Generalized EP elements in Banach *-algebras

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Generalized EP elements in Banach *-algebras

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Huanyin Chen,Marjan Sheibani^*

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Submitted:

06 October 2023

Posted:

09 October 2023

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Abstract

We introduce a new generalized inverse (i.e., generalized EP element) which is a natural generalization of EP and *-DMP elements in a Banach *-algebra. We present polar-like characterizations of generalized EP elements. The necessary and sufficient conditions under which the sum of two generalized EP elements is a generalized EP element are investigated. Finally, the generalized core-EP orders for generalized EP elements are characterized.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

1. Introduction

Let

A

be a Banach algebra with involution *. An element a in

A

has core inverse if there exists some

x \in A

such that

a x^{2} = x, {(a x)}^{*} = a x, a = x a^{2} .

Such x is unique if it exists and is denoted by

a^{#} .

The core inverse was extensively considered in the context of Banach algebras, e.g., [1,6,17,18,19,26].

An element a in

A

is EP (i.e., an EP element) if there exists some

x \in A

such that

a x^{2} = x, {(a x)}^{*} = x a, a = x a^{2} .

Evidently,

a \in A

is EP if and only if there exists

x \in A

such that

a^{2} x = a, a x = x a, {(a x)}^{*} = a x

if and only if there exists

x \in A

such that

a x^{2} = x, {(x a)}^{*} = x a, x a^{2} = a

if and only if

a \in A^{#}

and

{(a a^{#})}^{*} = a a^{#}

([2,16,23,24,25,27]). Here,

a \in A

has group inverse provided that there exists

x \in A

such that

a x^{2} = x, a x = x a, a = x a^{2} .

Such x is unique if exists, denoted by

a^{#}

, and called the group inverse of a.

An element a in a Banach *-algebra

A

is *-DMP (i.e., *-DMP element) if there exist

m \in N

and

x \in A

such that

a x^{2} = x, {(a x)}^{*} = x a = a x, a^{m} = x a^{m + 1} .

As is well known,

a \in A

is *-DMP if and only if

a^{m} \in A

is EP for some

m \in N

(see [6,11,13]). In [21], Mosic and Djordjevic introduced and studied the gDMP inverse for a Hilbert space operator using its generalized Drazin inverse and its Moore-Penrose inverse.

The motivation of this paper is to introduce and study a new kind of generalized inverse as a natural generalization of EP and *-DMP elements mentioned above. Let

A^{q n i l} = {x \in A ∣ lim_{n \to \infty} ‖ x^{n} ‖^{\frac{1}{n}} = 0} .

As is well known,

x \in A^{q n i l}

if and only if

1 + λ x \in A

is invertible for any

λ \in C

Definition 1.1.

An element

a \in A

is generalized EP (i.e., generalized EP element) if there exist

x, y \in A

such that

a = x + y, x^{*} y = y x = 0, x \in A i s E P, y \in A^{q n i l} .

Recall that

a \in A

has g-Drazin inverse (i.e., generalized Drazin inverse) if there exists

x \in A

such that

a x^{2} = x, a x = x a, a - a^{2} x \in A^{q n i l} .

Such x is unique, if exists, and denote it by

a^{d}

. AS it is well known, a has g-Drazin inverse if and only if a has quasi-polar property, i.e., there exists an idempotent

p \in A

such that

a + p \in A^{- 1}

and

a p \in A^{q n i l}

(see [3]). In Section 2, we investigate polar-like characterizations of generalized EP elements. We prove that

a \in A

is generalized EP if and only if there exists a projection

p \in A

(i.e.,

p^{2} = p = p^{*}

)such that

a + p \in A^{- 1}, a p = p a

and

a p \in A^{q n i l}

In Section 3, we are concerned with additive properties of generalized EP elements. The necessary and sufficient conditions under which the sum of two generalized EP elements is a generalized EP element are investigated by using orthogonal and commuting perturbations.

An element

a \in A

has generalized core-EP inverse if there exists

x \in A

such that

x = a x^{2}, {(a x)}^{*} = a x, lim_{n \to \infty} | | a^{n} - x a^{n + 1} {| |}^{\frac{1}{n}} = 0 .

The preceding x is unique if it exists, and denoted by

a^{d}

. Let

a, b \in A

have generalized core-EP inverses. Recall that

a \leq^{d} b

a a^{d} = b a^{d}

and

a^{d} a = a^{d} b

. We refer the reader to [4] for properties of generalized core-EP inverses in a Banach *-algebra. Finally, in Section 4, the generalized core EP-orders for generalized EP elements in a Banach *-algebra are characterized. The properties of core-EP orders are thereby extended to wider cases.

Throughout the paper, all Banach *-algebras are complex with an identity. An element p in

A

is a projection provided that

p^{2} = p = p^{*}

. We use

A^{d}

and

A^{e}

to denote the sets of all generalized core-EP invertible and generalized EP elements in

A

. The commutant of

a \in R

is defined by

c o m m (a) = {x \in R | x a = a x}

. The double commutant of

a \in R

is defined by

c o m m^{2} (a) = {x \in R | x y = y x f o r a l l y \in c o m m (a)}

2. Polar-like Characterizations

In this section, we present a polar-like property for EP elements in a Banach *-algebra. The related characterize of EP elements are thereby derived. We begin with

Lemma 2.1.

Let

a \in A

. Then the following are equivalent:

(1): $a \in A^{e}$ .
(2): There exists $x \in c o m m (a)$ such that

$a x^{2} = x, {(a x)}^{*} = a x, a - x a^{2} \in A^{q n i l} .$
(3): There exists $x \in A$ such that

$a x^{2} = x, {(a x)}^{*} = x a, a - x a^{2} \in A^{q n i l} .$

Proof.

(1) \Rightarrow (2)

By hypothesis, there exist

z, y \in A

such that

a = z + y, z^{*} y = y z = 0, z \in A i s E P, y \in A^{q n i l} .

Set

x = z^{#}

. Then

z z^{#} = z^{#} z

by [27]. Hence

z^{#} y = z^{#} z z^{#} y = z^{#} {(z z^{#})}^{*} y = z^{#} {(z^{#})}^{*} (z^{*} y) = 0

. We check that

a x = (z + y) z^{#} = z z^{#}, x a = z^{#} (z + y) = z^{#} z; a x^{2} = (z + y) {(z^{#})}^{2} = z {(z^{#})}^{2} = z^{#} = x, x a^{2} = z^{#} {(z + y)}^{2} = z^{#} z^{2} = z .

Therefore

{(a x)}^{*} = {(z z^{#})}^{*} = z z^{#} = a x, a x = z z^{#} = z^{#} z = x a, x a^{2} - a = z - a = - y \in A^{q n i l} .

(2) \Rightarrow (3)

This is obvious.

(3) \Rightarrow (1)

By hypotheses, we have

z \in A

such that

a z^{2} = z, {(a z)}^{*} = z a, a - z a^{2} \in A^{q n i l} .

Then

{(z a)}^{*} = {({(a z)}^{*})}^{*} = a z

. In view of [25],

a z = z a

. Set

x = a z a

and

y = a - a z a .

We claim that x is EP. Evidently, we verify that

z x^{2} = z a (z a^{2} z) a = z a^{2} z a = a z a = x, x z^{2} = a z a z^{2} = a z^{2} = z, x z = a z a z = z a z a = z x, {(x z)}^{*} = {(a z a z)}^{*} = {(a z)}^{*} = a z = (a z a) z = x z .

