Research Article On the Idempotent Solutions of a Kind of Operator Equations

International Scholarly Research Network ISRN Applied Mathematics Volume 0, Article ID 84665, pages doi:0.540/0/84665 Research Article On the Idempotent Solutions of a Kind of Operator Equations Chun Yuan Deng School of Mathematics Science, South China Normal University, Guangzhou 5063, China Correspondence should be addressed to Chun Yuan Deng, cy-deng@63.net Received March 0; Accepted 9 April 0 Academic Editors: F. Messine and F. Tadeo Copyright q 0 Chun Yuan Deng. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. This paper provides some relations between the idempotent operators and the solutions to operator equations ABA A and BAB B.. Introduction Let H be a complex Hilbert space. Denote by B H the Banach algebra of all bounded linear operators on H. ForA, B B H,ifA and B satisfy the relations ABA A, BAB B,. we say the pair of A, B is the solution to..in, Vidav has investigated the self-adjoint solutions to. and showed that the pair of A, B is self-adjoint solution to. if and only if there exists unique idempotent operator P such that A PP and B P P.In, Rakočević gave another proof of this result by using some propertiesof generalizedinverses. In 3, Schmoeger generalized the Vidav s result concerning. by using some properties of Drazin inverses. The aim of this paper is to investigate some connections between idempotent operators and the solutions to.. We prove main results as follows. A and B are idempotent solution to. if and only if there exist idempotent operators P and Q satisfying. such that A PQ and B QP. If A is generalized Drazin invertible such that A π B I A π 0. Then A and B satisfy. if and only if A P N and B P N,whereN and

ISRN Applied Mathematics N are arbitrary quasinilpotent elements satisfying., P and P are arbitrary idempotent elements satisfying R P R P and P i c N j, i, j,. Before proving the main results in this paper, let us introduce some notations and terminology which are used in the later. For T B H, wedenotebyr T, N T, σ p 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 P P. P is called an orthogonal projection if P P P,whereP denotes the adjoint of P. AnoperatorA B H is unitary if AA A A I. A is positive if Ax, x 0forallx Handits unique positive square root is denoted by A /. For a closed subspace K of H,T K denotes the restriction of T on K and P K denotes the orthogonal projection onto K. The generalized Drazin inverse see 4, 5 is the element T d B H such that TT d T d T, T d TT d T d, T T T d is quasinilpotent.. It is clear T d T if T B H is invertible. If T is generalized Drazin invertible, then the spectral idempotent T π of T corresponding to {0} is given by T π I TT d. The operator matrix form of T with respect to the space decomposition H N T π R T π is given by T T T,whereT is invertible and T is quasinilpotent.. Some Lemmas To prove the main results, some lemmas are needed. Lemma. see 6, Lemma.. Let P be an idempotent in B H. Then there exists an invertible operator S B H such that SPS is an orthogonal projection. Lemma. see 7, Theorem.. Let P, Q B H with P P and Q Q Q.IfR P R Q, thenp P I is always invertible and Q P P P I P P I P.. Lemma.3 see 8, Remark... Let A, B B H. Thenσ AB \{0} σ BA \{0}. ( Lemma.4 see 9, 0. Let A B H have the matrix form A A A A A ).ThenA 0 if andonlyifa ii 0, i,, A A and there exists a contraction operator D such that A A / DA/. Lemma.5. Let A, B B H with ABA A and BAB B.Then AB k A k B AB k, A k B l A k l B AB k l. for all nonnegative integer k, l.

ISRN Applied Mathematics 3 Proof. The conditions ABA A and BAB B imply that AB ABAB A B AB. Now suppose AB k A k B AB k holds for nonnegative integer k m. Then, for k m, we have AB m AB AB m A B AB m A AB m A m B AA m B A B m AB B m AB m..3 Hence AB k A k B AB k and A k B l A k AB l A k A l B A k l B AB k l for all nonnegative integer k, l. An element T B H whose spectrum σ T consists of the set {0} is said to be quasinilpotent 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 A m 0, then A is m-nilpotent element. For the quasi-nilpotent operator, we have the following results. Lemma.6. Let A and B satisfy.. ThenA 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 A B 0. Proof. Because σ A {0} σ ABA {0} σ A B {0} σ AB {0} σ BAB {0} σ B {0}, it follows that A is quasinilpotent if and only if B is quasinilpotent. By Lemma.5, if there is a nonnegative integer m suchthata m 0, then B m B m BAB BA m B 0..4 Similarly we can show that A n 0ifB n 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 areinvertible. From the condition AB BA, weobtaina I B A ABA 0,B I B B BAB 0. It follows A 0andB 0. Lemma.7. Let A and B satisfy.. Then for every integer k, ( σ A ) ( σ B ), ( ) ( ) σ A k B σ B k A..5 Proof. Since A B AB by Lemma.5, ( ) ( σ A B {0} σ ABA {0} σ A ) {0}, ( σ AB ) ( {0} σ BAB {0} σ B ) {0}..6

