The Group Inverse of the Combinations of Two Idempotent Operators

Abstract

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

1. Introduction

Let ℋ be a complex Hilbert space. Denote by ℬ(ℋ) the Banach algebra of all bounded linear operators on ℋ. ℛ(𝑇) and 𝒩(𝑇) represent the range and the null space of 𝑇, respectively. The identity onto a Hilbert space ℋ is denoted by 𝐼_ℋ or 𝐼 if there does not exist confusion. For 𝑇∈ℬ(ℋ), the group inverse [1] of 𝑇 is the unique element 𝑇^#∈ℬ(ℋ) such that 𝑇𝑇^#=𝑇^#𝑇,𝑇^#𝑇𝑇^#=𝑇^#,𝑇=𝑇𝑇^#𝑇. (1)𝑇^# exists if and only if 𝑇 has finite ascent and descent such that 𝑖(𝑇)=asc(𝑇)=desc(𝑇)≤1 [2]. When ind(𝑇)=0, the group inverse reduces to the standard inverse; that is, 𝑇^#=𝑇⁻¹. In particular, 𝑎^#=𝑎⁻¹ if 𝑎≠0 and 𝑎^#=0 if 𝑎=0 for a scalar 𝑎. One of the most important applications of group inverses is to derive some closed-form formulas for general solutions to operator equations. An operator 𝑃∈ℬ(ℋ) is said to be idempotent if 𝑃²=𝑃. If 𝑇 is group invertible, then ℛ(𝑇) is closed and the spectral idempotent 𝑇^𝜋 is given by 𝑇^𝜋=𝐼−𝑇𝑇^#. The operator matrix form of 𝑇 with respect to the space decomposition ℋ=𝒩(𝑇^𝜋)⊕ℛ(𝑇^𝜋) is given by 𝑇=𝑇₁⊕0, where 𝑇₁ is invertible on 𝒩(𝑇^𝜋) [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 𝐶^∗-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 𝐶^∗-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 𝑃,𝑄 and their products 𝑃𝑄, 𝑄𝑃, 𝑃𝑄𝑃, 𝑄𝑃𝑄, (𝑃𝑄)², and (𝑄𝑃)². 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 (𝑃𝑄)²=𝑃(𝑄𝑃)². A previous study of the group invertibility of two idempotents was made under the conditions (𝑃𝑄)²=(𝑄𝑃)² or 𝑃𝑄=𝑄𝑃 or 𝑃𝑄𝑃=𝑄𝑃𝑄 or (𝑃𝑄)²=0. It is clear that these conditions are special cases of our results.

2. Main Results and Proofs

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

Lemma 1 (see [12]). Let 𝑃 and 𝑄 be two idempotents.(i)𝑃−𝑄 is invertible if and only if 𝐼−𝑃𝑄 and 𝑃+𝑄−𝑃𝑄 are invertible.(ii)If 𝑃−𝑄 is group invertible, then 𝐼−𝑃𝑄 and 𝑃+𝑄−𝑃𝑄 are group invertible.

Proof. Since the properties in the lemma are similarly invariant, without loss of generality, we can assume that 𝑃 is an orthogonal projection. In this case, 𝑃 and 𝑄 have the operator matrix representations as follows 𝑄𝑃=𝐼000,𝑄=¹𝑄₂𝑄₃𝑄₄ (2) with respect to the space decomposition ℋ=ℛ(𝑃)⊕𝒩(𝑃), respectively. Since 𝑄²=𝑄, 𝑄²₁+𝑄₂𝑄₃𝑄₁𝑄₂+𝑄₂𝑄₄𝑄₃𝑄₁+𝑄₄𝑄₃𝑄₃𝑄₂+𝑄²₄=𝑄₁𝑄₂𝑄₃𝑄₄. (3) So 𝐼−𝑃𝑄=𝐼−𝑄₁−𝑄₂,𝑄0𝐼𝑃+𝑄−𝑃𝑄=𝐼0₃𝑄₄,(𝑃−𝑄)²=𝐼−𝑄₁00𝑄₄. (4) It is clear that item (i) holds, and if 𝑃−𝑄 is group invertible, then (𝑃−𝑄)² is group invertible, which implies that 𝐼−𝑄₁ and 𝑄₄ are group invertible.