Therefore

x \in A

is EP.

By hypothesis,

a (1 - z a) \in A^{q n i l}

. By virtue of Cline’s formula (see [3]),

y = a - z a^{2} \in A^{q n i l}

. Moreover, we see that

x^{*} y = {(a z a)}^{*} (1 - a z) a = a^{*} {(a z)}^{*} (1 - a z) a = a^{*} (a z) (1 - a z) a = 0, y x = (a - a z a) a z a = a (a - z a^{2}) z a = 0 .

This completes the proof. □

We are ready to prove:

Theorem 2.2.

Let

a \in A

. Then the following are equivalent:

(1): $a \in A^{e}$ .
(2): There exists a projection $p \in c o m m (a)$ such that $a + p \in A^{- 1}$ and $a p \in A^{q n i l}$ .

Proof.

(1) \Rightarrow (2)

In view of Lemma 2.1, there exists

x \in c o m m (a)

such that

a x^{2} = x, {(a x)}^{*} = a x, a - x a^{2} \in A^{q n i l} .

Let

p = 1 - a x

. Then

p = p^{2} \in c o m m (a)

and

a + p = a + 1 - a x

. We check that

(a + 1 - a x) (x + 1 - a x) = a x + a (1 - a x) + (1 - a x) x + (1 - a x) = 1

. Likewise,

(x + 1 - a x) (a + 1 - a x) = 1

. Then

a + p \in A^{- 1}

. Additionally,

a p = a - a^{2} x \in A^{q n i l}

, as desired.

(2) \Rightarrow (1)

By hypothesis, there exists an idempotent

p \in c o m m (a)

such that

a + p \in A^{- 1}

and

a p \in A^{q n i l}

. Set

x = (1 - p) {(a + p)}^{- 1}

. Then

x \in c o m m (a)

and

a x^{2} = a (1 - p) {(a + p)}^{- 2} = (1 - p) (a + p) {(a + p)}^{- 2} = (1 - p) {(a + p)}^{- 1} = x

. Moreover, we verify that

a - a^{2} x = a - a^{2} (1 - p) {(a + p)}^{- 1} = a - {(a + p)}^{2} (1 - p) {(a - p)}^{- 1} = a - (a + p) (1 - p) = a p \in A^{q n i l},

as desired. □

Corollary 2.3.

Every generalized EP element in a Banach *-algebra is the sum of three invertible elements.

Proof.

Let

a \in A^{e}

. In view of Theorem 2.2, we have

p^{2} = p = p^{*} \in A

such that

u : = a + p \in A^{- 1}

. Then

a = u - p

. Obviously,

- p = \frac{1 - 2 p}{2} - \frac{1}{2}

. It is easy to verify that

{(\frac{1 - 2 p}{2})}^{2} = \frac{1}{4},

and so

{(\frac{1 - 2 p}{2})}^{- 1} = 2 (1 - 2 p) .

Therefore

a = u + \frac{1 - 2 p}{2} - \frac{1}{2}

, as desired. □

Theorem 2.4.

Every Hermitan periodic element (i.e.,

a^{*} = a

and

a^{k} = a^{k}

for some distinct

k, l \in N

) in a Banach *-algebra is generalized EP.

Proof.

Assume that

a^{*} = a

and

a^{k} = a^{k}

for some positive integers

k, l (k > l)

. Then

a^{l} = a^{k} = a^{(k - l) + l} = \dots = a^{l (k - l) + l}

, and so

a^{l} = {(a^{l})}^{k - l + 1}

. Choose

m = l (k - l)

. Then

a - a^{m + 1} \in A^{n i l}

. According to the Dirichlet Theorem, there exists a prime k such that

k = s m + 1

for some

s \in N

. One easily checks that

\begin{matrix} a - a a^{m} \in A^{n i l}, \\ a a^{m} - a a^{2 m} = (a - a^{m + 1}) a^{m} \in A^{n i l}, \\ a a^{2 m} - a a^{3 m} = (a - a^{m + 1}) a^{2 m} \in A^{n i l}, \\ ⋮ \\ a a^{(s - 1) m} - a a^{s m} = (a - a^{m + 1}) a^{(s - 1) m} \in A^{n i l} . \end{matrix}

Therefore

a - a^{k} = a - a^{s m + 1} = \sum_{i = 0}^{s - 1} (a a^{i m} - a a^{(i + 1) m} \in A^{n i l}

. Set

n = k - 1

. Then

a - a^{n + 1} \in A^{n i l}

In view of [9, Theorem 3.5], there exists some

e^{n + 1} = e \in Z [a, \frac{1}{n}]

such that

w : = a - e \in Z [\frac{1}{n}] (a^{n + 1} - a) \subseteq A^{n i l} .

Thus,

e a = a e

. Let

q = e^{n - 1}

. Then

q^{2} = e^{2 n - 1} = e^{n + 1} e^{n - 2} = e^{n - 1} = q \in c o m m (a)

. Since

a = a^{*}

, we see that

q = q^{*}

. Since

a - e \in A^{n i l}

and

e a = a e

, we see that

a^{n - 1} - q \in A^{n i l}

. Set

p = 1 - q

. Then

p^{2} = p^{*} = p \in c o m m (a)

. One easily checks that

a^{n - 1} + p^{n - 1} = a^{n - 1} + p = (a^{n - 1} - q) + 1 \in A^{- 1} .

Since

n - 1 = k

is prime, we see that

a + p \in A^{- 1}

. Moreover, we have that

\begin{matrix} a^{n - 1} p & = & a^{n - 1} (1 - q) = [a^{n - 1} - q] (1 - q) \\ \in & A^{n i l} . \end{matrix}

Hence

{(a p)}^{n - 1} \in A^{n i l}

, and so

a p \in A^{n i l}

. Therefore

a \in A^{e}

by Theorem 2.2. □

We are ready to prove:

Theorem 2.5.

Let

a \in A

. Then the following are equivalent:

(1): $a \in A^{e}$ .
(2): There exists $x \in c o m m (a)$ such that

${(a x)}^{*} = a x, a - a^{2} x \in A^{q n i l} .$

Proof.

(1) \Rightarrow (2)

This is obvious by Lemma 2.1.

(2) \Rightarrow (1)

By hypothesis, there exists some

x \in c o m m (a)

such that

{(a x)}^{*} = a x, a - a^{2} x \in A^{q n i l}

. Set

s = x a x

. Then

z \in c o m m (a)

. We check that

\begin{matrix} a - a^{2} s & = & a - a x a x a \\ = & (1 + a x) (a - a^{2} x) \\ \in & A^{q n i l}, \\ s - s^{2} a & = & x a x - x a x a x a x \\ = & x (a - a^{2} x) x + x a x (a - a^{2} x) x \\ \in & A^{q n i l} . \end{matrix}

a s - {(a s)}^{2} = (a - a^{2} s) s \in A^{q n i l} .

Let

q = a s - {(a s)}^{2}, z = - \frac{1}{2} \sum_{k = 1}^{\infty} (\begin{matrix} \frac{1}{2} \\ k \end{matrix}) {(4 q {(4 q - 1)}^{- 1})}^{k}, e = a s - (2 a s - 1) z .

