# On the Idempotent Solutions of a Kind of Operator Equations

Authors: Chun Yuan Deng
Journal: ISRN Applied Mathematics
 (2011)

---

## Abstract

This paper provides some relations between the idempotent operators and the solutions to operator equationsABA=A2 and BAB=B2.


---

## Body

## 1. Introduction

Letℋ be a complex Hilbert space. Denote by ℬ(ℋ) the Banach algebra of all bounded linear operators on ℋ. For A,B∈ℬ(ℋ), if A and B satisfy the relations(1.1)ABA=A2,BAB=B2,
we say the pair of (A,B) is the solution to (1.1). In [1], Vidav has investigated the self-adjoint solutions to (1.1) and showed that the pair of (A,B) is self-adjoint solution to (1.1) if and only if there exists unique idempotent operator P such that A=PP* and B=P*P. In [2], Rakočević gave another proof of this result by using some properties of generalized inverses. In [3], Schmoeger generalized the Vidav's result concerning (1.1) by using some properties of Drazin inverses. The aim of this paper is to investigate some connections between idempotent operators and the solutions to (1.1). We prove main results as follows.(1)
A and B are idempotent solution to (1.1) if and only if there exist idempotent operators P and Q satisfying (1.1) such that A=PQ and B=QP.(2)
IfA is generalized Drazin invertible such that AπB(I-Aπ)=0. Then A and B satisfy (1.1) if and only if A=P1+N1 and B=P2+N2, where N1 and N2 are arbitrary quasinilpotent elements satisfying (1.1), P1 and P2 are arbitrary idempotent elements satisfying ℛ(P1)=ℛ(P2) and Pi⊥cNj, i,j=1,2.Before proving the main results in this paper, let us introduce some notations and terminology which are used in the later. ForT∈ℬ(ℋ), we denote by ℛ(T), 𝒩(T), σp(T) and σ(T) the range, the null space, the point spectrum, and the spectrum of T, respectively. An operator P∈ℬ(ℋ) is said to be idempotent if P2=P. P is called an orthogonal projection if P=P2=P*, where P* denotes the adjoint of P. An operator A∈ℬ(ℋ) is unitary if AA*=A*A=I. A is positive if (Ax,x)≥0 for all x∈ℋ and its unique positive square root is denoted by A1/2. For a closed subspace 𝒦 of ℋ,T|𝒦 denotes the restriction of T on 𝒦 and P𝒦 denotes the orthogonal projection onto 𝒦. The generalized Drazin inverse (see [4, 5]) is the element Td∈ℬ(ℋ) such that(1.2)TTd=TdT,TdTTd=Td,T-T2Tdisquasinilpotent.
It is clear Td=T-1 if T∈ℬ(ℋ) is invertible. If T is generalized Drazin invertible, then the spectral idempotent Tπ of T corresponding to {0} is given by Tπ=I-TTd. The operator matrix form of T with respect to the space decomposition ℋ=𝒩(Tπ)⊕ℛ(Tπ) is given by T=T1⊕T2, where T1 is invertible and T2 is quasinilpotent.


