Abstract
This paper provides some relations between the idempotent operators and the solutions to operator equations and .
1. Introduction
Let be a complex Hilbert space. Denote by the Banach algebra of all bounded linear operators on . For , if and satisfy the relations we say the pair of is the solution to (1.1). In [1], Vidav has investigated the self-adjoint solutions to (1.1) and showed that the pair of is self-adjoint solution to (1.1) if and only if there exists unique idempotent operator such that and . In [2], RakoΔeviΔ 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) and are idempotent solution to (1.1) if and only if there exist idempotent operators and satisfying (1.1) such that and .(2)If is generalized Drazin invertible such that . Then and satisfy (1.1) if and only if and , where and are arbitrary quasinilpotent elements satisfying (1.1), and are arbitrary idempotent elements satisfying and , .
Before proving the main results in this paper, let us introduce some notations and terminology which are used in the later. For , we denote by , , and the range, the null space, the point spectrum, and the spectrum of , respectively. An operator is said to be idempotent if . is called an orthogonal projection if , where denotes the adjoint of . An operator is unitary if . is positive if for all and its unique positive square root is denoted by . For a closed subspace of , denotes the restriction of on and denotes the orthogonal projection onto . The generalized Drazin inverse (see [4, 5]) is the element such that It is clear if is invertible. If is generalized Drazin invertible, then the spectral idempotent of corresponding to is given by . The operator matrix form of with respect to the space decomposition is given by , where is invertible and is quasinilpotent.
2. Some Lemmas
To prove the main results, some lemmas are needed.
Lemma 2.1 (see [6, Lemma 1.1]). Let be an idempotent in . Then there exists an invertible operator such that is an orthogonal projection.
Lemma 2.2 (see [7, Theorem 2.1]). Let with and . If , then is always invertible and
Lemma 2.3 (see [8, Remark 1.2.1]). Let . Then .
Lemma 2.4 (see [9, 10]). Let have the matrix form . Then if and only if , , and there exists a contraction operator such that .
Lemma 2.5. Let with and . Then for all nonnegative integer .
Proof. The conditions and imply that . Now suppose holds for nonnegative integer . Then, for , we have Hence and for all nonnegative integer .
An element whose spectrum consists of the set is said to be quasi-nilpotent [8]. It is clear that is quasi-nilpotent if and only if the spectral radius . In particular, if there exists a positive integer such that , then is -nilpotent element. For the quasi-nilpotent operator, we have the following results.
Lemma 2.6. Let and satisfy (1.1). Then is quasinilpotent if and only if is quasinilpotent. In particular, is nilpotent if and only if is nilpotent; if is quasinilpotent and , then .
Proof. Because , it follows that is quasinilpotent if and only if is quasinilpotent.
By Lemma 2.5, if there is a nonnegative integer such that , then Similarly we can show that if . Hence is nilpotent if and only if is nilpotent.
If is quasinilpotent, then is quasinilpotent, so and are invertible. From the condition , we obtain . It follows and .
Lemma 2.7. Let and satisfy (1.1). Then for every integer ,
Proof. Since by Lemma 2.5, Note that is invertible if and only if is invertible. We get . Next, if , then from we obtain , that is, is invertible. It follows that because and , so . Now, Hence, for every integer .
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) and are idempotent solution to (1.1).(b)There exist idempotent operators and satisfying (1.1) such that
Proof. Clearly, we only needs prove that (a) implies (b). Since and are idempotent operators, and , without loss of generality, we can assume that one of and is orthogonal projection by Lemma 2.1. For example, assume that is an orthogonal projection. From , we obtain . Since is an orthogonal projection and , we have and . By Lemma 2.4, and can be written in the forms of with respect to the space decomposition , respectively, where , , , and the entries omitted are zero. It is easy to see that as operators on , are injective positive contractions, and is a contraction from into by Lemma 2.4. Since is an orthogonal projection, that is, Comparing both sides of the above equation and observing that self-adjoint operators , , are injective, by a straightforward computation we obtain Hence where 0 and 1 are not in , is unitary from onto (see [11] and Lemma 1 in [12]). Denote by . A direct computation shows that and if and only if and . Since and injective self-adjoint operators, we obtain and . Moreover, we can show when and . Hence, where is a contraction on , 0 and 1 are not in , is unitary from onto , are arbitrary, and . Let Then we can deduce that idempotent operators and satisfy (1.1), and .
