠ , we have (aΠ abΠ −aΠ bbΠ +wwΠ )k ∈ J(A ) for integer k > k1 +k2 +k3 . Hence, (a−b)k −(a−b)k+1 x ∈
J(A ) for some integer k.
Therefore, a − b ∈ A pD and
(a − b)‡ = w‡ + a‡
∞
X
i=0
(ba‡ )i bΠ − aΠ
∞
X
(b‡ a)i b‡ .
i=0
This completes the proof.
Acknowledgments
The first author is grateful to China Scholarship Council for supporting him to purse his further study
with Professor Pedro Patrı́cio in University of Minho, Portugal. We would like to express our sincere thanks
to the referee for his/her careful reading and valuable remarks that improved the presentation of our work
tremendously.
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