4 ISRN Applied Mathematics Note that A is invertible if and only if B is invertible. We get σ A σ B.Next,if0 / σ AB, thenfrom AB A B we obtain AB A, thatis,a is invertible. It follows that A B I because ABA A and BAB B,soσ AB σ BA. Now, ( ) ( σ A k B σ AB k) { } λ k : λ σ AB, ( ) ( σ B k A σ BA k) { } μ k : μ σ BA..7 Hence, σ A k B σ B k A for every integer k. 3. Idempotent Solutions In this section, we will show that the solutions to. have a closed connection with the idempotent operators. Our main results are as follows. Theorem 3.. The following assertions are equivalent. a A and B are idempotent solution to.. b There exist idempotent operators P and Q satisfying. such that A PQ, B QP. 3. Proof. Clearly, we only needs prove that a implies b. SinceA and B are idempotent operators, ABA A and BAB B, without loss of generality, we can assume that one of A and B is orthogonal projection by Lemma.. For example, assume that B is an orthogonal projection. From ABA A, weobtainn B R A 0. Since B is an orthogonal projection and BAB B, wehaveba B B and N I R A B R A 0. By Lemma.4, A and B can be written in the forms of I 0 P 3 P 4 I I P 3 P 4 A 0, B 0 Q Q / DQ / Q / D Q / Q, 3. 0 with respect to the space decomposition H 4 i H i, respectively, where H N I R A B R A, H R A H, H 4 N B R A, H 3 R A H 4 and the entries omitted are zero. It is easy to see that Q i as operators on H i,i,, are injective positive contractions, and D is a contraction from H 3 into H by Lemma.4. SinceB is an orthogonal projection, Q Q / Q / D Q / Q DQ / Q Q / Q / D Q / Q DQ /, 3.3

ISRN Applied Mathematics 5 that is, Q/ DQ D Q / Q 3/ DQ / Q / Q / D Q 3/ Q 3/ D Q / Q Q DQ 3/ Q/ D Q DQ / Q Q / Q / D Q / Q DQ /. 3.4 Comparing both sides of the above equation and observing that self-adjoint operators Q i, I Q i, i, are injective, by a straightforward computation we obtain Q D I Q D, DD I, D D I. 3.5 Hence B I Q Q / I Q / D 0, D I Q / Q / D I Q D 3.6 where 0 and are not in σ p Q, D is unitary from H 3 onto H see andlemmain. DenotebyABA T ij i,j 4. A direct computation shows that T P 3 D Q / I Q /, T Q P 3 D Q / I Q /, 3.7 and ABA A if and only if T 0andT I. SinceQ and I Q injective self-adjoint operators, we obtain P 3 0andP 3 D Q / I Q /.Moreover,wecanshowBAB B when P 3 0andP 3 D Q / I Q /.Hence, I 0 0 P 4 I I P 3 P 4 A 0, B 0 Q Q / I Q / D D I Q / Q / D I Q D 0, 3.8 where Q is a contraction on H, 0 and are not in σ p Q, D is unitary from H 3 onto H, P i4 B H 4, H i,i, are arbitrary, P 3 B H 3, H and P 3 D Q / I Q /.Let ( ) I P3 P I 0, 0 0 I P 4 Q Q / I Q / D Q P 4 Q D I Q / Q / D I Q D D I Q / Q / 0 P 4. 3.9