As for 𝑛-idempotents, we have the following decomposition.

Lemma 2 (see [21, Theorem 2.3]). Let 𝐴∈ℬ(ℋ). Then 𝐴^𝑛=𝐴 if and only if(i)𝜎(𝐴)⊆{0,𝑒^{𝑖(2𝑘𝜋/(𝑛−1))}∶0≤𝑘≤𝑛−2};(ii)there exists a resolution set {𝐸_𝜆∶𝜆∈𝜎(𝐴)} of the identity 𝐼 and an invertible operator 𝑆 such that 𝑆𝐴𝑆⁻¹=_{𝜆∈𝜎(𝐴)}⊕𝜆𝐸_𝜆, (5)where ⊕ denotes the orthogonal direct sum, 𝐸_𝜆 is an orthogonal projection with ∑_{𝜆∈𝜎(𝐴)}𝐸_𝜆=𝐼, and 𝐸_𝜆𝐸_𝜇=𝐸_𝜇𝐸_𝜆=0 if 𝜆, 𝜇∈𝜎(𝐴), 𝜆≠𝜇.

Lemma 3. Let 𝑀=^𝐴𝐵_𝐶𝐷∈ℬ(ℋ⊕𝒦) be such that 𝐴∈ℬ(ℋ) is invertible. Then 𝑀 is invertible if and only if the Schur complement 𝑆=𝐷−𝐶𝐴⁻¹𝐵 is invertible.

Lemma 4 (see [1, Theorem 7.7.3]). Let 𝑀=^𝐴𝐶_0𝐷∈ℬ(ℋ⊕𝒦) be such that 𝐴^# exists. Then 𝑀^# exists if and only if 𝐷^# exists and 𝐴^𝜋𝐶𝐷^𝜋=0. In this case, 𝐴𝐶0𝐷^#=𝐴^#𝑌0𝐷^#, (6) where 𝑌=(𝐴^#)²𝐶𝐷^𝜋+𝐴^𝜋𝐶(𝐷^#)²−𝐴^#𝐶𝐷^#.

Lemma 5 (see [19, Theorem 3.1]). Let 𝑀=^𝐴𝐵_𝐶𝐷∈ℬ(ℋ⊕𝒦) be such that 𝐴∈ℬ(ℋ) is invertible and the Schur complement 𝑆=𝐷−𝐶𝐴⁻¹𝐵 is group invertible. Then 𝑀 is group invertible if and only if 𝑅=𝐴²+𝐵𝑆^𝜋𝐶 is invertible.

As we know, an operator 𝐴∈ℬ(ℋ) is said to be involutory if 𝐴²=𝐼, to be anti-idempotent if 𝐴²=−𝐴, and to be tripotent if 𝐴³=𝐴. 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 𝑃,𝑄 satisfy 𝑃³=𝑃, 𝑄³=𝑄, and 𝑃𝑄=𝑄𝑃. For any scalar 𝑎,𝑏,𝑐,𝑑,𝑒,𝑓,𝑔,ℎ, and 𝑖, let Φ=𝑎𝐼+𝑏𝑃+𝑐𝑄+𝑑𝑃²+𝑒𝑄²+𝑓𝑃𝑄+𝑔𝑃²𝑄+ℎ𝑃𝑄²+𝑖(𝑃𝑄)². (7)(i)If 𝜆₁⋯𝜆₉≠0, then Φ is invertible. In this case, Φ⁻¹=∑⁹_𝑖=1𝜆_𝑖⁻¹𝐸_𝑖.(ii)Φ is always group invertible and Φ^#=∑⁹_𝑖=1𝜆^#_𝑖𝐸_𝑖.
The 𝐸_𝑖 and 𝜆_𝑖,𝑖=1,2,…,9, in items (𝑖) and (𝑖𝑖) are defined as

