Abstract

This paper provides some relations between the idempotent operators and the solutions to operator equations 𝐴𝐡𝐴=𝐴2 and 𝐡𝐴𝐡=𝐡2.

1. Introduction

Let β„‹ be a complex Hilbert space. Denote by ℬ(β„‹) the Banach algebra of all bounded linear operators on β„‹. For 𝐴,π΅βˆˆβ„¬(β„‹), if 𝐴 and 𝐡 satisfy the relations𝐴𝐡𝐴=𝐴2,𝐡𝐴𝐡=𝐡2,(1.1) we say the pair of (𝐴,𝐡) 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,𝐡) is self-adjoint solution to (1.1) if and only if there exists unique idempotent operator 𝑃 such that 𝐴=π‘ƒπ‘ƒβˆ— and 𝐡=π‘ƒβˆ—π‘ƒ. In [2], RakočeviΔ‡ 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 𝐡 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 𝐡=𝑄𝑃.(2)If 𝐴 is generalized Drazin invertible such that π΄πœ‹π΅(πΌβˆ’π΄πœ‹)=0. Then 𝐴 and 𝐡 satisfy (1.1) if and only if 𝐴=𝑃1+𝑁1 and 𝐡=𝑃2+𝑁2, where 𝑁1 and 𝑁2 are arbitrary quasinilpotent elements satisfying (1.1), 𝑃1 and 𝑃2 are arbitrary idempotent elements satisfying β„›(𝑃1)=β„›(𝑃2) and π‘ƒπ‘–βŸ‚π‘π‘π‘—, 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. For π‘‡βˆˆβ„¬(β„‹), we denote by β„›(𝑇), 𝒩(𝑇), πœŽπ‘(𝑇) and 𝜎(𝑇) the range, the null space, the point spectrum, and the spectrum of 𝑇, respectively. An operator π‘ƒβˆˆβ„¬(β„‹) is said to be idempotent if 𝑃2=𝑃. 𝑃 is called an orthogonal projection if 𝑃=𝑃2=π‘ƒβˆ—, where π‘ƒβˆ— denotes the adjoint of 𝑃. An operator π΄βˆˆβ„¬(β„‹) is unitary if π΄π΄βˆ—=π΄βˆ—π΄=𝐼. 𝐴 is positive if (𝐴π‘₯,π‘₯)β‰₯0 for all π‘₯βˆˆβ„‹ and its unique positive square root is denoted by 𝐴1/2. For a closed subspace 𝒦 of β„‹,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 π‘‡π‘‘βˆˆβ„¬(β„‹) such that𝑇𝑇𝑑=𝑇𝑑𝑇,𝑇𝑑𝑇𝑇𝑑=𝑇𝑑,π‘‡βˆ’π‘‡2𝑇𝑑isquasinilpotent.(1.2) It is clear 𝑇𝑑=π‘‡βˆ’1 if π‘‡βˆˆβ„¬(β„‹) is invertible. If 𝑇 is generalized Drazin invertible, then the spectral idempotent π‘‡πœ‹ of 𝑇 corresponding to {0} is given by π‘‡πœ‹=πΌβˆ’π‘‡π‘‡π‘‘. The operator matrix form of 𝑇 with respect to the space decomposition β„‹=𝒩(π‘‡πœ‹)βŠ•β„›(π‘‡πœ‹) is given by 𝑇=𝑇1βŠ•π‘‡2, 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]). Let 𝑃 be an idempotent in ℬ(β„‹). Then there exists an invertible operator π‘†βˆˆβ„¬(β„‹) such that SPSβˆ’1 is an orthogonal projection.

Lemma 2.2 (see [7, Theorem 2.1]). Let 𝑃,π‘„βˆˆβ„¬(β„‹) with 𝑃=𝑃2 and 𝑄=𝑄2=π‘„βˆ—. If β„›(𝑃)=β„›(𝑄), then 𝑃+π‘ƒβˆ—βˆ’πΌ is always invertible and 𝑄=𝑃𝑃+π‘ƒβˆ—ξ€Έβˆ’πΌβˆ’1=𝑃+π‘ƒβˆ—ξ€Έβˆ’πΌβˆ’1π‘ƒβˆ—.(2.1)

Lemma 2.3 (see [8, Remark 1.2.1]). Let 𝐴,π΅βˆˆβ„¬(β„‹). Then 𝜎(𝐴𝐡)⧡{0}=𝜎(𝐡𝐴)⧡{0}.