As in the proof of [3], we have an idempotent

e \in c o m m^{2} (a z)

such that

a z - e \in A^{q n i l}

. We easily check that

(a + 1 - a s) ((s + 1 - a s) = 1 + (a - a^{2} s) (1 - s) + (s - s^{2} a) .

Hence,

\begin{matrix} a + 1 - e & = & (a + 1 - a s) + (a s - e) \in A^{- 1}, \\ a (1 - e) & = & (a - a^{2} s) + a (a s - e) \in A^{q n i l} . \end{matrix}

Since

a \in c o m m (a s)

, we have

e a = a e

. As

{(a x)}^{*} = a x

, we see that

{(a s)}^{*} = a s

. This implies that

q^{*} = q

, and then

z^{*} = z

. Therefore

e^{*} = e

. In light of Theorem 2.2,

a \in A^{e}

. □

Theorem 2.6.

Let

a \in A

. Then the following are equivalent:

(1): $a \in A^{e}$ .
(2): There exists a projection $p \in c o m m (a)$ such that

$1 - p \in a A a a n d p a p \in A^{q n i l} .$
(3): There exists a projection $p \in c o m m (a)$ such that

$p a p \in {(p A p)}^{q n i l} a n d (1 - p) a (1 - p) \in {((1 - p) A (1 - p))}^{- 1} .$

Proof.

(1) \Rightarrow (2)

By virtue of Theorem 2.2, there exists a projection

p \in c o m m (a)

such that

u : = a + p \in A^{- 1}

and

a p \in A^{q n i l}

. By using Cline’s formula, we have

p a p \in A^{q n i l}

. Moreover,

(1 - p) a = (1 - p) u

, and then

u^{- 1} (1 - p) a = 1 - p

. Similarly,

a (1 - p) u^{- 1} = 1 - p

. Accordingly,

1 - p = a (1 - p) u^{- 2} (1 - p) a \in a A a

, as required.

(2) \Rightarrow (3)

p a p \in A^{q n i l}

, we see that

p a p \in {(e A e)}^{q n i l}

. Write

1 - e = a r a

for some

r \in A

. Hence,

(1 - p) a (1 - p) (1 - p) r a (1 - p) = 1 - p = (1 - p) a r (1 - p) a (1 - p)

. Therefore

(1 - p) a (1 - p) \in {((1 - p) A (1 - p))}^{- 1}

, as desired.

(3) \Rightarrow (1)

By using Cline’s formula, we have

p a, a p \in A^{q n i l}

. Since

(1 - p) a (1 - p) \in {((1 - p) A (1 - p))}^{- 1}

, we have some

b \in A

such that

(1 - p) a (1 - p) b (1 - p) = (1 - p) b (1 - p) a (1 - p) = 1 - p

. This implies that

(a + p) ((1 - p) b (1 - p) + p) = a (1 - p) b (1 - p) + a p + p = 1 + a p \in A^{- 1} .

Analogously,

((1 - p) b (1 - p) + p) (a + p) = 1 + p a \in A^{q n i l}

. Accordingly,

a + p \in A^{q n i l}

and

a p \in A^{q n i l}

, thus yielding the result. □

Corollary 2.7.

Let

a \in A

. Then the following are equivalent:

(1): $a \in A^{e}$ .
(2): There exists a projection $p \in c o m m (a)$ such that

$1 - p \in a A ⋂ A a a n d p a p \in A^{q n i l} .$

Proof.

(1) \Rightarrow (2)

This is clear by Theorem 2.6 as

a A a \subseteq a A ⋂ A a

(2) \Rightarrow (1)

By hypothesis, there exists a projection

p \in c o m m (a)

such that

1 - p \in a A ⋂ A a a n d p a p \in A^{q n i l} .

Write

1 - p = a x = y a

for some

x, y \in A

. Then

1 - p = a x = a x (1 - p) = a x y z \in a A a

. This completes the proof by Theorem 2.6. □

An element

a \in A

has generalized core-EP inverse if there exists

x \in A

such that

x = a x^{2}, {(a x)}^{*} = a x, lim_{n \to \infty} | | a^{n} - x a^{n + 1} {| |}^{\frac{1}{n}} = 0 .

the preceding x is unique if it exists, and denoted by

a^{d}

. We use

A^{d}

to stand for the set of all generalized core-EP invertible element a in

A

. We refer the reader to [4] for more properties of the generalized core-EP inverse in Banach *-algebra. We say that a has dual generalized core-EP inverse if

a^{*} \in A^{d}

denote

a_{d} = {({(a^{*})}^{d})}^{*} .

We now derive

Theorem 2.8.

Let

a \in A

. Then the following are equivalent:

(1): $a \in A^{e}$ .
(2): $a \in A^{d}$ and $a^{π}$ is a projection.
(3): $a \in A^{d}$ and $a^{d} = a^{d}$ .
(4): $a \in A^{d} ⋂ A_{d}$ and $a^{d} = a_{d}$ .

Proof.

(1) \Rightarrow (2)

This is obvious by Lemma 2.1.

(2) \Rightarrow (1)

Since

a \in A^{d}

, there exists

x \in c o m m (a)

such that

a x^{2} = x, a - a^{2} x \in A^{q n i l}

. By hypothesis,

a^{π} = 1 - a x

is a projection. Hence,

{(a x)}^{*} = a x

. In view of Lemma 2.1,

a \in A^{e}

(1) \Rightarrow (3)

Since every EP element has core inverse, it follows by [4] that

a \in A^{d}

; hence,

a \in A^{d}

. By the uniqueness of the g-Drazin inverse of a, we have

a^{d} = a^{d}

(3) \Rightarrow (1)

Since

a \in A^{d}

, there exists

x \in A

such that

a x^{2} = x, {(a x)}^{*} = a x, lim_{n \to \infty} | | a^{n} - x a^{n + 1} {| |}^{\frac{1}{n}} = 0 .

By hypothesis,

x = a^{d}

, nd then

x \in c o m m (a)

. This implies that

{lim}_{n \to \infty} | | {(a - x a^{2})}^{n} {| |}^{\frac{1}{n}} = 0,

and so

a - x a^{2} \in A^{q n i l}

. Therefore

a \in A^{e}

by Theorem 2.5.

(1) \Rightarrow (4)

By the discussion above,

a \in A^{d}

and

a^{d} = a^{d}

. Dually, we have

a_{d} = {({(a^{*})}^{d})}^{*} = a^{d} = a^{d}

, as desired.

(4) \Rightarrow (1)

Since

a \in A^{d}

, it follows by [4, Theorem 1.2] that there exist

x, y \in A

such that

a = x + y, x^{*} y = y x = 0, x \in A^{#}, y \in A^{q n i l} .

By hypothesis,

x^{#} = a^{d} = a_{d} {({(a^{*})}^{d})}^{*}

. Hence,

{[a^{*} {(x^{#})}^{*}]}^{*} = a^{*} {(x^{#})}^{*}

. This implies that

{(x^{#} a)}^{*} = x^{#} a

. We easily check that

\begin{matrix} x^{#} a & = & x^{#} (x + y) \\ = & x^{#} x + x^{#} x x^{#} y \\ = & x^{#} x + x^{#} {(x^{#})}^{*} (x^{*} y) \\ = & x^{#} x . \end{matrix}

Then

{(x^{#} x)}^{*} = x^{#} x

. In view of [27, Theorem 2.2],

x \in A^{#}

. Accordingly,

a \in A^{e}

, as asserted. □

Corollary 2.9.

