Maps preserving the Douglas solution of operator equations
aa r X i v : . [ m a t h . F A ] F e b MAPS PRESERVING THE DOUGLAS SOLUTION OFOPERATOR EQUATIONS
ZSIGMOND TARCSAY
Abstract.
We consider bijective maps φ on the full operator algebra B ( H )of an infinite dimensional Hilbert space with the property that, for every A, B, X ∈ B ( H ), X is the Douglas solution of the equation A = BX if andonly if Y = φ ( X ) is the Douglas solution of the equation φ ( A ) = φ ( B ) Y . Weprove that those maps are implemented by a unitary or anti-unitary map U ,i.e., φ ( A ) = UAU ∗ . Introduction
Operator equations of the form(1.1) BX = A arise in many problems in engineering, physics, and statistics. In [3] R. G. Douglasconsidered the problem in the context of Hilbert space operators. He establishedthe following characterisation of solvability: Douglas’ Factorization Theorem.
Let H be a Hilbert space and let A, B ∈ B ( H ) .Then the following statements are equivalent: (i) the operator equation BX = A has a solution X ∈ B ( H ) , (ii) there exists λ > s.t. k A ∗ x k ≤ λ k B ∗ x k for all x ∈ H , (iii) we have the range inclusion ran A ⊆ ran B .In any case, there is a unique operator X = D such that ( D
1) ran D ⊆ [ker B ] ⊥ , ( D
2) ker D = ker A , ( D k D k = inf { λ > k A ∗ x k ≤ λ k B ∗ x k ( ∀ x ∈ H ) } . In fact, condition ( D
1) automatically implies both ( D
2) and ( D D satisfying conditions ( D D
3) is called the
Douglas solution (or re-duced solution ) of (1.1).There are many important objects in operator theory, including the Moore-Penrose pseudoinverse [2], the parallel sum [7] or the Schur complement [13], which
Mathematics Subject Classification.
Primary: 15A86; 47B49 Secondary: 47A62.
Key words and phrases.
Douglas theorem, operator equation, preserver problem.The corresponding author Zs. Tarcsay was supported by DAAD-TEMPUS Cooperation Project“Harmonic Analysis and Extremal Problems” (grant no. 308015), by the J´anos Bolyai ResearchScholarship of the Hungarian Academy of Sciences, and by the ´UNKP–20-5-ELTE-185 New Na-tional Excellence Program of the Ministry for Innovation and Technology. “Application DomainSpecific Highly Reliable IT Solutions” project has been implemented with the support providedfrom the National Research, Development and Innovation Fund of Hungary, financed under theThematic Excellence Programme TKP2020-NKA-06 (National Challenges Subprogramme) fund-ing scheme. can be defined as the Douglas solution of a suitably posed operator equation. Gen-eralisations of Douglas’ factorisation theorem to Banach spaces [1,4], locally convexspaces [12], Hilbert C ∗ -modules [9] and unbounded operators and relations [8, 11]are also available.In this paper we consider the following nonlinear preserver problem: Let φ : B ( H ) → B ( H ) be a bijective map with the property that, for every triple A, B, X of bounded operators in B ( H ), X is the Douglas solution of the equation A = BX if and only if Y = φ ( X ) is the Douglas solution of the equation φ ( A ) = φ ( B ) Y . (Shortly, we say in that case that φ preserves the Douglas solution in bothdirections.) In this note we describe the form all such transformation φ . Our resultshows that the structure of those mappings is quite rigid, namely, every Douglassolution preserving map φ is of the form φ ( A ) = U AU ∗ , A ∈ B ( H ) , for a fixed unitary or anti-unitary map U .We stress that the only constraint concerns the dimension of the underlyingHilbert space, and no algebraic or topological assumptions like linearity or conti-nuity on φ are imposed. On the contrary: these properties follow from the intrinsicstructure of such a transformation.2. The main theorem
Before we state and prove our main result, let us fix some notation. Throughout, H denotes an infinite dimensional complex Hilbert space, and B ( H ) stands for the C ∗ -algebra of all bounded, linear operators A : H → H . The kernel and rangespaces of an operator A ∈ B ( H ) are denoted by ker A and ran A , respectively. Thecollection of the ranges of all bounded operators is denoted by Lat( H ), that is,Lat( H ) := { ran A : A ∈ B ( H ) } . Note that Lat( H ) is not identical with the class of all linear submanifolds of H (cf.[7]). For the set of one-dimensional subspaces of H we use the symbol Lat ( H ):Lat ( H ) := { C e : e ∈ H , e = 0 } . For given two vectors e, f ∈ H , we define the one-rank operator e ⊗ f by( e ⊗ f )( x ) := h x , f i e, x ∈ H . If M ∈
Lat( H ) is a closed subspace of H then we denote by P M the orthogonalprojection onto M . If M is of the form M = C e for some e ∈ H then we set P e := P M . Finally, the set of invertible operators (that is, of those operators whichdo not have 0 in their spectrum) is denoted by GL( H ). Theorem.
