# The Group Inverse of the Combinations of Two Idempotent Operators

Authors: Shunqin Wang; Chunyuan Deng
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

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


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## Body

## 1. Introduction

Letℋ 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 Iℋ 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
(1)TT#=T#T,T#TT#=T#,T=TT#T.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 Tπ is given by Tπ=I-TT#. The operator matrix form of T with respect to the space decomposition ℋ=𝒩(Tπ)⊕ℛ(Tπ) is given by T=T1⊕0, where T1 is invertible on 𝒩(Tπ) [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 idempotentsP,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. LetP 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]).
LetP and Q be two idempotents.(i)
P-Q is invertible if and only if I-PQ and P+Q-PQ are invertible.(ii)
IfP-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 thatP is an orthogonal projection. In this case, P and Q have the operator matrix representations as follows
(2)P=(I000),Q=(Q1Q2Q3Q4)
with respect to the space decomposition ℋ=ℛ(P)⊕𝒩(P), respectively. Since Q2=Q,
(3)(Q12+Q2Q3Q1Q2+Q2Q4Q3Q1+Q4Q3Q3Q2+Q42)=(Q1Q2Q3Q4).
So
(4)I-PQ=(I-Q1-Q20I),P+Q-PQ=(I0Q3Q4),(P-Q)2=(I-Q100Q4).
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 forn-idempotents, we have the following decomposition.Lemma 2 (see [21, Theorem 2.3]).
LetA∈ℬ(ℋ). Then An=A if and only if(i)
σ(A)⊆{0,ei(2kπ/(n-1)):0≤k≤n-2};(ii)
there exists a resolution set{Eλ:λ∈σ(A)} of the identity I and an invertible operator S such that
(5)SAS-1=∑λ∈σ(A)‍⊕λEλ,where ⊕ denotes the orthogonal direct sum, Eλ is an orthogonal projection with ∑λ∈σ(A)‍Eλ=I, and EλEμ=EμEλ=0 if λ, μ∈σ(A), λ≠μ.Lemma 3.
LetM=(ABCD)∈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]).
LetM=(AC0D)∈ℬ(ℋ⊕𝒦) be such that A# exists. Then M# exists if and only if D# exists and AπCDπ=0. In this case,
(6)(AC0D)#=(A#Y0D#),
where Y=(A#)2CDπ+AπC(D#)2-A#CD#.Lemma 5 (see [19, Theorem 3.1]).
LetM=(ABCD)∈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+BSπC is invertible.As we know, an operatorA∈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.
LetP,Q satisfy P3=P, Q3=Q, and PQ=QP. For any scalar a,b,c,d,e,f,g,h, and i, let
(7)Φ=aI+bP+cQ+dP2+eQ2+fPQ+gP2Q+hPQ2+i(PQ)2.(i)
Ifλ1⋯λ9≠0, then Φ is invertible. In this case, Φ-1=∑i=19λi-1Ei.(ii)
Φ is always group invertible and Φ#=∑i=19λi#Ei.
TheEi and λi,i=1,2,…,9, in items (i) and (ii) are defined as(8)
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.Proof.
SinceP3=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 Qi3=Qi,i=1,2,3. In a similar way, Qi,i=1,2,3, can be written as Qi=Si-1[I⊕-I⊕0]Si,i=1,2,3. Let S=(S1⊕S2⊕S3)S0. Now, we get
(9)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.
Let Ei and λi,i=1,2,…,9, be defined as in (8). By (9), SEiS-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, ∑i=19Ei=I, Ei2=Ei, and EiEj=EjEi=0,i≠j,i,j=1,2,…,9. We get Φ=aI+bP+cQ+dP2+eQ2+fPQ+gP2Q+hPQ2+i(PQ)2=∑i=19λiEi. Hence, if λ1⋯λ9≠0, Φ is invertible and Φ-1=∑i=19λi-1Ei. If λ1⋯λ9=0, Φ is group invertible and Φ#=∑i=19λi#Ei.The matrix case of Theorem6 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.
