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

Transcription

1 EXPLICIT SOLUTION TO MODULAR OPERATOR EQUATION T XS SX T A M MOHAMMADZADEH KARIZAKI M HASSANI AND SS DRAGOMIR Abstract In this paper by using some block operator matrix techniques we find explicit solution of the operator equation T XS SX T A in the general setting of the adjointable operators between Hilbert C -modules Furthermore we solve the operator equation T XS SX T A when ran(t) + ran(s) is closed 1 Introduction and preliminaries The equation T XS SX T 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(t) When T 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 T XS SX T A for adjointable Hilbert module operators than those with closed ranges Furthermore we solve the operator equation T XS SX T A when ran(t) + ran(s) is closed Throughout 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 T L(X Y) the range and the null space of T are denoted by ran(t) and ker(t ) respectively In case X Y L(X X ) which we abbreviate to L(X ) is a C -algebra The identity operator on X is denoted by 1 X or 1 if there is no ambiguity Theorem 11 5 Theorem 32 Suppose that T L(X Y) has closed range Then 2000 Mathematics Subject Classification Primary 47A62; Secondary 15A24 46L08 Key words and phrases Operator equation Moore-Penrose inverse Hilbert C -module *Corresponding author 1

2 2 MOHAMMADZADEH KARIZAKI HASSANI AND DRAGOMIR ker(t ) is orthogonally complemented in X with complement ran(t ) ran(t) is orthogonally complemented in Y with complement ker(t ) The map T 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 The Moore-Penrose inverse of T denoted by T is the unique operator T L(X Y) satisfying the following conditions T T T T T T T T (T T ) T T (T T ) T T It is well-known that T exists if and only if ran(t) is closed and in this case (T ) (T ) Let T L(X Y) be a closed range then T T is the orthogonal projection from Y onto ran(t) and T T is the orthogonal projection from X onto ran(t ) Projection in the sense that they are self adjoint idempotent operators A matrix form of a bounded adjointable operator T 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 T can be written as the following 2 2 matrix (11) T T 1 T 2 T 3 T 4 where T 1 L(M N ) T 2 L(M N ) T 3 L(M N ) and T 4 L(M N ) Note that P M denotes the projection corresponding to M In fact T 1 P N T P M T 2 P N T (1 P M ) T 3 (1 P N )T P M and T 4 (1 P N )T (1 P M ) Lemma 12 ( see 4 Corollary 12) Suppose that T L(X Y) has closed range Then T has the following matrix decomposition with respect to the orthogonal decompositions of closed submodules X ran(t ) ker(t) and Y ran(t) ker(t ) T 1 0 ran(t ) ran(t) T ker(t ) ker(t )

3 EXPLICIT SOLUTION TO MODULAR OPERATOR EQUATION T XS SX T A 3 where T 1 is invertible Moreover T T ran(t) ker(t ) ran(t ) ker(t ) 2 Solutions to T XS SX T A In this section we will study the operator equation T XS SX T A in the general context of the Hilbert C -modules First in the following theorem we solve to the operator equation T XS SX T A in the case when S and T are invertible operators Theorem 21 Let X Y Z be Hilbert A-modules S L(X Y) and T L(Z Y) be invertible operators and A L(Y) Then the following statements are equivalent (a) There exists a solution X L(X Z) to the operator equation T XS SX T A (b) A A If (a) or (b) is satisfied then any solution to Eq (21) T XS SX T A X L(X Z) has the form (22) X 1 2 T 1 A(S ) 1 + T 1 Z(S ) 1 where Z L(Y) satisfy Z Z Proof (a) (b) Obvious (b) (a) Note that if A A then X 1 2 T 1 A(S ) 1 + T 1 Z(S ) 1 is a solution to Eq (21) The following sentences state this claim T ( 1 2 T 1 A(S ) 1 + T 1 Z(S ) 1 )S S( 1 2 S 1 A (T ) 1 + S 1 Z (T ) 1 )T 1 2 (T T 1 A(S ) 1 S SS 1 A (T ) 1 T ) + T T 1 Z(S ) 1 S SS 1 Z (T ) 1 T A + Z Z A