Let H be an infinite dimensional Hilbert space. A bijective map φ : B ( H ) → B ( H ) preserves the Douglas solution in both directions if and only if thereexists a unitary or anti-unitary operator U : H → H such that (2.1) φ ( A ) = U AU ∗ , A ∈ B ( H ) . Proof.
We start by observing that the bijective map φ : B ( H ) → B ( H ) preservesthe Douglas solution in both directions if and only if A = BX ran X ⊆ [ker B ] ⊥ (cid:27) ⇐⇒ (cid:26) φ ( A ) = φ ( B ) φ ( X )ran φ ( X ) ⊆ [ker φ ( B )] ⊥ (2.2) APS PRESERVING DOUGLAS SOLUTION 3 for all
A, B, X ∈ B ( H ). From this we infer that φ satisfies(2.3) φ ( BD ) = φ ( B ) φ ( D ) , ∀ B, D ∈ B ( H ) , ran D ⊆ [ker B ] ⊥ . Claim 1. ran( B ) = H ⇐⇒ ran( φ ( B )) = H , ( B ∈ B ( H ))Indeed, if ran B = H then equation (1.1) is solvable for every A ∈ B ( H ), andtherefore the same is true for equation φ ( B ) Y = φ ( A ). Since φ is bijective, thisimplies ran φ ( B ) = H . The converse direction is proved similarly. Claim 2. ker B = { } ⇐⇒ ker φ ( B ) = { } , ( B ∈ B ( H )) . For if ker B = { } , then the Douglas solution of BX = B is X = I , so by (2.2)we get φ ( B ) = φ ( B ) φ ( I ) , and ran φ ( I ) ⊆ [ker φ ( B )] ⊥ . Since φ ( I ) is surjective, we infer that ker φ ( B ) = { } . This proves Claim 2.We see from Claims 1 & 2 and equation (2.3) that the restriction φ | GL( H ) of φ onto the group of invertible elements GL( H ) is in fact an automorphism of GL( H ).Hence, by [10, Theorem 3.1], there exists a linear or conjugate linear topologicalisomorphism S : H → H such that either(2.4) φ ( X ) = SXS − , ∀ X ∈ GL( H ) , or(2.5) φ ( X ) = ( SX − S − ) ∗ , ∀ X ∈ GL( H ) . In Claim 4 we shall demonstrate that (2.5) may not happen. To do so we first needthe following.
Claim 3. φ maps orthogonal projections into orthogonal projections. Indeed, if P = P = P ∗ then ran P = [ker P ] ⊥ and therefore (2.3) implies φ ( P ) = φ ( P · P ) = φ ( P ) φ ( P ) , hence φ ( P ) is an idempotent. Furthermore, the Douglas solution of the equation P = P X is just X = P , hence the Douglas solution of φ ( P ) = φ ( P ) Y is Y = φ ( P ), which in turn implies ran φ ( P ) ⊆ [ker φ ( P )] ⊥ . Thus φ ( P ) is an orthogonalprojection. Claim 4. (2.5) cannot hold.
For assume towards a contradiction that it does. Let ( e n ) n ∈ N be an orthonormalsequence in H , and take a self-adjoint operator A ∈ B ( H ) such that Ae n = n e n .Let us denote by P n := e n ⊗ e n the orthogonal projection onto C e n . Then we have AP n = n P n and ran P n ⊆ [ker A ] ⊥ . From (2.3) and (2.5) we thus obtain φ ( A ) φ ( P n ) = φ ( AP n ) = φ ( n IP n ) = φ ( n I ) φ ( P n ) = nφ ( P n ) . Since φ ( P n ) is a non-zero orthogonal projection, this means that every integer n isan eigenvalue of φ ( A ), that is impossible. This proves Claim 4.So we see now that φ acts on GL( H ) by (2.4). Next we observe that φ preservesrange inclusion in both directions, i.e.,ran A ⊆ ran B ⇐⇒ ran φ ( A ) ⊆ ran φ ( B ) , A, B ∈ B ( H ) , ZS. TARCSAY which is obvious by the very definition of Douglas solution preserving maps. Hence φ induces a bijective map Φ : Lat( H ) → Lat( H ) on the set of operator ranges bythe correspondence Φ(ran A ) := ran φ ( A ) , A ∈ B ( H ) . Claim 5.