Theorem 3.1 shows that the arbitrary pair of idempotent solution can be written as , with idempotent operators and satisfying (1.1). Next, we discuss the uniqueness of the idempotent solution to (1.1).
Theorem 3.2. Let be given idempotent. Then (1.1) has unique idempotent solution if and only if . In this case, satisfy and .
Proof. Suppose that the pair being the idempotent solution to (1.1). By the proof of Theorem 3.1, if , the idempotent solution is not unique because and are arbitrary elements; if , then and have the form Since is injection, is unitary and , so and . Hence, we obtain that is uniquely determined and . Therefore the idempotent solution is unique and and .
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) and are self-adjoint solution to (1.1).(b)There is an idempotent operator such that
Proof. (b) implies (a) is clear. Now, suppose that (a) holds. From we have , so . The spectral mapping theorem gives Thus, for , we have and therefore . is closed since 0 is not the accumulation point of . Hence has the matrix form according to the space decomposition , where is invertible. Similarly, we can derive that . By Lemma 2.4, can be written as with and . From , we have . From , we have since the square root of a positive operator is unique. Define . Then ,
The next characterizations of the solutions to (1.1) are clear.
Corollary 3.4.
(a) For arbitrary idempotent operators and , , are the solution to (1.1).
(b) If is an idempotent operator satisfying , then is one solution to (1.1) if and only if there exists a square-zero operator such that and .
(c) If is an orthogonal projection, then self-adjoint operator satisfies (1.1) if and only if .
Proof. (a) See Theorem 2.2 in [3].
(b) By simultaneous similarity transformations, and can be written as and since . From , we obtain . From , we obtain . Select . Then , and .
(c) We use the notations from Theorem 3.3. If is an orthogonal projection, then 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 and are said to be c-orthogonal, denoted by , whenever and . The next result is a generalization of Theorems 2.2 and 3.2 in [6], where the same problems have been considered for ind and ind.
Theorem 4.1. Let be generalized Drazin invertible such that . Then where and are arbitrary quasinilpotent elements satisfying (1.1), and are arbitrary idempotent elements satisfying and , .
Proof. Let us consider the matrix representation of and relative to the . We have
where is invertible and is quasinilpotent. From , we have
From and (4.3), we have
From , we have . Now, it follows from (4.2), (4.3) and (4.4) that
with
Hence, is quasinilpotent by Lemma 2.6. Select
Then and are idempotent operators and . and are quasinilpotent operators satisfying (1.1) and .
For the proof of sufficiency observe that leads to . Straightforward calculations show that and .
We also prove the next result which can be seen as one corollary of Theorem 4.1.
Corollary 4.2. Let be generalized Drazin invertible such that . Then where and are 2-nilpotent operators satisfying , and are arbitrary idempotent elements satisfying and .
Proof. The sufficiency is clear. For the proof of the necessity, let and have the matrix representation as (4.2). From and , we know that and satisfy (4.3) and (4.4). The condition implies that It follows that because and is invertible by (4.3) and (4.4). Also, (4.2), (4.3), and (4.4) imply is quasinilpotent and It follows immediately that by Lemma 2.6. Now, we obtain with Select Then and are idempotent operators and . and satisfying , and .
If we assume that instead of the condition , we will get a much simpler expression for and .
Corollary 4.3. Let be generalized Drazin invertible such that . Then where and are 2-nilpotent operators satisfying , is arbitrary idempotent element satisfying and .
Proof. Similar to the proof of Theorem 4.1, Corollary 4.2. If , then and have the matrix representations with and , so, by Corollary 4.2, the result is clear.
Remark 4.4. (1) Let denote the spectrum radius of operator . In Corollaries 4.2 and 4.3, since nilpotent operators and are commutative, we get
that is, and are nilpotent. Hence, is a nilpotent operator and can be decomposed as the -orthogonality sum of an idempotent operator and a 2-nilpotent operator.
(2) Let have the Drazin inverse . Then, by (4.2), can be written as , where is invertible and is quasi-nilpotent (see also [4, 5]). If , then
where is the Moore-Penrose inverse of and is the group inverse. In fact, if , then because the self-adjoint quasi-nilpotent operator must be zero. Hence ,.
(3) If self-adjoint operators and satisfy (1.1), then ,
where is the orthogonal projection on . In fact, let , then . Select
Similar to the proof of Theorem 4.1, we have
This shows that since . If , then . Hence . If , then , so . From , we have , so since . Hence . 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).