Lemma 2.4 (see [9, 10]). Let π΄βˆˆβ„¬(𝐻) have the matrix form 𝐴=𝐴11𝐴12𝐴21𝐴22. Then 𝐴β‰₯0 if and only if 𝐴𝑖𝑖β‰₯0, 𝑖=1,2, 𝐴21=π΄βˆ—12 and there exists a contraction operator 𝐷 such that 𝐴12=𝐴1/211𝐷𝐴1/222.

Lemma 2.5. Let 𝐴,π΅βˆˆβ„¬(β„‹) with 𝐴𝐡𝐴=𝐴2 and 𝐡𝐴𝐡=𝐡2. Then (𝐴𝐡)π‘˜=π΄π‘˜π΅=π΄π΅π‘˜,π΄π‘˜π΅π‘™=π΄π‘˜+π‘™βˆ’1𝐡=π΄π΅π‘˜+π‘™βˆ’1(2.2) for all nonnegative integer π‘˜,𝑙β‰₯1.

Proof. The conditions 𝐴𝐡𝐴=𝐴2 and 𝐡𝐴𝐡=𝐡2 imply that (𝐴𝐡)2=𝐴𝐡𝐴𝐡=𝐴2𝐡=𝐴𝐡2. Now suppose (𝐴𝐡)π‘˜=π΄π‘˜π΅=π΄π΅π‘˜ holds for nonnegative integer 2β‰€π‘˜β‰€π‘š. Then, for π‘˜=π‘š+1, we have (𝐴𝐡)π‘š+1=(𝐴𝐡)2(𝐴𝐡)π‘šβˆ’1=𝐴2𝐡(𝐴𝐡)π‘šβˆ’1=𝐴(𝐴𝐡)π‘š=π΄π‘š+1𝐡=π΄π΄π‘šπ΅=𝐴2π΅π‘š=𝐴𝐡2π΅π‘šβˆ’1=π΄π΅π‘š+1.(2.3) Hence (𝐴𝐡)π‘˜=π΄π‘˜π΅=π΄π΅π‘˜ and π΄π‘˜π΅π‘™=π΄π‘˜βˆ’1𝐴𝐡𝑙=π΄π‘˜βˆ’1𝐴𝑙𝐡=π΄π‘˜+π‘™βˆ’1𝐡=π΄π΅π‘˜+π‘™βˆ’1 for all nonnegative integer π‘˜,𝑙β‰₯1.

An element π‘‡βˆˆβ„¬(β„‹) 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{|πœ†|βˆΆπœ†βˆˆπœŽ(𝑇)}=0. In particular, if there exists a positive integer π‘š such that π΄π‘š=0, then 𝐴 is π‘š-nilpotent element. For the quasi-nilpotent operator, we have the following results.

Lemma 2.6. Let 𝐴 and 𝐡 satisfy (1.1). Then 𝐴 is quasinilpotent if and only if 𝐡 is quasinilpotent. In particular, 𝐴 is nilpotent if and only if 𝐡 is nilpotent; if 𝐴 is quasinilpotent and 𝐴𝐡=𝐡𝐴, then 𝐴2=𝐡2=0.

Proof. Because 𝜎(𝐴2)βˆͺ{0}=𝜎(𝐴𝐡𝐴)βˆͺ{0}=𝜎(𝐴2𝐡)βˆͺ{0}=𝜎(𝐴𝐡2)βˆͺ{0}=𝜎(𝐡𝐴𝐡)βˆͺ{0}=𝜎(𝐡2)βˆͺ{0}, it follows that 𝐴 is quasinilpotent if and only if 𝐡 is quasinilpotent.

By Lemma 2.5, if there is a nonnegative integer π‘šβ‰₯1 such that π΄π‘š=0, then π΅π‘š+1=π΅π‘šβˆ’1𝐡𝐴𝐡=π΅π΄π‘šπ΅=0.(2.4) Similarly we can show that 𝐴𝑛+1=0 if 𝐡𝑛=0. Hence 𝐴 is nilpotent if and only if 𝐡 is nilpotent.

If 𝐴 is quasinilpotent, then 𝐡 is quasinilpotent, so πΌβˆ’π΄ and πΌβˆ’π΅ are invertible. From the condition 𝐴𝐡=𝐡𝐴, we obtain 𝐴2(πΌβˆ’π΅)=𝐴2βˆ’π΄π΅π΄=0,𝐡2(πΌβˆ’π΅)=𝐡2βˆ’π΅π΄π΅=0. It follows 𝐴2=0 and 𝐡2=0.