The restriction Φ | Lat ( H ) of Φ onto Lat ( H ) is a bijection of Lat ( H ) . Let B ∈ B ( H ) be a rank one operator and take two non-zero vectors e , e ∈ ran Φ( B ). Let E , E ∈ B ( H ) be rank one operators such that ran E i = e i , thenthe equations Φ( B ) Y = E i , i = 1 ,
2, are both solvable and therefore, setting C i := φ − ( E i ), equations BX = C i are solvable too. Since B has rank one, it follows thatran C = ran C = ran B and therefore ran E = ran E , hence e and e are linearlydependent. A similar argument applied to φ − shows that Φ : Lat ( H ) → Lat ( H )is a bijection. This proves Claim 5.It is clear that for all non-zero vectors e, f, g ∈ H one has C g ⊆ C e + C f ⇐⇒ Φ( C g ) ⊆ Φ( C e ) + Φ( C f ) , hence Φ is a projectivity in the sense of [5]. The fundamental theorem of projectivegeometry (see e.g. [5]) implies that Φ is implemented by a semi-linear bijection T : H → H in the sense thatΦ( C e ) = C · T e, e ∈ H , e = 0 . Moreover, T is unique up to a scalar multiplier. Take now an M ∈
Lat( H ), thenΦ( M ) = Φ (cid:18) [ e ∈M C e (cid:19) = [ e ∈M Φ( C e ) = [ e ∈M C · T e = T ( M ) , whence it follows that(2.6) ran φ ( A ) = ran T A, A ∈ B ( H ) . By [6, Theorem 1] we infer that T : H → H is either a linear or a conjugate lineartopological isomorphism.
Claim 6.
We claim that S = λT for some non-zero scalar λ . Let us assume first that both
S, T are linear. In this case, φ ( λA ) = λφ ( A ) forevery A ∈ B ( H ) and λ ∈ C . By (2.6), φ ( P x ) = P T x , x ∈ H , where P x denotes the orthogonal projection onto C x . Hence φ ( XP x ) = φ ( X ) φ ( P x ) = SXS − P T x , X ∈ GL( H ) , x ∈ H . Consequently, for every fixed X ∈ GL( H ), C · T Xx = ran(
T XP x ) = ran φ ( XP x ) = C · SXS − T x, ∀ x ∈ H . The uniqueness part of the fundamental theorem of projective geometry implies
T X = λ X SXS − T, X ∈ GL( H ) , where λ X ∈ C is a scalar depending on X . After rearranging this yields S − T X = λ X XS − T , whence, by setting R := S − T , we obtain that(2.7) RX = λ X XR, ∀ X ∈ GL( H ) . APS PRESERVING DOUGLAS SOLUTION 5
It is easy to check that the map X λ X is continuous with λ I = 1, λ XY = λ X λ Y and λ αX = λ X , (0 = α ∈ C ). If X, Y, X + Y ∈ GL( H ) then also λ X + Y ( X + Y ) = λ X X + λ Y Y, whence we obtain [ λ X + Y − λ X ] I = [ λ Y − λ X + Y ] Y X − . Therefore, if
Y X − / ∈ C I , then λ ( X ) = λ ( X + Y ) = λ ( Y ). For every X ∈ GL( H )with | α | < k X k − we have I + αX ∈ GL( H ) and so λ X = λ αX = λ I + αX = λ I = 1 . According to (2.7) this means that R commutes with every X ∈ GL( H ) and there-fore R ∈ C I . Consequently, S = αT for some (non-zero) α .The same argument applies when S and T are both conjugate linear.Assume now that S is linear and T is conjugate linear, or conversely, that S isconjugate linear and T is linear. Just like above we conclude that (2.7) holds truewith the conjugate linear operator R = S − T . Then we have the following identitieson the spectra: σ ( X ∗ ) = σ ( RXR − ) = λ X · σ ( X ) , ∀ X ∈ GL( H )But this cannot hold for every X ∈ GL( H ) (take e.g. an X with spectrum σ ( X ) = { , i, − i } ), so this is a contradiction. The proof of Claim 6 is therefore complete.We may and we will therefore assume that S = T . Claim 7. T is unitary or anti-unitary Let P be an orthogonal projection, then 2 P − I is unitary such that (2 P − I ) P = P . Hence φ ( P ) = φ (2 P − I ) φ ( P ) = (2 T P T − − I ) φ ( P ) , whence we get(2.8) φ ( P ) = T P T − . In particular, for e ∈ H , k e k = 1, we have(2.9) φ ( e ⊗ e ) = T e ⊗ ( T − ) ∗ e. Furthermore, φ ( e ⊗ e ) = φ ( P e ) = P T e = 1 k T e k T e ⊗ T e, whence ( T − ) ∗ e = k T e k T e , and therefore