## 2. Some Lemmas

To prove the main results, some lemmas are needed.Lemma 2.1 (see [6, Lemma 1.1]).
LetP be an idempotent in ℬ(ℋ). Then there exists an invertible operator S∈ℬ(ℋ) such that SPS-1 is an orthogonal projection.Lemma 2.2 (see [7, Theorem 2.1]).
LetP,Q∈ℬ(ℋ) with P=P2 and Q=Q2=Q*. If ℛ(P)=ℛ(Q), then P+P*-I is always invertible and
(2.1)Q=P(P+P*-I)-1=(P+P*-I)-1P*.Lemma 2.3 (see [8, Remark 1.2.1]).
LetA,B∈ℬ(ℋ). Then σ(AB)∖{0}=σ(BA)∖{0}.Lemma 2.4 (see [9, 10]).
LetA∈ℬ(H) have the matrix form A=(A11A12A21A22). Then A≥0 if and only if Aii≥0, i=1,2, A21=A12* and there exists a contraction operator D such that A12=A111/2DA221/2.Lemma 2.5.
LetA,B∈ℬ(ℋ) with ABA=A2 and BAB=B2. Then
(2.2)(AB)k=AkB=ABk,AkBl=Ak+l-1B=ABk+l-1
for all nonnegative integer k,l≥1.Proof.
The conditionsABA=A2 and BAB=B2 imply that (AB)2=ABAB=A2B=AB2. Now suppose (AB)k=AkB=ABk holds for nonnegative integer 2≤k≤m. Then, for k=m+1, we have
(2.3)(AB)m+1=(AB)2(AB)m-1=A2B(AB)m-1=A(AB)m=Am+1B=AAmB=A2Bm=AB2Bm-1=ABm+1.
Hence (AB)k=AkB=ABk and AkBl=Ak-1ABl=Ak-1AlB=Ak+l-1B=ABk+l-1 for all nonnegative integer k,l≥1.An elementT∈ℬ(ℋ) whose spectrum σ(T) consists of the set {0} is said to be quasi-nilpotent [8]. It is clear that T is quasi-nilpotent if and only if the spectral radius γ(T)=sup{|λ|:λ∈σ(T)}=0. In particular, if there exists a positive integer m such that Am=0, then A is m-nilpotent element. For the quasi-nilpotent operator, we have the following results.Lemma 2.6.
LetA and B satisfy (1.1). Then A is quasinilpotent if and only if B is quasinilpotent. In particular, A is nilpotent if and only if B is nilpotent; if A is quasinilpotent and AB=BA, then A2=B2=0.Proof.
Becauseσ(A2)∪{0}=σ(ABA)∪{0}=σ(A2B)∪{0}=σ(AB2)∪{0}=σ(BAB)∪{0}=σ(B2)∪{0}, it follows that A is quasinilpotent if and only if B is quasinilpotent.By Lemma2.5, if there is a nonnegative integer m≥1 such that Am=0, then (2.4)Bm+1=Bm-1BAB=BAmB=0.
Similarly we can show that An+1=0 if Bn=0. Hence A is nilpotent if and only if B is nilpotent.IfA is  quasinilpotent, then B is quasinilpotent, so I-A and I-B are invertible. From the condition AB=BA, we obtain A2(I-B)=A2-ABA=0,B2(I-B)=B2-BAB=0. It follows A2=0 and B2=0.Lemma 2.7.
LetA and B satisfy (1.1). Then for every integer k≥1,
(2.5)σ(A2)=σ(B2),σ(AkB)=σ(BkA).Proof.
SinceA2B=AB2 by Lemma 2.5,
(2.6)σ(A2B)∪{0}=σ(ABA)∪{0}=σ(A2)∪{0},σ(AB2)∪{0}=σ(BAB)∪{0}=σ(B2)∪{0}.
Note that A is invertible if and only if B is invertible. We get σ(A2)=σ(B2). Next, if 0∉σ(AB), then from (AB)2=A2B we obtain AB=A, that is, A is invertible. It follows that A=B=I because ABA=A2 and BAB=B2, so σ(AB)=σ(BA). Now,
(2.7)σ(AkB)=σ((AB)k)={λk:λ∈σ(AB)},σ(BkA)=σ((BA)k)={μk:μ∈σ(BA)}.
Hence, σ(AkB)=σ(BkA) for every integer k≥1.


