Generalized core-EP inverse for triangular operator matrices

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Generalized core-EP inverse for triangular operator matrices

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Huanyin Chen^*

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Submitted:

01 August 2024

Posted:

04 August 2024

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Abstract

We investigate the existence and representation of the generalized core-EP inverse of some triangular matrices over a Banach algebra. Further, the general representations of the generalized core-EP inverse of a triangular matrix over a $C^*$-algebra are presented. As applications, the generalized core-EP inverses of some block operator matrices over Hilbert spaces are given.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

MSC: 15A09; 16U90; 16W10

1. Introduction

Let

A

be a Banach *-algebra. An element

a \in A

has Drazin inverse provided that there exists

x \in A

such that

a x^{2} = x, a x = x a, x a^{k + 1} = a^{k},

where k is the index of a (denoted by

i n d (a)

), i.e., the smallest k such that the previous equations are satisfied. Such x is unique if exists, denoted by

a^{D}

, and called the Drazin inverse of a. We say that

a \in A

has group inverse x if

i n d (a) = 1

, i.e., there exists a unique

x \in A

such that

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

We denote the group inverse x by

a^{#}

. Evidently, a square complex matrix A has group inverse if and only if

r a n k (A) = r a n k (A^{2})

An element

a \in A

has core-EP inverse (i.e., pseudo core inverse) if there exist

x \in A

and

k \in N

such that

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

where the smallest k is the index of a (denoted by

i (a)

). If such x exists, it is unique, and denote it by

a^{D}

. We say that

a \in A

has core inverse x if

i (a) = 1

, i.e., there exists a unique

x \in A

such that

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

We denote the core inverse x by

a^{#}

. As is well known, an element

a \in A

has core inverse x if and only if

a = a x a, x A = a A, A x = A a^{*} .

As a natural generalization of core-EP invertibility, the authors introduced the generalized core-EP inverse in a Banach algebra

A

with an involution *. An element

a \in A

has generalized core-EP inverse if 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 .

If such x exists, it is unique, and denote it by

a^{d}

The generalized inverses mentioned above are powerful tools in linear algebra and operator algebra for dealing with matrices and operators that do not have a traditional inverse. They are used in various applications and provide a means to find solutions to linear systems and has applications across various scientific and engineering disciplines. Recently, many authors have studied them from many different views, e.g., [2,4,5,7,8,9,10,11,13,17,20,21,24,25].

Recall that

a \in A

has 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} .

Here,

A^{q n i l} = {a \in A ∣ 1 + λ a \in A^{- 1}} .

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

a^{d}

. We use

A^{d}, A^{#}

and

A^{d}

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

A

, respectively. If a and x satisfy the equations

a = a x a

and

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

, then x is called

(1, 3)

-inverse of a and is denoted by

a^{(1, 3)}

. We use

A^{(1, 3)}

to stand for sets of all

(1, 3)

-invertible elements in

A

. We list several characterizations of generalized core-EP inverse.

Theorem 1.1.(see [3,6])Let

A

be a Banach *-algebra, and let

a \in A

. Then the following are equivalent:

(1): $a \in A^{d}$ .
(2): There exist $x, y \in A$ such that

$a = x + y, x^{*} y = y x = 0, x \in A^{#}, y \in A^{q n i l} .$
(3): There exists a projection $p \in A$ such that

$a + p \in A^{- 1}, p a = p a p \in A^{q n i l} .$
(4): $x a x = x, i m (x) = i m (x^{*}) = i m (a^{d})$ .
(5): $a \in A^{d}$ and $a^{d} \in A^{#}$ . In this case, $a^{d} = {(a^{d})}^{2} {(a^{d})}^{#} .$
(6): $a \in A^{d}$ and $a^{d} \in A^{(1, 3)}$ . In this case, $a^{d} = {(a^{d})}^{2} {(a^{d})}^{(1, 3)} .$
(7): $a \in A^{d}$ and there exists a projection $q \in A$ such that $a^{d} A = q A$ . In this case, $a^{d} = a^{d} q .$

The motivation of this paper is to investigate the generalized core-EP inverse for the triangular matrices over a Banach *-algebra.

In Section 2, we establish necessary and sufficient conditions under which the block operator triangular matrix

(\begin{matrix} a & b \\ 0 & d \end{matrix})

over a Banach algebra has the generalized core-EP inverse with upper triangular form.

C^{*}

-algebra is a Banach algebra equipped with an involution operation * that satisfies satisfies the

C^{*}

-identity:

| | x^{*} x | | = {| | x | |}^{2}

for all

x \in A

. In Section 3, we particularly investigate the generalized core-EP inverse of a triangular block operator matrices over a

C^{*}

-algebra. We prove that every triangular operator matrix

(\begin{matrix} a & b \\ 0 & d \end{matrix})

over a

C^{*}

-algebra with generalized core-EP invertible diagonal entries has the generalized core-EP inverse and its representation of generalized core-EP inverse is presented.

The set of all bounded linear operators on a Hilbert space H, denoted

B (H)

, forms a

C^{*}

-algebra with the operator norm and the adjoint operation. Lex X and Y be Hilbert spaces. We use

B (X, Y)

to stand for the set of all bounded linear operators from X to Y. Finally, in Section 4, we apply our results and study the generalized core-EP inverse for the block operator matrix

M = (\begin{matrix} A & B \\ C & D \end{matrix}),

where

A \in B {(X)}^{d}, B \in B (X, Y), C \in B (Y, X), D \in B {(Y)}^{d}

. Here, M is a linear operator on Hilbert space

X \oplus Y

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

p \in A

is a projection if

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

A^{D}, A^{d}

and

A^{n i l}

denote the sets of all Drazin, generalized core-EP invertible and nilpotent elements in

A

respectively. Let

a \in A^{d}

. We use

a^{π}

to stand for the spectral idempotent

a^{π} = 1 - a a^{d}

2. Triangular Operator Matrices over Banach *-Algebras

Let

A

be a Banach *-algebra. Then

M_{2} (A)

is a Banach *-algebra with *-transpose as the involution. We come now to generalized EP-inverse of a triangular matrix over

A

. To prove the main results, some lemmas are needed. We begin with

Lemma 2.1.

Let

a \in A^{d}

and

b \in A

. Then the following are equivalent:

(1): $(1 - a^{d} a) b = 0$ .
(2): $(1 - a a^{d}) b = 0$ .
(3): $a^{π} b = 0$ .

Proof.

(1) \Rightarrow (3)

Since

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

, we have

b = a^{d} a b

. In view of ????,

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

. Thus,

(1 - a a^{d}) b = (1 - a a^{d}) {(a^{d})}^{2} {(a^{d})}^{#} a b = 0 .

(3) \Rightarrow (2)

Since

a^{d} = {(a^{d})}^{2} a = a^{d} [a^{d} {(a^{d})}^{#} a^{d}] a = [{(a^{d})}^{2} {(a^{d})}^{#}] a a^{d} = a^{d} a a^{d}

. Then

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

; and so

(1 - a a^{d}) b = (1 - a a^{d}) a^{d} a^{2} a^{d} b = 0

, as required.