Lemma 2.7. Let 𝐴 and 𝐡 satisfy (1.1). Then for every integer π‘˜β‰₯1, πœŽξ€·π΄2𝐡=𝜎2𝐴,πœŽπ‘˜π΅ξ€Έξ€·π΅=πœŽπ‘˜π΄ξ€Έ.(2.5)

Proof. Since 𝐴2𝐡=𝐴𝐡2 by Lemma 2.5, πœŽξ€·π΄2𝐡𝐴βˆͺ{0}=𝜎(𝐴𝐡𝐴)βˆͺ{0}=𝜎2ξ€ΈπœŽξ€·βˆͺ{0},𝐴𝐡2𝐡βˆͺ{0}=𝜎(𝐡𝐴𝐡)βˆͺ{0}=𝜎2ξ€Έβˆͺ{0}.(2.6) Note that 𝐴 is invertible if and only if 𝐡 is invertible. We get 𝜎(𝐴2)=𝜎(𝐡2). Next, if 0βˆ‰πœŽ(𝐴𝐡), then from (𝐴𝐡)2=𝐴2𝐡 we obtain 𝐴𝐡=𝐴, that is, 𝐴 is invertible. It follows that 𝐴=𝐡=𝐼 because 𝐴𝐡𝐴=𝐴2 and 𝐡𝐴𝐡=𝐡2, so 𝜎(𝐴𝐡)=𝜎(𝐡𝐴). Now, πœŽξ€·π΄π‘˜π΅ξ€Έξ€·=𝜎(𝐴𝐡)π‘˜ξ€Έ=ξ€½πœ†π‘˜ξ€Ύ,πœŽξ€·π΅βˆΆπœ†βˆˆπœŽ(𝐴𝐡)π‘˜π΄ξ€Έξ€·=𝜎(𝐡𝐴)π‘˜ξ€Έ=ξ€½πœ‡π‘˜ξ€Ύ.βˆΆπœ‡βˆˆπœŽ(𝐡𝐴)(2.7) Hence, 𝜎(π΄π‘˜π΅)=𝜎(π΅π‘˜π΄) for every integer π‘˜β‰₯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 𝐡 are idempotent solution to (1.1).(b)There exist idempotent operators 𝑃 and 𝑄 satisfying (1.1) such that 𝐴=𝑃𝑄,𝐡=𝑄𝑃.(3.1)