Let

A \in B (X)

. Then the following are equivalent:

(1): $A \in B {(X)}^{e}$ .
(2): $A \in B {(X)}^{d}$ and $A A^{d} A \in B (X)$ is EP.
(3): $A \in B {(X)}^{d}$ and $A^{d} A^{*} A A^{π} = 0$ .

Proof.

(1) \Rightarrow (2)

In view of Theorem 2.8,

A \in B {(X)}^{d}

. By hypothesis, there exist

X, Y \in B (X)

such that

A = X + Y, X^{*} Y = Y X = 0, X \in B (X) i s E P, Y \in B {(X)}^{q n i l} .

In view of [24, Corollary 5],

X = A A^{d} A

. Hence,

A A^{d} A \in B (X)

is EP.

(2) \Rightarrow (1)

In view of [24, Theorem 4], there exist unique

X, Y \in B (X)

such that

A = X + Y, X^{*} Y = 0, Y X = 0, X \in B {(X)}^{#}, Y \in B {(X)}^{q n i l} .

Explicitly,

X = A A^{d} A

. By hypothesis, X is

E P

. Therefore

A \in B {(X)}^{e}

(1) \Rightarrow (3)

Clearly,

A \in B {(X)}^{d}

. By hypothesis, there exist

X, Y \in B (X)

such that

A = X + Y, X^{*} Y = Y X = 0, X \in B (X) i s E P, Y \in B {(X)}^{q n i l} .

By virtue of [24, Corollary 1],

X = A^{2} A^{d}

. Therefore

X^{*} Y = {(A^{2} A^{d})}^{*} (A - A^{2} A^{d}) = 0 .

Hence,

A A^{d} A^{*} A (I - A A^{d}) = 0 .

A^{d} A^{*} A A^{π} = 0 .

(3) \Rightarrow (1)

In view of [24, Theorem 1], there exist unique

X, Y \in B (X)

such that

A = X + Y, Y X = 0, X \in B (X) i s E P, Y \in B {(X)}^{q n i l} .

Explicitly,

X = A^{2} A^{d}

. By hypothesis,

X^{*} Y = 0

. Therefore

A \in B {(X)}^{e}

. □

3. Additive Properties

In this section, we are concerned with additive properties of generalized EP elements. Let

a, p^{2} = p \in A

. Then a has the Pierce decomposition relative to p, and we denote it by

{(\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix})}_{p}

. We now derive

Lemma 3.1.

Let p be a projection,

a \in {(p A p)}^{d}

and

x = {(\begin{matrix} a & b \\ 0 & d \end{matrix})}_{p} .

Then

x \in A)

is generalized EP if and only if

a, d \in A

are generalized EP and

\begin{matrix} \sum_{n = 0}^{\infty} a^{n} a^{π} b {(d^{d})}^{n + 2} = 0, \\ \sum_{n = 0}^{\infty} {(a^{d})}^{n + 2} b d^{n} d^{π} = 0 . \end{matrix}

Proof.

⟹ Since

x \in A^{e}

, it follows by [4, Theorem 1.2] that

x^{d} \in A^{#}

. In this case,

{(x^{d})}^{#} = {(x^{e})}^{d}

and

x^{e} = {(x^{d})}^{2} {(x^{d})}^{#}

. In view of Theorem 2.8,

x^{e} = x^{d}

. By virtue of [8, Theorem 2.1], we can write

x^{e} = (\begin{matrix} * & * \\ 0 & * \end{matrix}),

and so

{(x^{e})}^{d} = (\begin{matrix} * & * \\ 0 & * \end{matrix}) .

This implies that

{(x^{d})}^{#} = (\begin{matrix} * & * \\ 0 & * \end{matrix}) .

Obviously, we have

x^{d} = {(\begin{matrix} a^{d} & z \\ 0 & d^{d} \end{matrix})}_{p},

where

z = \sum_{n = 0}^{\infty} {(a^{d})}^{n + 2} b d^{n} d^{π} + \sum_{n = 0}^{\infty} a^{n} a^{π} b {(d^{d})}^{n + 2} - a^{d} b d^{d} .

In light of [26, Theorem 2.5],

a^{d}, d^{d} \in A^{#}

and

{(a^{d})}^{π} z = 0

. This implies that

a^{π} z = 0 .

Since

x^{d} \in A_{#}

, then

{(x^{d})}^{*} {\in A)}^{#}

. It follows by [26, Theorem 2.5] that

{({(d^{*})}^{d})}^{π} z^{*} = 0

, and so

z {(d^{d})}^{π} = 0

, i.e.,

z d^{π} = 0

. We easily check that

\begin{matrix} a^{π} z & = & \sum_{n = 0}^{\infty} a^{n} a^{π} b {(d^{d})}^{n + 2}, \\ z d^{π} & = & \sum_{n = 0}^{\infty} {(a^{d})}^{n + 2} b d^{n} d^{π} . \end{matrix}

Thus, we have

\begin{matrix} \sum_{n = 0}^{\infty} a^{n} a^{π} b {(d^{d})}^{n + 2} = 0, \\ \sum_{n = 0}^{\infty} {(a^{d})}^{n + 2} b d^{n} d^{π} = 0 . \end{matrix}

In view of Theorem 2.8,

a, d \in A^{e}

, as desired.

⟸ Since

a, d \in A

are generalized EP, it follows by Theorem 2.8 that

a^{d}, d^{d} \in A^{#} ⋂ A_{#}

. In view of [8],

x \in A^{d}

and

x^{d} = (\begin{matrix} a^{d} & z \\ 0 & d^{d} \end{matrix}),

where

\begin{matrix} z & = & \sum_{n = 0}^{\infty} {(a^{d})}^{n + 2} b d^{n} d^{π} + \sum_{n = 0}^{\infty} a^{n} a^{π} b {(d^{d})}^{n + 2} - a^{d} b d^{d} \\ = & - a^{d} b d^{d} . \end{matrix}

We easily check that

a^{π} z = 0

and

z d^{π} = 0

. In light of [26, Theorem 2.5],

x^{d} \in A^{#}

. In this case,

{(x^{d})}^{#} = (\begin{matrix} {(a^{d})}^{#} & - {(a^{d})}^{#} z {(d^{d})}^{#} \\ 0 & {(d^{d})}^{#} \end{matrix}) .

Thus, we have

\begin{matrix} x^{d} = {(x^{d})}^{2} {(x^{d})}^{#} \\ = & {(\begin{matrix} a^{d} & - a^{d} b d^{d} \\ 0 & d^{d} \end{matrix})}^{2} (\begin{matrix} {(a^{d})}^{#} & {(a^{d})}^{#} a^{d} b d^{d} {(d^{d})}^{#} \\ 0 & {(d^{d})}^{#} \end{matrix}) \\ = & (\begin{matrix} {(a^{d})}^{2} & - {(a^{d})}^{2} b d^{d} - a^{d} b {(d^{d})}^{2} \\ 0 & {(d^{d})}^{2} \end{matrix}) (\begin{matrix} {(a^{d})}^{#} & {(a^{d})}^{#} a^{d} b d^{d} {(d^{d})}^{#} \\ 0 & {(d^{d})}^{#} \end{matrix}) \\ = & (\begin{matrix} a^{d} & - a^{d} b d^{d} \\ 0 & d^{d} \end{matrix}) . \end{matrix}

This implies that

x^{d} = x^{d}

. According to Theorem 2.8,

x \in A

is generalized EP. □

Lemma 3.2.