(2) \Rightarrow (1)

Since

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

, we get

b = a a^{d} b

. Therefore

(1 - a^{d} a) b = (1 - a^{d}) a a^{d} b = (a - a^{d} a^{2}) a^{d} b = 0

, as asserted. □

Let

A

be a Banach *-algebra. Then

M_{2} (A)

is a Banach *-algebra with *-transpose as the involution. We come now to generalized EP-inverse of a triangular matrix over

A

Lemma 2.2.

Let

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

(1): If $a, d \in A^{d}$ , then $x \in M_{2} {(A)}^{d}$ and $x^{d} = (\begin{matrix} a^{d} & z \\ 0 & d^{d} \end{matrix})$ , where

$z = \sum_{i = 0}^{\infty} {(a^{d})}^{i + 2} b d^{i} d^{π} + \sum_{i = 0}^{\infty} a^{i} a^{π} b {(d^{d})}^{i + 2} - a^{d} b d^{d} .$
(2): If $a, d \in A^{#}$ and $a^{π} b = 0$ , then $x \in M_{2} {(A)}^{#}$ and

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

Proof.

See [26][Lemma 2.1] and [23][Theorem 2.5]. □

We are ready to prove:

Theorem 2.3.

Let

x = (\begin{matrix} a & b \\ 0 & d \end{matrix}) \in M_{2} (A)

with

a, d \in A^{d}

. Then the following are equivalent:

(1): $x \in M_{2} (A)$ has upper triangular generalized core-EP inverse.
(2): $a, d \in A^{d}$ and

$\sum_{i = 0}^{\infty} a^{i} a^{π} b {(d^{d})}^{i + 2} = 0 .$

In this case,

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

where

z = - a^{d} b d^{d}

Proof.

By virtue of Lemma 2.2, we have

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

where

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

(1) \Rightarrow (2)

By virtue of Theorem 1.1,

x^{d}

has core inverse and that

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

Hence,

{(x^{d})}^{#} = x^{2} [{(x^{d})}^{2} {(x^{d})}^{#}],

and so

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

is a upper triangular matrix. Write

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

Then

x^{d} {({(x^{d})}^{#})}^{2} = {(x^{d})}^{#}, {(x^{d})}^{#} {(x^{d})}^{2} = x^{d}, {(x^{d} {(x^{d})}^{#})}^{*} = x^{d} {(x^{d})}^{#} .

This implies that

a^{d} α^{2} = α, α {(a^{d})}^{2} = a^{d}, {(a^{d} α)}^{*} = a^{d} α .

Hence,

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

By using Theorem 1.1 again,

a \in A^{d}

. Likewise,

d \in A^{d}

. In view of Lemma 2.2,

{(a^{d})}^{π} s = 0

. This implies that

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

Therefore

\sum_{i = 0}^{\infty} a^{i} a^{π} b {(d^{d})}^{i + 2} = 0,

as asserted.

(2) \Rightarrow (1)

By hypothesis, we get

{(a^{d})}^{π} s = (1 - a^{d} a^{2} a^{d}) s = a^{π} s = a^{π} [\sum_{i = 0}^{\infty} {(a^{d})}^{i + 2} b d^{i} d^{π} - a^{d} b d^{d}] = 0 .

Then it follows by Lemma 2.2 that

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

where

t = - {(a^{d})}^{#} s {(d^{d})}^{#} .

Hence,

t = - {(a^{d})}^{#} [\sum_{i = 0}^{\infty} {(a^{d})}^{i + 2} b d^{i} d^{π} - a^{d} b d^{d}] {(d^{d})}^{#} = {(a^{d})}^{#} a^{d} b d^{d} {(d^{d})}^{#} .

Then we have

{(x^{d})}^{2} = (\begin{matrix} {(a^{d})}^{2} & w \\ 0 & {(d^{d})}^{2} \end{matrix}),

where

w = \sum_{i = 0}^{\infty} {(a^{d})}^{i + 3} b d^{i} d^{π} - {(a^{d})}^{2} b d^{d} - a^{d} b {(d^{d})}^{2} .

Therefore

\begin{matrix} x^{d} & = & {(x^{d})}^{2} {(x^{d})}^{#} \\ = & (\begin{matrix} {(a^{d})}^{2} & w \\ 0 & {(d^{d})}^{2} \end{matrix}) (\begin{matrix} {(a^{d})}^{#} & t \\ 0 & {(d^{d})}^{#} \end{matrix}) \\ = & (\begin{matrix} a^{d} & z \\ 0 & d^{d} \end{matrix}), \end{matrix}

where

\begin{matrix} z & = & {(a^{d})}^{2} t + w {(d^{d})}^{#} \\ = & {(a^{d})}^{2} [{(a^{d})}^{#} a^{d} b d^{d} {(d^{d})}^{#}] - [{(a^{d})}^{2} b d^{d} + a^{d} b {(d^{d})}^{2}] {(d^{d})}^{#} \\ = & {(a^{d})}^{2} b d^{d} {(d^{d})}^{#} - a^{d} (a^{d} b + b d^{d}) d^{d} {(d^{d})}^{#} \\ = & {(a^{d})}^{2} b d^{d} {(d^{d})}^{#} - {(a^{d})}^{2} b d^{d} {(d^{d})}^{#} - a^{d} [b {(d^{d})}^{2} {(d^{d})}^{#}] \\ = & - a^{d} b d^{d} \end{matrix}

This completes the proof. □

Corollary 2.4.

Let

α = (\begin{matrix} a & b \\ 0 & d \end{matrix}) \in M_{2} (A)

with

a, d \in A^{d}

. If

a^{π} b d^{d} = 0,

then

α \in M_{2} {(A)}^{d}

and

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

Proof.

Since

a^{π} b d^{d} = 0,

it follows by Theorem 1.1 that

a^{π} b {(d^{d})}^{2} {(d^{d})}^{#} = 0

; hence,

a^{π} b d^{d} = [a^{π} b {(d^{d})}^{2} {(d^{d})}^{#}] b^{d} b = 0 .

By using Lemma 2.1, we have

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

, and so

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

. Then

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

In light of Theorem 2.3,

α^{d} = (\begin{matrix} a^{d} & - a^{d} b d^{d} \\ 0 & d^{d} \end{matrix}),

as asserted. □

It is very hard to determine the core-EP inverse of a triangular complex matrix (see [10]). As a consequence of Theorem 2.3, we now derive the following.

Corollary 2.5.

Let

M = (\begin{matrix} A & B \\ 0 & D \end{matrix})

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

. If

\sum_{i = 0}^{i (A)} A^{i} A^{π} B {(D^{D})}^{i + 2} = 0,

then

M^{D} = (\begin{matrix} A^{D} & Z \\ 0 & D^{D} \end{matrix}),

where

Z = - A^{D} B D^{D} .

Proof.

Since the generalized core-EP inverse and generalized core-EP inverse coincide with each other for a complex matrix, we obtain the result by Theorem 2.3. □

Corollary 2.6.

