Math. Nachr. 194 (1998), 205-224

The Point Spectrum of Unitary Dilations in Kre'in Spaces

I3y DIRK TEMME of Amsterdam

(Received October 30, 1995)

Abstract. For a bicontractive operator T on a Kreh space the connections between its eigenvalues illid eigenstructure and the eigenvalues and eigenstructure of its minimal unitary dilation U are tdudied. For eigenvalues on the unit circle of T in general only part of the eigenspace of T will return ILS an eigenspace of U and the corresponding eigenvalue will be a singular critical point of U.

0. Introduction

Contractive operators in Hilbert spaces and their unitary dilations are well under- stood. In spaces with an indefinite metric the classical dilation theory can be continued to a large extend, but also new features arise. For instance, in general a contractive operator in a space with an indefinite metric cannot be written as the sum of a unitary operator and a completely - non - unitary operator, an eigenvector may be perpendic- ular to itself and the norms of spectral projections may be arbitrarily large.

Spectral properties of unitary dilations in Pontryagin and Kreh spaces have been considered in the papers [BDS], [CG], and [DLS]. Recently, detailed information about eigenvalues, Jordan chains and their behaviour in the indefinite inner product for dissipative and contractive operators has been obtained. See [AL] and [RT]. In this paper, this knowledge will be used to identify some relations between spectral properties of a bicontraction on a Krein space and spectral properties of its unitary dilation.

As a motivation the following example may be presented. Consider the linear oper- ator U on L2(T) @ C 2 represented by the matrix

1991 Mathematics Subject Classification. Primary: 47B50; Secondary: 47A10, 47A20. Keywords and phrases. Indefinite metric, unitary dilations, point spectrum.

206 Math. Nachr. 194 (19DH)

Here M, denotes multiplication of a function in &(T) with the variable z , and ( . , . ) denotes the standard Hilbert space scalar product on Lz(T). The number & in t.11t1

right upper entry is meant to be the constant function on T with value 4. Equip i.1w space with the indefinite scalar product [ . I . ] given by

This scalar product makes the space Lz(T) @ C 2 into a Pontryagin space. Wil.11 respect to this indefinite scalar product the operator U is unitary. The spectrum of 1 I is

.(V) = TI and the algebraic eigenspace corresponding to the eigenvalue 1 is

On the other hand, consider the contractive operator

on CC with the inner product [ . , given by

Its spectrum contains only the eigenvalue 1 with corresponding algebraic eigenspaw

The operator U is a minimal unitary dilation of T. The number 1 is a so-callvct singular critical point of U. It is typical that the algebraic eigenspace of U at 1 11iih

smaller dimension than the algebraic eigenspace of T at 1. The aim of this paper i:, to show that it can be foreseen from the contraction whether such a singular critiviil point will appear in the spectrum of the minimal unitary dilation.

Section 1 contains some basic material on indefinite scalar product spaces and O I I

unitary dilations. In Section 2 a question on whether or not the span of the algebIilIi eigenspaces in the eigenvalues X and 1/x of a contraction will be nondegeneratci 1 4

considered. In the next sections it is described which eigenvalues of a bicontrac1,ivi. linear operator will appear in the spectrum of the unitary dilation, and a recipr Ioi computing the Jordan chains of the unitary dilation from the chains of the origiiiiil operator is given. Eigenvalues outside the unit circle will be treated in Sectioii 3 , eigenvalues inside in Section 4. Eigenvalues on the unit circle will be considered iii

Section 5 . Here the eigenstructure of the contraction is rich and interesting and Ic;i(Iti to regular and singular critical points for the unitary dilation.

R(T, (1)) = C 2 .

'Ibirime, Point Spectrum of Unitary Dilations 207

1. Preliminaries

(:onsider a linear vector space, equipped with a scalar product [ - , - 1 that satisfies 1 1 1 1 % common rules of linearity and symmetry. That is, for all vectors 2, y, z in the wlmce and A, p E Q: we have [y, z] = [z, y ] and [Xz + py, 21 = A[%, z] + p[y, z ] . The ac.nlar product is allowed to be indefinite in the sense that a vector z can be positive, IivKative or neutral ([z, z] > 0, < 0 or = 0 respectively). A subspace is called positive, iionnegative, neutral, nonpositive, negative if all its nonzero vectors have that property. 0 1 1 any nonnegative or nonpositive subspace M the Schwarz inequality holds, i. e.,

-

IoI all 2, y E M . Vectors 2, y are called orthogonal if [z, y ] = 0 and subsets MI N are orlhogonal ((MINI = 0, M I n/) if [z,y] = 0 for all z E M , y E N . The orthogonal mnpanion M I of M is the subspace containing all vectors that are orthogonal to M . ' 1 ' 1 ~ scalar product is called nondegenerate if for any vector z # 0 there exists a vector v such that [z,y] # 0. A subspace M is called degenerate if M contains a nonzero vector z such that z I M , and it is called nondegenerate otherwise. The isotropic part M o of M is the subspace M o = M r l M I . A pair of neutral subspaces MI N Ihat satisfies M' nn/ = M nN' = (0) is called skewly linked, which we denote by

Let K be a vector space with a nondegenerate scalar product [ . , a ] as above. The npace K: is called a Krein space if there exists a negative subspace Ic- such that both Ic- quipped with the scalar product -[ . , a ] and (L)' equipped with [ . , e ] are Hilbert npaces. In such a case the dimension of the subspace Ic- is uniquely determined. If dim K- = 0, then K is a Hilbert space, and if d i m L is finite, then K is called a I'ontryagin space. The number dim K- is called the rank of negativity of the space A'. Denote K+ = ( K - ) l . The space K admits the fundamental decomposition K = h'+ [+I K - . Here + denotes a direct sum and the brackets express the orthogonality. Any z E K equals x+ + z- for some uniquely determined z+ E Ic+, z- E K - . The iiorm 11 . (JK on K induced by this fundamental decomposition is defined by 11~11: = I$:+ , z+] - [z- , z-1. For more details about these spaces see [AI], [Bo], [IKL].