𝐸₁=14𝑃+𝑃²𝑄+𝑄²,𝐸₂1=−4𝑃+𝑃²𝑄−𝑄²,𝐸₃=12𝑃+𝑃²𝐼−𝑄²,𝐸₄1=−4𝑃−𝑃²𝑄+𝑄²,𝐸₅=14𝑃−𝑃²𝑄−𝑄²,𝐸₆1=−2𝑃−𝑃²𝐼−𝑄²,𝐸₇=12𝐼−𝑃²𝑄+𝑄²,𝐸₈1=−2𝐼−𝑃²𝑄−𝑄²,𝐸₉=𝐼−𝑃²𝐼−𝑄²,𝜆₁𝜆=𝑎+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔+ℎ+𝑖,₂𝜆=𝑎+𝑏−𝑐+𝑑+𝑒−𝑓−𝑔+ℎ+𝑖,₄𝜆=𝑎−𝑏+𝑐+𝑑+𝑒−𝑓+𝑔−ℎ+𝑖,₅𝜆=𝑎−𝑏−𝑐+𝑑+𝑒+𝑓−𝑔−ℎ+𝑖,₃=𝑎+𝑏+𝑑,𝜆₆𝜆=𝑎−𝑏+𝑑,₇=𝑎+𝑐+𝑒,𝜆₈=𝑎−𝑐+𝑒,𝜆₉=𝑎. (8)

Proof. Since 𝑃³=𝑃, by Lemma 2, 𝜎(𝑃)⊆{0,1,−1} and there exists an invertible operator 𝑆₀ such that 𝑃=𝑆₀⁻¹[𝐼⊕−𝐼⊕0]𝑆₀. We consider a partition 𝑄 conforming with 𝑃. Since 𝑃𝑄=𝑄𝑃, 𝑄 can be written as 𝑄=𝑆₀⁻¹[𝑄₁⊕𝑄₂⊕𝑄₃]𝑆₀, where 𝑄³_𝑖=𝑄_𝑖,𝑖=1,2,3. In a similar way, 𝑄_𝑖,𝑖=1,2,3, can be written as 𝑄_𝑖=𝑆_𝑖⁻¹[𝐼⊕−𝐼⊕0]𝑆_𝑖,𝑖=1,2,3. Let 𝑆=(𝑆₁⊕𝑆₂⊕𝑆₃)𝑆₀. Now, we get 𝑃=𝑆⁻¹[𝐼⊕𝐼⊕𝐼⊕−𝐼⊕−𝐼⊕−𝐼⊕0⊕0⊕0]𝑆,𝑄=𝑆⁻¹[𝐼⊕−𝐼⊕0⊕𝐼⊕−𝐼⊕0⊕𝐼⊕−𝐼⊕0]𝑆. (9) Let 𝐸_𝑖 and 𝜆_𝑖,𝑖=1,2,…,9, be defined as in (8). By (9), 𝑆𝐸_𝑖𝑆⁻¹ is a diagonal block matrix such that the 𝑖th diagonal element is the identity 𝐼 and the remaining diagonal elements are 0, 𝑖=1,2,…,9. Moreover, ∑⁹_𝑖=1𝐸_𝑖=𝐼, 𝐸²_𝑖=𝐸_𝑖, and 𝐸_𝑖𝐸_𝑗=𝐸_𝑗𝐸_𝑖=0,𝑖≠𝑗,𝑖,𝑗=1,2,…,9. We get Φ=𝑎𝐼+𝑏𝑃+𝑐𝑄+𝑑𝑃²+𝑒𝑄²+𝑓𝑃𝑄+𝑔𝑃²𝑄+ℎ𝑃𝑄²+𝑖(𝑃𝑄)²=∑⁹_𝑖=1𝜆_𝑖𝐸_𝑖. Hence, if 𝜆₁⋯𝜆₉≠0, Φ is invertible and Φ⁻¹=∑⁹_𝑖=1𝜆_𝑖⁻¹𝐸_𝑖. If 𝜆₁⋯𝜆₉=0, Φ is group invertible and Φ^#=∑⁹_𝑖=1𝜆^#_𝑖𝐸_𝑖.

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 𝑃,𝑄∈ℬ(ℋ) is noncommutative. Next, let 𝑃²=𝑃, 𝑄²=𝑄, and (𝑃𝑄)²=𝑃(𝑄𝑃)². It is easy to verify that this condition includes some specific cases: (i)𝑃𝑄𝑃=0, (ii)(𝑃𝑄)²=0 (see [14]),(iii)𝑃𝑄𝑃=𝑄, (iv)𝑃𝑄𝑃=𝑃𝑄, (v)𝑃𝑄𝑃=𝑄𝑃𝑄, (vi)𝑃𝑄=𝑄𝑃 (see [17]),(vii) (𝑃𝑄)²=(𝑄𝑃)² (see [14]).Applying Lemmas 3 and 4, we get the following main result.