Let

M = (\begin{matrix} A & B \\ 0 & D \end{matrix})

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

. If A is invertible, then

M^{D} = (\begin{matrix} A^{- 1} & - A^{- 1} B D^{D} \\ 0 & D^{D} \end{matrix}) .

Proof.

Straightforward. □

The condition "

x^{d} \in M_{2} (A)

is upper triangular" in Theorem 2.3 is necessary as the following shows.

Example 2.7.

Let σ and τ be linear operators, acting on separable Hilbert space

l_{2} (N)

with the conjugate adjoint as an involution, defined as follows respectively:

\begin{matrix} σ (x_{1}, x_{2}, x_{3}, x_{4}, \dots) & = & (0, x_{1}, x_{2}, x_{3}, \dots), \\ τ (x_{1}, x_{2}, x_{3}, x_{4}, \dots) & = & (x_{2}, x_{3}, x_{4}, x_{5}, \dots) . \end{matrix}

Then

τ σ = 1

. Take

M = (\begin{matrix} σ & 1 - σ τ \\ 0 & τ \end{matrix})

. Then

M^{d} = M^{- 1} = (\begin{matrix} τ & 0 \\ 1 - σ τ & σ \end{matrix}) .

In this case, M is upper triangular matrix, but its generalized core-EP inverse is lower triangular.

3. Triangular Matrices with $C^{*}$ -Algebra Entries

The aim of this section is to investigate the generalized core-EP inverse of triangular matrices over a

C^{*}

-algebra. Throughout this section,

A

is always a

C^{*}

-algebra. We start by

Lemma 3.1.

Let

A

be a

C^{*}

-algebra and let

a \in A^{#} ⋂ A^{†}

. Then

a^{†} a a^{#} = a^{†} .

Proof.

Since

(1 - a a^{d}) a^{d} = 0

, by virtue of [23][Lemma 2.4], we have

(1 - a a^{#}) a^{d} = 0

. This implies that

a^{†} {[(1 - a a^{#}) a^{d} (a^{2} a^{†})]}^{*} = 0

, and so

a^{†} {[(1 - a a^{#}) (a a^{†})]}^{*} = 0

. Hence,

a^{†} {(a a^{†})}^{*} {(1 - a a^{#})}^{*} = 0

. Therefore

a^{†} (1 - a a^{#}) = 0

, as required. □

Set

e_{a} = 1 - a a^{†}

and

f_{d} = 1 - d^{†} d

. Then we derive

Lemma 3.2.

Let

A

be a

C^{*}

-algebra and let

a, d \in A^{#} ⋂ A^{†}

. Then

d d^{#} [1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})] = [1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})] d d^{#} .

Proof.

It is easy to check that

{(e_{a} b d^{†} d d^{#})}^{*} e_{a} b d^{†} = {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}) = {(e_{a} b d^{†})}^{*} (e_{a} b d^{†} d d^{#}) .

Then

d d^{#} {(e_{a} b d^{†})}^{*} e_{a} b d^{†} = {(e_{a} b d^{†})}^{*} e_{a} b d^{†} d d^{#} .

Therefore

d d^{#} [1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})] = [1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})] d d^{#},

as asserted. □

In [15], Li and Du investigate the core inverse of a triangular block complex matrix. We now extend Li and Du’s result to block operator matrices over a

C^{*}

-algebra by a new route.

Lemma 3.3.

Let

A

be a

C^{*}

-algebra and let

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

. If

a, d \in A^{#}, a^{π} b d^{π}

= 0

and

e_{a} b f_{d} = 0

, then

x \in A^{#}

. In this case,

x^{#} = (\begin{matrix} α & β \\ γ & δ \end{matrix})

, where

\begin{matrix} α & = & a^{#} + [a^{π} b d^{†} - a^{#} b] d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*}, \\ β & = & (a^{π} b d^{†} - a^{#} b) d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1}, \\ γ & = & d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*}, \\ δ & = & d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} . \end{matrix}

Proof.

Since

a \in A^{#}

, by virtue of [23][Lemma 2.1], it has group inverse, and so a is regular. As

A

is a

C^{*}

-algebra, it follows by [14][Theorem 2.8] that

a \in A^{†}

. Likewise,

d \in A^{†}

. Since every

C^{*}

-algebra has the symmetry property, we have

1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}) \in A^{- 1}

Let

z = (\begin{matrix} α & β \\ γ & δ \end{matrix})

, where

α, β, γ

and

δ

as defined above.

Claim 1.

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

Since

e_{a} b f_{d} = 0

, by virtue of Lemma 3.1, we have

\begin{matrix} (1 - a a^{#}) b d^{#} & = & (1 - a a^{†} a a^{#}) b d^{#} \\ = & (1 - a a^{†}) b d^{#} \\ = & (1 - a a^{†}) b d^{†} d d^{#} \\ = & e_{a} b d^{†} . \end{matrix}

Hence,

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

Therefore

\begin{matrix} a α + b γ \\ = & a a^{#} + (1 - a a^{#}) b d^{#} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} {(e_{a} b d^{†})}^{*} \\ = & a a^{#} + e_{a} b d^{†} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} {(e_{a} b d^{†})}^{*} . \end{matrix}

Hence,

{(a α + b γ)}^{*} = a α + b γ .

By virtue of Lemma 3.1, we have

\begin{matrix} d d^{#} {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}) & = & {(e_{a} b d^{†} d d^{#})}^{*} (e_{a} b d^{†}) \\ = & {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}) \\ = & {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}) d d^{#} . \end{matrix}

Hence,

\begin{matrix} d d^{#} [1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})] & = & d d^{#} + d d^{#} {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}) \\ = & d d^{#} + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}) d d^{#} \\ = & [1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})] d d^{#} . \end{matrix}

Thus, we derive that

d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} = {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} d d^{#} .

Since

d δ = d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1},

we have

{(d δ)}^{*} = d δ .

In view of Lemma 3.2, we verify that

\begin{matrix} a β + b δ \\ = & - a a^{#} b d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} + b d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} \\ = & [1 - a a^{#}] b d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} \\ = & [1 - a a^{#}] b (d^{#} d d^{#}) {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} \\ = & [1 - a a^{#}] b d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} d d^{#} \\ = & e_{a} b d^{†} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} d d^{#} \\ = & e_{a} b d^{†} (d d^{#}) {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} \\ = & e_{a} b d^{†} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} \end{matrix}

Therefore

\begin{matrix} {(a β + b δ)}^{*} \\ = & {[(1 - a a^{#}) b d^{#} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} d d^{#}]}^{*} \\ = & d d^{#} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} {[(1 - a a^{#}) b d^{#}]}^{*} \\ = & d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} \\ = & {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} \\ = & d γ . \end{matrix}

We compute that

\begin{matrix} x z & = & (\begin{matrix} a & b \\ 0 & d \end{matrix}) (\begin{matrix} α & β \\ γ & δ \end{matrix}) \\ = & (\begin{matrix} a α + b γ & a β + b δ \\ d γ & d δ \end{matrix}) \end{matrix}

Therefore

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

Claim 2.

x z^{2} = z

We compute that

\begin{matrix} (1 - x z) z \\ = & (\begin{matrix} 1 - (a α + b γ) & - (a β + b δ) \\ - d γ & 1 - d δ \end{matrix}) (\begin{matrix} α & β \\ γ & δ \end{matrix}) \\ = & (\begin{matrix} [1 - (a α + b γ)] α - (a β + b δ) γ & [1 - (a α + b γ)] β - (a β + b δ) δ \\ - d γ α + (1 - d δ) γ & - d γ β + (1 - d δ) δ \end{matrix}) . \end{matrix}

Obviously, we have

a α + b γ = a a^{#} + (a β + b δ) {(e_{a} b δ^{†})}^{*} .