Proof. Clearly, we only needs prove that (a) implies (b). Since 𝐴 and 𝐡 are idempotent operators, 𝐴𝐡𝐴=𝐴2 and 𝐡𝐴𝐡=𝐡2, without loss of generality, we can assume that one of 𝐴 and 𝐡 is orthogonal projection by Lemma 2.1. For example, assume that 𝐡 is an orthogonal projection. From 𝐴𝐡𝐴=𝐴, we obtain 𝒩(𝐡|β„›(𝐴))=0. Since 𝐡 is an orthogonal projection and 𝐡𝐴𝐡=𝐡, we have π΅π΄βˆ—π΅=𝐡 and 𝒩(𝐼|β„›(𝐴)βŸ‚βˆ’π΅|β„›(𝐴)βŸ‚)=0. By Lemma 2.4, 𝐴 and 𝐡 can be written in the forms of βŽ›βŽœβŽœβŽœβŽœβŽœβŽπ΄=𝐼0𝑃13𝑃14𝐼𝑃23𝑃2400βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ βŽ›βŽœβŽœβŽœβŽœβŽœβŽπΌπ‘„,𝐡=1𝑄11/2𝐷𝑄21/2𝑄21/2π·βˆ—π‘„11/2𝑄20⎞⎟⎟⎟⎟⎟⎠,(3.2) with respect to the space decomposition βˆ‘β„‹=4𝑖=1βŠ•β„‹π‘–, respectively, where β„‹1=𝒩(𝐼ℛ(𝐴)βˆ’π΅|β„›(𝐴)), β„‹2=β„›(𝐴)βŠ–β„‹1, β„‹4=𝒩(𝐡|β„›(𝐴)βŸ‚), β„‹3=β„›(𝐴)βŸ‚βŠ–β„‹4 and the entries omitted are zero. It is easy to see that 𝑄𝑖 as operators on β„‹1+𝑖,𝑖=1,2, are injective positive contractions, and 𝐷 is a contraction from β„‹3 into β„‹2 by Lemma 2.4. Since 𝐡 is an orthogonal projection, 𝑄1𝑄11/2𝐷𝑄21/2𝑄21/2π·βˆ—π‘„11/2𝑄2ξƒͺ2=𝑄1𝑄11/2𝐷𝑄21/2𝑄21/2π·βˆ—π‘„11/2𝑄2ξƒͺ,(3.3) that is, 𝑄21+𝑄11/2𝐷𝑄2π·βˆ—π‘„11/2𝑄13/2𝐷𝑄21/2+𝑄11/2𝐷𝑄23/2𝑄21/2π·βˆ—π‘„13/2+𝑄23/2π·βˆ—π‘„11/2𝑄22+𝑄21/2π·βˆ—π‘„1𝐷𝑄21/2ξƒͺ=𝑄1𝑄11/2𝐷𝑄21/2𝑄21/2π·βˆ—π‘„11/2𝑄2ξƒͺ.(3.4) Comparing both sides of the above equation and observing that self-adjoint operators 𝑄𝑖, πΌβˆ’π‘„π‘–, 𝑖=1,2 are injective, by a straightforward computation we obtain 𝑄2=π·βˆ—ξ€·πΌβˆ’π‘„1𝐷,π·π·βˆ—=𝐼,π·βˆ—π·=𝐼.(3.5) Hence βŽ›βŽœβŽœβŽπ‘„π΅=πΌβŠ•1𝑄11/2ξ€·πΌβˆ’π‘„1ξ€Έ1/2π·π·βˆ—ξ€·πΌβˆ’π‘„1ξ€Έ1/2𝑄11/2π·βˆ—ξ€·πΌβˆ’π‘„1ξ€Έπ·βŽžβŽŸβŽŸβŽ βŠ•0,(3.6) where 0 and 1 are not in πœŽπ‘(𝑄1), 𝐷 is unitary from β„‹3 onto β„‹2 (see [11] and Lemma 1 in [12]). Denote by 𝐴𝐡𝐴=(𝑇𝑖𝑗)1≀𝑖,𝑗≀4. A direct computation shows that 𝑇12=𝑃13π·βˆ—π‘„11/2ξ€·πΌβˆ’π‘„1ξ€Έ1/2,𝑇22=𝑄1+𝑃23π·βˆ—π‘„11/2ξ€·πΌβˆ’π‘„1ξ€Έ1/2,(3.7) and 𝐴𝐡𝐴=𝐴 if and only if 𝑇12=0 and 𝑇22=𝐼. Since 𝑄1 and πΌβˆ’π‘„1 injective self-adjoint operators, we obtain 𝑃13=0 and 𝑃23π·βˆ—π‘„11/2=(πΌβˆ’π‘„1)1/2. Moreover, we can show 𝐡𝐴𝐡=𝐡 when 𝑃13=0 and 𝑃23π·βˆ—π‘„11/2=(πΌβˆ’π‘„1)1/2. Hence, βŽ›βŽœβŽœβŽœβŽœβŽœβŽπ΄=𝐼00𝑃14𝐼𝑃23𝑃2400βŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽ βŽ›βŽœβŽœβŽœβŽœβŽœβŽπΌπ‘„,𝐡=1𝑄11/2ξ€·πΌβˆ’π‘„1ξ€Έ1/2π·π·βˆ—ξ€·πΌβˆ’π‘„1ξ€Έ1/2𝑄11/2π·βˆ—ξ€·πΌβˆ’π‘„1𝐷0⎞⎟⎟⎟⎟⎟⎠,(3.8) where 𝑄1 is a contraction on β„‹2, 0 and 1 are not in πœŽπ‘(𝑄1), 𝐷 is unitary from β„‹3 onto β„‹2, 𝑃𝑖4βˆˆβ„¬(β„‹4,ℋ𝑖),𝑖=1,2 are arbitrary, 𝑃23βˆˆβ„¬(β„‹3,β„‹2) and 𝑃23π·βˆ—π‘„11/2=(πΌβˆ’π‘„1)1/2. Let 𝑃=πΌβŠ•πΌπ‘ƒ23ξƒͺβŽ›βŽœβŽœβŽœβŽœβŽœβŽ00βŠ•0,𝑄=𝐼𝑃14𝑄1𝑄11/2ξ€·πΌβˆ’π‘„1ξ€Έ1/2𝐷𝑄1𝑃24π·βˆ—ξ€·πΌβˆ’π‘„1ξ€Έ1/2𝑄11/2π·βˆ—ξ€·πΌβˆ’π‘„1ξ€Έπ·π·βˆ—ξ€·πΌβˆ’π‘„1ξ€Έ1/2𝑄11/2𝑃240⎞⎟⎟⎟⎟⎟⎠.(3.9) Then we can deduce that idempotent operators 𝑃 and 𝑄 satisfy (1.1), 𝑃𝑄=𝐴 and 𝑄𝑃=𝐡.

