On the Decomposition of Families of Quasinormal Operators

Opuscula Math. 33, no. 3 (2013), 419–438 http://dx.doi.org/10.7494/OpMath.2013.33.3.419 Opuscula Mathematica

ON THE DECOMPOSITION OF FAMILIES OF QUASINORMAL OPERATORS

Zbigniew Burdak

Communicated by Aurelian Gheondea

Abstract. The canonical injective decomposition of a jointly quasinormal family of operators is given. Relations between the decomposition of a quasinormal operator T and the decomposition of a partial isometry in the polar decomposition of T are described. The decomposition of pairs of commuting quasinormal partial isometries and its applications to pairs of commuting quasinormal operators is shown. Examples are given.

Keywords: multiple canonical decomposition, quasinormal operators, partial isometry.

Mathematics Subject Classiﬁcation: 47B20, 47A45.

1. INTRODUCTION

Let L(H) be the algebra of all bounded linear operators on a complex Hilbert space H. If T L(H), then T ∗ stands for the adjoint of T. By (T ) and (T ) we denote the ∈ N R kernel and the range space of T respectively. By a subspace we always understand a closed subspace. The orthogonal complement of a subspace H H is denoted by H⊥ or H H . The commutant of T L(H) 0 ⊂ 0 0 ∈ denoted by T 0 is the algebra of all operators commuting with T. A subspace H H 0 ⊂ is T hyperinvariant, when it is invariant for every S T 0. An orthogonal projection ∈ onto H is denoted by P . A subspace H reduces operator T L(H) (or is reduc- 0 H0 0 ∈ ing for T ) if and only if P T 0. An operator is called completely non unitary (non H0 ∈ normal, non isometric etc.) if there is no non trivial subspace reducing it to a unitary operator (normal operator, isometry). Such an operator is also called pure. Let denote some property of an operator T L(H) (like being unitary, normal, W ∈ isometry etc.) If there is a decomposition H = H H such that H ,H reduce 1 ⊕ 2 1 2 operator T and T has the property while T is pure, then |H1 W |H2 T = T T |H1 ⊕ |H2 c AGH University of Science and Technology Press, Krakow 2013 419

420 Zbigniew Burdak is called a canonical decomposition of T. The classical example of a canonical W decomposition is the Wold decomposition [11]. Wold showed that any isometry can be decomposed into a unitary and a completely non unitary operators. An isometry n S L(H) is called a unilateral shift if H = S ( (S∗)). A completely non ∈ n 0 N unitary isometry is a unilateral shift. ≥ L An operator T L(H) is called quasinormal, if it is normal on (T ) (i.e. ∈ R TT ∗T = T ∗TT ). The class of quasinormal operators has been introduced by Brown in [2]. Quasinormal operators are subnormal. Every quasinormal operator T can be decomposed to N S A, where N is normal, S is a unilateral shift and A is a positive, ⊕ ⊗ injective operator. For a quasinormal operator T holds (T ) (T ∗). Thus (T ) N ⊂ N N reduces a quasinormal operator. Recall that T = √T T. An operator W is a partial | | ∗ isometry, if the restriction W (W ) is an isometry. Every T L(H) has the polar |N ⊥ ∈ decomposition T = W T , where W is a partial isometry. Operator T is quasinormal | | if and only if W T = T W. Recall after [7] that a family of operators L(H) is | | | | A ⊂ called jointly quasinormal, if ST ∗T = T ∗TS holds for all S, T . ∈ A Assume that, for some property , there are canonical decompositions of W W operators T ,T L(H). If there is a decomposition H = H H H H , 1 2 ∈ 11 ⊕ 12 ⊕ 21 ⊕ 22 where each H reduces operators T ,T for i, j = 1, 2, such that: T has the ij 1 2 1|Hij property for i = 1, is pure for i = 2 and T has the property for j = 1 and W 2|Hij W is pure for j = 2, then T = T T T T for k = 1, 2 k k|H11 ⊕ k|H12 ⊕ k|H21 ⊕ k|H22 is called a multiple canonical decomposition. Multiple canonical decompositions W can be deﬁned in a similar way for families of operators. Recall that operators T ,T L(H) doubly commute if T ,T ∗ T 0. Recall after [4] that a family of doubly 1 2 ∈ 1 1 ∈ 2 commuting operators has a multiple canonical decomposition if each operator in the family has a (single) canonical decomposition. However, the doubly commutativity assumption is rather strong. Only a normal operator can doubly commute with itself. On the other hand, a pair T,T has a multiple canonical decomposition if T has a canonical decomposition. In the present paper we show that, although there does not need to exist a normal canonical decomposition of a jointly quasinormal family of operators, there is an injective canonical decomposition. We also give a generalization of this decomposition to a pair of commuting quasinormal partial isometries. The generalization is not a canonical decomposition. In the last paragraph we give some applications of this decomposition to arbitrary pairs of commuting quasinormal operators.