Then we compute that

\begin{matrix} [1 - (a α + b γ)] α \\ = & [1 - a a^{#} - (a β + b δ) {(e_{a} b δ^{†})}^{*}] [a^{#} + (a^{π} b d^{†} - a^{#} b) γ] \\ = & [1 - a a^{#} - (a β + b δ) {(e_{a} b δ^{†})}^{*}] [(a^{π} b d^{†} - a^{#} b) γ] \\ = & [1 - a a^{#} - (a β + b δ) {(e_{a} b δ^{†})}^{*}] [(a^{π} b d^{†}] γ \\ = & [1 - a a^{#} - (a β + b δ) {(e_{a} b δ^{†})}^{*}] (e_{a} b d^{†}) γ \\ = & e_{a} b d^{†} - (a β + b δ) {(e_{a} b δ^{†})}^{*} (e_{a} b d^{†}) γ \\ = & e_{a} b d^{†} - e_{a} b d^{†} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b δ^{†})}^{*} (e_{a} b d^{†}) γ \\ = & e_{a} b d^{†} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} [1 + {(e_{a} b d^{†})}^{*} e_{a} b d^{†} - {(e_{a} b δ^{†})}^{*} e_{a} b d^{†}] γ \\ = & e_{a} b d^{†} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} γ \\ = & (a β + b δ) γ; \end{matrix}

hence,

[1 - (a α + b γ)] α - (a β + b δ) γ = 0 .

Analogously, we derive that

\begin{matrix} [1 - (a α + b γ)] β \\ = & [1 - a a^{#} - (a β + b δ) {(e_{a} b δ^{†})}^{*}] [(a^{π} b d^{†} - a^{#} b) d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} \\ = & [1 - a a^{#} - (a β + b δ) {(e_{a} b δ^{†})}^{*}] [(a^{π} b d^{†} - a^{#} b) δ \\ = & (a β + b δ) δ; \end{matrix}

hence,

[1 - (a α + b γ)] β - (a β + b δ) δ = 0 .

Also we check that

\begin{matrix} - d γ α + (1 - d δ) γ \\ = & - d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} [a^{#} + [a^{π} b d^{†} - a^{#} b] \\ d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*}] \\ + & [1 - d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1}] d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} \\ = & - {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} [a^{π} b d^{†} - a^{#} b] d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*}] \\ + & [1 - d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1}] d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} \\ = & - {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} a^{π} b d^{†} d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*}] \\ + & [1 - {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1}] d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} \\ = & {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} [- {(e_{a} b d^{†})}^{*} a^{π} b d^{†} + {(e_{a} b d^{†})}^{*} e_{a} b d^{†}] \\ d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*}] \\ = & {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} [{(e_{a} b d^{†})}^{*} e_{a} (1 - a^{π}) b d^{†}] d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*}] \\ = & 0 . \end{matrix}

Furthermore, we verify that

\begin{matrix} - d γ β + (1 - d δ) δ \\ = & - d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} [(a^{π} b d^{†} - a^{#} b) d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1}] \\ + & [1 - d d^{#} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1}] d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} \\ = & - {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} (a^{π} b d^{†} - a^{#} b) d^{#} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} \\ + & [1 - {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1}] d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} \\ = & - {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} a^{π} b d^{†} d^{#} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} \\ + & [1 - {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1}] d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} \\ = & {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} [- {(e_{a} b d^{†})}^{*} a^{π} b d^{†} + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})] d^{#} \\ {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} \\ = & {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} [- {(e_{a} b d^{†})}^{*} a^{π} b d^{†} + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})] d^{#} \\ {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} \\ = & {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} e_{a} (1 - a^{π}) b d^{†} + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})] d^{#} \\ {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} \\ = & 0 . \end{matrix}

Therefore

x z^{2} = z

Claim 3.

x z x = x

\begin{matrix} (1 - x z) x \\ = & (\begin{matrix} 1 - (a α + b γ) & - (a β + b δ) \\ - d γ & 1 - d δ \end{matrix}) (\begin{matrix} a & b \\ 0 & d \end{matrix}) \\ = & (\begin{matrix} [1 - (a α + b γ)] a & [1 - (a α + b γ)] b - (a β + b δ) d \\ - d γ a γ & - d γ b + (1 - d δ) d \end{matrix}) . \end{matrix}

Obviously, we have

{(e_{a})}^{*} a = (1 - a a^{†}) a = 0 .

Then

\begin{matrix} [1 - (a α + b γ)] a \\ = & [1 - a a^{#} - e_{a} b d^{†} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} {(e_{a} b d^{†})}^{*}] a \\ = & [(1 - a a^{#}) a - e_{a} b d^{†} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} {(e_{a} b d^{†})}^{*} {(e_{a})}^{*} a)] \\ = & 0 . \end{matrix}

Similarly, we have

\begin{matrix} - d γ a & = & - d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} a \\ = & - d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} {(e_{a})}^{*} a \\ = & 0 . \end{matrix}

Clearly,

(1 - a a^{#}) b = (1 - a a^{#} a a^{†}) b = e_{a} b = e_{a} b (f_{d} + d^{†} d) = e_{a} b d^{d a g} d

. Then we have

\begin{matrix} [1 - (a α + b γ)] b \\ = & [1 - a a^{#} - e_{a} b d^{†} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} {(e_{a} b d^{†})}^{*}] b \\ = & e_{a} b - e_{a} b d^{†} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} {(e_{a} b d^{†})}^{*} (e_{a} b) \\ = & e_{a} b d^{†} d - e_{a} b d^{†} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} {(e_{a} b d^{†})}^{*} (e_{a} b d^{d a g}) d \\ = & e_{a} b d^{†} [1 - {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} {(e_{a} b d^{†})}^{*} (e_{a} b d^{d a g})] d \\ = & e_{a} b d^{†} {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1} [1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}) - {(e_{a} b d^{†})}^{*} (e_{a} b d^{d a g})] d \\ = & e_{a} b d^{†} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} d \\ = & (a β + b δ) d . \end{matrix}

Finally, we verify that

\begin{matrix} d γ b & = & d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} b \\ = & d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} {(e_{a})}^{*} b \\ = & d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} {(e_{a} b d^{†})}^{*} e_{a} b (f_{d} + d^{†} d) \\ = & d d^{#} [1 - {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1}] [{(e_{a} b d^{†})}^{*} (e_{a} b d^{†})] d \\ = & d d^{#} [1 - {(1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†}))}^{- 1}] d \\ = & d d^{#} d - d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} d \\ = & d - d d^{#} {[1 + {(e_{a} b d^{†})}^{*} (e_{a} b d^{†})]}^{- 1} d \\ = & d (1 - δ d) \\ = & (1 - d δ) d . \end{matrix}

Therefore

x z x = x

. In light of [24][Theorem 3.3],

x \in A^{#}

and

x^{#} = z

, as asserted. □

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

Theorem 3.4.