Theorem 3.1 shows that the arbitrary pair of idempotent solution (𝐴,𝐡) can be written as 𝐴=𝑃𝑄, 𝐡=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. Let 𝐡 be given idempotent. Then (1.1) has unique idempotent solution 𝐴 if and only if 𝒩(𝐡|β„›(𝐴)βŸ‚)=0. In this case, 𝐴,𝐡 satisfy 𝐴𝐡=𝐴 and 𝐡𝐴=𝐡.

Proof. Suppose that the pair (𝐴,𝐡) being the idempotent solution to (1.1). By the proof of Theorem 3.1, if β„‹4=𝒩(𝐡|β„›(𝐴)βŸ‚)β‰ 0, the idempotent solution 𝐴 is not unique because 𝑃14 and 𝑃24 are arbitrary elements; if β„‹4=𝒩(𝐡|β„›(𝐴)βŸ‚)=0, then 𝐴 and 𝐡 have the form βŽ›βŽœβŽœβŽœβŽπ΄=𝐼00𝐼𝑃230βŽžβŽŸβŽŸβŽŸβŽ βŽ›βŽœβŽœβŽœβŽœβŽπΌπ‘„,𝐡=1𝑄11/2ξ€·πΌβˆ’π‘„1ξ€Έ1/2π·π·βˆ—ξ€·πΌβˆ’π‘„1ξ€Έ1/2𝑄11/2π·βˆ—ξ€·πΌβˆ’π‘„1ξ€Έπ·βŽžβŽŸβŽŸβŽŸβŽŸβŽ .(3.10) Since 𝑄1=π‘„βˆ—1 is injection, 𝐷 is unitary and 𝑃23π·βˆ—π‘„11/2=(πΌβˆ’π‘„1)1/2, so 𝑄11/2π·π‘ƒβˆ—23=(πΌβˆ’π‘„1)1/2 and π‘ƒβˆ—23=π·βˆ—π‘„1βˆ’1/2(πΌβˆ’π‘„1)1/2. Hence, we obtain that 𝑃23 is uniquely determined and 𝑃23=𝑄1βˆ’1/2(πΌβˆ’π‘„1)1/2𝐷. Therefore the idempotent solution 𝐴 is unique and 𝐴𝐡=𝐴 and 𝐡𝐴=𝐡.

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 𝐡 are self-adjoint solution to (1.1).(b)There is an idempotent operator 𝑃 such that 𝐴=π‘ƒπ‘ƒβˆ—,𝐡=π‘ƒβˆ—π‘ƒ.(3.11)

Proof. (b) implies (a) is clear. Now, suppose that (a) holds. From 𝐴(π΅βˆ’πΌ)2𝐴=𝐴𝐡2π΄βˆ’2𝐴𝐡𝐴+𝐴2=𝐴2π΅π΄βˆ’π΄2=𝐴3βˆ’π΄2,(3.12) we have 𝐴3βˆ’π΄2=𝐴(π΅βˆ’πΌ)(𝐴(π΅βˆ’πΌ))βˆ—β‰₯0, so 𝜎(𝐴3βˆ’π΄2)βŠ‚[0,∞). The spectral mapping theorem gives πœ†3βˆ’πœ†2β‰₯0,βˆ€πœ†βˆˆπœŽ(𝐴)⧡{0}.(3.13) Thus, for βˆ€πœ†βˆˆπœŽ(𝐴)⧡{0}, we have πœ†β‰₯1 and therefore 𝐴β‰₯0. β„›(𝐴) is closed since 0 is not the accumulation point of 𝜎(𝐴). Hence 𝐴 has the matrix form 𝐴=𝐴1βŠ•0 according to the space decomposition β„‹=β„›(𝐴)βŠ•β„›βŸ‚(𝐴), where 𝐴1 is invertible. Similarly, we can derive that 𝐡β‰₯0. By Lemma 2.4, 𝐡 can be written as 𝐡=𝐡1𝐡2π΅βˆ—2𝐡4 with 𝐡1β‰₯0 and 𝐡4β‰₯0. From 𝐴𝐡𝐴=𝐴2, we have 𝐡1=𝐼. From 𝐡𝐴𝐡=𝐡2, we have 𝐼+𝐡2π΅βˆ—2=𝐴1,π΅βˆ—2𝐡2+𝐡24=π΅βˆ—2𝐴1𝐡2ξƒ°βŸΉπ΅24=ξ€·π΅βˆ—2𝐡2ξ€Έ2⟹𝐡4=π΅βˆ—2𝐡2,(3.14) since the square root of a positive operator is unique. Define 𝑃=𝐼𝐡200ξ€Έ. Then 𝑃2=𝑃, π‘ƒπ‘ƒβˆ—=𝐼+𝐡2π΅βˆ—20ξƒͺ00=𝐴,π‘ƒβˆ—ξƒ©π‘ƒ=𝐼𝐡2π΅βˆ—2π΅βˆ—2𝐡2ξƒͺ=𝐡.(3.15)

