#### Abstract

We present some inverses and group inverses results for linear combinations of two idempotents and their products.

#### 1. Introduction

Let be a complex Hilbert space. Denote by the Banach algebra of all bounded linear operators on . and represent the range and the null space of , respectively. The identity onto a Hilbert space is denoted by or if there does not exist confusion. For , the group inverse [1] of is the unique element such that exists if and only if has finite ascent and descent such that [2]. When ind, the group inverse reduces to the standard inverse; that is, . In particular, if and if for a scalar . One of the most important applications of group inverses is to derive some closed-form formulas for general solutions to operator equations. An operator is said to be idempotent if . If is group invertible, then is closed and the spectral idempotent is given by . The operator matrix form of with respect to the space decomposition is given by , where is invertible on [1].

Idempotents are a type of simplest operators. Various expressions or equalities consisting of idempotents occur in operator theory and its applications. Some previous work on linear combinations of idempotents in statistics can be found in [3]. There have been several papers devoted to the invertibility of a linear combination of two idempotent operators in a Hilbert space or in a -algebra. In [4], Buckholtz studied the idempotency of the difference of two operators in a Hilbert space. Du and Li in [5] had established the spectral characterization of generalized projections. In [6], the invertibility of the difference of two orthogonal projectors in a -algebra was studied. Li in [7] had investigated how to express Moore-Penrose inverses of products and differences. In [8], J. K. Baksalary and O. M. Baksalary discussed the invertibility of a linear combination of idempotent matrices. This paper was improved by Koliha and Rakočević [9] by showing that the rank of a linear combination of two idempotents is constant. Du et al. [10] extended the conclusion on an infinite-dimensional Hilbert space.

The purpose of this note is to characterize the invertibility and the group invertibility of the linear combinations of idempotents and their products , , , , , and . These linear combinations were studied by some authors in recent years [2, 4, 6–9, 11–18]. Some formulas for the group inverse of a sum of two bounded operators under some conditions were given (see [7, 11, 14, 19, 20]). Here, we will find group invertibility for a linear combination of two idempotents under the condition . A previous study of the group invertibility of two idempotents was made under the conditions or or or . It is clear that these conditions are special cases of our results.

#### 2. Main Results and Proofs

We start by discussing some lemmas. Let and be two idempotents. Now, we consider the invertibility of . This problem is the subject of Buckholtz’s papers [4] and Koliha and Rakočević’s paper [12].

Lemma 1 (see [12]). * Let and be two idempotents.*(i)* is invertible if and only if and are invertible.*(ii)*If is group invertible, then and are group invertible. *

* Proof. *Since the properties in the lemma are similarly invariant, without loss of generality, we can assume that is an orthogonal projection. In this case, and have the operator matrix representations as follows
with respect to the space decomposition , respectively. Since ,
So
It is clear that item (i) holds, and if is group invertible, then is group invertible, which implies that and are group invertible.

As for -idempotents, we have the following decomposition.

Lemma 2 (see [21, Theorem 2.3]). * Let . Then if and only if*(i)*;*(ii)*there exists a resolution set of the identity and an invertible operator such that
**where denotes the orthogonal direct sum, is an orthogonal projection with , and if , , . *

Lemma 3. *Let be such that is invertible. Then is invertible if and only if the Schur complement is invertible. *

Lemma 4 (see [1, Theorem ]). * Let be such that exists. Then exists if and only if exists and . In this case,
**
where .*

Lemma 5 (see [19, Theorem 3.1]). * Let be such that is invertible and the Schur complement is group invertible. Then is group invertible if and only if is invertible. *

As we know, an operator is said to be involutory if , to be anti-idempotent if , and to be tripotent if . Obviously, involutory, idempotent, and anti-idempotent are special cases of tripotent. For linear combinations of two commutative tripotents and their products, we have the following result.

Theorem 6. *Let satisfy , , and . For any scalar , and , let
*(i)*If , then is invertible. In this case, .*(ii)* is always group invertible and .** The and , in items and are defined as*

* Proof. * Since , by Lemma 2, and there exists an invertible operator such that . We consider a partition conforming with . Since , can be written as , where . In a similar way, , can be written as . Let . Now, we get
Let and , be defined as in (8). By (9), is a diagonal block matrix such that the th diagonal element is the identity and the remaining diagonal elements are , . Moreover, , , and . We get . Hence, if , is invertible and . If , is group invertible and .

The matrix case of Theorem 6 was first investigated by Tian [17]. The commutative relations ensure that idempotents (or tripotents) have simple block matrix forms. It is natural to ask whether this kind of combinational properties still hold when a pair of idempotents is noncommutative. Next, let , , and . It is easy to verify that this condition includes some specific cases: (i), (ii) (see [14]),(iii), (iv), (v), (vi) (see [17]),(vii) (see [14]).Applying Lemmas 3 and 4, we get the following main result.

Theorem 7. *Let and be two idempotents such that . For any scalar and with , and , let
*(i)* is invertible if and only if is invertible.*(ii)* is always group invertible.*

* Proof. *Let and have the forms as in (2). Then
If , then . Moreover, by (3), and . These imply that , , and . So and can be rewritten as block matrix forms as
with respect to the space decomposition , respectively. From , by (12), we deduce that
By (2) and (12) we get as an operator on which can be represented as operator matrix form: DenoteSince , then and are invertible. The Schur complement has the form
Hence, by Lemma 3, is invertible if and only if (see (12)) is invertible if and only if (see (2)) is invertible, which is equivalent to that
is invertible.

If the idempotent operator is not invertible, by Lemma 4, is group invertible:
where the omitted element can be got by Lemma 4. Note that and are invertible. By Lemma 5, is group invertible.

If , we get the following main result.

Theorem 8. *Let and be two idempotents such that . For any scalar , and with , and , let be defined as in (10). Then*(i)* is invertible if and only if is invertible;*(ii)*if any one of and is invertible, then is always group invertible.*

* Proof. *We use the notations from the proof of Theorem 7. By (14), if , then the upper left submatrix fails to be invertible. Perturb it a little. as an operator on can be written asDenote
The Schur complement of in (19) has the structure
Hence, by Lemma 3, is invertible if and only if idempotents and (see (12)) are invertible if and only if idempotents and are invertible, which is equivalent to the fact that (or ; see (4)) is invertible by Lemma 1.

If is invertible, then is invertible by (12). Since is idempotent, then . By Lemma 4, is group invertible:
If is invertible, then is invertible by (12). Since is idempotent, then . By Lemma 4, is group invertible:
So, if any one of and is invertible, then . By Lemma 5, is invertible. Hence is group invertible.

#### 3. Concluding Remarks

In Theorems 7 and 8, the inverse and the group inverse formulae can be obtained by using the results in Lemma 3 and [19, Theorem 3.1], respectively. This is a trivial and redundant work. If is group invertible and , by definition (1), it is also trivial to check Hence, is always group invertible. In particular, if is invertible (see [18, Theorem 3.1]), then From Theorems 7 and 8 we know that in (10) is always group invertible if . It seems very difficult to find the minimum requirements that guarantee that in (10) is group invertible, which could be the topic of some future research. Hence, we suggest the following question: what are the minimum requirements that guarantee that in (10) is group invertible?

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

Shunqin Wang is supported by the Basic and Advanced Research Program of Henan Science Committee (no. 102300410145) and the Research Award for Teachers in Nanyang Normal University (nynu200749). Chun Yuan Deng is supported by the National Natural Science Foundation of China under Grant 11171222 and the Doctoral Program of the Ministry of Education under Grant 20094407120001.