2. DECOMPOSITIONS OF A QUASINORMAL OPERATOR

By the model given by Brown a quasinormal operator has a normal canonical decomposition. Let T L(H) be a quasinormal operator. Recall after [10] that if T is ∈ additionally a contraction, then the subspace (I T ∗T ) is the maximal subspace N − reducing T to an isometry. Note that a subspace reduces operator T if and only if it reduces αT, for α C 0 . If α = 1, then (I T ∗T ) = (I (αT )∗(αT )) . Since αT ∈ \{ } | | 6 N − 6 N − On the decomposition of families of quasinormal operators 421

is not a contraction for suﬃciently large α , it is not known whether (I (αT )∗(αT )) | | N − reduces T. In this section the existence of the maximal subspace reducing a bounded quasinormal operator to an isometry will be proved. As a consequence, any eigenspace of T reduces operator T to an isometry weighted by the corresponding eigenvalue. | | Moreover, such a subspace is T hyperinvariant. | | Remark 2.1. Let A L(H) be a positive operator. For x (I A) we have ∈ ∈ N − x = Ax (A). Consequently, the subspace (I A) is orthogonal to (A). An ∈ R N − N arbitrary x (A A2) can be decomposed to x = (x Ax)+Ax (A) (I A). ∈ N − − ∈ N ⊕N − It follows that (A A2) (I A) (A). Since the reverse inclusion is obvious, N − ⊂ N − ⊕ N we have (A A2) = (I A) (A). N − N − ⊕ N For arbitrary x (I A2), we have ∈ N − 0 (A(x Ax), x Ax) = (Ax x, x Ax) = x Ax 2 0. ≤ − − − − −k − k ≤ Consequently, (I A2) = (I A). N − N − A decomposition of an operator T is called T hyperinvariant, when subspaces in the corresponding decomposition of the Hilbert space H = i Hi are T hyperinvariant. It follows that Hi and H Hi are T hyperinvariant and consequently subspaces Hi L reduce every operator in T 0. Proposition 2.2. Let T L(H) and T = √T T . Then the decomposition ∈ | | ∗ H = ( T T T ) (I T ) (T ) R | | − ∗ ⊕ N − | | ⊕ N is T hyperinvariant. | | Proof. Since the operator T T ∗T is self-adjoint, then | | −

H = ( T T T ) ( T T ∗T ). R | | − ∗ ⊕ N | | − By Remark 2.1, we obtain the decomposition

( T T ∗T ) = ( T ) (I T ). N | | − N | | ⊕ N − | | Obviously commutants of T and I T are equal. Since the kernel of an operator is a | | −| | hyperinvariant subspace, then (I T ) and ( T ) = (T ) are T hyperinvariant. N − | | N | | N | | The orthogonal complement of a subspace, which is hyperinvariant for a self-adjoint operator is also a hyperinvariant subspace. Thus ( T T T ) is T hyperinvariant R | | − ∗ | | as well. As a corollary we obtain the following decomposition. Proposition 2.3. Let T L(H) be a quasinormal operator. There is a decomposition ∈ 2 2 H = (T T T T ) (I T ∗T ) (T ), R ∗ − ∗ ⊕ N − ⊕ N where the subspaces are the maximal reducing operator T, such that:

(i) T (T T T 2T 2) is a completely non isometric, injective, quasinormal operator, |R ∗ − ∗ 422 Zbigniew Burdak