Let

a, b \in A

be generalized EP. If

a b = b a = a^{*} b = 0

, then

a + b

is generalized EP.

Proof.

In view of Theorem 2.8 ,

a, b \in A^{d}

a^{d} = a^{d}

and

b^{d} = b^{d}

. Since

a^{*} b = 0

, we have

b^{*} a = 0

. By virtue of [5], we have

{(a + b)}^{d} = a^{d} + b^{d}

. Clearly,

{(a + b)}^{d} = a^{d} + b^{d}

, and then

{(a + b)}^{d} = {(a + b)}^{d}

. By using Theorem 2.8 again,

a + b

is generalized EP. □

We come now to the demonstration for which this section has been developed.

Theorem 3.3.

Let

a, b, a^{π} b \in A

be generalized EP. If

a^{π} a b = a^{π} b a = a^{π} a^{*} b = 0

, then the following are equivalent:

(1): $a + b \in A$ is generalized EP.
(2): $(a + b) a a^{d} \in A$ is generalized EP

$\begin{matrix} \sum_{n = 0}^{\infty} {(a + b)}^{n} a a^{d} {(a + b)}^{π} a a^{d} b {(a^{π} b^{d})}^{n + 2} = 0, \\ \sum_{n = 0}^{\infty} {({(a + b)}^{d} a a^{d})}^{n + 2} b a^{π} (a^{n} + b^{n}) a^{π} b^{π} = 0 . \end{matrix}$

Proof.

Let

p = a a^{d}

. By hypothesis,

p^{π} b p = (1 - a a^{d}) b a a^{d} = (a^{π} b a) a^{d} = 0

. So we get

a = {(\begin{matrix} a_{1} & 0 \\ 0 & a_{4} \end{matrix})}_{p}, b = {(\begin{matrix} b_{1} & b_{2} \\ 0 & b_{4} \end{matrix})}_{p} .

Hence

a + b = {(\begin{matrix} a_{1} + b_{1} & b_{2} \\ 0 & a_{4} + b_{4} \end{matrix})}_{p} .

Here,

a_{1} = a^{2} a^{d}

and

b_{1} = a a^{d} b a a^{d} = b a a^{d}

. Then

a_{1} + b_{1} = (a + b) a a^{d} .

Moreover, we see that

\begin{matrix} {(a_{1} + b_{1})}^{i} & = & {(a + b)}^{i} a a^{d}, \\ {(a_{1} + b_{1})}^{d} & = & {(a + b)}^{d} a a^{d}, \\ {(a_{1} + b_{1})}^{π} & = & 1 - (a + b) {(a + b)}^{d} a a^{d} . \end{matrix}

Also we have

a_{4} = a a^{π}

and

b_{4} = a^{π} b a^{π} = a^{π} b

, and so

a_{4} + b_{4} = a^{π} a + a^{π} b .

Then we check that

\begin{matrix} {(a_{4} + b_{4})}^{i} & = & a^{π} (a^{i} + b^{i}), \\ {(a_{4} + b_{4})}^{d} & = & a^{π} b^{d}, \\ {(a_{4} + b_{4})}^{π} & = & 1 - a^{π} b b^{d} . \end{matrix}

Clearly,

a^{π} a

is generalized EP. By hypothesis,

a^{π} b

is generalized EP. Further, we see that

\begin{matrix} (a^{π} a) (a^{π} b) & = & 0, \\ (a^{π} b) (a^{π} a) & = & 0, \\ {(a^{π} a)}^{*} (a^{π} b) & = & 0 . \end{matrix}

In view of Lemma 3.2,

a_{4} + b_{4}

is generalized EP.

In light of Lemma 3.1,

a + b

is generalized EP if and only if

a_{1} + b_{1}

is generalized EP and

\begin{matrix} \sum_{n = 0}^{\infty} {(a_{1} + b_{1})}^{n} {(a_{1} + b_{1})}^{π} b_{2} {({(a_{4} + b_{4})}^{d})}^{n + 2} = 0, \\ \sum_{n = 0}^{\infty} {({(a_{1} + b_{1})}^{d})}^{n + 2} b_{2} {(a_{4} + b_{4})}^{n} {(a_{4} + b_{4})}^{π} = 0 . \end{matrix}

Therefore

a + b

is generalized EP if and only if

(a + b) a a^{d}

is generalized EP and

\begin{matrix} \sum_{n = 0}^{\infty} {(a + b)}^{n} a a^{d} {(a + b)}^{π} a a^{d} b {(a^{π} b^{d})}^{n + 2} = 0, \\ \sum_{n = 0}^{\infty} {({(a + b)}^{d} a a^{d})}^{n + 2} b a^{π} (a^{n} + b^{n}) a^{π} b^{π} = 0 . \end{matrix}

This completes the proof. □

Corollary 3.4.

Let

a, b, a^{π} b \in A

be EP. If

a^{π} b a = 0

, then the following are equivalent:

(1): $a + b \in A$ is generalized EP.
(2): $(a + b) a a^{#} \in A$ is generalized EP and

$\begin{matrix} \sum_{n = 0}^{\infty} {(a + b)}^{n} a a^{#} {(a + b)}^{π} a a^{#} b {(a^{π} b^{#})}^{n + 2} = 0, \\ \sum_{n = 0}^{\infty} {({(a + b)}^{d} a a^{#})}^{n + 2} b a^{π} (a^{n} + b^{n}) a^{π} b^{π} = 0 . \end{matrix}$

Proof.

Since

a \in A

is EP, we see that

a^{π} a = a^{π} a^{*} = 0

. Therefore we complete the proof by Theorem 3.3. □

Lemma 3.5.

Let

a, b \in A

be generalized EP. If

a b = b a

and

a^{*} b = b a^{*}

, then the following are equivalent:

(1): $a + b \in A$ is generalized EP.
(2): $1 + b a^{d} \in A$ is generalized EP.

Proof.

(1) \Rightarrow (2)

In view of [5, Theorem 3.4],

1 + a^{d} b \in A^{d}

and

{(1 + a^{d} b)}^{d} = a^{π} + a^{2} a^{d} {(a + b)}^{d} .

Then

\begin{matrix} (1 + a^{d} b) {(1 + a^{d} b)}^{d} & = & a^{π} + (1 + a^{d} b) a^{2} a^{d} {(a + b)}^{d} \\ = & a^{π} + a a^{d} (a + b) {(a + b)}^{d} \\ = & 1 - a a^{d} {(a + b)}^{π} . \end{matrix}

In view of Theorem 2.8,

a a^{d}

are

{(a + b)}^{π}

are projections. Then

(1 + a^{d} b) {(1 + a^{d} b)}^{d}

is a projection, and so is

{(1 + a^{d} b)}^{π}

. Therefore

1 + a^{d} b \in A

is generalized EP by Theorem 2.8. Since

a b = b a

, it follows by [3] that

a^{d} b = b a^{d}

, and so

1 + b a^{d}

is generalized EP.

(2) \Rightarrow (1)

Since

1 + a^{d} b = 1 + b a^{d} \in A

is generalized EP, it follows by Theorem 2.8 that

1 + a^{d} b \in A^{d}

and

{(1 + a^{d} b)}^{π}

is a projection. In view of [28, Theorem 3.3],

a + b \in A^{d}

and

{(a + b)}^{d} = {(1 + a^{d} b)}^{d} a^{d} + b^{d} {(1 + a a^{π} b^{d})}^{- 1} a^{π} .

