EXPLICIT SOLUTION TO MODULAR OPERATOR EQUATION T XS SX T = A

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1 Kragujevac Journal of Mathematics Volume 4(2) (216), Pages EXPLICI SOLUION O MODULAR OPERAOR EQUAION XS SX A M MOHAMMADZADEH KARIZAKI 1, M HASSANI 2, AND S S DRAGOMIR 3 Abstract In this paper, by using some block operator matrix techniques, we find explicit solution of the operator equation XS SX A in the general setting of the adjointable operators between Hilbert C -modules Furthermore, we solve the operator equation XS SX A, when ran() + ran(s) is closed 1 Introduction and Preliminaries he equation XS SX A was studied by Yuan 8 for finite matrices and Xu et al 7 generalized the results to Hilbert C -modules, under the condition that ran(s) is contained in ran() When equals an identity matrix or identity operator, this equation reduces to XS SX A, which was studied by Braden 1 for finite matrices, and Djordjevic 3 for the Hilbert space operators In this paper, by using block operator matrix techniques and properties of the Moore-Penrose inverse, we provide a new approach to the study of the equation XS SX A for adjointable Hilbert module operators than those with closed ranges Furthermore, we solve the operator equation XS SX A, when ran() + ran(s) is closed hroughout this paper, A is a C -algebra Let X and Y be two Hilbert A-modules, and L(X, Y) be the set of the adjointable operators from X to Y For any L(X, Y), the range and the null space of are denoted by ran() and ker( ) respectively In case X Y, L(X, X) which we abbreviate to L(X), is a C -algebra he identity operator on X is denoted by 1 X or 1 if there is no ambiguity heorem 11 5, heorem 32 Suppose that L(X, Y) has closed range hen ker( ) is orthogonally complemented in X, with complement ran( ); Key words and phrases Operator equation, Moore-Penrose inverse, Hilbert C -module 21 Mathematics Subject Classification Primary: 47A62 Secondary: 15A24, 46L8 Received: December 5, 215 Accepted: April 4,

2 EXPLICI SOLUION O MODULAR OPERAOR EQUAION XS SX A 281 ran() is orthogonally complemented in Y, with complement ker( ); he map L(Y, X) has closed range Xu and Sheng 6 showed that a bounded adjointable operator between two Hilbert A-modules admits a bounded Moore-Penrose inverse if and only if it has closed range he Moore-Penrose inverse of, denoted by, is the unique operator L(X, Y) satisfying the following conditions:,, ( ), ( ) It is well-known that exists if and only if ran() is closed, and in this case ( ) ( ) Let L(X, Y) be a closed range, then is the orthogonal projection from Y onto ran() and is the orthogonal projection from X onto ran( ) Projection, in the sense that they are self adjoint idempotent operators A matrix form of a bounded adjointable operator L(X, Y) can be induced by some natural decompositions of Hilbert C -modules Indeed, if M and N are closed orthogonally complemented submodules of X and Y, respectively, and X M M, Y N N, then can be written as the following 2 2 matrix 1 2, 3 4 where, 1 L(M, N), 2 L(M, N), 3 L(M, N ) and 4 L(M, N ) Note that P M denotes the projection corresponding to M In fact 1 P N P M, 2 P N (1 P M ), 3 (1 P N ) P M and 4 (1 P N ) (1 P M ) Lemma 11 (see 4, Corollary 12) Suppose that L(X, Y) has closed range hen has the following matrix decomposition with respect to the orthogonal decompositions of closed submodules X ran( ) ker() and Y ran() ker( ): 1 : where 1 is invertible Moreover 1 1 : ran( ) ker( ) ran() ker( ) ran() ker( ) ran( ) ker( ) Remark 11 Let X and Y be two Hilbert A-modules, we use the notation X Y to denote the direct sum of X and Y, which is also a Hilbert A-module whose A-valued inner product is given by ( ) ( ) x1 x2, x y 1 y 1, x 2 + y 1, y 2, 2 for ( x i ) X and y i Y, i 1, 2 o simplify the notation, we use x y to denote x X Y y,