(ii) T (I T T ) is an isometry, |N − ∗ (iii) T (T ) = 0. |N Moreover, each of these subspaces is T hyperinvariant. | | Proof. Applying Proposition 2.2 to the operator T ∗T we obtain the decomposition

2 H = (T T (T T ) ) (I T ∗T ) (T ∗T ). R ∗ − ∗ ⊕ N − ⊕ N

Since T ∗T = T ∗T , the decomposition is T ∗T hyperinvariant. Commutants of T ∗T | | and its square root T are equal. Thus the decomposition is also T hyperinvariant. | | 2 2 2 | | Since (T ) = (T ∗T ) and by quasinormality (T ∗T ) = T ∗ T the decomposition is N N equivalent to 2 2 H = (T T T T ) (I T ∗T ) (T ). R ∗ − ∗ ⊕ N − ⊕ N Since T is quasinormal, it commutes with T . Consequently, subspaces obtained in | | the decomposition reduce T. Since for an isometry T ∗T = I, then a subspace reducing T, reduces the operator to an isometry if and only if it is a subspace of (I T ∗T ). On the other N − hand we have shown that (I T ∗T ) reduces T. Thus it is the maximal subspace N − reducing T to an isometry. Since (T ) reduces T, it is the maximal subspace re- N ducing T to a zero operator. Since (T T T 2T 2) is the orthogonal complement R ∗ − ∗ of (T ) (I T ∗T ), it is the maximal subspace reducing T to a completely non N ⊕ N − isometric, injective, quasinormal operator. By the proposition above, we obtain the following decomposition. Theorem 2.4. Let T L(H) be a quasinormal operator, where H is a separable ∈ Hilbert space. There is a decomposition

2 H = (λ T T T ) (λ I T ∗T ), R | | − ∗ ⊕ N − λ Λ λ Λ \∈ M∈ where Λ is the set of all eigenvalues of T . The subspaces are the maximal reducing | | operator T such that: (i) T has no eigenvectors, ( T λT T ) | | λ Λ R | |− ∗ 1 ∈ (ii) λ− T (λ2I T T ) is an isometry for λ = 0, |N − ∗ 6 (iii) T (T ) = 0. |N Moreover, each of the subspaces is T hyperinvariant. | | Since H is separable, there is only countably many eigenvalues of T . Hence, in | | the decomposition above, there is countably many subspaces and the orthogonal sum can be used. Proof. Note that if (T ) = 0 , then λ = 0 is an eigenvalue of T and (T ) = ( T ) N 6 { } | | N N | | is a summand of the decomposition. Since T is quasinormal, then (T ) reduces T. N Obviously it is the maximal subspace reducing T to a zero operator. On the decomposition of families of quasinormal operators 423

1 For any α C 0 denote by λ = α . Proposition 2.3 applied to the operator ∈ \{ } 2 2| | αT shows that (I α T ∗T ) = (λ I T ∗T ) is a T hyperinvariant subspace N − | | N − | | 1 and reduces the operator αT to an isometry. Consequently, λ− T (λ2I T T ) = |N − ∗ 2 α T (λ I T ∗T ) is an isometry. Assume that for some subspace L H reducing | | |N − 1 1 1 ⊂ T the operator λ− T L is an isometry. Then (λ− T )∗(λ− T ) L = I L implies that 2 | 2 2 | | L (I λ− T ∗T ) = (λ I T ∗T ). Consequently, (λ I T ∗T ) is the maximal ⊂ N − N − 1 N − subspace reducing T such that λ− T is an isometry. 2 2 If (λ I T ∗T ) = 0, then λ is an eigenvalue of T . Subspaces (λ I T ∗T ) are N − 6 | | N − orthogonal, since they are diﬀerent eigenspaces of T . | | By Proposition 2.2 applied to the operator αT , we obtain

( α T α 2T T ) = H ( (I α T ) (αT )). (2.1) R | || | − | | ∗ N − | || | ⊕ N 2 2 2 2 2 Note that α T ∗T = α T ∗T and since T is quasinormal, (T ∗T ) = T ∗ T . Thus | | | | 2 by Proposition 2.2 applied to the operator α T ∗T , we obtain | |