LetP 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
(10)Γ=aP+bQ+cPQ+dQP+ePQP+fQPQ+g(PQ)2+h(QP)2.(i)
Γ is invertible if and only if P+Q-PQ is invertible.(ii)
Γ is always group invertible.Proof.
LetP and Q have the forms as in (2). Then
(11)(PQ)2=[(I000)(Q1Q2Q3Q4)]2=(Q12Q1Q200),(QP)2=[(Q1Q2Q3Q4)(I000)]2=(Q120Q3Q10).
If (PQ)2=P(QP)2, then Q1Q2=0. Moreover, by (3), Q2Q4=Q2 and Q3Q2+Q42=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
(12)P=(I0000I0000000000),Q=(0Q110Q210Q1200Q31Q32Q41Q420Q330I)
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
(13)Q122=Q12,Q412=Q41,Q41Q31=Q31,Q11Q12+Q21Q33=Q11,Q33Q12=0,Q31Q11+Q32Q12+Q41Q32+Q42Q33=Q32,Q41Q42+Q31Q21=0.
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: (14)Γ=aP+bQ+cPQ+dQP+ePQP+fQPQ+g(PQ)2+h(QP)2=(aI+(b+c+d+e)Q1+(f+g+h)Q12(b+c)Q2(b+d)Q3+(f+h)Q3Q1bQ4+fQ3Q2)=(aI(b+c+d+e)Q11+(f+g+h)Q11Q120(b+c)Q210aI+(b+c+d+e+f+g+h)Q1200(b+d)Q31(b+d)Q32+(f+h)(Q31Q11+Q32Q12)bQ41bQ42+fQ31Q210(b+d)Q330bI).Denote(15)A1=(aI(b+c+d+e)Q11+(f+g+h)Q11Q120aI+(b+c+d+e+f+g+h)Q12),B1=(0(b+c)Q2100),C1=((b+d)Q31(b+d)Q32+(f+h)(Q31Q11+Q32Q12)0(b+d)Q33),D1=(bQ41bQ42+fQ31Q210bI).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
(16)S=D1-C1A1-1B1=(bQ41bQ42+[f-a-1(b+c)(b+d)]Q31Q210bI).
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
(17)P+Q-PQ=(I0Q3Q4)
is invertible.
If the idempotent operatorQ41 is not invertible, by Lemma 4, S is group invertible:
(18)S#=(b-1Q41*0b-1I),Sπ=(I-Q41*00),
where the omitted element * can be got by Lemma 4. Note that B1Sπ=0 and A12+B1SπC1 are invertible. By Lemma 5, Γ is group invertible.Ifa+b+c+d+e+f+g+h=0, we get the following main result.Theorem 8.
LetP 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 ofI-PQ and P+Q-PQ is invertible, then Γ is always group invertible.Proof.
We use the notations from the proof of Theorem7. 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(19)Γ=(aI(b+c)Q210(b+c+d+e)Q11+(f+g+h)Q11Q120bI0(b+d)Q33(b+d)Q31bQ42+fQ31Q21bQ41(b+d)Q32+(f+h)(Q31Q11+Q32Q12)000a(I-Q12)).Denote
(20)A0=(aI(b+c)Q210bI),B0=(0(b+c+d+e)Q11+(f+g+h)Q11Q120(b+d)Q33),C0=((b+d)Q31bQ42+fQ31Q2100),D0=(bQ41(b+d)Q32+(f+h)(Q31Q11+Q32Q12)0a(I-Q12)).
The Schur complement of A0 in (19) has the structure
(21)S0=D0-C0A0-1B0=(bQ41*0a(I-Q12)).
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.
IfI-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:
(22)S0#=(b-1Q41*0a-1I),S0π=I-S0S0#=(I-Q41*00).
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:
(23)S0#=(b-1I*0a-1(I-Q12)),(24)S0π=I-S0S0#=(0*0Q12).
So, if any one of I-PQ and P+Q-PQ is invertible, then B0S0πC0=0. By Lemma 5, A02+B0S0πC0=A02 is invertible. Hence Γ is group invertible.


## 3. Concluding Remarks

In Theorems7 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
(25)(aP+bQ+cPQ)#=b-1(P-Q)#(P-Q)-b-1P(P-Q)#-a+cab(P-Q)#P+(a+b+c)#P(P-Q)π+a+b+cab(P-Q)#P(P-Q)#.
Hence, aP+bQ+cPQ is always group invertible. In particular, if P-Q is invertible (see [18, Theorem 3.1]), then
(26)(aP+bQ+cPQ)-1=-1bQ(P-Q)-1-a+cab(P-Q)-1P+a+b+cab(P-Q)-1P(P-Q)-1.
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?



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