## 3. Idempotent Solutions

In this section, we will show that the solutions to (1.1) have a closed connection with the idempotent operators. Our main results are as follows.Theorem 3.1.
The following assertions are equivalent.(a)
A and B are idempotent solution to (1.1).(b)
There exist idempotent operatorsP and Q satisfying (1.1) such that
(3.1)A=PQ,B=QP.Proof.
Clearly, we only needs prove that (a) implies (b). SinceA and B are idempotent operators, ABA=A2 and BAB=B2, without loss of generality, we can assume that one of A and B is orthogonal projection by Lemma 2.1. For example, assume that B is an orthogonal projection. From ABA=A, we obtain 𝒩(B|ℛ(A))=0. Since B is an orthogonal projection and BAB=B, we have BA*B=B and 𝒩(I|ℛ(A)⊥-B|ℛ(A)⊥)=0. By Lemma 2.4, A and B can be written in the forms of
(3.2)A=(I0P13P14IP23P2400),B=(IQ1Q11/2DQ21/2Q21/2D*Q11/2Q20),
with respect to the space decomposition ℋ=∑i=14⊕ℋi, respectively, where ℋ1=𝒩(Iℛ(A)-B|ℛ(A)), ℋ2=ℛ(A)⊖ℋ1, ℋ4=𝒩(B|ℛ(A)⊥), ℋ3=ℛ(A)⊥⊖ℋ4 and the entries omitted are zero. It is easy to see that Qi as operators on ℋ1+i,i=1,2, are injective positive contractions, and D is a contraction from ℋ3 into ℋ2 by Lemma 2.4. Since B is an orthogonal projection,
(3.3)(Q1Q11/2DQ21/2Q21/2D*Q11/2Q2)2=(Q1Q11/2DQ21/2Q21/2D*Q11/2Q2),
that is,
(3.4)(Q12+Q11/2DQ2D*Q11/2Q13/2DQ21/2+Q11/2DQ23/2Q21/2D*Q13/2+Q23/2D*Q11/2Q22+Q21/2D*Q1DQ21/2)=(Q1Q11/2DQ21/2Q21/2D*Q11/2Q2).
Comparing both sides of the above equation and observing that self-adjoint operators Qi, I-Qi, i=1,2 are injective, by a straightforward computation we obtain
(3.5)Q2=D*(I-Q1)D,DD*=I,D*D=I.
Hence
(3.6)B=I⊕(Q1Q11/2(I-Q1)1/2DD*(I-Q1)1/2Q11/2D*(I-Q1)D)⊕0,
where 0 and 1 are not in σp(Q1), D is unitary from ℋ3 onto ℋ2 (see [11] and Lemma 1 in [12]). Denote by ABA=(Tij)1≤i,j≤4. A direct computation shows that
(3.7)T12=P13D*Q11/2(I-Q1)1/2,T22=Q1+P23D*Q11/2(I-Q1)1/2,
and ABA=A if and only if T12=0 and T22=I. Since Q1 and I-Q1 injective self-adjoint operators, we obtain P13=0 and P23D*Q11/2=(I-Q1)1/2. Moreover, we can show BAB=B when P13=0 and P23D*Q11/2=(I-Q1)1/2. Hence,
(3.8)A=(I00P14IP23P2400),B=(IQ1Q11/2(I-Q1)1/2DD*(I-Q1)1/2Q11/2D*(I-Q1)D0),
where Q1 is a contraction on ℋ2, 0 and 1 are not in σp(Q1), D is unitary from ℋ3 onto ℋ2, Pi4∈ℬ(ℋ4,ℋi),i=1,2 are arbitrary, P23∈ℬ(ℋ3,ℋ2) and P23D*Q11/2=(I-Q1)1/2. Let
(3.9)P=I⊕(IP2300)⊕0,Q=(IP14Q1Q11/2(I-Q1)1/2DQ1P24D*(I-Q1)1/2Q11/2D*(I-Q1)DD*(I-Q1)1/2Q11/2P240).
Then we can deduce that idempotent operators P and Q satisfy (1.1), PQ=A and QP=B.Theorem3.1 shows that the arbitrary pair of idempotent solution (A,B) can be written as A=PQ, B=QP with idempotent operators P and Q satisfying (1.1). Next, we discuss the uniqueness of the idempotent solution to (1.1).Theorem 3.2.