4 4 MOHAMMADZADEH KARIZAKI HASSANI AND DRAGOMIR On the other hand let X be any solution to (21) Then X T 1 A(S ) 1 + T 1 SX T (S ) 1 We have X T 1 A(S ) 1 + T 1 SX T (S ) T 1 A(S ) T 1 A(S ) 1 + T 1 SX T (S ) T 1 A(S ) 1 + T 1 ( 1 2 A + SX T )(S ) 1 Taking Z 1 2 A + SX T we get Z Z Corollary 22 Suppose that Y Z are Hilbert A-modules and T L(Z Y) is invertible operator A L(Y) Then the following statements are equivalent (a) There exists a solution X L(Y Z) to the operator equation T X X T A (b) A A If (a) or (b) is satisfied then any solution to Eq (23) T X X T A X L(Y Z) has the form (24) X 1 2 T 1 A + T 1 Z where Z L(Y) satisfy Z Z In the following theorems we obtain explicit solutions to the operator equation (25) T XS SX T A via matrix form and complemented submodules Theorem 23 Suppose S L(Z Y) is an invertible operator and T L(Z Y) has closed range and A L(Y) Then the following statements are equivalent (a) There exists a solution X L(Z) to Eq (25) (b) A A and (1 T T )A(1 T T ) 0 If (a) or (b) is satisfied then any solution to Eq (25) has the form (26) X 1 2 T AT T (S ) 1 + T ZT T (S ) 1 + T A(1 T T )(S ) 1 + (1 T T )Y (S ) 1 where Z L(Y) satisfies T (Z Z )T 0 and Y L(Y Z) is arbitrary

5 EXPLICIT SOLUTION TO MODULAR OPERATOR EQUATION T XS SX T A 5 Proof (a) (b) Obviously A A Also (1 T T )A(1 T T ) (1 T T )(T XS SX T )(1 T T ) (T T T T )XS (1 T T ) (1 T T )SX (T T T T ) 0 (b) (a) Note that the condition (1 T T )A(1 T T ) 0 is equivalent to A AT T + T T A + T T AT T On the other hand since T (Z Z )T 0 then (Z Z )T ker(t ) ker(t ) Therefore T (Z Z )T 0 or equivalently T T ZT T T T Z (T ) T 0 Hence we have 1 2 T T AT T + T T ZT T + T T A(1 T T ) + T (1 T T )Y 1 2 T T A (T ) T T T Z (T ) T (1 T T )A (T ) T Y (1 T T )T AT T + T T A + T T AT T + T T ZT T T T Z (T ) T A That is any operator X of the form (26) is a solution to Eq (25) Now suppose that (27) X X 0 (S ) 1 X 0 L(Y Z) is a solution to Eq (25) We let (27) in Eq (25) Hence (25) get into the following equation (28) T X 0 X 0T A X 0 L(Y Z) Since T has closed range we have Z ran(t ) ker(t) and Y ran(t) ker(t ) Now T has the matrix form T 1 0 ran(t ) ran(t) T ker(t ) ker(t ) where T 1 is invertible On the other hand A A and (1 T T )A(1 T T ) 0 imply that A has the form A A 1 A 2 ran(t) ran(t) A 2 0 ker(t ) ker(t ) where A 1 A 1 Let X 0 have the form X 1 X 2 ran(t) X 0 X 3 X 4 ker(t ) ran(t ) ker(t )