Let A : K1 + K2 be a linear operator, with dense domain ?>(A), between Kre'in paces Kl and KZ with scalar product [ . , . I 1 and [ . , . ] 2 , respectively. The adjoint ,A* : K 2 --f K1 with domain ?>(A*) is defined as usual, and satisfies [Ax, y ] 2 = [z, A'yll for all z E D ( A ) , y E ?>(A*). The operator A is called contractive if [ A z , Az]z 5 [z, z]1 lor all z E D ( A ) , and isometric if [ A z , Az]z = [z, zI1 for all E ?>(A). The contraction 11 is called a bicontraction if A* is also a contraction. The operator A : K + K on the KreTn space K is called unitary if A*A = AA* = IK. The linear operator A : K + K with dense domain ?>(A) is called selfadjoint if A = A* and A is called dissipative if Im[Az,z] 2 0 for all z E D ( A ) . For Krein spaces K1 and K2 the npace L(Icl , K 2 ) consists of all bounded linear operators with full domain from K1 on I c 2 . If K1 = Ic2 = K we denote L(K) = L(K1,Ic2). It is known, see, e.g., IDR], Theorem 1.3.7, that any contraction T E L(K) on a Pontryagin space K is a hicontraction.

Let K be a Krein space and let T E L ( K ) . A unitary dilation of T is a unitary

M #N.

208 Math. Nachr. 194 (1998)

operator U on a Krein space K' 3 K such that

~ U " r x = Tnx and n(U*)"rz = (T*)"x

for all x E K . Here A : K' + K denotes the orthogonal projection and r = A* is tlw canonical embedding of K in K'. A unitary dilation U is called a minimal unitary dilation if, moreover,

K' = Span{ ..., U-'K, U°Kl U'K, . .. } .

For any bounded T on K there exists a minimal unitary dilation. If U on K' is II

minimal unitary dilation of a bicontractive T on K , then the ranks of negativity of A ' and of K' are equal. Moreover, if U1 on K1 and U2 on K2 both are minimal unitaty dilations of a bicontractive T E L(K), then iY1 and U2 are unitarily equivalent, sov [DR], Theorem 3.1.7. That is, there exists a bijective isometry 6 : K1 + I c 2 such thril 4-'U24 = U1, and rjx = x for all x E K. See [BDS] and [CG] and the references givcw there.

So the minimal unitary dilation of a bicontractive operator in L ( K ) is determincvl uniquely up to unitary equivalence. In this paper we will consider the special represcw tation called the Schaer form. We follow the construction that was used in [DR]. Lilt

T E L ( K ) be an operator on a Krein space K . Then there exist Krein spaces 31- will1 scalar product [ . , a ] - , 31+ with scalar product [ . , .]+ and operators A : Ic + 31 , B : 31+ + 31- and C : 31+ + Ic such that

AA*+BB' = I x - , TT* + CC* = IK , T A * + C B * = 0 , A*A+T*T = I K ,

B * B + C*C = I x + , B'A+C*T = 0 .

Given 31- and 31+ the norms 11 - 11- and 11 II+ are induced by some fundamentid decompositions of 31- and 31+, respectively. Make copies Icj = 'H- with scalar product [ , -]j = [. , -1- and norm 11 - Ilj = 11 . 11- for j = -1, -2 , . . . and Kj = 'H+ with scaliir product [ - , . ]j = [ . , .]+ and norm 11 - 1 1 , = 11 I]+ for j = 1 , 2 , . . . . Let

(1.2)

be the Krein space consisting of all vectors

Ic' = . * * @ K-2 @ K-1 @ [K] @ K1@ K z @ ...

x = ( . . . , 5-2, x-1, [ Z O ] , 21, 572, . . .IT with Xj E K j for j # 0 and xo E K: such that

The scalar product [ . , . ] ~ l on K' is given by

‘Ihtne, Point Spectrum of Unitary Dilations

u =

209

0

A

[TI 0

0

B

C 0

TIE entry with brackets always denotes the central entry acting from K: to K. This U Irc a. unitary dilation of T. The minimality of the dilation can be achieved by taking .4 and C such that Ker A’ = (0) and KerC = (0). If T E L(K) is a bicontraction, I.lien both I - T’T and I - TT’ are positive semidefinite on K. In that special case t . 1 1 ~ spaces 3c- and 3c+ are Hilbert spaces.

In this paper the operators xj : Ic’ + Kj for j # 0 and no : Ec’ + K: denote the orthogonal projections. The operators Tj = ~r; for j E P are the canonical embeddings. ‘NJC notions a(A), p(A), ap(A) and o, (A) express the spectrum, the resolvent set, the 1)oint spectrum and the continuous spectrum of the operator A, respectively. Let S l w some set in C . Then R(A, S) denotes the linear span of the algebraic eigenspaces of the operator A corresponding to eigenvalues in S n op(A). That is,

’ R(A,S) = span{Ker(A-XI)”(XESno, (A) ,n€N}

Assume X with 1x1 = 1 is an eigenvalue of the operator T. The eigenvalue X will I)(: called a (non)degenerate eigenvalue if R(T, {A}) is (non)degenerate. Note that, riccording to [L], Proposition 11.5.6, for a unitary operator U on a Pontryagin space t.110 notion of a (non)degenerate eigenvalue is equivalent to the notion of a singular (regular) critical point introduced there.