The next characterizations of the solutions to (1.1) are clear.

Corollary 3.4. (a) For arbitrary idempotent operators 𝑃 and 𝑄, 𝐴=𝑃𝑄, 𝐡=𝑄𝑃 are the solution to (1.1).
(b) If 𝐴 is an idempotent operator satisfying 𝐴𝐡=𝐡𝐴, then 𝐡 is one solution to (1.1) if and only if there exists a square-zero operator 𝑁0 such that 𝐡=𝐴+𝑁0 and 𝐴𝑁0=𝑁0𝐴=0.
(c) If 𝐴 is an orthogonal projection, then self-adjoint operator 𝐡 satisfies (1.1) if and only if 𝐴=𝐡.

Proof. (a) See Theorem 2.2 in [3].
(b) By simultaneous similarity transformations, 𝐴 and 𝐡 can be written as 𝐴=πΌβŠ•0 and 𝐡=𝑁21βŠ•π‘22 since 𝐴𝐡=𝐡𝐴. From 𝐴𝐡𝐴=𝐴2, we obtain 𝑁21=I. From 𝐡𝐴𝐡=𝐡2, we obtain 𝑁222=0. Select 0βŠ•π‘22=𝑁0. Then 𝑁20=0, 𝐡=𝐴+𝑁0 and 𝐴𝑁0=𝑁0𝐴=0.
(c) We use the notations from Theorem 3.3. If 𝐴 is an orthogonal projection, then 𝐴1=𝐼,𝐡2=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 operators 𝐴 and 𝐡 are said to be c-orthogonal, denoted by π΄βŸ‚π‘π΅, whenever 𝐴𝐡=0 and 𝐡𝐴=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(𝐡)≀1.

Theorem 4.1. Let 𝐴 be generalized Drazin invertible such that π΄πœ‹π΅(πΌβˆ’π΄πœ‹)=0. Then 𝐴and𝐡satisfy(1.1)iff𝐴=𝑃1+𝑁1,𝐡=𝑃2+𝑁2,(4.1) where 𝑁1 and 𝑁2 are arbitrary quasinilpotent elements satisfying (1.1), 𝑃1 and 𝑃2 are arbitrary idempotent elements satisfying β„›(𝑃1)=β„›(𝑃2) and π‘ƒπ‘–βŸ‚π‘π‘π‘—, 𝑖,𝑗=1,2.

Proof. Let us consider the matrix representation of 𝐴 and 𝐡 relative to the 𝑃=πΌβˆ’π΄πœ‹. We have 𝐴𝐴=100𝐴2ξƒͺ𝑃𝐡,𝐡=1𝐡2𝐡3𝐡4ξƒͺ𝑃,(4.2) where 𝐴1 is invertible and 𝐴2 is quasinilpotent. From 𝐴𝐡𝐴=𝐴2, we have 𝐡1=𝐼,𝐡2𝐴2=0,𝐴2𝐡3=0,𝐴2𝐡4𝐴2=𝐴22.(4.3) From 𝐡𝐴𝐡=𝐡2 and (4.3), we have 𝐼+𝐡2𝐡3=𝐴1,𝐡2+𝐡2𝐡4=𝐴1𝐡2,𝐡3+𝐡4𝐡3=𝐡3𝐴1,𝐡3𝐡2+𝐡24=𝐡3𝐴1𝐡2+𝐡4𝐴2𝐡4.(4.4) From π΄πœ‹π΅(πΌβˆ’π΄πœ‹)=0, we have 𝐡3=0. Now, it follows from (4.2), (4.3) and (4.4) that 𝐴=𝐼00𝐴2ξƒͺ𝑃,𝐡=𝐼𝐡20𝐡4ξƒͺ𝑃,(4.5) with 𝐴2𝐡4𝐴2=𝐴22,𝐡4𝐴2𝐡4=𝐡24,𝐡2𝐴2=0,𝐡2𝐡4=0.(4.6) Hence, 𝐡4 is quasinilpotent by Lemma 2.6. Select 𝑁1=0βŠ•π‘ƒπ΄2,𝑁2=0βŠ•π‘ƒπ΅4,𝑃1=πΌβŠ•π‘ƒ0,𝑃2=π΅βˆ’π‘2.(4.7) Then 𝑃1 and 𝑃2 are idempotent operators and β„›(𝑃1)=β„›(𝑃2). 𝑁1 and 𝑁2 are quasinilpotent operators satisfying (1.1) and π‘ƒπ‘–βŸ‚π‘π‘π‘—,𝑖,𝑗=1,2.
For the proof of sufficiency observe that β„›(𝑃1)=β„›(𝑃2) leads to 𝑃1𝑃2=𝑃2,𝑃2𝑃1=𝑃1. Straightforward calculations show that 𝐴𝐡𝐴=𝐴2 and 𝐡𝐴𝐡=𝐡2.