Since

(1 - a^{π} b b^{d}) (1 + a a^{π} b^{d}) = 1 - a^{π} b b^{d}

, we have

(1 - a^{π} b b^{d}) {(1 + a a^{π} b^{d})}^{- 1} = 1 - a^{π} b b^{d} .

Then we check that

\begin{matrix} (a + b) {(1 + b)}^{d} \\ = & a^{d} (a + b) {(1 + a^{d} b)}^{d} + (a + b) a^{π} b^{d} {(1 + a a^{π} b^{d})}^{- 1} \\ = & a a^{d} (1 + a^{d} b) {(1 + a^{d} b)}^{d} + (a a^{π} b^{d} + a^{π} b b^{d}) {(1 + a a^{π} b^{d})}^{- 1} \\ = & a a^{d} (1 + a^{d} b) {(1 + a^{d} b)}^{d} + 1 - (1 - a^{π} b b^{d}) {(1 + a a^{π} b^{d})}^{- 1} \\ = & a a^{d} (1 + a^{d} b) {(1 + a^{d} b)}^{d} + 1 - [1 - a^{π} b b^{d}] \\ = & a a^{d} (1 + a^{d} b) {(1 + a^{d} b)}^{d} + a^{π} b b^{d} . \end{matrix}

Therefore

\begin{matrix} {(a + b)}^{π} & = & 1 - a a^{d} (1 + a^{d} b) {(1 + a^{d} b)}^{d} - a^{π} b b^{d} \\ = & a a^{d} - a a^{d} (1 + a^{d} b) {(1 + a^{d} b)}^{d} + a^{π} b^{π} \\ = & a a^{d} {(1 + a^{d} b)}^{π} + a^{π} b^{π} . \end{matrix}

Hence,

{(a + b)}^{π}

is a projection. Accordingly,

a + b \in A

is generalized EP by Theorem 2.8. □

We are ready to prove:

Theorem 3.6.

Let

a, b, a^{π} b \in A

be generalized EP. If

a^{π} a b = a^{π} b a

and

a^{π} a^{*} b = a^{π} b a^{*}

, then the following are equivalent:

(1): $a + b \in A$ is generalized EP.
(2): $1 + b a^{d} \in A$ is generalized EP and

$\begin{matrix} \sum_{n = 0}^{\infty} {(a + b)}^{n} a a^{d} {(a + b)}^{π} a a^{d} b {(a^{π} b^{π})}^{n + 2} = 0, \\ \sum_{n = 0}^{\infty} {({(a + b)}^{d} a a^{d})}^{n + 2} b a^{π} (a^{n} + b^{n}) a^{π} b^{π} = 0 . \end{matrix}$

Proof.

Since

a^{π} a b = a^{π} b a

, we have

a (a^{π} b) = (a^{π} b) a

. In view of [3, Theorem 15.2.12],

a^{d} (a^{π} b) = (a^{π} b) a^{d}

. Hence,

a^{π} b a^{d} = 0

. Let

p = a a^{d}

. Then

p^{π} b p = (a^{π} b a^{d}) a = 0

, and then we have

a = {(\begin{matrix} a_{1} & 0 \\ 0 & a_{4} \end{matrix})}_{p}, b = {(\begin{matrix} b_{1} & b_{2} \\ 0 & b_{4} \end{matrix})}_{p} .

Thus,

a + b = {(\begin{matrix} a_{1} + b_{1} & b_{2} \\ 0 & a_{4} + b_{4} \end{matrix})}_{p} .

Here,

a_{1} = a^{2} a^{d}

and

b_{1} = a a^{d} b a a^{d} = (1 - a^{π}) b a a^{d} = b a a^{d}

. Then

a_{1} + b_{1} = (a + b) a a^{d} = (1 + b a^{d}) a^{2} a^{d} .

We verify that

\begin{matrix} {(a_{1} + b_{1})}^{i} & = & {(a + b)}^{i} a a^{d}, \\ {(a_{1} + b_{1})}^{d} & = & {(a + b)}^{d} a a^{d}, \\ {(a_{1} + b_{1})}^{π} & = & 1 - (a + b) {(a + b)}^{d} a a^{d} . \end{matrix}

Further, we have

a_{4} = a a^{π}

and

b_{4} = a^{π} b a^{π} = a^{π} b

; hence,

a_{4} + b_{4} = a^{π} a + a^{π} b .

Then we check that

\begin{matrix} {(a_{4} + b_{4})}^{i} & = & a^{π} (a^{i} + b^{i}), \\ {(a_{4} + b_{4})}^{d} & = & a^{π} b^{d}, \\ {(a_{4} + b_{4})}^{π} & = & 1 - a^{π} b b^{d} . \end{matrix}

Obviously,

a^{π} a

and

a^{π} b

are generalized EP. Furthermore, we have

\begin{matrix} (a^{π} a) (a^{π} b) & = & a^{π} a b = a^{π} b a = (a^{π} b) (a^{π} a), \\ {(a^{π} a)}^{*} (a^{π} b) & = & a^{π} a^{*} b = a^{π} b a^{*} = a^{π} b a^{π} a^{*} = (a^{π} b) {(a^{π} a)}^{*}, \\ 1 + {(a^{π} a)}^{d} (a^{π} b) & = & 1 h a s g e n e r a l i z e d c o r e - E P i n v e r s e . \end{matrix}

By virtue Lemma 3.5,

a_{4} + b_{4}

is generalized EP.

By virtue of Lemma 3.1,

a + b

is generalized EP if and only if

a_{1} + b_{1}

is generalized EP and

\begin{matrix} \sum_{n = 0}^{\infty} {(a_{1} + b_{1})}^{n} {(a_{1} + b_{1})}^{π} b_{2} {({(a_{4} + b_{4})}^{d})}^{n + 2} = 0, \\ \sum_{n = 0}^{\infty} {({(a_{1} + b_{1})}^{d})}^{n + 2} b_{2} {(a_{4} + b_{4})}^{n} {(a_{4} + b_{4})}^{π} = 0 . \end{matrix}

Claim 1. Assume that

1 + b a^{d}

is generalized EP. Then we see that

\begin{matrix} (b a^{d}) (a a^{d}) & = & b a^{d} = (1 - a^{π}) b a^{d} = (a a^{d}) (b a^{d}), \\ (b a^{d}) {(a a^{d})}^{*} & = & (b a^{d}) (a a^{d}) = (a a^{d}) (b a^{d}) = {(a a^{d})}^{*} (b a^{d}), \end{matrix}

and then

\begin{matrix} (1 + b a^{d}) (a a^{d}) & = & (a a^{d}) (1 + b a^{d}), \\ (1 + b a^{d}) {(a a^{d})}^{*} & = & {(a a^{d})}^{*} (1 + b a^{d}) . \end{matrix}

By virtue of Lemma 3.5,

a_{1} + b_{1} = (1 + b a^{d}) (a a^{d})

is generalized EP.

Claim 2. Assume that

a_{1} + b_{1} = (1 + b a^{d}) a a^{d}

is generalized EP. Obviously, we have

a^{π} (1 + b a^{d}) a a^{d} = {(a^{π})}^{*} (1 + b a^{d}) a a^{d} = (1 + b a^{d}) a a^{d} a^{π} = 0

. It follows by Lemma 3.2 that

1 + b a^{d} = a^{π} + (1 + b a^{d}) a a^{d}

is generalized EP.