2. Algebraic eigenspaces of a contraction

Consider a unitary operator U on a Krein space K’. It is known, see [Bo], The- orem 11.2.5, that for any eigenvalue X of U with 1x1 # 1 the algebraic eigenspace R(U, {A} ) is neutral. For a contraction T E L(K) on a Kreh space K the algebraic vigenspace R(T, (X}) is nonnegative if 1x1 < 1 and nonpositive if 1x1 > 1, see [AL], I’roposition 2.1. If U is unitary on a Pontryagin space, then, according to [IKL], Lemma 12.1, the subspace R ( U , {A, l/x}) is nondegenerate. In this section the latter Ntatement will be generalized for contractions on finite dimensional Pontryagin spaces. For contractions on infinite dimensional Pontryagin spaces and unitary operators on infinite dimensional Krein spaces counterexamples will be given.

210 Math. Nachr. 194 (19!)H)

For a contraction the number 0 also may be in the point spectrum and should IJU considered separately.

Proposition 2.1. Let T be a contraction on a Krein space K . Then R(T, (0)) ia II

Proof. Assume Tnx = 0. Then

positive subspace.

0 = [T"z,T"z] 5 a * * 5 [Tz,Tz] 5 [z , z] . Hence R(T, (0)) is nonnegative. Moreover, the neutral vectors in R(T, (0)) fonii a T-invariant subspace which is, in fact, the isotropic part R(T,(O))O. Assurricb x E R(T, (0 ) )O and Tx = 0. Then [(I - T*T)z,x] = 0. The operator I - T'T beiiiy, nonnegative it follows that ( I - T'T)z = 0. Hence x = 0. This proves the statement.

LI

Assume T is a contractive operator on a Kreh space K. Assume there is a numbcv E E T with E E p(T) U uC(T). Then, without loss of generality, we can assume thilI E = -1. The Cayley transform

A = i (IK - T) ( Ix +T)-' is a dissipative operator on K. That is,

Im [Az,z] 2 0 for all x E 'D(A) .

Note that if -1 E oc(T), then A is an unbounded operator with dense domain. Con- versely, the contraction T is reconstructed from A by

T = ( ~ I K - A ) ( i l ~ + A ) - ' .

The following relations hold:

1-X 1 + X

X E o(T) --.. i - E u ( A ) ;

i - X - E ~ ( 2 ' ) ; i + X X E a ( A )

Moreover, any finite dimensional invariant subspace M of the operator T is invariant for A , and conversely. See [Bo], Section 11.5.

The following theorem is due to AZIZOV, see [AI], Theorem 11.2.15.

Theorem 2.2. Assume A is a densely defined dissipative operator on a K r e h spacc K with domain V ( A ) . For any A -invariant subspace M the isotropic part M o is A -invariant.

' h i m e , Point Spectrum of Unitary Dilations 211

Proof . Consider the graph space G = K x K equipped with the indefinite inner

Afisume z is an element of the isotropic part M o of M . Then in particular [x, Az] = 0 riiid Im [Az,z] = 0. Hence the vector

G X )

is neutral in G, and isotropic in the nonnegative graph

That is,

F'rom (SA)' = GA* it follows that

I n other words x E D(A) n D(A') and Ax = A'z. Hence for all v E M we have

[Az,y] = [A'z,y] = [z,Ay] = 0

since Ay E M. 0

Corollary 2.3. Let T be a contractive operator on a Krein space K: and assume there is a number E E T with E E p(T)Uu,(T). For any X # 0 the subspaces R(T, {A})' and R(T , { A , 1/ 1))' are T -invariant.

Proof. Combine (2.1) with Theorem 2.2. 0

As a consequence we have

Theorem 2.4. Assume T is a contractive operator on a finite dimensional Pon- tryagin space K. For any X # 0 the subspace R(T , {A, l/x}) is nondegenerate.

Proof. From Corollary 2.3 we have that R ( T , { A , l / X } ) ' is T-invariant. Hence for any z E R ( T , { X l l / ~ } ) o we have that [z,z] = [Tz,Tz] = 0 and therefore [ ( I - T'T)z,x] = 0. Applying Schwarz inequality (1.1) on the positive semidefi- nite scalar product ( ( I - T'T) . , - 1 it follows that [(I - T'T)x, g] = 0 for all y E K:. Since [ a , . ] is nondegenerate it follows that ( I - T'T)z = 0. Hence R(T, {A, 1/x}) is contained in Ker ( I - T'T). Let p # XI 1/ be an element of u(T). Proposition 2.5 in [AL] yields

0

R(T, {A, l / X } ) O 1 RP, { P I ) *

212 Math. Nachr. 104 (ISM)

Hence R(T, { A , l /X} ) ' is orthogonal to the whole space K, and it necessarily ody contains the zero vector. C1

Corollary 2.5. Assume A is a dissipative opemtor on a finite dimensional Po71 tryagin space K. For any A the subspace R(A, { A , I}) is nondegenemte.