Let

A

be a

C^{*}

-algebra and

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

. If

a, d \in A^{d}

, then

x \in A^{d}

and

x^{d} = (\begin{matrix} {(a^{d})}^{2} α + w γ & {(a^{d})}^{2} β + w δ \\ {(d^{d})}^{2} γ & {(d^{d})}^{2} δ \end{matrix}),

where

\begin{matrix} w & = & \sum_{i = 0}^{\infty} {(a^{d})}^{i + 3} b d^{i} d^{π} + \sum_{i = 0}^{\infty} a^{i} a^{π} b {(d^{d})}^{i + 3} - {(a^{d})}^{2} b d^{d} - a^{d} b {(d^{d})}^{2}, \\ s & = & \sum_{i = 0}^{\infty} {(a^{d})}^{i + 2} b d^{i} d^{π} + \sum_{i = 0}^{\infty} a^{i} a^{π} b {(d^{d})}^{i + 2} - a^{d} b d^{d}, \\ α & = & a^{2} a^{d} + [a^{π} s {[d^{d}]}^{†} - a^{2} a^{d} s] d^{2} d^{d} {(1 + {(e_{a^{d}} s {[d^{d}]}^{†})}^{*} (e_{a^{d}} s {[d^{d}]}^{†}))}^{- 1} {(e_{a^{d}} s {[d^{d}]}^{†})}^{*}, \\ β & = & (a^{π} s {[d^{d}]}^{†} - a^{2} a^{d} s) d^{2} d^{d} {[1 + {(e_{a^{d}} s {[d^{d}]}^{†})}^{*} (e_{a^{d}} s {[d^{d}]}^{†})]}^{- 1}, \\ γ & = & d^{2} d^{d} {[1 + {(e_{a^{d}} s {[d^{d}]}^{†})}^{*} (e_{a^{d}} s {[d^{d}]}^{†})]}^{- 1} {(e_{a^{d}} s {[d^{d}]}^{†})}^{*}, \\ δ & = & d^{2} d^{d} {[1 + {(e_{a^{d}} s {[d^{d}]}^{†})}^{*} (e_{a^{d}} s {[d^{d}]}^{†})]}^{- 1} . \end{matrix}

Proof.

In view of Theorem 1.1,

a, d \in A^{d}

and

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

. By virtue of Lemma 2.2, we have

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

where

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

Since

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

, i.e.,

a^{d}

is regular. As

A

is a

C^{*}

-algebra, it follows by [14][Theorem 2.8],

a^{d} \in A^{†}

. Likewise,

d^{d} \in A^{†}

. Let

e_{a^{d}} = 1 - a^{d} {(a^{d})}^{†}

and

f_{d^{d}} = 1 - {[d^{d}]}^{†} d^{d}

. Then we check that

\begin{matrix} {(a^{d})}^{π} s {(d^{d})}^{π} & = & a^{π} [\sum_{i = 0}^{\infty} {(a^{d})}^{i + 2} b d^{i} d^{π} + \sum_{i = 0}^{\infty} a^{i} a^{π} b {(d^{d})}^{i + 2} - a^{d} b d^{d}] d^{π} \\ = & a^{π} [\sum_{i = 0}^{\infty} a^{i} a^{π} b {(d^{d})}^{i + 2}] d^{π} \\ = & 0 \end{matrix}

and

\begin{matrix} e_{a^{d}} s f_{d^{d}} & = & [1 - a^{d} {(a^{d})}^{†}] [\sum_{i = 0}^{\infty} {(a^{d})}^{i + 2} b d^{i} d^{π} + \sum_{i = 0}^{\infty} a^{i} a^{π} b {(d^{d})}^{i + 2} \\ - & a^{d} b d^{d}] [1 - {[d^{d}]}^{†} d^{d}] \\ = & [1 - a^{d} {(a^{d})}^{†}] [\sum_{i = 0}^{\infty} a^{i} a^{π} b {(d^{d})}^{i + 2}] [1 - {[d^{d}]}^{†} d^{d}] \\ = & 0 . \end{matrix}

It follows by Lemma 3.3 that

{(x^{d})}^{#} = (\begin{matrix} α & β \\ γ & δ \end{matrix}),

where

\begin{matrix} α & = & {[a^{d}]}^{#} + [a^{π} s {[d^{d}]}^{†} - {[a^{d}]}^{#} s] {[d^{d}]}^{#} {[1 + {(e_{a^{d}} s {[d^{d}]}^{†})}^{*} (e_{a^{d}} s {[d^{d}]}^{†})]}^{- 1} {(e_{a^{d}} s {[d^{d}]}^{†})}^{*}, \\ β & = & (a^{π} s {[d^{d}]}^{†} - {[a^{d}]}^{#} s) {[d^{d}]}^{#} {[1 + {(e_{a^{d}} s {[d^{d}]}^{†})}^{*} (e_{a^{d}} s {[d^{d}]}^{†})]}^{- 1}, \\ γ & = & {[d^{d}]}^{#} {[1 + {(e_{a^{d}} s {[d^{d}]}^{†})}^{*} (e_{a^{d}} s {[d^{d}]}^{†})]}^{- 1} {(e_{a^{d}} s {[d^{d}]}^{†})}^{*}, \\ δ & = & {[d^{d}]}^{#} {[1 + {(e_{a^{d}} s {[d^{d}]}^{†})}^{*} (e_{a^{d}} s {[d^{d}]}^{†})]}^{- 1} . \end{matrix}

Set

\begin{matrix} w & : = & a^{d} s + s d^{d} \\ = & \sum_{i = 0}^{\infty} {(a^{d})}^{i + 3} b d^{i} d^{π} + \sum_{i = 0}^{\infty} a^{i} a^{π} b {[d^{d}]}^{i + 3} - {(a^{d})}^{2} b d^{d} - a^{d} b {(d^{d})}^{2} . \end{matrix}

Accordingly,

\begin{matrix} x^{d} & = & {(x^{d})}^{2} {(x^{d})}^{#} \\ = & (\begin{matrix} {(a^{d})}^{2} & w \\ 0 & {(d^{d})}^{2} \end{matrix}) (\begin{matrix} α & β \\ γ & δ \end{matrix}) \\ = & (\begin{matrix} {(a^{d})}^{2} α + w γ & {(a^{d})}^{2} β + w δ \\ {(d^{d})}^{2} γ & {(d^{d})}^{2} δ \end{matrix}) . \end{matrix}

In view of Theorem 1.1,

a^{d} = {(a^{d})}^{2} {[a^{d}]}^{#}

. Hence,

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

Similarly, we have

{(d^{d})}^{#} = d^{2} d^{d} .