3 282 M MOHAMMADZADEH KARIZAKI, M HASSANI, AND S S DRAGOMIR Lemma 12 (see 2, Lemma 3) Suppose that X is a Hilbert A-module and S, S L(X) hen L(X X) has closed range if and only if ran() + ran(s) is closed, and S ( + SS ) S ( + SS ) Lemma 13 (see 2, Corollary 4) Suppose that X is a Hilbert A-module and S, L(X) hen L(X X) has closed range if and only if ran( S ) + ran(s ) is closed, and ( + S S) ( + S S) S S 2 Solutions to XS SX A In this section, we will study the operator equation XS SX A in the general context of the Hilbert C -modules First, in the following theorem we solve to the operator equation XS SX A, in the case when S and are invertible operators heorem 21 Let X, Y, Z be Hilbert A-modules, S L(X, Y) and L(Z, Y) be invertible operators and A L(Y) hen the following statements are equivalent: (a) here exists a solution X L(X, Z) to the operator equation XS SX A; (b) A A If (a) or (b) is satisfied, then any solution to (21) XS SX A, X L(X, Z), has the form where Z L(Y) satisfy Z Z X A(S ) Z(S ) 1, Proof (a) (b): Obvious (b) (a): Note that, if A A then X 1 1 A(S ) Z(S ) 1 is a solution 2 to (21) he following sentences state this claim ( A(S ) Z(S ) 1 ) S S ( 1 2 S 1 A ( ) 1 + S 1 Z ( ) ( 1 A(S ) 1 S SS 1 A ( ) 1 ) + 1 Z(S ) 1 S SS 1 Z ( ) 1 A + Z Z A )

4 EXPLICI SOLUION O MODULAR OPERAOR EQUAION XS SX A 283 On the other hand, let X be any solution to (21) hen X 1 A(S ) SX (S ) 1 We have X 1 A(S ) SX (S ) A(S ) A(S ) SX (S ) A(S ) ( 1 2 A + SX )(S ) 1 aking Z 1 2 A + SX, we get Z Z Corollary 21 Suppose that Y, Z are Hilbert A-modules and L(Z, Y) is invertible operator, A L(Y) hen the following statements are equivalent: (a) here exists a solution X L(Y, Z) to the operator equation X X A; (b) A A If (a) or (b) is satisfied, then any solution to has the form where Z L(Y) satisfy Z Z X X A, X L(Y, Z), X A + 1 Z, In the following theorems we obtain explicit solutions to the operator equation (22) XS SX A, via matrix form and complemented submodules heorem 22 Suppose S L(Z, Y) is an invertible operator and L(Z, Y) has closed range and A L(Y) hen the following statements are equivalent: (a) here exists a solution X L(Z) to (22); (b) A A and (1 )A(1 ) If (a) or (b) is satisfied, then any solution to (22) has the form (23) X 1 2 A (S ) 1 + Z (S ) 1 + A(1 )(S ) 1 + (1 )Y (S ) 1, where Z L(Y) satisfies (Z Z ), and Y L(Y, Z) is arbitrary Proof (a) (b): Obviously, A A Also, (1 )A(1 ) (1 )( XS SX )(1 ) ( )XS (1 ) (1 )SX ( ) (b) (a): Note that the condition (1 )A(1 ) is equivalent to A A + A A On the other hand, since (Z Z ), then (Z Z ) ker( ) ker( ) herefore (Z Z )