2 4 2 2 2 2 ( α T T α T T ) = H ( (I α T ∗T ) ( α T ∗T )). (2.2) R | | ∗ − | | ∗ N − | | ⊕ N | | 2 Since α = 0, then ( α T ∗T ) = (αT ) = (T ). By Remark 2.1 applied to A = 6 N | | N 2 N 2 2 α T , we obtain (I α T ) = (I α T ) = (I α T ∗T ). Consequently, | || | N − | || | N − | | | | N − | | the right hand sides of equalities (2.1) and (2.2) are equal. It follows that

( α 2T T α 4T 2T 2) = ( α T α 2T T ) = (λ T T T ). R | | ∗ − | | ∗ R | || | − | | ∗ R | | − ∗ By the equality above and Proposition 2.3 applied to the operator αT , we have that

2 H (λ I T ∗T ) = (λ T T T ) (T ). N − R | | − ∗ ⊕ N Note that either (T ) is 0 or is an eigenspace of T . Thus N { } | | 2 H (λ I T ∗T ) = (λ T T T ). N − R | | − ∗ λ Λ λ Λ M∈ \∈ It is a T hyperinvariant subspace, since it is an intersection of such subspaces. By | | the construction above, the orthogonal complement of λ Λ (λ T T ∗T ) is a sub- ∈ R | | − space generated by all eigenvectors of T . Thus (λ T T T ) is the maximal | | λ Λ RT | | − ∗ subspace reducing T such that T has no eigenvalues.∈ | | T

3. DECOMPOSITIONS OF SOME FAMILIES OF QUASINORMAL OPERATORS

In [4] it has been proved that a family of doubly commuting quasinormal operators has a multiple normal canonical decomposition. The results do not extend to a jointly quasinormal family – Example 1 in [9]. We can obtain the following decomposition. 424 Zbigniew Burdak

Theorem 3.1. Let Ti i Z L(H) be a family of jointly quasinormal operators on { } ∈ ⊂ a separable Hilbert space H, where Z Z is ﬁnite or inﬁnite. Denote by Λ the set of ⊂ all sequences αi i Z such that αi is an eigenvalue of Ti or αi = for i Z. There { } ∈ | | ∞ ∈ is a decomposition H = Hα α Λ M∈ into subspaces reducing the family Ti i Z , where for every α Λ and i Z { } ∈ ∈ ∈

(i) Ti Hα = 0 for αi = 0, |1 (ii) α− T is an isometry for α (0, ), i i|Hα i ∈ ∞ (iii) T is such that T has no eigenvectors for α = . i|Hα | i| i ∞ i i H = H H i T Proof. Denote by j Ji λ the decomposition of the operator i given ∞ ⊕ ∈ j i by Theorem 2.4, where λj j Ji are all eigenvalues of Ti for i Z. Precisely, { }L∈ | | ∈ i i i 2 i i H = (T ),H i = ((λ ) I T T ) H = (λ T T T ). 0 i λ j i∗ i and j Ji j i i∗ i Note N j N − ∞ ∈ R | | − i that Λ is the cartesian product of λj j Ji for i Z. Since decomposi- { } ∈ ∪ {∞} T ∈ tions are T hyperinvariant for every i Z and the family is jointly quasinormal, | i| ∈ each subspace in each decomposition reduces the whole family Ti i Z . Thus sub- i { } ∈ spaces Hα = i Z Hαi reduce the family, since they are intersections of such sub- ∈ spaces. By the construction above, the subspaces Hα have suitable properties and T H = α Λ Hα. ∈ BesideL a normal canonical decomposition of a quasinormal operator there is also a canonical decomposition into an injective operator and a zero operator. Obviously a zero operator is normal. Therefore an injective decomposition can be understand as a partial result compared to the normal decomposition. A jointly quasinormal family need not have a multiple normal canonical decomposition. However, as a corollary of Theorem 3.1, a jointly quasinormal family of operators has a multiple injective canonical decomposition.