Theorem 7. Let 𝑃 and 𝑄 be two idempotents such that (𝑃𝑄)²=𝑃(𝑄𝑃)². For any scalar 𝑎,𝑏,𝑐,𝑑,𝑒,𝑓,𝑔 and ℎ with 𝑎≠0,𝑏≠0, and 𝑎+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔+ℎ≠0, let ()Γ=𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔𝑃𝑄²+ℎ(𝑄𝑃)². (10)(i)Γ is invertible if and only if 𝑃+𝑄−𝑃𝑄 is invertible.(ii)Γ is always group invertible.

Proof. Let 𝑃 and 𝑄 have the forms as in (2). Then (𝑃𝑄)²=𝐼0𝑄00¹𝑄₂𝑄₃𝑄₄²=𝑄²₁𝑄₁𝑄₂,()00𝑄𝑃²=𝑄₁𝑄₂𝑄₃𝑄₄𝐼000²=𝑄²₁0𝑄₃𝑄₁0. (11) If (𝑃𝑄)²=𝑃(𝑄𝑃)², then 𝑄₁𝑄₂=0. Moreover, by (3), 𝑄₂𝑄₄=𝑄₂ and 𝑄₃𝑄₂+𝑄²₄=𝑄₄. These imply that ℛ(𝑄₂)⊂𝒩(𝑄₁), ℛ(𝐼−𝑄₄)⊂𝒩(𝑄₂), and 𝑄₃ℛ(𝑄₂)⊂ℛ(𝐼−𝑄₄). So 𝑃 and 𝑄 can be rewritten as 4×4 block matrix forms as 𝑃=𝐼0000𝐼0000000000,𝑄=0𝑄₁₁0𝑄₂₁0𝑄₁₂𝑄00₃₁𝑄₃₂𝑄₄₁𝑄₄₂0𝑄₃₃0𝐼 (12) with respect to the space decomposition ℋ=ℛ(𝑄₂)⊕[ℛ(𝑃)⊖ℛ(𝑄₂)]⊕ℛ(𝐼−𝑄₄)⊕[𝒩(𝑃)⊖ℛ(𝐼−𝑄₄)], respectively. From 𝑄²=𝑄, by (12), we deduce that 𝑄²₁₂=𝑄₁₂,𝑄²₄₁=𝑄₄₁,𝑄₄₁𝑄₃₁=𝑄₃₁,𝑄₁₁𝑄₁₂+𝑄₂₁𝑄₃₃=𝑄₁₁,𝑄₃₃𝑄₁₂𝑄=0,₃₁𝑄₁₁+𝑄₃₂𝑄₁₂+𝑄₄₁𝑄₃₂+𝑄₄₂𝑄₃₃=𝑄₃₂,𝑄₄₁𝑄₄₂+𝑄₃₁𝑄₂₁=0. (13) By (2) and (12) we get Γ as an operator on ℛ(𝑄₂)⊕(ℛ(𝑃)⊖ℛ(𝑄₂))⊕ℛ(𝐼−𝑄₄)⊕(𝒩(𝑃)⊖ℛ(𝐼−𝑄₄)) which can be represented as 4×4 operator matrix form: Γ=𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)²+ℎ(𝑄𝑃)²=𝑎𝐼+(𝑏+𝑐+𝑑+𝑒)𝑄₁+𝑓+𝑔+ℎ𝑄²₁(𝑏+𝑐)𝑄₂(𝑏+𝑑)𝑄₃+𝑓+ℎ𝑄₃𝑄₁𝑏𝑄₄+𝑓𝑄₃𝑄₂=𝑎𝐼(𝑏+𝑐+𝑑+𝑒)𝑄₁₁+𝑓+𝑔+ℎ𝑄₁₁𝑄₁₂0(𝑏+𝑐)𝑄₂₁0𝑎𝐼+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔+ℎ𝑄₁₂()𝑄00𝑏+𝑑₃₁(𝑏+𝑑)𝑄₃₂+𝑓+ℎ𝑄₃₁𝑄₁₁+𝑄₃₂𝑄₁₂𝑏𝑄₄₁𝑏𝑄₄₂+𝑓𝑄₃₁𝑄₂₁0(𝑏+𝑑)𝑄₃₃.0𝑏𝐼 (14)Denote𝐴₁=𝑎𝐼(𝑏+𝑐+𝑑+𝑒)𝑄₁₁+𝑓+𝑔+ℎ𝑄₁₁𝑄₁₂0𝑎𝐼+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔+ℎ𝑄₁₂,𝐵₁=0(𝑏+𝑐)𝑄₂₁,𝐶00₁=(𝑏+𝑑)𝑄₃₁(𝑏+𝑑)𝑄₃₂+𝑓+ℎ𝑄₃₁𝑄₁₁+𝑄₃₂𝑄₁₂0(𝑏+𝑑)𝑄₃₃,𝐷₁=𝑏𝑄₄₁𝑏𝑄₄₂+𝑓𝑄₃₁𝑄₂₁.0𝑏𝐼 (15)Since 𝑎+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔+ℎ≠0, then 𝑎𝐼+(𝑏+𝑐+𝑑+𝑒+𝑓+𝑔+ℎ)𝑄₁₂ and 𝐴₁ are invertible. The Schur complement has the form 𝑆=𝐷₁−𝐶₁𝐴₁⁻¹𝐵₁=𝑏𝑄₄₁𝑏𝑄₄₂+𝑓−𝑎⁻¹(𝑏+𝑐)(𝑏+𝑑)𝑄₃₁𝑄₂₁.0𝑏𝐼 (16) Hence, by Lemma 3, Γ is invertible if and only if 𝑄₄₁ (see (12)) is invertible if and only if 𝑄₄ (see (2)) is invertible, which is equivalent to that 𝑄𝑃+𝑄−𝑃𝑄=𝐼0₃𝑄₄ (17) is invertible.
If the idempotent operator 𝑄₄₁ is not invertible, by Lemma 4, 𝑆 is group invertible: 𝑆^#=𝑏⁻¹𝑄₄₁∗0𝑏⁻¹𝐼,𝑆^𝜋=𝐼−𝑄₄₁∗,00 (18) where the omitted element ∗ can be got by Lemma 4. Note that 𝐵₁𝑆^𝜋=0 and 𝐴²₁+𝐵₁𝑆^𝜋𝐶₁ are invertible. By Lemma 5, Γ is group invertible.