Thus, we conclude that

a_{1} + b_{1}

is generalized EP if and only if so is

1 + b a^{d}

. Therefore

a + b

is generalized EP if and only if

1 + b a^{d}

is generalized EP and

\begin{matrix} \sum_{n = 0}^{\infty} {(a + b)}^{n} a a^{d} {(a + b)}^{π} a a^{d} b {(a^{π} b^{π})}^{n + 2} = 0, \\ \sum_{n = 0}^{\infty} {({(a + b)}^{d} a a^{d})}^{n + 2} b a^{π} (a^{n} + b^{n}) a^{π} b^{π} = 0 . \end{matrix}

□

Corollary 3.7.

Let

a, b \in A

be EP. If

a b = b a

and

a^{*} b = b a^{*}

, then the following are equivalent:

(1): $a + b \in A$ is generalized EP.
(2): $1 + b a^{#} \in A$ is generalized EP.

Proof.

This is obvious by Theorem 3.6. □

4. Generalized Core-EP Orders

This section is devoted to the generalized core-EP orders involved in generalized EP elements. We now extend [13, Theorem 4.4] as follows.

Theorem 4.1.

a \in A^{e}, b \in A^{d}

. Then the following are equivalent:

(1): $a \leq^{d} b$ .
(2): $a^{d} \leq^{d} b^{d}$ and $a^{d} a = a^{d} b$ .
(3): $a^{d} b^{d} = b^{d} a^{d}$ and $a^{d} a = a^{d} b$ .

Proof.

(1) \Rightarrow (3)

By hypothesis, we have

a^{d} a = a^{d} b, a a^{d} = b a^{d} .

Since

a \in A^{e}

, it follows by Theorem 2.8 that

a^{d} = a^{d}

, and then

a a^{d} b = a a^{d} a = b a^{d} a = b a^{d} a = b a a^{d} = b a a^{d} .

Moreover,

{(a a^{d})}^{*} b = b {(a a^{d})}^{*}

, and so

a a^{d} b^{*} = b^{*} a a^{d} .

In light of [5, Lemma 3.2],

a a^{d} b^{d} = b^{d} a a^{d} .

Therefore

\begin{matrix} | | a^{d} - a a^{d} b^{d} | | \\ = & | | a^{k} {(a^{d})}^{k + 1} - b^{d} b^{k + 1} {(a^{d})}^{k + 1} | | \\ \leq & | | a^{k} {(a^{d})}^{k + 1} - b^{k} {(a^{d})}^{k + 1} | | + | | b^{k} - b^{d} b^{k + 1} | | | | a^{d})^{k + 1} | | \\ = & | | b^{k} - b^{d} b^{k + 1} | | | | a^{d})^{k + 1} | | . \end{matrix}

Since

lim_{k \to \infty} | | b^{k} - b^{d} b^{k + 1} {| |}^{\frac{1}{k}} = 0,

we deduce that

lim_{k \to \infty} | | a^{d} - a a^{d} b^{d} {| |}^{\frac{1}{k}} = 0 .

Then

a a^{d} b^{d} = a^{d};

hence,

a^{d} [a a^{d} b^{d}] = {(a^{d})}^{2} .

Accordingly,

a^{d} b^{d} = {(a^{d})}^{2} = a^{d} a^{d} = b^{d} a^{d} = b^{d} a^{d} .

(3) \Rightarrow (1)

Since

a, b \in A^{e}

, then

a^{d} = a^{d}

and

b^{d} = b^{d}

. We verify that

\begin{matrix} b a^{d} & = & b {(a^{d})}^{2} a = b {(a^{d})}^{2} b = b {(a^{d})}^{k + 1} b^{k}, \\ b b^{d} a a^{d} & = & b b^{d} {(a^{d})}^{k + 1} b^{k + 1} = b {(a^{d})}^{k + 1} b^{d} b^{k + 1}, \\ a a^{d} & = & a^{d} a = a^{d} b = {(a^{d})}^{k} b^{k}, \\ b b^{d} a a^{d} & = & b^{d} {(a^{d})}^{k} b^{k + 1} = {(a^{d})}^{k} b^{d} b^{k + 1} . \end{matrix}

Then

| | b a^{d} - b b^{d} a a^{d} {| |}^{\frac{1}{k}} \leq {| | b | |}^{\frac{1}{k}} | | a^{d} {| |}^{1 + \frac{1}{k}} | | b^{k} - b^{d} b^{k + 1} {| |}^{\frac{1}{k}} .

Since

{lim}_{k \to \infty} | | b^{k} - b^{d} b^{k + 1} {| |}^{\frac{1}{k}} = 0

, we have

lim_{k \to \infty} | | b a^{d} - b b^{d} a a^{d} {| |}^{\frac{1}{k}} = 0 .

This implies that

b a^{d} = b b^{d} a a^{d}

. Likewise,

a a^{d} = b b^{d} a a^{d}

. Therefore

b a^{d} = b b^{d} a a^{d} = a a^{d}

, as required.

(2) \Rightarrow (3)

By hypothesis, we have

{(a^{d})}^{d} a^{d} = {(a^{d})}^{d} b^{d}

. In view of [5, Theorem 3.5],

{(a^{d})}^{d} = a^{2} a^{d}

. Then

a^{2} a^{d} a^{d} = a^{2} a^{d} b^{d} .

Hence,

a a^{d} = a^{2} a^{d} b^{d} .

On the other hand,

a^{d} {(a^{d})}^{d} = b^{d} {(a^{d})}^{d}

. Then

a^{d} a^{2} a^{d} = b^{d} a^{2} a^{d} .

This implies that

a a^{d} = b^{d} a^{2} a^{d} .

Therefore

\begin{matrix} a^{d} b^{d} & = & {(a^{d})}^{2} (a^{2} a^{d} b^{d}) \\ = & {(a^{d})}^{2} (a a^{d}) = {(a^{d})}^{2} = a a^{d} {(a^{d})}^{2} \\ = & (b^{d} a^{2} a^{d}) {(a^{d})}^{2} = b^{d} a^{d}, \end{matrix}

as desired.

(3) \Rightarrow (2)

In view of [5, Theorem 3.5],

{(a^{d})}^{d} = a^{2} a^{d}

. Then we check that

\begin{matrix} {(a^{d})}^{d} a^{d} & = & a^{2} {(a^{d})}^{2} = a^{2} (a a^{d}) {(a^{d})}^{2} = a^{2} b {(a^{d})}^{3} = a^{2} b (a a^{d}) {(a^{d})}^{3} \\ = & a^{2} b^{2} {(a^{d})}^{4} = \dots = a^{2} b^{k} {(a^{d})}^{k + 2}, \\ {(a^{d})}^{d} b^{d} & = & a^{2} a^{d} b^{d} = a^{2} b^{d} a^{d} = a^{2} b^{d} a {(a^{d})}^{2} = a^{2} b^{d} b^{k + 1} {(a^{d})}^{k + 2} \end{matrix}

Hence,

\begin{matrix} | | {(a^{d})}^{d} a^{d} - a^{d})^{d} b^{d} {| |}^{\frac{1}{k}} \\ \leq & | | a^{2} {| |}^{\frac{1}{k}} | | b^{k} - b^{d} b^{k + 1} {| |}^{\frac{1}{k}} | | a^{d} {| |}^{1 + \frac{2}{k}} . \end{matrix}

Since

{lim}_{k \to \infty} | | b^{k} - b^{d} b^{k + 1} {| |}^{\frac{1}{k}} = 0

, we deduce that

lim_{k \to \infty} | | {(a^{d})}^{d} a^{d} - a^{d})^{d} b^{d} {| |}^{\frac{1}{k}} = 0 .