5 284 M MOHAMMADZADEH KARIZAKI, M HASSANI, AND S S DRAGOMIR Hence we have 1 2 A + Z + A(1 ) + (1 )Y 1 2 A ( ) Z ( ) (1 )A ( ) Y (1 ) A + A A + Z Z ( ) A hat is, any operator X of the form (23) is a solution to (22) Now, suppose that (24) X X (S ) 1, X L(Y, Z), is a solution to (22) We let (24) in (22) Hence (22) get into the following equation (25) X X A, X L(Y, Z) Since has closed range, we have Z ran( ) ker() and Y ran() ker( ) Now, has the matrix form 1 ran( : ) ran() ker( ) ker(, ) where 1 is invertible On the other hand, A A and (1 )A(1 ) imply that A has the form A1 A A 2 ran() ran() A : 2 ker( ) ker(, ) where A 1 A 1 Let X have the form X1 X X 2 ran() : X 3 X 4 ker( ) ran( ) ker( ) Now by using matrix form for operators, X and A, we have 1 X1 X 2 X 1 X3 1 A1 A X 3 X 4 X2 X4 2 A 2 or equivalently herefore and 1 X 1 X X 2 X X 1 X 1 1 A 1, (26) 1 X 2 A 2 A1 A 2 A 2 Now, we obtain X 1 and X 2 By Lemma 11, 1 is invertible Hence, Corollary 21 implies that X A Z 1,,

6 EXPLICI SOLUION O MODULAR OPERAOR EQUAION XS SX A 285 where Z 1 L(ran()) satisfy Z 1 Z 1 Now, multiplying 1 1 from the left to (26), we get (27) X A 2 1 Hence A Z A 2 is a solution to (25), such that X X 3 X 3, X 4 are arbitrary operators Let 4 Y1 Y Y 2 ran() ran( : X 3 X 4 ker( ), ) ker( ) and and hen Z Z1 Z 2 ran() : Z 3 Z 4 ker( ) A A 1 A(1 ) 1 A 2 ran() ker( ), Z 1 1 Z 1, (1 )Y X 3 X 4 Consequently, X 1 A + Z + A(1 ) + (1 )Y, where 2 Z L(Y) satisfies (Z Z ), and Y L(Y, Z) is arbitrary In the following theorem we obtain explicit solution to the operator equation XS SX A when (1 P ran(s) ) and S have closed ranges heorem 23 Suppose that X, Y, Z are Hilbert A-modules, S L(X, Y), L(Z, Y) and A L(Y) and such that (1 SS ) and S have closed ranges If the equation (28) XS SX A, X L(X, Z), is solvable, then X S A (29) X S S Proof aking H : Z X Y Y he operator H has closed range, since let {z n x n } be sequence chosen in Z X, {z n }, {x n } be sequences chosen in Z and X, respectively such that (z n ) + S(x n ) y for some y Y hen (1 SS ) (z n ) (1 SS )( (z n ) + S(x n )) (1 SS )(y) Since ran((1 SS )) is assumed to be closed, (1 SS )(y) (1 SS ) (z 1 ) for some z 1 Z It follows that y (z 1 ) ker(1 SS ) ran(s), hence y (z 1 ) + S(x) for X some x X herefore H has closed range Let Y X : Z X Z X

7 286 M MOHAMMADZADEH KARIZAKI, M HASSANI, AND S S DRAGOMIR and H S : Y Y Z X and B (28) get into (21) HY H B A : Y Y Y Y, therefore Since H has closed range and (28) has solution, then (21) has solution, therefore with multiplying HH on the left and multiply H (H ) on the right, we have Y H B(H ) he proof of the following remark is the same as in the matrix case Remark 21 Let L(Y, Z) and S L(X, Y) have closed ranges, and A L(X, Z) hen the equation (211) XS A, X L(Y), has a solution if and only if AS S A In which case, any solution of (211) has the form X AS Now, we solve to the operator equation XS SX A in the case when ran() + ran(s) is closed heorem 24 Suppose that X is a Hilbert A-module, S,, A L(X) such that ran() + ran(s) is closed hen the following statements are equivalent: (a) here exists a solution X L(X) to the operator equation XS SX A; (b) A A and P ran( +SS )AP ran( +SS ) A If (a) or (b) is satisfied, then any solution to (212) XS SX A, X L(X), has the form (213) X ( + SS ) A( + SS ) S Proof (a) (b) Suppose that (212) has a solution X L(X) A A On the other hand, (212) get into S X A (214) X S hen obviously Since (212) is solvable, then (214) is solvable Since ran() + ran(s) is closed, then S Lemma 12 implies that exists Hence, by Remark 21 we have S S A A S S