Therefore we verify the formulas of

α, β, γ

and

δ

mentioned before. □

Corollary 3.5.

Let

A

be a

C^{*}

-algebra and

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

. If

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

, then

x \in A^{d}

and

x^{d} = (\begin{matrix} s d^{d} γ & s d^{d} δ \\ {(d^{d})}^{2} γ & {(d^{d})}^{2} δ \end{matrix}),

where

\begin{matrix} s & = & \sum_{i = 0}^{\infty} a^{i} b {(d^{d})}^{i + 2}, \\ α & = & s {[d^{d}]}^{†} d^{2} d^{d} {(1 + {(s {[d^{d}]}^{†})}^{*} (s {[d^{d}]}^{†}))}^{- 1} {(s {[d^{d}]}^{†})}^{*}, \\ β & = & (s {[d^{d}]}^{†}) d^{2} d^{d} {[1 + {(s {[d^{d}]}^{†})}^{*} (s {[d^{d}]}^{†})]}^{- 1}, \\ γ & = & d^{2} d^{d} {[1 + {(s {[d^{d}]}^{†})}^{*} (s {[d^{d}]}^{†})]}^{- 1} {(s {[d^{d}]}^{†})}^{*}, \\ δ & = & d^{2} d^{d} {[1 + {(s {[d^{d}]}^{†})}^{*} (s {[d^{d}]}^{†})]}^{- 1} . \end{matrix}

Proof.

Since

a \in A^{q n i l}

, we see that

a^{d} = 0

, and therefore we obtain the result by Theorem 3.4. □

Corollary 3.6.

Let

A

be a

C^{*}

-algebra and

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

. If

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

, then

x \in A^{d}

and

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

Proof.

Since

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

d^{d} = 0

, we complete the proof by Theorem 3.4. □

Since the algebra

C^{n \times n}

of all

n \times n

complex matrices is a

C^{*}

-algebra, the following result gives a simpler formula to compute the core-EP inverse of a block triangular complex matrices (see [10]).

Corollary 3.7.

Let

A

be a

C^{*}

-algebra and

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

. If

a, d \in A^{D}

, then

x \in A^{D}

and

x^{D} = (\begin{matrix} {(a^{D})}^{2} α + w γ & {(a^{D})}^{2} β + w δ \\ {(d^{D})}^{2} γ & {(d^{D})}^{2} δ \end{matrix}),

where

\begin{matrix} w & = & \sum_{i = 0}^{m} {(a^{D})}^{i + 3} b d^{i} d^{π} + \sum_{i = 0}^{m} a^{i} a^{π} b {(d^{D})}^{i + 3} - {(a^{D})}^{2} b d^{D} - a^{D} b {(d^{D})}^{2}, \\ s & = & \sum_{i = 0}^{m} {(a^{D})}^{i + 2} b d^{i} d^{π} + \sum_{i = 0}^{m} a^{i} a^{π} b {(d^{D})}^{i + 2} - a^{d} b d^{D}, \\ α & = & a^{2} a^{D} + [a^{π} s {[d^{D}]}^{†} - a^{2} a^{D} s] d^{2} d^{D} {(1 + {(e_{a^{D}} s {[d^{D}]}^{†})}^{*} (e_{a^{D}} s {[d^{D}]}^{†}))}^{- 1} \\ {(e_{a^{D}} s {[d^{D}]}^{†})}^{*}, \\ β & = & (a^{π} s {[d^{D}]}^{†} - a^{2} a^{D} s) d^{2} d^{D} {[1 + {(e_{a^{D}} s {[d^{D}]}^{†})}^{*} (e_{a^{D}} s {[d^{D}]}^{†})]}^{- 1}, \\ γ & = & d^{2} d^{D} {[1 + {(e_{a^{D}} s {[d^{D}]}^{†})}^{*} (e_{a^{D}} s {[d^{D}]}^{†})]}^{- 1} {(e_{a^{D}} s {[d^{D}]}^{†})}^{*}, \\ δ & = & d^{2} d^{D} {[1 + {(e_{a^{D}} s {[d^{D}]}^{†})}^{*} (e_{a^{D}} s {[d^{D}]}^{†})]}^{- 1}, \\ m & = & max {i n d (a), i n d (d)} . \end{matrix}

Proof.

Since

a, d \in A^{D}

, it follows by [3][Corollary 3.4] that

a, d \in A^{d} ⋂ A^{D}

. By virtue of Theorem 3.4,

x \in A^{d}

. Clearly,

x \in A^{D}

. By using [3][Corollary 3.4] again,

x \in A^{D}

and

x^{D} = x^{d}

, as required. □

4. Applications

Lex X and Y be Hilbert spaces, and let

M = (\begin{matrix} A & B \\ C & D \end{matrix}),

where

A \in B {(X)}^{d}, B \in B (X, Y), C \in B (Y, X), D \in B {(Y)}^{d}

. Choose

p = (\begin{matrix} I_{X} & 0 \\ 0 & I_{Y} \end{matrix}) .

Then M can be regarded as the Pierce matrix

{(\begin{matrix} p M p & p M p^{π} \\ p^{π} M p & p^{π} M p^{π} \end{matrix})}_{p} .

Here, every subblock matrices can be seen as the bounded linear operators on Hilbert space

X \oplus Y

. Throughout this section, without loss the generality, we consider M as the block operator matrix in a specifical case

X = Y

. In this case,

B (X \oplus X)

is indeed a

C^{*}

-algebra. The following lemma is crucial.

Lemma 4.1.

Let

a \in A^{d}

and

b \in A^{q n i l}

. If

a^{*} b = 0

and

b a = 0

, then

a + b \in A^{d}

. In this case,

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

Proof.

Since

a \in A^{d}

, by virtue of Theorem 1.1, there exist

x \in A^{#}

and

y \in A^{q n i l}

such that

a = x + y, x^{*} y = 0, y x = 0

. As in the proof of [3],

x = a a^{d} a

and

y = a - a a^{d} a

. Then

a = x + (y + b)

. Since

b y = b (a - a a^{d} a) = 0

, it follows by [26][Lemma 2.10] that

y + b \in A^{q n i l}

. We directly verify that

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

In light of Theorem 1.1,

a + b \in A^{d}

. In this case,

{(a + b)}^{d} = x^{#} = a^{d},

as asserted. □

Theorem 4.2.