This implies that

{(a^{d})}^{d} a^{d} = {(a^{d})}^{d} b^{d} .

On the other hand, we have

\begin{matrix} a^{d} {(a^{d})}^{d} & = & a^{d} a^{2} a^{d} = (a^{d} a) a a^{d} \\ = & b^{d} a^{2} a^{d} = b^{d} {(a^{d})}^{d} . \end{matrix}

Thus,

a^{d} {(a^{d})}^{d} = b^{d} {(a^{d})}^{d} .

This completes the proof. □

The core-EP order for core-EP inverse of complex matrices was studied in [25, Theorem 4.2]. As an immediate consequence of Theorem 4.1, we give an alternative characterization of core-EP order for core-EP inverses as follows.

Corollary 4.2.

Let

A, B \in C^{n \times n}

. Then the following are equivalent:

(1): $A \leq^{D} B$ .
(2): $A^{D} \leq^{D} B^{D}$ and $A^{D} A = A^{D} B$ .
(3): $A^{D} B^{D} = B^{D} A^{D}$ and $A^{D} A = A^{D} B$ .

Theorem 4.3.

Let

a \in A^{e}, b \in A^{d}

. If

a \leq^{d} b

, then the following are equivalent:

(1): $b \in A^{e}$ .
(2): $b (1 - a a^{d}) \in A^{e}$ .

Proof.

Since

a \leq^{d} b

, we have that

a^{d} a = a^{d} b

and

a a^{d} = b a^{d}

. Then

a a^{d} b = a a^{d} a = b a^{d} a = b a a^{d} .

(1) \Rightarrow (2)

Since

b (1 - a a^{d}) = (1 - a a^{d}) b

, it follows by [28, Theorem 3.1] that

\begin{matrix} {[b (1 - a a^{d})]}^{d} & = & b^{d} (1 - a a^{d}) = b^{d} - b^{d} a a^{d} = b^{d} - {(b a a^{d})}^{d} \\ = & b^{d} - {(a a^{d} b)}^{d} = b^{d} - {(a a^{d} a)}^{d} = b^{d} - a^{d} . \end{matrix}

We verify that

\begin{matrix} [b (1 - a a^{d})] [b^{d} - a^{d}] = b b^{d} - a a^{d}; \\ {[(b (1 - a a^{d})) (b^{d} - a^{d})]}^{*} = [b (1 - a a^{d})] [b^{d} - a^{d}], \end{matrix}

Moreover, we check that

\begin{matrix} {(b (1 - a a^{d}))}^{n} - (b^{d} - a^{d}) {(b (1 - a a^{d}))}^{n + 1} \\ = & b^{n} (1 - a a^{d}) - (b^{d} - a^{d}) b^{n + 1} (1 - a a^{d}) \\ = & b^{n} (1 - a a^{d}) - b^{d} b^{n + 1} (1 - a a^{d}) \\ = & (b^{n} - b^{d} b^{n + 1}) (1 - a a^{d}) . \end{matrix}

Since

{lim}_{n \to \infty} | | b^{n} - b^{d} b^{n + 1} {| |}^{\frac{1}{n}} = 0

, we deduce that

lim_{n \to \infty} | | {(b (1 - a a^{d}))}^{n} - (b^{d} - a^{d}) {(b (1 - a a^{d}))}^{n + 1} {| |}^{\frac{1}{n}} = 0 .

Therefore

{[b (1 - a a^{d})]}^{d} = b^{d} - a^{d} = {[b (1 - a a^{d})]}^{d} .

By virtue of Theorem 2.8,

b (1 - a a^{d})

is generalized EP.

(2) \Rightarrow (1)

Obviously,

b = x + y

, where

x = b a a^{d}

and

y = b (1 - a a^{d})

Claim 1.

x \in A

are generalized EP. We directly verify that

x^{d} = a^{d} = a^{d} = x^{d}

. In view of Theorem 2.8,

x \in A^{e}

Claim 2.

x y = y x = x^{*} y = 0

. Since

a a^{d} a = b a a^{d}

, we have that

a^{*} a a^{d} = a a^{d} b^{*}

. Then we verify that

\begin{matrix} x y & = & b a a^{d} b (1 - a a^{d}) = b a a^{d} a (1 - a a^{d}) = 0, \\ y x & = & b (1 - a a^{d}) b a^{d} a = b (1 - a a^{d}) a a^{d} a = 0, \\ x^{*} y & = & a a^{d} b^{*} b (1 - a a^{d}) = a^{*} a a^{d} b (1 - a a^{d}) \\ = & a^{*} a a^{d} a (1 - a a^{d}) = 0 . \end{matrix}

In light of Lemma 3.2,

b = x + y \in A

is generalized EP, as asserted. □

As an immediate consequence, we now improve [13, Theorem 4.5] as follows.

Corollary 4.4.

Let

a \in A^{e}, b \in A^{D}

. If

a \leq^{d} b

, then the following are equivalent:

(1): $b \in A$ is *-DMP.
(2): $b (1 - a a^{d}) \in A$ is *-DMP.

Proof.

(1) \Rightarrow (2)

By virtue of Theorem 2.8,

b (1 - a a^{d}) \in A^{e}

and

{[b (1 - a a^{d})]}^{d} = {[b (1 - a a^{d})]}^{d}

. Since

b \in A

is *-DMP, it follows by Theorem 2.8 that

b \in A^{D}

. Since

a \leq^{d} b

, we have

a a^{d} b = a a^{d} b = a a^{d} a = b a^{d} a = b a a^{d} .

In view of [28, Theorem 3.1],

a a^{d} b \in A^{D}

. Obviously,

a a^{d} b a^{π} = b a a^{d} a^{π} = 0

. Set

p = a a^{d}

. Then

b = {(\begin{matrix} a a^{d} b & 0 \\ a^{π} b a a^{d} & b a^{π} \end{matrix})}_{p} .

By virtue of [8, Theorem 2.1],

b a^{π} \in A^{D}

. Hence,

{[b (1 - a a^{d})]}^{d} = {[b (1 - a a^{d})]}^{D} .

Therefore

{[(b (1 - a a^{d})) {(b (1 - a a^{d}))}^{D}]}^{*} = (b (1 - a a^{d})) {(b (1 - a a^{d}))}^{D} .

Accordingly,

b (1 - a a^{d}) \in A

is *-DMP.

(2) \Rightarrow (1)

In view of Theorem 4.3,

b \in A^{e}

. Then

b \in A^{d}

and

b^{d} = b^{d}

. Since

b \in A^{D}

, it follows by [10, Theorem 2.3] that

b \in A^{D}

. Therefore

b^{d} = b^{D}

. Then

{(b b^{D})}^{*} = b b^{D}

. This implies that

b \in A

is *-DMP by [13, Lemma 2.2]. □

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Generalized EP elements in Banach *-algebras

Abstract

1. Introduction

2. Polar-like Characterizations

3. Additive Properties

4. Generalized Core-EP Orders

References

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