8 EXPLICI SOLUION O MODULAR OPERAOR EQUAION XS SX A 287 By applying Lemma 12, Corollary 13 are shown that S S A S S S ( + SS ) A( + SS ) A( + SS ) S S ( + SS ) ( + SS )( + SS ) A( + SS ) ( + SS ) Pran( +SS )AP ran(( +SS ) ) Pran( +SS )AP ran( +SS ) A S hat is, P ran( +SS )AP ran( +SS ) A (b) (a): If P ran( +SS )AP ran( +SS ) A, then we have Pran( +SS )AP ran(( +SS ) ) Pran( +SS )AP ran( +SS ) A, or equivalently ( + SS )( + SS ) A( + SS ) ( + SS ) S S A S S A herefore, Remark 21 implies that (214) is solvable and hence (212) is solvable Now, by applying Remark 21 and Lemma 12, Corollary 13 imply that X X S A S ( + SS ) A ( + SS ) ( + SS ) S S ( + SS ) ( + SS ) A( + SS ) ( + SS ) A( + SS ) S S ( + SS ) A( + SS ) S ( + SS ) A( + SS ) S

9 288 M MOHAMMADZADEH KARIZAKI, M HASSANI, AND S S DRAGOMIR herefore ( + SS ) A( + SS ) S ( + SS ) A( + SS ) S Consequently, X has the form (213) Using exactly similar arguments, we obtain the following analogue of heorem 21, in which to (21) is replaced by (215) XS + SX A All results of this section can be rewritten for to (215), considering the following theorem heorem 25 Let X, Y, Z be Hilbert A-modules, S L(X, Y) and L(Z, Y) be invertible operators and A L(Y) hen the following statements are equivalent: (a) here exists a solution X L(X, Z) to the operator equation XS +SX A (b) A A If (a) or (b) is satisfied, then any solution to has the form where Z L(Y) satisfy Z Z XS + SX A, X L(X, Z) X A(S ) 1 1 Z(S ) 1, References 1 H Braden, he equations A X ± X A B, SIAM J Matrix Anal Appl 2 (1998), CY Deng, H K Du, Representations of the Moore-Penrose inverse for a class of 2 by 2 block operator valued partial matrices, Linear Multilinear Algebra, 58 (21), D S Djordjevic, Explicit solution of the operator equation A X + X A B, J Comput Appl Math 2(27), M Mohammadzadeh Karizaki, M Hassani, M Amyari and M Khosravi, Operator matrix of Moore-Penrose inverse operators on Hilbert C -modules, Colloq Math 14 (215), E C Lance, Hilbert C -Modules, LMS Lecture Note Series 21, Cambridge Univ Press, Q Xu, L Sheng, Positive semi-definite matrices of adjointable operators on Hilbert C*-modules, Linear Algebra Appl 428 (28), Q Xu, L Sheng, Y Gu, he solutions to some operator equations, Linear Algebra Appl 429 (28), Y Yuan, Solvability for a class of matrix equation and its applications, J Nanjing Univ (Math Biquart) 18 (21),

10 EXPLICI SOLUION O MODULAR OPERAOR EQUAION XS SX A University of orbat-e-heydarieh, Iran address: mohammadzadehkarizaki@gmailcom 2 Department of Mathematics, Mashhad Branch, Islamic Azad University, Mashhad, Iran address: mhassanimath@gmailcom 3 Mathematics, College of Engineering & Science Victoria University, PO Box 14428, Melbourne City, MC 81, Australia address: SeverDragomir@vueduau