6 ISRN Applied Mathematics Then we can deduce that idempotent operators P and Q satisfy., PQ A and QP B. Theorem 3. 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..next,wediscussthe uniqueness of the idempotent solution to.. Theorem 3.. Let B be given idempotent. Then. 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.. By the proof of Theorem 3.,ifH 4 N B R A / 0, the idempotent solution A is not unique because P 4 and P 4 are arbitrary elements; if H 4 N B R A 0, then A and B have the form I 0 0 A I P 3, 0 I B Q Q / I Q / D. D I Q / Q / D I Q D 3.0 Since Q Q is injection, D is unitary and P 3D Q / I Q /,soq / DP3 I Q / and P3 D Q / I Q /.Hence,weobtainthatP 3 is uniquely determined and P 3 Q / I Q / D. Therefore the idempotent solution A is unique and AB A and BA B. The following result was first given by Vidav. We give an alternative short proof. Theorem 3.3 see,theorem. The following assertions are equivalent. a A and B are self-adjoint solution to.. b There is an idempotent operator P such that A PP, B P P. 3. Proof. b implies a is clear. Now, suppose that a holds. From A B I A AB A ABA A A BA A A 3 A, 3. we have A 3 A A B I A B I 0, so σ A 3 A 0,. The spectral mapping theorem gives λ 3 λ 0, λ σ A \ {0}. 3.3 Thus, for λ σ A \{0},wehaveλ and therefore A 0. R A is closed since 0 is not the accumulation point of σ A. HenceA has the matrix form A A 0 according to the space decomposition H R A R A,whereA is invertible. Similarly, we can derive that B 0.

ISRN Applied Mathematics 7 ( ) By Lemma.4, B can be written as B B B B B with B 4 0andB 4 0. From ABA A,we have B I.FromBAB B,wehave I B B A }, B B B 4 B A B 4 B ( B B ) B4 B B, 3.4 since the square root of a positive operator is unique. Define P ( IB ) 0 0.ThenP P, ( I B B PP 0 ) A, 0 0 ( ) I B P P B B B B. 3.5 The next characterizations of the solutions to. are clear. Corollary 3.4. a For arbitrary idempotent operators P and Q, A PQ, B QP are the solution to.. b If A is an idempotent operator satisfying AB BA, thenb is one solution to. if and only if there exists a square-zero operator N 0 such that B A N 0 and AN 0 N 0 A 0. c If A is an orthogonal projection, then self-adjoint operator B satisfies. ifandonlyif A B. Proof. a See Theorem. in 3. b By simultaneous similarity transformations, A and B canbewrittenasa I 0 and B N N since AB BA. FromABA A,weobtainN I. From BAB B,we obtain N 0. Select 0 N N 0.ThenN 0 0, B A N 0 and AN 0 N 0 A 0. c We use the notations from Theorem 3.3.IfA is an orthogonal projection, then A I, B 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 B are said to be c-orthogonal, denoted by A c B, whenever AB 0and BA 0. The next result is a generalization of Theorems. and 3. in 6, wherethesame problems have been considered for ind A andind B. Theorem 4.. Let A be generalized Drazin invertible such that A π B I A π 0.Then A and B satisfy. iff A P N,B P N, 4. where N and N are arbitrary quasinilpotent elements satisfying., P and P are arbitrary idempotent elements satisfying R P R P and P i c N j, i, j,.

8 ISRN Applied Mathematics Proof. Let us consider the matrix representation of A and B relative to the P I A π.we have ( ) ( ) A 0 B B A, B, 4. 0 A B 3 B 4 where A is invertible and A is quasinilpotent. From ABA A,wehave P B I, B A 0, A B 3 0, A B 4 A A. 4.3 P From BAB B and 4.3,wehave I B B 3 A, B B B 4 A B, B 3 B 4 B 3 B 3 A, 4.4 B 3 B B 4 B 3A B B 4 A B 4. From A π B I A π 0, we have B 3 0. Now, it follows from 4., 4.3 and 4.4 that ( ) ( ) I 0 I B A, B, 4.5 0 A 0 B P 4 P with A B 4 A A, B 4A B 4 B 4, B A 0, B B 4 0. 4.6 Hence, B 4 is quasinilpotent by Lemma.6. Select N 0 P A, N 0 P B 4, P I P 0, P B N. 4.7 Then P and P are idempotent operators and R P R P. N and N are quasinilpotent operators satisfying. and P i c N j,i,j,. For the proof of sufficiency observe that R P R P leads to P P P,P P P. Straightforward calculations show that ABA A and BAB B. We also prove the next result which can be seen as one corollary of Theorem 4.. Corollary 4.. Let A be generalized Drazin invertible such that A π AB A π BA. Then A and B satisfy. iff A P N,B P N, 4.8 where N and N are -nilpotent operators satisfying N N N N, P and P are arbitrary idempotent elements satisfying R P R P and P i c N j,i,j,.