T x = k T x k k x k ( T − ) ∗ x, x ∈ H , x = 0 . This yields that x ⊥ y implies T x ⊥ T y for every pair of non-zero vectors x, y ∈ H .An easy calculation shows that k T e k = k T f k =: α for every unit vectors e, f ∈ H , e ⊥ f . It follows therefore that U := α − T is either a unitary or an anti-unitaryoperator.We may therefore assume that U = T . Then(2.10) φ ( e ⊗ e ) = U e ⊗ U e, e ∈ H , k e k = 1 . ZS. TARCSAY
Let now e, f ∈ H be unit vectors, then φ ( e ⊗ f ) is of rank one, namely, ran φ ( e ⊗ f ) = C · U e . Consequently, φ ( e ⊗ f ) = U e ⊗ u e,f for some u e,f ∈ H (depending on thevectors e, f ). Since we have e ⊗ e = ( e ⊗ f )( f ⊗ e ) , ran( f ⊗ e ) ⊆ [ker( e ⊗ f )] ⊥ , from (2.3) it follows that U e ⊗ U e = (
U e ⊗ u e,f )( U f ⊗ u f,e ) = h U f , u e,f i · ( U e ⊗ u f,e ) , consequently u f,e ∈ C · U e . In particular, φ ( e ⊗ f ) = λ e,f · U e ⊗ U f, for some λ e,f ∈ C . Consider now a unitary operator V ∈ B ( H ) such that V e = f .Then f ⊗ f = V ( e ⊗ f ), whence it follows that U f ⊗ U f = φ ( V ) φ ( e ⊗ f ) = U V U − ( λ e,f U e ⊗ U f ) = λ e,f · U f ⊗ U f, and therefore λ e,f = 1. Consequently, φ ( e ⊗ f ) = U e ⊗ U f for every pair of unitvectors e, f ∈ H . This implies(2.11) φ ( x ⊗ y ) = U x ⊗ U y, x, y ∈ H . To conclude the proof, consider an arbitrary operator A ∈ B ( H ). Take first aunit vector e ∈ [ker A ] ⊥ then the identities φ ( Ae ⊗ e ) = φ ( A ) ⊗ φ ( e ⊗ e ) and (2.11)imply U Ae ⊗ U e = φ ( A ) U e ⊗ U e.
Hence
U Ae = φ ( A ) U e . This proves that(2.12) φ ( A ) U x = U Ax, ∀ x ∈ [ker A ] ⊥ . Denote by P the orthogonal projection onto [ker A ] ⊥ . Since the Douglas solutionof A = AX is X = P , the Douglas solution of φ ( A ) = φ ( A ) Y is in turn Y = φ ( P ). Here, φ ( P ) coincides with the orthogonal projection onto U (ker A ⊥ ). Thuswe conclude thatker φ ( A ) = ker φ ( P ) = [ U (ker A ⊥ )] ⊥ = U (ker A ) . In particular,(2.13) 0 = φ ( A ) U x = U Ax, ∀ x ∈ ker A. From identities (2.12) and (2.13) it follows that φ satisfies (2.1). (cid:3) Remark.
The above proof relies on tools which depend on the infinite-dimensionalityof H . It remains an intriguing open question whether the statement of the Theoremremains true in the finite dimensional case. Acknowledgement.
The author is grateful to Gy¨orgy P. Geh´er for many en-lightening conversations about the contents of the paper. He is also extremely grate-ful to Bence Horv´ath for carefully reading the paper and for his valuable commentswhich much improved the exposition of the paper.
APS PRESERVING DOUGLAS SOLUTION 7
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Zs. Tarcsay, Department of Applied Analysis and Computational Mathematics, E¨otv¨osLor´and University, P´azm´any P´eter s´et´any 1/c., Budapest H-1117, Hungary, and Alfr´edR´enyi Institute of Mathematics, Re´altanoda utca 13-15., Budapest H-1053, Hungary
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