C A = 0, C B = 0

and

D^{*} C = 0

, then M has generalized core-EP inverse. In this case,

M^{d} = (\begin{matrix} {(A^{d})}^{2} Λ + W Γ & {(A^{d})}^{2} Σ + W Δ \\ {(D^{d})}^{2} Γ & {(D^{d})}^{2} Δ \end{matrix}),

where

\begin{matrix} W & = & \sum_{i = 0}^{\infty} {(A^{d})}^{i + 3} B D^{i} D^{π} + \sum_{i = 0}^{\infty} A^{i} A^{π} B {(D^{d})}^{i + 3} - {(A^{d})}^{2} B D^{d} - A^{d} B {(D^{d})}^{2}, \\ S & = & \sum_{i = 0}^{\infty} {(A^{d})}^{i + 2} B D^{i} D^{π} + \sum_{i = 0}^{\infty} A^{i} A^{π} B {(D^{d})}^{i + 2} - A^{d} B D^{d}, \\ Λ & = & A^{2} A^{d} + [A^{π} S {[D^{d}]}^{†} - A^{2} A^{d} S] D^{2} D^{d} {(I + {(e_{A^{d}} S {[D^{d}]}^{†})}^{*} (e_{A^{d}} S {[D^{d}]}^{†}))}^{- 1} \\ {(e_{A^{d}} S {[D^{d}]}^{†})}^{*}, \\ Σ & = & (A^{π} S {[D^{d}]}^{†} - A^{2} A^{d} S) D^{2} D^{d} {[I + {(e_{A^{d}} S {[D^{d}]}^{†})}^{*} (e_{A^{d}} S {[D^{d}]}^{†})]}^{- 1}, \\ Γ & = & D^{2} D^{d} {[I + {(e_{A^{d}} S {[D^{d}]}^{†})}^{*} (e_{A^{d}} S {[D^{d}]}^{†})]}^{- 1} {(e_{A^{d}} S {[D^{d}]}^{†})}^{*}, \\ Δ & = & D^{2} D^{d} {[I + {(e_{A^{d}} S {[D^{d}]}^{†})}^{*} (e_{A^{d}} S {[D^{d}]}^{†})]}^{- 1} . \end{matrix}

Proof.

Write

M = P + Q

, where

P = (\begin{matrix} A & B \\ 0 & D \end{matrix}), Q = (\begin{matrix} 0 & 0 \\ C & 0 \end{matrix}) .

We easily check that

\begin{matrix} P^{*} Q & = & (\begin{matrix} A^{*} & 0 \\ B^{*} & D^{*} \end{matrix}) (\begin{matrix} 0 & 0 \\ C & 0 \end{matrix}) = (\begin{matrix} 0 & 0 \\ D^{*} C & 0 \end{matrix}) = 0, \\ Q P & = & (\begin{matrix} 0 & 0 \\ C & 0 \end{matrix}) (\begin{matrix} A & B \\ 0 & D \end{matrix}) = (\begin{matrix} 0 & 0 \\ C A & C B \end{matrix}) = 0 \end{matrix}

Since A and D have generalized core-EP inverses, it follows by Theorem 3.4 that P has generalized core-EP inverse, and that

P^{d} = (\begin{matrix} {(A^{d})}^{2} Λ + W Γ & {(A^{d})}^{2} Σ + W Δ \\ {(D^{d})}^{2} Γ & {(D^{d})}^{2} Δ \end{matrix}),

where

W, S, Λ, Σ, Γ, Δ

as defined before. Obviously, Q is nilpotent, and so it is quasinilpotent. According to Lemma 4.1,

M^{d} = P^{d}

, required. □

Corollary 4.3.

B D = 0, B C = 0

and

A^{*} B = 0

, then M has generalized core-EP inverse. In this case,

M^{d} = (\begin{matrix} {(A^{d})}^{2} Δ & {(A^{d})}^{2} Γ \\ {(D^{d})}^{2} Σ + W Δ & {(D^{d})}^{2} Λ + W Γ \end{matrix}),

where

\begin{matrix} W & = & \sum_{i = 0}^{\infty} {(D^{d})}^{i + 3} C A^{i} A^{π} + \sum_{i = 0}^{\infty} D^{i} D^{π} C {(A^{d})}^{i + 3} - {(D^{d})}^{2} C A^{d} - D^{d} C {(A^{d})}^{2}, \\ S & = & \sum_{i = 0}^{\infty} {(D^{d})}^{i + 2} C A^{i} A^{π} + \sum_{i = 0}^{\infty} D^{i} D^{π} C {(A^{d})}^{i + 2} - D^{d} C A^{d}, \\ Λ & = & D^{2} D^{d} + [D^{π} S {[A^{d}]}^{†} - D^{2} D^{d} S] A^{2} A^{d} {(I + {(e_{D^{d}} S {[A^{d}]}^{†})}^{*} (e_{D^{d}} S {[A^{d}]}^{†}))}^{- 1} \\ {(e_{D^{d}} S {[A^{d}]}^{†})}^{*}, \\ Σ & = & (D^{π} S {[A^{d}]}^{†} - D^{2} D^{d} S) A^{2} A^{d} {[I + {(e_{D^{d}} S {[A^{d}]}^{†})}^{*} (e_{D^{d}} S {[A^{d}]}^{†})]}^{- 1}, \\ Γ & = & A^{2} A^{d} {[I + {(e_{D^{d}} S {[A^{d}]}^{†})}^{*} (e_{D^{d}} S {[A^{d}]}^{†})]}^{- 1} {(e_{D^{d}} S {[A^{d}]}^{†})}^{*}, \\ Δ & = & A^{2} A^{d} {[I + {(e_{D^{d}} S {[A^{d}]}^{†})}^{*} (e_{D^{d}} S {[A^{d}]}^{†})]}^{- 1} . \end{matrix}

Proof.

Obviously,

(\begin{matrix} 0 & I \\ I & 0 \end{matrix}) M (\begin{matrix} 0 & I \\ I & 0 \end{matrix}) = (\begin{matrix} D & C \\ B & A \end{matrix})

. Applying Theorem 4.2 to the matrix

(\begin{matrix} D & C \\ B & A \end{matrix})

, we see that

(\begin{matrix} D & C \\ B & A \end{matrix})

has generalized core-EP inverse. This implies that M has generalized core-EP inverse. Additionally,

M^{d} = (\begin{matrix} 0 & I \\ I & 0 \end{matrix}) {(\begin{matrix} D & C \\ B & A \end{matrix})}^{d} (\begin{matrix} 0 & I \\ I & 0 \end{matrix}) .

Therefore we complete the proof by Theorem ???. □

Lemma 4.4.

Let

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

. If

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

and

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

, then

a + b \in A^{d}

. In this case,

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

Proof.

Since

a b = b a

, it follows by [26][Theorem 2.3] that

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

. Likewise, we have

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

. Since

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

, by using [26][Theorem 2.3] again,

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

. Then

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

. Hence,

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

. In view of Theorem 1.1 and [7],

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

. We directly verify that

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

This implies that

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

. We easily verify that

b^{d} (1 + a b^{d}) = (1 + a b^{d}) b^{d}, {(b^{d})}^{*} (1 + a b^{d}) = (1 + a b^{d}) {(b^{d})}^{*} .

Then we derive that

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

According to [7][Theorem 3.5],

{(a + b)}^{d} \in A^{#}

and

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

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

□

Theorem 4.5.