Theorem 3.2. Let Ti i Z L(H) be a family of jointly quasinormal operators on a { } ∈ ⊂ separable Hilbert space H, where Z Z is ﬁnite or inﬁnite. There is a decomposition ⊂

H = Hα α 0,1 Z ∈{M} into subspaces reducing the family Ti i Z , where for every α Λ and i Z { } ∈ ∈ ∈ (i) T = 0 for α = 0, i|Hα i (ii) T is injective for α = 1. i|Hα i There is a natural question of a decomposition with weaker than a joint quasinormality assumption. By the following Remarks 3.3 and 3.5, we can describe an interesting subclass of quasinormal operators. Remark 3.3. Let H reduces a quasinormal operator T. Let T = W T be the polar 0 | | decomposition, where W is a partial isometry. Obviously H and H H reduce T ∗T 0 0 On the decomposition of families of quasinormal operators 425

which is equivalent to the commutativity of PH and PH H with T ∗T. Consequently 0 0 PH ,PH H commute with T and subspaces H0,H H0 are T invariant. Note that 0 0 | | | | ( T ) = (T ) = (W ). Thus for any x H we have W ∗x ( T ). It follows that N | | N N ∈ ⊥ N | | T ∗x = T W ∗x = 0 if and only if W ∗x = 0. Thus (W ∗) = (T ∗) (T ). It follows | | N N ⊃ N that W x, W ∗x are orthogonal to (T ) = ( T ). Since ( T ) is a hyperinvariant N N | | N | | subspace of T , it reduces PH H . Thus for any x H0 we have | | 0 ∈

0 = PH H T x = PH H T W x = T PH H W x 0 0 | | | | 0 and consequently

PH H0 W x = P ( T )PH H0 W x = PH H0 P ( T )W x = 0. N | | N | |

Similarly, PH H W ∗x = 0. Since x has been taken arbitrary, it follows that H0 0 reduces W. By Remark 3.3 subspaces in any decomposition of a quasinormal operator reduce also a partial isometry in the polar decompositions of the operator. In this sense a decomposition of a quasinormal operator generates the corresponding decomposition of a partial isometry. It can be shown that subspaces in the Wold decomposition of an isometric part of a partial isometry reduce a whole quasinormal operator. In general, a subspace reducing a partial isometry in the polar decomposition of a quasinormal operator does not need to reduce the quasinormal operator.

Example 3.4. Let H = n 0 Cen Cfn, where en, fn are orthonormal vectors for ≥ ⊕ n 0. Deﬁne T en = 2en+1, T fn = fn+1. Then W en = en+1, W fn = fn+1. A closed ≥ L subspace generated by e + f for n 0 reduces W but does not reduce T. n n ≥ Partial isometries which are obtained in the polar decomposition of quasinormal operators are also interesting because they are quasinormal. Remark 3.5. Let T L(H) be a quasinormal operator and T = W T denotes the ∈ | | polar decomposition. Then W is quasinormal.

Proof. For a partial isometry, we have WW ∗W = W. Thus we need to show that 2 W ∗W = W. It is trivial for vectors from (W ) = ( T ) = (T ). Take arbitrary x N N | | N such that x = T y for some y H. Since (T ) (T ∗) for a quasinormal operator, | | ∈ N ⊂ N then (T ) H (T ) and W (T ) is an isometry. Consequently, W ∗WT = T and R ⊂ N |R 2 2 W ∗W x = W ∗W T y = W ∗W T y = T y = W T y = W x. | | | | 2 Thus we have W ∗W = W on a dense set in H (W ). N Quasinormal partial isometries forms an interesting subclass of quasinormal operators. It is easy to verify that the injective decomposition of a quasinormal operator corresponds to the decomposition of a partial isometry in the polar decomposition to an isometry and a zero operator. We restrict our consideration to pairs of commuting quasinormal partial isometries. Note some properties of such pairs. 426 Zbigniew Burdak

Lemma 3.6. For T ,T L(H) commuting quasinormal partial isometries hold: 1 2 ∈ (i) (T T ) = (T ) T ∗T ( (T T )) for i = 1, 2, N 1 2 N i ⊕ i i N 1 2 (ii) T ∗T ( (T T )) (T ∗),T ∗T ( (T T )) (T ∗), 1 1 N 1 2 ⊂ N 2 2 2 N 1 2 ⊂ N 1 (iii) (T T ) (T ∗T ∗). N 1 2 ⊂ N 1 2