Proof. Combine (2.2) with Theorem 2.4. 1 1

A different proof of the statement in Corollary 2.5 for real A appears in [Ba]. Note that Theorem 2.4 holds in particular for X on the unit circle T. In that case X :

1/ and R(T, { A } ) = R(T, {A , l/x}) is nondegenerate. If K is infinite dimensional, however, even for unitary U the root manifold R ( U , { A } ) for A E ap(V) n T can Iw degenerate as well.

On the other hand, in the beginning of this section it was mentioned that Thco- rem 2.4 also holds for unitary U on an infinite dimensional Pontryagin space K'. Thv theorem, however, does not generalize to contractions on infinite dimensional Pontrya- gin spaces. Indeed, consider the operator T on K represented by the matrix

and equip K: with the indefinite metric given by the Gram matrix

G =

' 0 1 0 0

1 2 0 0 * . .

0 0 1 0 * . .

0 0 0 1 * .

Then T is contractive and R(T, {1/2, 2)) = R(T, (2)) = span {e l } is neutral.

R ( U , { 1/2,2}) is degenerate, consider the operator In order to construct an example of a unitary operator U for which the subspaw

T =

2 1 0

0 2 1 * .

0 0 2 . .

'l'imme, Point Spectrum of Unitary Dilations 213

011 the Hilbert space 3-1 = L2 with scalar product ( . , . ). Let IC = 3-1 @ 3-1 be the Krein Npace with the indefinite scalar product defined by

Then the operator

u = ( T 0 T-• 0 )

is unitary in K and R(T, {1/2, 2)) = R(T, (2)) is neutral.

3. Spectrum outside the unit circle

Let K be a Kreh space. Let T E C(K) be a bicontraction and let U be the minimal unitary dilation of T on the Krein space K' 3 K . Without loss of generality it will be assumed that U and K' have the form of (1.4) and (1.2). The aim of this section is 1.0 show that the point spectra of T and U outside the unit circle T coincide and that Ifhere is a 1 - 1 relation between the corresponding Jordan chains of T and U . In this section 1x1 > 1.

Lemma 3.1. Let 1x1 > 1. Let yl , . . . , pm E K' be a Jordan chain of U at A. Let (3.1) z k = nopk for k = 1, ... , m e Then zl, . . . , zm E K is a Jordan chain ofT at A. In particular, q,(U)\D C ap(T)\b find roR(& { X I ) c R(T, { X I ) .

P roof . Let yo = 0 E K'. Clearly r j p o = 0 for j = 1 ,2 , . . . . Fix k E { O , l , . . . , m - 1). We have (U - X)yk+l = pk. Assume r j p k = 0 for j = 1 , 2 , . . . . From the special form of U and the square summability of the coefficients of yk+l it follows that njyk+l = 0 for j = 1 ,2 , . . . . By induction r j y l = . = r j y m = 0 for all j = 1,2 , . . . . Using induction it is also proved that nopl, . . . , r0ym # 0. Using the special form of U it is immediately clear that (T- X ) r o v k + l = noyk for k = 0 , 1 , . . . , rn - 1. Hence X E op(T)

0

Conversely, the vectors in a Jordan chain of U corresponding to eigenvalues outside the unit circle can be explicitly constructed, iteratively, in terms of the vectors in a Jordan chain of T . Let q,. . . , x, E K be a Jordan chain of T at X with 1x1 > 1. Let 20 = 0 E K. We need the operator V on K' represented by the matrix

and nOylr . . . , nay, E K is a Jordan chain of T at A.

v = . . . i / x 2 11x3 o ... 1/x 1 / X 2 0 . * .

0 1/x 0 ... 0 [O] ...

214 Math. Nachr. 194 (190H)

The nonzero entries are not scalars but multiples of the identity on K-. Note that \ ’ is of the form

where W has spectrum o(W) = { z I Iz - XI 5 1). Hence the operator V is bourltlvtl since 1x1 > 1. Let yo = 0 E K’. Define

(3.3) vk+l = TOzk+l -t VT-iAzk+i - Vvk

for k = 0,1,. . . ,m - 1.

Lemma 3.2. Let T E L(K) be a bicontmction on the Krer’n space K. Let U br its minimal unitary dilation. Let 1x1 > 1. Let 21 , . . . , z, E K be a Jordan chain I)/

T at A. Then y1,. . . ,urn given by (3.3) is a Jordan chain of U at A. In particulw, op(T) \ B C a,(U) \B.

Proof. It suffices to check that, for k = O , l , . . . ,m - 1,

+ + ... 1 0

0 1 0

0 [O] ...

= Y k .

From (3.3) it is clear that vk+l # 0 since zk+l # 0. CJ

Starting with a Jordan chain zl , . . . , x, E K of T at X Lemma 3.2 gives a corw sponding Jordan chain yl, . . . , ym E K’ of U at A. From the construction in Lemma 3.2 it is clear that applying Lemma 3.1 on the chain y1,. . . ,urn results in the origiriiil chain 2 1 , . . . , x,. If we start with a chain y1,. . . ,ym E K‘ of U at X and apply fils1 Lemma 3.1 and next Lemma 3.2 we also end up with the original chain yl , . . . , y,,, .

Theorem 3.3. Let K be a Krein space and let T E C(K) be a bicontmction. Lr/ U be its minimal unitary dilation. Let 1x1 > 1. The formula (3.1) defines a I I correspondence between the Jordan chains of U at A. and the Jordan chains of T at X The transformation from the chain 2 1 , . . . , x, t o the chain yl . . . , y, given by (3.3) is its inverse. I n particular, op(T) \ = o p ( U ) \ B and .rroR(U, { A } ) = R(T, { A ) ) . Moreover, the partial multiplicities of T and U at X coincide.