LetB be given idempotent. Then (1.1) has unique idempotent solution A if and only if 𝒩(B|ℛ(A)⊥)=0. In this case, A,B satisfy AB=A and BA=B.Proof.
Suppose that the pair(A,B) being the idempotent solution to (1.1). By the proof of Theorem 3.1, if ℋ4=𝒩(B|ℛ(A)⊥)≠0, the idempotent solution A is not unique because P14 and P24 are arbitrary elements; if ℋ4=𝒩(B|ℛ(A)⊥)=0, then A and B have the form
(3.10)A=(I00IP230),B=(IQ1Q11/2(I-Q1)1/2DD*(I-Q1)1/2Q11/2D*(I-Q1)D).
Since Q1=Q1* is injection, D is unitary and P23D*Q11/2=(I-Q1)1/2, so Q11/2DP23*=(I-Q1)1/2 and P23*=D*Q1-1/2(I-Q1)1/2. Hence, we obtain that P23 is uniquely determined and P23=Q1-1/2(I-Q1)1/2D. Therefore the idempotent solution A is unique and AB=A and BA=B.The following result was first given by Vidav [1]. We give an alternative short proof.Theorem 3.3 (see [1, Theorem 2]).
The following assertions are equivalent.(a)
A and B are self-adjoint solution to (1.1).(b)
There is an idempotent operatorP such that
(3.11)A=PP*,B=P*P.Proof.
(b) implies (a) is clear. Now, suppose that (a) holds. From(3.12)A(B-I)2A=AB2A-2ABA+A2=A2BA-A2=A3-A2,
we have A3-A2=A(B-I)(A(B-I))*≥0, so σ(A3-A2)⊂[0,∞). The spectral mapping theorem gives
(3.13)λ3-λ2≥0,∀λ∈σ(A)∖{0}.
Thus, for ∀λ∈σ(A)∖{0}, we have λ≥1 and therefore A≥0. ℛ(A) is closed since 0 is not the accumulation point of σ(A). Hence A has the matrix form A=A1⊕0 according to the space decomposition ℋ=ℛ(A)⊕ℛ⊥(A), where A1 is invertible. Similarly, we can derive that B≥0. By Lemma 2.4, B can be written as B=(B1B2B2*B4) with B1≥0 and B4≥0. From ABA=A2, we have B1=I. From BAB=B2, we have
(3.14)I+B2B2*=A1,B2*B2+B42=B2*A1B2}⟹B42=(B2*B2)2⟹B4=B2*B2,
since the square root of a positive operator is unique. Define P=(IB200). Then P2=P,
(3.15)PP*=(I+B2B2*000)=A,P*P=(IB2B2*B2*B2)=B.The next characterizations of the solutions to (1.1) are clear.Corollary 3.4.
(a) For arbitrary idempotent operatorsP and Q, A=PQ, B=QP are the solution to (1.1).
(b) IfA is an idempotent operator satisfying AB=BA, then B is one solution to (1.1) if and only if there exists a square-zero operator N0 such that B=A+N0 and AN0=N0A=0.
(c) IfA is an orthogonal projection, then self-adjoint operator B satisfies (1.1) if and only if A=B.Proof.
(a) See Theorem 2.2 in [3].
(b) By simultaneous similarity transformations,A and B can be written as A=I⊕0 and B=N21⊕N22 since AB=BA. From ABA=A2, we obtain N21=I. From BAB=B2, we obtain N222=0. Select 0⊕N22=N0. Then N02=0, B=A+N0 and AN0=N0A=0.
(c) We use the notations from Theorem3.3. If A is an orthogonal projection, then A1=I,B2=0 in the proof of Theorem 3.3, so the result is a direct corollary of Theorem 3.3.