ISRN Applied Mathematics 9 Proof. The sufficiency is clear. For the proof of the necessity, let A and B have the matrix representation as 4.. FromABA A and BAB B, we know that A i,i, andb i,i,, 3, 4satisfy 4.3 and 4.4. The condition A π AB A π BA implies that A B 3 B 3 A, A B 4 B 4 A. 4.9 It follows that B 3 0becauseA B 3 0andA is invertible by 4.3 and 4.4. Also, 4., 4.3,and 4.4 imply A is quasinilpotent and A A B 4 A, B 4 B 4A B 4, B 4 A A B 4. 4.0 It follows immediately that A B 4 0byLemma.6.Now,weobtain ( ) ( ) I 0 I B A, B, 4. 0 A 0 B P 4 P with B B 4 0, B A 0, A B 4 B 4 A, A B 4 0. 4. Select N 0 P A, N 0 P B 4, P I P 0, P B N. 4.3 Then P and P are idempotent operators and R P R P. N 0andN P i c N j, i, j, andn N N N. 0 satisfying If we assume that AB BA instead of the condition A π AB A π BA, we will get a much simpler expression for A and B. Corollary 4.3. Let A be generalized Drazin invertible such that AB BA. Then A and B satisfy. iff A P N,B P N, 4.4 where N and N are -nilpotent operators satisfying N N element satisfying P c N and P c N. N N, P is arbitrary idempotent Proof. Similar to the proof of Theorem 4., Corollary 4.. IfAB BA, thena and B have the matrix representations A I P A, B I P B 4, 4.5 with A B 4 0andB 4A A B 4,so,byCorollary 4., the result is clear.

0 ISRN Applied Mathematics Remark 4.4. Let γ T denote the spectrum radius of operator T. In Corollaries 4. and 4.3, since nilpotent operators A and B 4 are commutative, we get γ A B 4 γ A γ B 4 0, γ B 4 A γ B 4 γ A 0, 4.6 that is, A B 4 and B 4 A 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 -nilpotent operator. Let A have the Drazin inverse A d. Then, by 4., A can be written as A A A, where A is invertible and A is quasi-nilpotent see also 4, 5. IfA A,then A d A # A A 0, 4.7 where A is the Moore-Penrose inverse of A and A # is the group inverse. In fact, if A A, then A 0 because the self-adjoint quasi-nilpotent operator must be zero. Hence A A 0,A d A # A A 0. 3 If self-adjoint operators A and B satisfy., then A, AB BA iff A iff A B P N A, 4.8 where P N A is the orthogonal projection on N A.Infact,letH N A N A, then A A 0. Select ( ) B B B B B. 4.9 4 Similar to the proof of Theorem 4., wehave B I, A I B B, B A B B A B B 4. 4.0 This shows that A sincea I B B I.If A, then A I, B 0,B 4 0. Hence A B P N A.IfAB BA, thenb 0, so A I 0,B I B 4.FromBAB B,wehave B 4 0, so B 4 0sinceB 4 B 4.HenceA B P N A. These results see Theorem 4. 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. 0094407000.

ISRN Applied Mathematics References I. Vidav, On idempotent operators in a Hilbert space, Publications de l Institut Mathématique. Nouvelle Série, vol. 4, no. 8, pp. 57 63, 964. V. Rakočević, A note on a theorem of I. Vidav, Publications de l Institut Mathématique. Nouvelle Série, vol. 68, no. 8, pp. 05 07, 000. 3 C. Schmoeger, On the operator equations ABA A and BAB B, Publications de l Institut Mathématique. Nouvelle Série, vol. 78, no. 9, pp. 7 33, 005. 4 A. Ben-Israel and T. N. E. Greville, Generalized Inverses: Theory and Applications, Wiley-Interscience, New York, NY, USA, 974. 5 A. Ben-Israel and T. N. E. Greville, Generalized Inverses: Theory and Applications, Springer, New York, NY, USA, nd edition, 00. 6 C. Y. Deng, The Drazin inverses of sum and difference of idempotents, Linear Algebra and Its Applications, vol. 430, no. 4, pp. 8 9, 009. 7 J. J. Koliha and V. Rakocevic, Range projections and the Moore-Penrose inverse in rings with involution, Linear and Multilinear Algebra, vol. 55, no., pp. 03, 007. 8 G. J. Murphy, C -Algebra and Operator Theory, Academic Press, New York, NY, USA, 990. 9 H. K. Du, Operator matrix forms of positive operator matrices, Chinese Quarterly Mathematics, vol. 7, pp. 9, 99. 0 D. T. Kato, Perturbation Theory for Linear Operators, Springer, New York, NY, USA, 966. P. R. Halmos, Two subspaces, Transactions of the American Mathematical Society, vol. 44, pp. 38 389, 969. C. Y. Deng and H. Du, A new characterization of the closedness of the sum of two subspaces, Acta Mathematica Scientia, vol. 8, no., pp. 7 3, 008.

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