Proof. Let 2 1 , . . . ,zrn E k: be a Jordan chain of T at A. Let yo = 0 E K’ and 1 ~ 1 , y1,. . . , y, be defined by (3.3). Then 7rOVk = xk for k = 1,. . . , m.

' I \ ~ I I I I I I ( ~ , Point Spectrum of Unitary Dilations 215

I)y induction it follows that

for j = 1 ,2 , . . . . Indeed this translates to

yk+l = 7 0 8 0 2 / k + l + VT-IATOW+I - vyk .

I h c e pi = yk for k = 1,. . . , m. 0

From Lemma 11.1 in (IKL] it is known that for a contraction T on a Pontryagin space K: the set o(T) \ b consists of finitely many eigenvalues only, with finite dimensional iilgebraic eigenspaces. Moreover, for A E o(T) with 1x1 > 1 the algebraic eigenspace R(T, {A}) E K is a nonpositive subspace (see [IKL], Theorem 11.2).

The following statement also follows from Proposition 2.1 in [AL].

Corollary 3.4. Let K be a Krein space. Let T E L(K) be a contraction. Let 1x1 > 1. Then R(T, { A } ) is a nonpositive subspace.

Proof. Assume q, . . . ,zm is a Jordan chain of T at A. Construct the Krein space K' and the minimal unitary dilation U of T as in (1.2) and (1.4). Since T is a contraction the space Xft- = K-1 = K-2 = s - . , needed in the construction of K', is a Hilbert space with scalar product ( . , . ). Use Lemma 3.2 to construct the Jordan chain ~1 , .. . , y,,, E K' of U at A. From Theorem 11.2.5 in [Bo] we know that yl , .. . y, are

Moreover, we have Xk = noyk. Hence neutral vectors. From (3.3) it follows that njpl = . . . = xjj2/, = 0 for all j = 1,2,. . . .

-m

[Zkr Zk]K = [Ykr?/k]K' - (njykinjgk) 5 0 . j = - 1

216 Math. Nachr. 194 (If IUR)

Hence x k is a nonpositive vector and the same holds for any vector z E R(T, { A ) ) , I I

Assume T E C(K) is a bicontraction on the KreTn space K. Let

TA = %(?-,{A))

for some A E up(%), 1x1 > 1. Consider the neutral part

N = {z E R(T, {A} ) I [.,.I = 0)

of R(T, {A} ) . For all z E N we have

0 = [ z , ~ ] 5 [ ( T X ) - ~ Z , (TA)-'z] 5 0 .

Hence (TA)-'N = N and TAN = N . It follows that the set JV is a T-invariiiiil subspace contained in Ker(I - T'T) = KerA, where A is the operator that appears ill the minimal unitary dilation U of T . This subspace N is the part of R(T, { A } ) 1.h i appears, after lifting to the big space K', unchanged in R ( U , {A) ) .

Corollary 3.5. Let K be a Krein space, and let T E C(K) be a bicontmction. /,v/ U on K' be its minimal unitary dilation. Let 1x1 > 1 and let

JV = {z E R(T, {A) ) I [z, z] = 0) .

Then the subspace roN E K' is U - invariant and

4. Spectrum inside the unit circle

The connections from the previous section between eigenvalues of T and of U (10 not hold for eigenvalues inside the unit circle. However, if T E L(K) and U on Ic' is i i

minimal unitary dilation of T , then U' is a minimal unitary dilation of 2''. Moreovca., if A with (XI < 1 is an eigenvalue of U , then 1 / X is an eigenvalue of U'. The resull,s of the previous section can be applied to the eigenvalue 1/X of the operators U' airit T'. It turns out that the Jordan chains of U at X can be explicitly constructed froiii the chains of T' at 1/X.

The following lemmas and the corollary describe the connection between the Joi,. dan chains corresponding to the eigenvalue X # 0 of a unitary operator U , and t , l i i ~

Jordan chains of U' at 1/X. The statements are well-known from spectral mappilit: theory, see e. g. (GGK], Theorem 1.3.3. Nevertheless, in order to have an explicit 1 I correspondence, and for completeness, we choose to give the proofs also.

Lemma 4.1. (See also (LT], Theorem 9.4.4.) Let U be a unitary operator on (1

Krein space. Assume yl,. . . , y,,, is a Jordan chain of U at A. For k = 1,. . . ,m, let

'jtwiiie, Point Spectrum of Unitary Dilations 217

I 1 11icri yi , . . . , &, i s a Jordan chain of U' at 1 / X .

Proof . Let yo = y; = 0. From

( u - X ) Y k = Yk-1

1 1 follows that (1 - X u * ) u k = U I Y k - 1

Ily induction it follows that

{or k = 1,. . . ,m. Then

- - Yl-1.

Lemma 4.2. For X # 0 let A,(X) be the m x m matrix with entries

if k > 1 ;

Then ( A m ( X ) ) - l = A m ( l / X ) f o r all X # 0.