We also prove the next result which can be seen as one corollary of Theorem 4.1.

Corollary 4.2. Let 𝐴 be generalized Drazin invertible such that π΄πœ‹π΄π΅=π΄πœ‹π΅π΄. Then 𝐴and𝐡satisfy(1.1)iff𝐴=𝑃1+𝑁1,𝐡=𝑃2+𝑁2,(4.8) where 𝑁1 and 𝑁2 are 2-nilpotent operators satisfying 𝑁1𝑁2=𝑁2𝑁1, 𝑃1 and 𝑃2 are arbitrary idempotent elements satisfying β„›(𝑃1)=β„›(𝑃2) and π‘ƒπ‘–βŸ‚π‘π‘π‘—,𝑖,𝑗=1,2.

Proof. The sufficiency is clear. For the proof of the necessity, let 𝐴 and 𝐡 have the matrix representation as (4.2). From 𝐴𝐡𝐴=𝐴2 and 𝐡𝐴𝐡=𝐡2, we know that 𝐴𝑖,𝑖=1,2 and 𝐡𝑖,𝑖=1,2,3,4 satisfy (4.3) and (4.4). The condition π΄πœ‹π΄π΅=π΄πœ‹π΅π΄ implies that 𝐴2𝐡3=𝐡3𝐴1,𝐴2𝐡4=𝐡4𝐴2.(4.9) It follows that 𝐡3=0 because 𝐴2𝐡3=0 and 𝐴1 is invertible by (4.3) and (4.4). Also, (4.2), (4.3), and (4.4) imply 𝐴2 is quasinilpotent and 𝐴22=𝐴2𝐡4𝐴2,𝐡24=𝐡4𝐴2𝐡4,𝐡4𝐴2=𝐴2𝐡4.(4.10) It follows immediately that 𝐴22=𝐡24=0 by Lemma 2.6. Now, we obtain 𝐴=𝐼00𝐴2ξƒͺ𝑃,𝐡=𝐼𝐡20𝐡4ξƒͺ𝑃,(4.11) with 𝐡2𝐡4=0,𝐡2𝐴2=0,𝐴2𝐡4=𝐡4𝐴2,𝐴22=𝐡24=0.(4.12) Select 𝑁1=0βŠ•π‘ƒπ΄2,𝑁2=0βŠ•π‘ƒπ΅4,𝑃1=πΌβŠ•π‘ƒ0,𝑃2=π΅βˆ’π‘2.(4.13) Then 𝑃1 and 𝑃2 are idempotent operators and β„›(𝑃1)=β„›(𝑃2). 𝑁21=0 and 𝑁22=0 satisfying π‘ƒπ‘–βŸ‚π‘π‘π‘—, 𝑖,𝑗=1,2 and 𝑁1𝑁2=𝑁2𝑁1.

If we assume that 𝐴𝐡=𝐡𝐴 instead of the condition π΄πœ‹π΄π΅=π΄πœ‹π΅π΄, we will get a much simpler expression for 𝐴 and 𝐡.