6 6 MOHAMMADZADEH KARIZAKI HASSANI AND DRAGOMIR Now by using matrix form for operators T X 0 and A we have T 1 0 X 1 X 2 X1 X3 T1 0 A 1 A 2 X 3 X 4 X2 X4 A 2 0 or equivalently T1X1 X 1T 1 T 1 X 2 X 2T 1 0 Therefore A 1 A 2 A 2 0 (29) (210) T 1 X 1 X 1T 1 A 1 T 1 X 2 A 2 Now we obtain X 1 and X 2 By Lemma 11 T 1 is invertible Hence Corollary 22 implies that X T 1 1 A 1 + T 1 1 Z 1 where Z 1 L(ran(T)) satisfy Z 1 Z 1 Now multiplying T 1 1 from the left to Eq (210) we get (211) X 2 T1 1 A Hence T 1 1 A 1 + T1 1 Z 1 T1 1 A 2 is a solution to Eq (28) such that X 3 X 4 X 3 X 4 are arbitrary operators Let Y 1 Y 2 ran(t) ran(t ) Y X 3 X 4 ker(t ) ker(t ) and and Then Z Z 1 Z 2 Z 3 Z T AT T 2 T 1 1 A 1 0 T A(1 T T ) 0 T 1 A 2 ran(t) ker(t ) T ZT T (1 T T )Y ran(t) ker(t ) T 1 1 Z 1 0 X 3 X 4 Consequently X T AT T +T ZT T +T A(1 T T )+(1 T T )Y where Z L(Y) satisfies T (Z Z )T 0 and Y L(Y Z) is arbitrary

7 EXPLICIT SOLUTION TO MODULAR OPERATOR EQUATION T XS SX T A 7 Remark 24 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 ( ) ( ) x 1 y 1 x 2 y 2 x 1 x 2 + y 1 y 2 for x i ( X and ) y i Y i 1 2 To simplify the notation we use x y to x denote X Y y In the following theorem we obtain explicit solution to the operator equation T XS SX T A when (1 P ran(s) )T and S have closed ranges Theorem 25 Suppose that X Y Z are Hilbert A-modules S L(X Y) T L(Z Y) and A L(Y) and such that (1 SS )T and S have closed ranges If the equation (212) T XS SX T A X L(X Z) is solvable then 0 X T S A 0 T 0 (213) X 0 S 0 T S Proof Taking H Z X Y Y The 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 T (z n ) + S(x n ) y for some y Y Then (1 SS )T (z n ) (1 SS )(T (z n ) + S(x n )) (1 SS )(y) Since ran((1 SS )T) is assumed to be closed (1 SS )(y) (1 SS )T (z 1 ) for some z 1 Z It follows that y T (x 1 ) ker(1 SS ) ran(s) hence y T (z 1 ) + S(x) for some x X Therefore H has closed range Let Y 0 X Z X Z X and H T 0 Y Y Z X and X 0 S 0 A 0 B Y Y Y Y therefore Eq (212) get into (214) HY H B

8 8 MOHAMMADZADEH KARIZAKI HASSANI AND DRAGOMIR Since H has closed range and equation (212) has solution then equation (214) has solution therefore with multiplying HH on the left and multiply H (H ) on the right we have Y H B(H ) The proof of the following remark is the same as in the matrix case Remark 26 Let T L(Y Z) and S L(X Y) have closed ranges and A L(X Z) Then the equation (215) T XS A X L(Y) has a solution if and only if (216) T T AS S A In which case any solution of Eq (215) has the form (217) X T AS Now we solve to the operator equation T XS SX T A in the case when ran(t) + ran(s) is closed Theorem 27 Suppose that X is a Hilbert A-module S T A L(X ) such that ran(t) + ran(s) is closed Then the following statements are equivalent (a) There exists a solution X L(X ) to the operator equation T XS SX T A (b) A A and P ran(tt +SS )AP ran(tt +SS ) A If (a) or (b) is satisfied then any solution to Eq (218) T XS SX T A X L(X ) has the form (219) X T (T T + SS ) A(T T + SS ) S Proof (a) (b) Suppose that Eq (218) has a solution X L(X ) Then obviously A A On the other hand equation (218) get into T S 0 X T 0 A 0 (220) X 0 S 0