If 𝑎+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔+ℎ=0, we get the following main result.

Theorem 8. Let 𝑃 and 𝑄 be two idempotents such that (𝑃𝑄)²=𝑃(𝑄𝑃)². For any scalar 𝑎,𝑏,𝑐,𝑑,𝑒,𝑓,𝑔, and ℎ with 𝑎≠0,𝑏≠0, and 𝑎+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔+ℎ=0, let Γ be defined as in (10). Then(i)Γ is invertible if and only if 𝑃−𝑄 is invertible;(ii)if any one of 𝐼−𝑃𝑄 and 𝑃+𝑄−𝑃𝑄 is invertible, then Γ is always group invertible.

Proof. We use the notations from the proof of Theorem 7. By (14), if 𝑎+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔+ℎ=0, then the upper left submatrix 𝐴₁ fails to be invertible. Perturb it a little. Γ as an operator on ℛ(𝑄₂)⊕(𝒩(𝑃)⊖ℛ(𝐼−𝑄₄))⊕ℛ(𝐼−𝑄₄)⊕(ℛ(𝑃)⊖ℛ(𝑄₂)) can be written as()𝑄Γ=𝑎𝐼𝑏+𝑐₂₁0(𝑏+𝑐+𝑑+𝑒)𝑄₁₁+𝑓+𝑔+ℎ𝑄₁₁𝑄₁₂0𝑏𝐼0(𝑏+𝑑)𝑄₃₃(𝑏+𝑑)𝑄₃₁𝑏𝑄₄₂+𝑓𝑄₃₁𝑄₂₁𝑏𝑄₄₁(𝑏+𝑑)𝑄₃₂+𝑓+ℎ𝑄₃₁𝑄₁₁+𝑄₃₂𝑄₁₂000𝑎𝐼−𝑄₁₂. (19)Denote 𝐴₀=𝑎𝐼(𝑏+𝑐)𝑄₂₁,𝐵0𝑏𝐼₀=0(𝑏+𝑐+𝑑+𝑒)𝑄₁₁+𝑓+𝑔+ℎ𝑄₁₁𝑄₁₂0(𝑏+𝑑)𝑄₃₃,𝐶₀=(𝑏+𝑑)𝑄₃₁𝑏𝑄₄₂+𝑓𝑄₃₁𝑄₂₁,𝐷00₀=𝑏𝑄₄₁(𝑏+𝑑)𝑄₃₂+𝑓+ℎ𝑄₃₁𝑄₁₁+𝑄₃₂𝑄₁₂0𝑎𝐼−𝑄₁₂. (20) The Schur complement of 𝐴₀ in (19) has the structure 𝑆₀=𝐷₀−𝐶₀𝐴₀⁻¹𝐵₀=𝑏𝑄₄₁∗0𝑎𝐼−𝑄₁₂. (21) Hence, by Lemma 3, Γ is invertible if and only if idempotents 𝑄₄₁ and 𝐼−𝑄₁₂ (see (12)) are invertible if and only if idempotents 𝑄₄ and 𝐼−𝑄₁ are invertible, which is equivalent to the fact that 𝑃−𝑄 (or (𝑃−𝑄)²; see (4)) is invertible by Lemma 1.
If 𝐼−𝑃𝑄 is invertible, then 𝐼−𝑄₁₂ is invertible by (12). Since 𝐼−𝑄₁₂ is idempotent, then 𝑄₁₂=0. By Lemma 4, 𝑆₀ is group invertible: 𝑆^#₀=𝑏⁻¹𝑄₄₁∗0𝑎⁻¹𝐼,𝑆^𝜋₀=𝐼−𝑆₀𝑆^#₀=𝐼−𝑄₄₁∗.00 (22) If 𝑃+𝑄−𝑃𝑄 is invertible, then 𝑄₄₁ is invertible by (12). Since 𝑄₄₁ is idempotent, then 𝑄₄₁=𝐼. By Lemma 4, 𝑆₀ is group invertible: 𝑆^#₀=𝑏⁻¹𝐼∗0𝑎⁻¹𝐼−𝑄₁₂S_𝜋0,=𝐼−𝑆₀𝑆^#₀=0∗0𝑄₁₂. (23)(24) So, if any one of 𝐼−𝑃𝑄 and 𝑃+𝑄−𝑃𝑄 is invertible, then 𝐵₀𝑆^𝜋₀𝐶₀=0. By Lemma 5, 𝐴²₀+𝐵₀𝑆^𝜋₀𝐶₀=𝐴²₀ is invertible. Hence Γ 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 𝑃−𝑄 is group invertible and 𝑎,𝑏≠0, by definition (1), it is also trivial to check (𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄)^#=𝑏⁻¹(𝑃−𝑄)^#(𝑃−𝑄)−𝑏⁻¹𝑃(𝑃−𝑄)^#−𝑎+𝑐()𝑎𝑏𝑃−𝑄^#𝑃+(𝑎+𝑏+𝑐)^#𝑃(𝑃−𝑄)^𝜋+𝑎+𝑏+𝑐()𝑎𝑏𝑃−𝑄^#𝑃(𝑃−𝑄)^#. (25) Hence, 𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄 is always group invertible. In particular, if 𝑃−𝑄 is invertible (see [18, Theorem 3.1]), then (𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄)⁻¹1=−𝑏𝑄(𝑃−𝑄)⁻¹−𝑎+𝑐()𝑎𝑏𝑃−𝑄⁻¹𝑃+𝑎+𝑏+𝑐()𝑎𝑏𝑃−𝑄⁻¹𝑃(𝑃−𝑄)⁻¹. (26) From Theorems 7 and 8 we know that Γ in (10) is always group invertible if (𝑃𝑄)²=𝑃(𝑄𝑃)². It seems very difficult to find the minimum requirements that guarantee that Γ 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 Γ 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.