Corollary 4.3. Let 𝐴 be generalized Drazin invertible such that 𝐴𝐡=𝐡𝐴. Then 𝐴and𝐡satisfy(1.1)iff𝐴=𝑃+𝑁1,𝐡=𝑃+𝑁2,(4.14) where 𝑁1 and 𝑁2 are 2-nilpotent operators satisfying 𝑁1𝑁2=𝑁2𝑁1, 𝑃 is arbitrary idempotent element satisfying π‘ƒβŸ‚π‘π‘1 and π‘ƒβŸ‚π‘π‘2.

Proof. Similar to the proof of Theorem 4.1, Corollary 4.2. If 𝐴𝐡=𝐡𝐴, then 𝐴 and 𝐡 have the matrix representations 𝐴=πΌβŠ•π‘ƒπ΄2,𝐡=πΌβŠ•π‘ƒπ΅4,(4.15) with 𝐴22=𝐡24=0 and 𝐡4𝐴2=𝐴2𝐡4, so, by Corollary 4.2, the result is clear.

Remark 4.4. (1) Let 𝛾(𝑇) denote the spectrum radius of operator 𝑇. In Corollaries 4.2 and 4.3, since nilpotent operators 𝐴2 and 𝐡4 are commutative, we get 𝛾𝐴2βˆ’π΅4𝐴≀𝛾2ξ€Έξ€·+π›Ύβˆ’π΅4𝐡=0,𝛾4𝐴2𝐡≀𝛾4𝛾𝐴2ξ€Έ=0,(4.16) that is, 𝐴2βˆ’π΅4 and 𝐡4𝐴2 are nilpotent. Hence, π΄βˆ’π΅ is a nilpotent operator and 𝐡𝐴 can be decomposed as the 𝑐-orthogonality sum of an idempotent operator and a 2-nilpotent operator.
(2) Let 𝐴 have the Drazin inverse 𝐴𝑑. Then, by (4.2), 𝐴 can be written as 𝐴=𝐴1βŠ•π΄2, where 𝐴1 is invertible and 𝐴2 is quasi-nilpotent (see also [4, 5]). If 𝐴=π΄βˆ—, then𝐴𝑑=𝐴#=𝐴+=𝐴1βˆ’1βŠ•0,(4.17) where 𝐴+ is the Moore-Penrose inverse of 𝐴 and 𝐴# is the group inverse. In fact, if 𝐴=π΄βˆ—, then 𝐴2=0 because the self-adjoint quasi-nilpotent operator must be zero. Hence 𝐴=𝐴1βŠ•0,𝐴𝑑=𝐴#=𝐴+=𝐴1βˆ’1βŠ•0.
(3) If self-adjoint operators 𝐴 and 𝐡 satisfy (1.1), then ‖𝐴‖β‰₯1,𝐴𝐡=𝐡𝐴iff‖𝐴‖=1iff𝐴=𝐡=𝑃𝒩(𝐴)βŸ‚,(4.18) where 𝑃𝒩(𝐴)βŸ‚ is the orthogonal projection on 𝒩(𝐴)βŸ‚. In fact, let β„‹=𝒩(𝐴)βŸ‚βŠ•π’©(𝐴), then 𝐴=𝐴1βŠ•0. Select 𝐡𝐡=1𝐡2π΅βˆ—2𝐡4ξƒͺ.(4.19) Similar to the proof of Theorem 4.1, we have 𝐡1=𝐼,𝐴1=𝐼+𝐡2π΅βˆ—2,π΅βˆ—2𝐴1𝐡2=π΅βˆ—2𝐴1𝐡2+𝐡24.(4.20) This shows that ‖𝐴‖β‰₯1 since 𝐴1=𝐼+𝐡2π΅βˆ—2β‰₯𝐼. If ‖𝐴‖=1, then 𝐴1=𝐼,𝐡2=0,𝐡4=0. Hence 𝐴=𝐡=𝑃𝒩(𝐴)βŸ‚. If 𝐴𝐡=𝐡𝐴, then 𝐡2=0, so 𝐴=πΌβŠ•0,𝐡=πΌβŠ•π΅4. From 𝐡𝐴𝐡=𝐡2, we have 𝐡24=0, so 𝐡4=0 since 𝐡4=π΅βˆ—4. Hence 𝐴=𝐡=𝑃𝒩(𝐴)βŸ‚. These results (see Theorem 4.2 in [6]) can be seen as the particular case of Corollary 4.3.

Acknowledgments

The author would like to thank the anonymous referees for their careful reading, very detailed comments, and many constructive suggestions which greatly improved the paper. C. Y. Deng is supported by a Grant from the Ph.D. Programs Foundation of Ministry of Education of China (no. 20094407120001).