## 4. The Perturbation of the Solutions

The operatorsA and B are said to be c-orthogonal, denoted by A⊥cB, whenever AB=0 and BA=0. The next result is a generalization of Theorems 2.2 and 3.2 in [6], where the same problems have been considered for ind(A)≤1 and ind(B)≤1.Theorem 4.1.
LetA be generalized Drazin invertible such that AπB(I-Aπ)=0. Then
(4.1)AandBsatisfy(1.1)iffA=P1+N1,B=P2+N2,
where N1 and N2 are arbitrary quasinilpotent elements satisfying (1.1), P1 and P2 are arbitrary idempotent elements satisfying ℛ(P1)=ℛ(P2) and Pi⊥cNj, i,j=1,2.Proof.
Let us consider the matrix representation ofA and B relative to the P=I-Aπ. We have
(4.2)A=(A100A2)P,B=(B1B2B3B4)P,
where A1 is invertible and A2 is quasinilpotent. From ABA=A2, we have
(4.3)B1=I,B2A2=0,A2B3=0,A2B4A2=A22.
From BAB=B2 and (4.3), we have
(4.4)I+B2B3=A1,B2+B2B4=A1B2,B3+B4B3=B3A1,B3B2+B42=B3A1B2+B4A2B4.
From AπB(I-Aπ)=0, we have B3=0. Now, it follows from (4.2), (4.3) and (4.4) that
(4.5)A=(I00A2)P,B=(IB20B4)P,
with
(4.6)A2B4A2=A22,B4A2B4=B42,B2A2=0,B2B4=0.
Hence, B4 is quasinilpotent by Lemma 2.6. Select
(4.7)N1=0⊕PA2,N2=0⊕PB4,P1=I⊕P0,P2=B-N2.
Then P1 and P2 are idempotent operators and ℛ(P1)=ℛ(P2). N1 and N2 are quasinilpotent operators satisfying (1.1) and Pi⊥cNj,i,j=1,2.
For the proof of sufficiency observe thatℛ(P1)=ℛ(P2) leads to P1P2=P2,P2P1=P1. Straightforward calculations show that ABA=A2 and BAB=B2.We also prove the next result which can be seen as one corollary of Theorem4.1.Corollary 4.2.
LetA be generalized Drazin invertible such that AπAB=AπBA. Then
(4.8)AandBsatisfy(1.1)iffA=P1+N1,B=P2+N2,
where N1 and N2 are 2-nilpotent operators satisfying N1N2=N2N1, P1 and P2 are arbitrary idempotent elements satisfying ℛ(P1)=ℛ(P2) and Pi⊥cNj,i,j=1,2.Proof.
The sufficiency is clear. For the proof of the necessity, letA and B have the matrix representation as (4.2). From ABA=A2 and BAB=B2, we know that Ai,i=1,2 and Bi,i=1,2,3,4 satisfy (4.3) and (4.4). The condition AπAB=AπBA implies that
(4.9)A2B3=B3A1,A2B4=B4A2.
It follows that B3=0 because A2B3=0 and A1 is invertible by (4.3) and (4.4). Also, (4.2), (4.3), and (4.4) imply A2 is quasinilpotent and
(4.10)A22=A2B4A2,B42=B4A2B4,B4A2=A2B4.
It follows immediately that A22=B42=0 by Lemma 2.6. Now, we obtain
(4.11)A=(I00A2)P,B=(IB20B4)P,
with
(4.12)B2B4=0,B2A2=0,A2B4=B4A2,A22=B42=0.
Select
(4.13)N1=0⊕PA2,N2=0⊕PB4,P1=I⊕P0,P2=B-N2.
Then P1 and P2 are idempotent operators and ℛ(P1)=ℛ(P2). N12=0 and N22=0 satisfying Pi⊥cNj, i,j=1,2 and N1N2=N2N1.If we assume thatAB=BA instead of the condition AπAB=AπBA, we will get a much simpler expression for A and B.Corollary 4.3.
LetA be generalized Drazin invertible such that AB=BA. Then
(4.14)AandBsatisfy(1.1)iffA=P+N1,B=P+N2,
where N1 and N2 are 2-nilpotent operators satisfying N1N2=N2N1, P is arbitrary idempotent element satisfying P⊥cN1 and P⊥cN2.Proof.
Similar to the proof of Theorem4.1, Corollary 4.2. If AB=BA, then A and B have the matrix representations
(4.15)A=I⊕PA2,B=I⊕PB4,
with A22=B42=0 and B4A2=A2B4, so, by Corollary 4.2, the result is clear.Remark 4.4.
(1) Letγ(T) denote the spectrum radius of operator T. In Corollaries 4.2 and 4.3, since nilpotent operators A2 and B4 are commutative, we get
(4.16)γ(A2-B4)≤γ(A2)+γ(-B4)=0,γ(B4A2)≤γ(B4)γ(A2)=0,
that is, A2-B4 and B4A2 are nilpotent. Hence, A-B is a nilpotent operator and BA can be decomposed as the c-orthogonality sum of an idempotent operator and a 2-nilpotent operator.
(2) LetA have the Drazin inverse Ad. Then, by (4.2), A can be written as A=A1⊕A2, where A1 is invertible and A2 is quasi-nilpotent (see also [4, 5]). If A=A*, then(4.17)Ad=A#=A+=A1-1⊕0,
where A+ is the Moore-Penrose inverse of A and A# is the group inverse. In fact, if A=A*, then A2=0 because the self-adjoint quasi-nilpotent operator must be zero. Hence A=A1⊕0,Ad=A#=A+=A1-1⊕0.
(3) If self-adjoint operatorsA and B satisfy (1.1), then ∥A∥≥1,(4.18)AB=BAiff‖A‖=1iffA=B=PN(A)⊥,
where P𝒩(A)⊥ is the orthogonal projection on 𝒩(A)⊥. In fact, let ℋ=𝒩(A)⊥⊕𝒩(A), then A=A1⊕0. Select
(4.19)B=(B1B2B2*B4).
Similar to the proof of Theorem 4.1, we have
(4.20)B1=I,A1=I+B2B2*,B2*A1B2=B2*A1B2+B42.
This shows that ∥A∥≥1 since A1=I+B2B2*≥I. If ∥A∥=1, then A1=I,B2=0,B4=0. Hence A=B=P𝒩(A)⊥. If AB=BA, then B2=0, so A=I⊕0,B=I⊕B4. From BAB=B2, we have B42=0, so B4=0 since B4=B4*. Hence A=B=P𝒩(A)⊥. These results (see Theorem 4.2 in [6]) can be seen as the particular case of Corollary 4.3.



---

*Source: 846165-2011-06-08.xml*