Proof . For k = 1,. . . ,m let

218 Math. Nachr. 194 (19!)H)

Applying Lemma 4.1 once more on the chain y;, . . . , y k i t follows that the chiiili yr, . . . , y$ is another Jordan chain of U at A. The columns of the matrix Am(l/X) ;III*

the vectors y:,. . . , y: written with respect to the basis {yi, . . . , &}, and the coluirirlfi of Am(A) are the vectors yi , .. . ,yA written with respect to the basis {yl, . . . , ym}. 1 1 follows that the columns of Am(l/A)Am(A) are the vectors yr, . . . ,y$ written wil.11 respect to the basis ( ~ 1 , . . . ,grn}. The chains y l , . . . ,vrn and p r , . . . ,& both ;LIY~

Jordan chains of U at A. Hence Am(l/A)Am(A) is an upper triangular Toeplitz matrix Computing the entries on the first row gives

m

(Am (l/A)Am(A))ik = (Am(1IA))lj (Am(A))jk j=1

1 if k = 1 ; 0 if k = 2 , . . . , m .

Hence Am(l/X)Am(X) = I. From reversing the role of X and l / X it follows t h d . 0

From the fact that the columns of Am(l/X)Am(X) are the vectors y:,. . . ,g; writteii

Am(X)Am(l/X) = I as well.

with respect to the basis { y l , . . . , ym}, it follows that y i = yk for k = 1 , . . . ,m.

Corollary 4.3. The relations (4.1) and (4.2) establish a 1 - 1 relation between t h Jordan chains of U at A and the Jordan chains of U' at 1 / A .

For the Jordan chains of U' at 1/X we can apply the results of the previous section. In the rest of this section 1x1 < 1. Let K be a KreK space and let T E L(K) be il

bicontraction. The space K' and the unitary dilation U on K' have the form (1.2) a r i d

(1.4), respectively. Let V' : K' + K' be the operator represented by the matrix

. . . [O] 0

. . . 0 x 0 ... 0 x 2 x ...

0 A3 x2 * * .

. .

The nonzero entries in V' are multiples of the identity on K+. The operator V' is bounded since 1x1 < 1.

' I ~ I I Z I E , Point Spectrum of Unitary Dilations 219

Assume z1 , . . . , xm E K is a Jordan chain of T' at 1 / A . Let gh = 0 E K'. Define

( 4 3) = TOxk+l + v ' ~ l C ' ~ ~ + l - v 'g ;

l o i k = 0,1,. . . , m - 1. According to Lemma 3.2 the vectors ui , .. . , gk form a Jordan i*Imiri of U' at 1 / A . Next, let

(4 .4 )

for k = 1,. . . , m. Then, according to Lemma 4.1, the vectors 91,. . . , gm form a Jordan drain of U at A.

Conversely, assume that 91,. . . , y,,, is a Jordan chain of U at A. Let

(4 .5) k k - 1

j=1 3 -

for k = 1 , . . . , m. Then, according to Lemma 4.1, the vectors yi, . . . , y h form a Jordan drain of U' at 1 / A . Next, let

(4.6) x k = TOY; I

for k = 1 , . . . , m. Then, according to Lemma 3.1, the vectors x1 , .. . , xm form a Jordan chain of T' at 1 / A .

Theorem 4.4. Let K be a K r e h space and let T E L(K) be a bicontmction. Let U on K' be its minimal unitary dilation. Let 1A1 < 1. The above transformation from the chain X I , . . . , xm E K to the chain gl , .. . , vm E Ic' given by (4.3) and (4.4) and llie transformation from the chain yl , . . . , 9, E K' to the chain 2 1 , . . . , xm E Ic given hy (4.5) and (4.6) are each others inverses and they establish a 1 - 1 correspondence bctween the set of Jordan chains ojT' at 1 / A and the set of Jordan chains of U at A. 171 particular op(U) r l D = (op(T*) \ ID) . - -1

Proof . Combine Corollary 4.3 with Theorem 3.3. 0

If T E L(K) is a bicontraction on the Krein space K, then for an eigenvalue A with ( A ( < 1 it is known, see (AL], Proposition 2.1, that the algebraic eigenspace R(T, {A}) is a nonnegative subspace. It may happen that A is also an eigenvalue of U . Indeed, if 1 / A is an eigenvalue of T' , then Theorem 4.4 yields that A is an eigenvalue of U . In that case the question arises what R(T, { A } ) and R(U, {A}) will have in common. F'rom (,he following proposition it turns out that if R(T, {A}) is a strictly positive subspace, h e n R ( U , {A}) has nothing to do with R(T, {A}). If the subspace R(T, {A}) contains ;I nontrivial neutral subspace, however, then the lifting of this neutral part returns in the subspace R ( U , {A}).

Proposition 4.5. Let K be a Krein space and let T E L(K) be a bicontmction. Let U on K' be the minimal unitary dilation of T . Let M be some subspace of R(T, { A } ) with 1x1 < 1. Then TOM C R ( U , { A } ) if and only i f M is neutml.

220 Math. Nachr. 194 (I!,UN)

Proof . Recall that R(U,{A} ) isaneutralsubspaceofK'. Hence,ifToM c R ( U , { A ) )

Conversely, assume M C R(T, {A} ) is neutral. For all 2 E R(T, {A}) we have then TOM is neutral and necessarily M is a neutral subspace of K.

0 5 [Tz,Ts] 5 [z,2].