B C = 0, C B = 0, C A = D C, C A^{*} = D^{*} C

and

D^{*} C A = D C A^{*}

, then M has generalized core-EP inverse. In this case,

M^{d} = (\begin{matrix} α & β \\ γ & δ \end{matrix}),

where

\begin{matrix} α & = & {(A^{d})}^{2} Λ + W Γ, \\ β & = & {(A^{d})}^{2} Σ + W Δ, \\ γ & = & - C A^{d} [{(A^{d})}^{2} Λ + W Γ] + (I - C S) {(D^{d})}^{2} Γ, \\ δ & = & - C A^{d} [{(A^{d})}^{2} Σ + W Δ] + (I - C S) {(D^{d})}^{2} Δ \end{matrix}

and

W, S, Λ, Σ, Γ

and Δ constructed as in Theorem 4.2.

Proof.

Write

M = P + Q

, where

P = (\begin{matrix} A & B \\ 0 & D \end{matrix}), Q = (\begin{matrix} 0 & 0 \\ C & 0 \end{matrix}) .

Then

P^{d} = (\begin{matrix} A^{d} & S \\ 0 & D^{d} \end{matrix}),

where

S = \sum_{i = 0}^{\infty} {(A^{d})}^{i + 2} B D^{i} D^{π} + \sum_{i = 0}^{\infty} A^{i} A^{π} B {(D^{d})}^{i + 2} - A^{d} B D^{d} .

Hence, we have

\begin{matrix} I - Q P^{d} & = & I - (\begin{matrix} 0 & 0 \\ C & 0 \end{matrix}) (\begin{matrix} A^{d} & S \\ 0 & D^{d} \end{matrix}) \\ = & (\begin{matrix} I & 0 \\ - C A^{d} & I - C S \end{matrix}) . \end{matrix}

We easily check that

\begin{matrix} P Q & = & (\begin{matrix} A & B \\ 0 & D \end{matrix}) (\begin{matrix} 0 & 0 \\ C & 0 \end{matrix}) = (\begin{matrix} B C & 0 \\ D C & 0 \end{matrix}) \\ = & (\begin{matrix} 0 & 0 \\ C A & C B \end{matrix}) = (\begin{matrix} 0 & 0 \\ C & 0 \end{matrix}) (\begin{matrix} A & B \\ 0 & D \end{matrix}) = Q P, \\ P^{*} Q & = & (\begin{matrix} A^{*} & 0 \\ B^{*} & D^{*} \end{matrix}) (\begin{matrix} 0 & 0 \\ C & 0 \end{matrix}) = (\begin{matrix} 0 & 0 \\ D^{*} C & 0 \end{matrix}) \\ = & (\begin{matrix} 0 & 0 \\ C A^{*} & 0 \end{matrix}) = (\begin{matrix} 0 & 0 \\ C & 0 \end{matrix}) (\begin{matrix} A^{*} & 0 \\ B^{*} & D^{*} \end{matrix}) = Q P, \\ P^{*} (Q P) & = & (\begin{matrix} A^{*} & 0 \\ B^{*} & D^{*} \end{matrix}) (\begin{matrix} 0 & 0 \\ C A & C B \end{matrix}) = (\begin{matrix} 0 & 0 \\ D^{*} C A & D^{*} C B \end{matrix}) \\ = & (\begin{matrix} B C A^{*} & 0 \\ D C A^{*} & 0 \end{matrix}) = (\begin{matrix} B C & 0 \\ D C & 0 \end{matrix}) (\begin{matrix} A^{*} & 0 \\ B^{*} & D^{*} \end{matrix}) = (P Q) P^{*} . \end{matrix}

Since A and D have generalized core-EP inverses, it follows by Theorem 3.4 that P has generalized core-EP inverse, and that

P^{d} = (\begin{matrix} {(A^{d})}^{2} Λ + W Γ & {(A^{d})}^{2} Σ + W Δ \\ {(D^{d})}^{2} Γ & {(D^{d})}^{2} Δ \end{matrix}),

where

W, S, Λ, Σ, Γ, Δ

as defined before. Obviously,

Q^{2} = 0

, and so it is quasinilpotent. According to Lemma 4.4,

\begin{matrix} M^{d} & = & {(I + Q P^{d})}^{- 1} P^{d} \\ = & (I - Q P^{d}) P^{d} \\ = & (\begin{matrix} I & 0 \\ - C A^{d} & I - C S \end{matrix}) (\begin{matrix} {(A^{d})}^{2} Λ + W Γ & {(A^{d})}^{2} Σ + W Δ \\ {(D^{d})}^{2} Γ & {(D^{d})}^{2} Δ \end{matrix}) \\ = & (\begin{matrix} α & β \\ γ & δ \end{matrix}), \end{matrix}

where

\begin{matrix} α & = & {(A^{d})}^{2} Λ + W Γ, \\ β & = & {(A^{d})}^{2} Σ + W Δ, \\ γ & = & - C A^{d} [{(A^{d})}^{2} Λ + W Γ] + (I - C S) {(D^{d})}^{2} Γ, \\ δ & = & - C A^{d} [{(A^{d})}^{2} Σ + W Δ] + (I - C S) {(D^{d})}^{2} Δ . \end{matrix}

as required. □

Corollary 4.6.

B C = 0, C B = 0, A B = B D, A^{*} B = B D^{*}

and

A^{*} B D = A B D^{*}

, then M has generalized core-EP inverse. In this case,

M^{d} = (\begin{matrix} α & β \\ γ & δ \end{matrix}),

where

\begin{matrix} α & = & - B D^{d} [{(D^{d})}^{2} Σ + W Δ] + (I - B S) {(A^{d})}^{2} Δ, \\ β & = & - B D^{d} [{(D^{d})}^{2} Λ + W Γ] + (I - B S) {(A^{d})}^{2} Γ, \\ γ & = & {(D^{d})}^{2} Σ + W Δ, \\ δ & = & {(D^{d})}^{2} Λ + W Γ \end{matrix}

and

W, S, Λ, Σ, Γ

and Δ constructed as in Corollary 4.3.

Proof.

Applying Theorem 4.5 to the matrix

(\begin{matrix} D & C \\ B & A \end{matrix})

, we see that it has generalized core-EP inverse. Analogously to Corollary 4.3, we have

M^{d} = (\begin{matrix} 0 & I \\ I & 0 \end{matrix}) {(\begin{matrix} D & C \\ B & A \end{matrix})}^{d} (\begin{matrix} 0 & I \\ I & 0 \end{matrix}) .

Therefore we obtain the result by Theorem 4.5. □

Conflicts of Interest

The authors declare there is no conflicts of interest.

Data Availability Statement

The data used to support the findings of this study are included within the article.

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Generalized core-EP inverse for triangular operator matrices

Abstract

1. Introduction

2. Triangular Operator Matrices over Banach *-Algebras

3. Triangular Matrices with C * -Algebra Entries

4. Applications

Conflicts of Interest

Data Availability Statement

References

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3. Triangular Matrices with $C^{*}$ -Algebra Entries