9 EXPLICIT SOLUTION TO MODULAR OPERATOR EQUATION T XS SX T A 9 Since Eq (218) is solvable then Eq (220) is solvable Since ran(t) + ran(s) T S is closed then 2 Lemma 4 implies that exists Hence by Remark 215 we have T S T S A 0 T 0 T 0 A 0 S 0 S 0 By applying 2 Lemma 4 Corollary 5 are shown that T S T S A 0 T 0 T 0 S 0 S 0 T S T (T T + SS ) 0 A(T T + SS ) T S (T T + SS ) 0 (T T + SS )(T T + SS ) A(T T + SS ) (T T + SS ) 0 A(T T + SS ) S T 0 S 0 P ran(tt +SS )AP ran((tt +SS ) ) 0 P ran(tt +SS )AP ran(tt +SS ) 0 A 0 That is P ran(tt +SS )AP ran(tt +SS ) A (b) (a) If P ran(tt +SS )AP ran(tt +SS ) A then we have P ran(tt +SS )AP ran((tt +SS ) ) 0 P ran(tt +SS )AP ran(tt +SS ) 0 A 0 or equivalently (T T + SS )(T T + SS ) A(T T + SS ) (T T + SS ) 0 T S T S A 0 T 0 T 0 A 0 S 0 S 0 Therefore Remark 215 implies that Eq (218) is solvable (220) is solvable and hence Eq

10 10 MOHAMMADZADEH KARIZAKI HASSANI AND DRAGOMIR Now by applying Remark 215 and 2 Lemma 4 Corollary 5 imply that 0 X T S A 0 T 0 X 0 S 0 T (T T + SS ) 0 A 0 (T T + SS ) T (T T + SS ) S S (T T + SS ) 0 T (T T + SS ) A(T T + SS ) T T (T T + SS ) A(T T + SS ) S S (T T + SS ) A(T T + SS ) T S (T T + SS ) A(T T + SS ) S Therefore T (T T + SS ) A(T T + SS ) T S (T T + SS ) A(T T + SS ) S 0 Consequently X has the form (219) Using exactly similar arguments we obtain the following analogue of Theorem 21 in which to Eq (21) is replaced by (221) T XS + SX T A All results of this section can be rewritten for to Eq (221) considering the following theorem Theorem 28 Let X Y Z be Hilbert A-modules S L(X Y) and T L(Z Y) be invertible operators and A L(Y) Then the following statements are equivalent (a) There exists a solution X L(X Z) to the operator equation T XS + SX T A (b) A A If (a) or (b) is satisfied then any solution to Eq (222) T XS + SX T A X L(X Z) has the form where Z L(Y) satisfy Z Z X 1 2 T 1 A(S ) 1 T 1 Z(S ) 1

11 EXPLICIT SOLUTION TO MODULAR OPERATOR EQUATION T XS SX T A 11 References 1 H Braden The equations A T X ± X T A B SIAM J Matrix Anal Appl 20 (1998) CY Deng H K Du Representations of the MoorePenrose inverse for a class of 2 by 2 block operator valued partial matrices Linear and Multilinear Algebran 58 (2010) D S Djordjevic Explicit solution of the operator equation A X+X A B J Comput Appl Math 200(2007) M Mohammadzadeh Karizaki M Hassani M Amyari and M Khosravi Operator matrix of Moore-Penrose inverse operators on Hilbert C -modules Colloq Math 140 (2015) E C Lance Hilbert C -Modules LMS Lecture Note Series 210 Cambridge Univ Press Q Xu L Sheng Positive semi-definite matrices of adjointable operators on Hilbert C*-modules Linear Algebra Appl 428 (2008) Q Xu L Sheng Y Gu The solutions to some operator equations Linear Algebra Appl 429 (2008) Y Yuan Solvability for a class of matrix equation and its applications JNanjing Univ (Math Biquart) 18 (2001) Mehdi Mohammadzadeh Karizaki Department of Mathematics Mashhad Branch Islamic Azad University Mashhad Iran address mohammadzadehkarizaki@gmailcom Mahmoud Hassani Department of Mathematics Mashhad Branch Islamic Azad University Mashhad Iran address mhassanimath@gmailcom hassani@mshdiauacir Silvestru Seve Dragomir Mathematics College of Engineering & Science Victoria University PO Box Melbourne City MC 8001 Australia address SeverDragomir@vueduau