It follows that the set of neutral vectors of R(T, { A } ) forms a T-invariant subsp;iw Hence (T - A)JM is neutral for j = 0,1,2, . . . . Hence (T - A)JM C Ker(IK - T'7') KerA for j = 0,1,2, . . . . For all z E KerA we have (V - A)TOZ = TO(T - X)z. FoI i l i1y

z E M there exists a number m such that (T - A)mx = 0. Hence

(u - A)mT02 = (u - X)m-lTO(T - A)Z = . . . = 70(T - = 0,

and TOX E R ( U , {A}) . I1

5 . Spectrum on the unit circle

The previous sections show that, roughly speaking, eigenvalues outside the U I H I

circle and the corresponding algebraic eigenspaces return completely in the minintril unitary dilation, while the algebraic eigenspaces for eigenvalues inside the unit circ.lv may disappear. The present section shows that for eigenvalues on the unit circlc II

mixture of these two possibilities occurs. Let T E L(K) be a contraction on the Krein space K. Let U on K' be its miii-

imal unitary dilation. It was pointed out in [AL], Remark 2, that on the subspacct Ker(1- T'T) the contraction T coincides with U in the following sense

TOTX = VTOZ for all z E Ker(I - T 'T) ,

where TO : K + K' denotes the canonical embedding. In general, however, the subs pi^(^^^ Ker(I - T'T) is not T-invariant. Therefore we consider the maximal T-invariaiil subspace contained in Ker ( I - T'T).

Lemma 5.1. There is a unique maximal T -invariant subspace M contained aii

Ker(I - T'T) .

Proof . Assume M I , M2 are maximal T-invariant subspaces contained i t i

Ker(1 - T'T) . Let z E MI \ M2. Then Tkx E M1 for all k E IN u {0}, and y + T'z E Ker ( I - T'T) for all y E Mz. So M2 +span (2, Tx, . . .} is a T -invariaiit subspace contained in Ker(I - T'T). This contradicts the maximality of M t . Hencsc,

In [AL], Proposition 1.5, it is shown that for each Jordan chain 2 1 , . . . ,zn of 'I'

Mi = M 2 . r1

corresponding to an eigenvalue A on %, the subspace

span {xi ,.. . , Zm} ,

with m = [(n + 1)/2], is contained in Ker(I - T'T) (this statement also follows, foi finite dimensional K, by applying Cayley transformation on Theorem 2.1 in [RT]).

' I ~ W I I I C , Point Spectrum of Unitary Dilations 221

11, follows that span (21,. . . , x , } is contained in M . In particular, eigenvectors of T c-orrcsponding to eigenvalues on T are always in M .

'I'hc following theorem says that, after lifting it, the subspace M is the part of R (T , { A ) ) that forms the algebraic eigenspace of the minimal unitary dilation U of T I I I H far as eigenvalues on the unit circle 7' are concerned. Recall that in the minimal iiiiitary dilation U there appears the operator A : Ic -+ 3t- that satisfies ZK - T'T = ,4',4 and KerA* = (0). If T is a contraction, then 3t- is a Hilbert space.

Theorem 5 .2 . Let T E L(K) be a bicontmction on the K d n space X . Let U on K' be its minimal unitary dilation. Let M be the maximal T - invariant subspace rwntained in Ker(I - T'T). Let X E %. Zf y l , . . . ,pn E K' i s a Jordan chain of l I at A, then y k = 70x1; for k = 1 , . . . , n , where 21, . . . , X , E M C Ic is a Jordan chin of TIM at A . If X I , . . . , x,, E M C K is a Jordan chain of TIM at A then '10x1 , . . . , TOG, E K' i s a Jordan chain of U at A. In particular op(U) n% = ap(T) n%, a d R ( U , { A } ) = TO(M n R(T, { A } ) ) .

Proof . Without loss of generality we may assume that Ic' and U have the form (1.2) imd (1.4), respectively.

Assume A E op(U) with /A1 = 1. Assume 511,. . . ,y, E K' is a Jordan chain of U at A . Let go = 0 E K'. Clearly K j g o = 0 for all j E Z \ (0). We have (U - A ) V ~ + ~ = gk for k = 0,1,. . . , n - 1. Fix k E (0, 1,. . . ,n - 1). Assume njyk = 0 for all j E Z \ (0). Ilom the special form of U and the square summability of the coefficients of y k + l it

for all j E z \ (0). In other words p k = qnopk for k = 1,. . . ,n. Let xj = nopj for k = 1,. . . , n. Since y l , . . . ,yn # 0 E Ic' it follows that 21,. . . , x,, # 0 E X. Using the special form of U it is clear that (T-A)zk+l = x k and Axk+l = 0 for k = 0,1,. . . , n-1. 11, follows that A E op(T) and 21,. . . , x , E Ic is a Jordan chain of T at A, contained in I<er(Z - T'T) = Ker(A*A). Hence 2 1 , . . . , zn E M .

Conversely, assume X I , . . . , 2, E M C K is a Jordan chain of TIM at A. Since A'A = Z - T'T 2 0 it follows that 21,. . . ,z, E KerA. Let xo = 0 E Ic. We have (T - A ) z k + l = xk for k = 0,1, . , . , n - 1. From the special form of U it follows Ishat (V - X)702k+l = 70x1; for k = O , l , . ..,?I - 1. Clearly 70x0 = 0 E K' and 7021,. . . , 7 o X n # 0. 0

l'ollows that njyk+l = 0 for all j E Z \ (0). By induction njyl = = K j y n = 0

Assume X with (XI = 1 is an eigenvalue of the contraction T. Recall that the eigen- value X will be called a (non)degenerate eigenvalue of T if R(T, { A } ) is (non)degenerate. Note that, due to [L], Proposition 11.5.6, for a unitary operator U on a Pontryagin space the notion of a degenerate eigenvalue is equivalent to the notion of a singular critical point introduced there.

In Section 2 it was shown that for an eigenvalue A on the unit circle the algebraic eigenspace R(T, { A } ) of a contraction T on a finite dimensional Pontryagin space is nondegenerate. The subspace M, however, even in the finite dimensional case, in general will be degenerate. If M is degenerate, then, according to Theorem 5.2, the corresponding algebraic eigenspace R ( U , {A)) of the minimal unitary dilation U will be degenerate. In other words, A is a degenerate eigenvalue of U.

222 Math. Nachr. 194 (19!)H)

Theorem 5.3. Le t T E t(K) be a bicontractive operator o n the Krer'n space K. f ,r/ U o n K' be i ts minimal unitary di lat ion. Let M be the maximal T - i nva r ian t subspmw contained in Ker(1 - T'T). Assume A E a,(T) wi th 1x1 = 1. Consider the follouwy statements:

(i) A is a nondegenemte eigenvalue of U ; (ii) M n R(T, { A } ) is nondegenernte;

(iii) R(T, { A } ) C M; (iv) Tla(T,(x)) i s isometric;

(v) R ( U , { A } ) = TOR(T1 {A} ) . The statements (i) and (ii) are equivalent. equivalent. The statements (i) and (ii) imply the statements (iii), (iv) and (v).

The statements (iii), (iv) and (v) wv

Proof. (i) (ii): From Theorem 5.2 it follows that

(ii) * (iii): Let M I = R(T, {A}) n M and let M2 = ( M 1 ) l n R(T, (A}) . Any x E R(T, { A } ) can be written as x = rnl +m2 with rnl E M1 and m2 E M f . It followh that m2 = x-ml E R(T, { A } ) n M f = M z . In order to show that M2 is T-invariaiit, take an arbitrary m2 E M2. For all ml E M 1 we have ml E M E Ker(I - T'T) Hence

[Trnl,Tm2] = [rnl,m2] = 0 .

So Tmz I TM1 = M I and M2 is T-invariant. However, any eigenvector of TIM:, necessarily is in M I . Hence Mz = {O), and M I = R(T, {A}) .

(iii) (v): Follows from Theorem 5.2. (iii) (iv): Follows from M C Ker(1- T'T). 0

Corollary 5.4. If li is f in i te dimensional, then the statements (i) - (v) are equiva- lent.

Proof. If K is finite dimensional, then, according to Theorem 2.4, the subspaw c1 R(T, { A } ) is nondegenerate. Then the implication (v) + (i) holds trivially.

Corollary 5.5. Let T be a contractive operator o n a Pontryagin space II,, and l d U o n ll; be i ts min ima l un i ta ry di lat ion. Assume A E u(T) with 1x1 = 1. Then X is ti

singular cr i t ical point of U if and only if M n R(T, { A } ) is degenerate.

Proof. Combine Theorem 5.3 with [L], Proposition 11.5.6. C1

If A is a degenerate eigenvalue of the bicontraction T, then the number

dim R(T, {A})'

gives the degree of degeneracy of A. In taking the minimal unitary dilation, the degrw of degeneracy can only grow. Indeed,

' h n m e , Point Spectrum of Unitary Dilations 223

Corollary 5.6. If U is the minimal unitary dilation of the bicontmction T and X E C T ~ ( T ) n %, then

dimR(U, {A})' 2 dimR(T, {A})'.

Proof. Recall from Corollary 2.3 that R(T, {A})' is T-invariant. It follows that 0

If we deal with a degenerate eigenvalue A E T, then the subspaces R(U, {A}) and M n R(T, { A } ) have a nontrivial isotropic part. The subspaces that are skewly linked 1.0 these isotropic parts are also connected in a natural way.

R(T, {A})' E Ker(I - T'T) and ToR(T, {A})' C R(U, {A})'.

Proposition 5.7. Assume T E L(K) i s bicontmctive on the Krein space K . Let U on lit be its minimal unitary dilation. Let M be the maximal T - invariant subspace contained in Ker (I - T'T). If M n R(T, {A}) is degenerate and (M n R(T,{A}) ) '#M' for some M' C li then R ( U , { A } ) o # ~ O M ' . If R ( U , {A})' #N' for some N' c K t , then ( M n R(T, {A}))' #TON'.

Proof. According to Theorem 5.2 we have R ( U , {A}) = TO(M n R(T, {A})) . Hence R(U, {A})' = (T'(M n R(T, {A})))' = Q((M n R(T, {A}))o). This proves both Elatements. 0

It is interesting to consider the latter proposition in the case that K is finite dimen- sional. Then for any A E a(T) n T the algebraic eigenspace R(T, { A } ) is nondegenerate. If M n R(T, {A}) is degenerate, then, according to [Bo], Lemma 1.10.7, its isotropic part (/vl n R(T, {A}))' finds a skewly linked subspace M' within R(T, {A}). Lift- ing the subspaces to the big space li' we get the subspace TOMI being skewly linked to the isotropic part of R(U, {A}). However, this subspace TOM' has no relation to R(Ul {A}).

Acknowledgements

I would like to thank Prof. HEINZ LANCER for giving me the opportunity to visit him in Vienna and for our discussions on the topic of this paper during my stay there and during Iris visits to Amsterdam.

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[L]

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P. Nieuwlandstmat 77' 1099 XM Amaterrlam The Netherlands e -mail: d.temmeBiguip.nl