Proof. Since for partial isometries Ti∗Ti = P (Ti) , then by inclusions (Ti) N ⊥ N ⊂ (T T ) for i = 1, 2 follows the decomposition in (i). N 1 2 We will show the ﬁrst inclusion in (ii). By inclusions T ( (T T )) (T ) 1 N 1 2 ⊂ N 2 ⊂ (T ∗) (T ∗T ∗), it follows that T ∗T ( (T T )) (T ∗). N 2 ⊂ N 1 2 1 1 N 1 2 ⊂ N 2 To show (iii) note that by a combination of properties of a quasinormal operator 2 and a partial isometry we have that Ti∗Ti = TiTi∗Ti = Ti for i = 1, 2. For adjoints we 2 2 2 obtain that Ti∗ = Ti∗ Ti. Therefore T1∗T2∗ = T1∗(T2∗) T2 = (T2∗) T1∗T2. It follows that 2 (T T ) (T ∗T ) ((T ∗) T ∗T ) = (T ∗T ∗). N 1 2 ⊂ N 1 2 ⊂ N 2 1 2 N 1 2 Denote by

H = L H : (T T ) L and L reduces T ,T (3.1) ni { ⊂ N 1 2 ⊂ 1 2} \ and H = H H . (3.2) Is ni Since (T T ) = (T ) T ∗T ( (T T )) (T ) + T ∗( (T )) and (T ) N 1 2 N 1 ⊕ 2 2 N 1 2 ⊂ N 1 2 N 1 N i ⊂ (T T ) for i = 1, 2, then H is the minimal subspace reducing both operators N 1 2 ni and containing (T ) and (T ). Thus H is the maximal subspace reducing the N 1 N 2 Is pair T ,T to isometries. Restrictions T ,T form a completely non isometric 1 2 1|Hni 2|Hni pair. The commuting isometries are being studied by many authors (see [1, 3, 8]). In the following part we consider a pair of commuting quasinormal partial isometries and describe how one of them acts between the kernel and the isometric part of the other one. Obviously H = (T ) (T ) reduces both operators. Consider H H . 00 N 1 ∩ N 2 ni 00 Recall after [6] that a pair of isometries V1,V2 is called compatible if P (V n) commutes R 1 with P (V m) for every n, m Z+. It can be shown that isometries are compatible if R 2 ∈ and only if

n n m n m (V ∗ ) = ( (V ∗ ) (V ∗ )) ( (V ∗ ) (V )) N 1 N 1 ∩ N 2 ⊕ N 1 ∩ R 2 for every n, m Z+. We will follow the idea of compatibility but decompose kernels ∈ of quasinormal operators instead of kernels of theirs adjoints.

Definition 3.7. We call a pair of commuting quasinormal operators T ,T L(H) 1 2 ∈ q-compatible, if (T ) ( (T ) (T )) is orthogonal to (T ) ( (T ) (T )). N 1 N 1 ∩ N 2 N 2 N 1 ∩ N 2 n Since (T ) reduces a quasinormal operator T , we have (T ) = (T ) for n Z+. N N N ∈ Therefore in the definition of q-compatibility we do not concern powers of the operator. To avoid misunderstanding in case of isometries which are quasinormal we can not call the defined property simply compatibility. On the decomposition of families of quasinormal operators 427

Remark 3.8. If there is the maximal subspace L H reducing the pair T ,T and ⊂ 1 2 such that ( (T ) L) ( (T ) L), N 1 ∩ ⊥ N 2 ∩ then each subspace L0 reducing operators T1,T2 to isometries or one to an isometry and the other one to a zero operator is a subspace of L. Indeed, in such case (T ) (T ) because at least one of the kernels is 0 . Since L reduces N 1|L0 ⊥ N 2|L0 { } 0 both operators (T ) = (T ) L for i = 1, 2. Then by the maximality of L N i|L0 N i ∩ 0 follows that L L. 0 ⊂ Our aim is to ﬁnd the maximal subspace L described in Remark 3.8. Decompose

(T ) (H H ) = G F , N i ∩ ni 00 i ⊕ i where

G = (T ) T ∗T ( (T T )),G = (T ) T ∗T ( (T T )) (3.3) 1 N 1 ∩ 2 2 N 1 2 2 N 2 ∩ 1 1 N 1 2 and F = (T ) (G H ) for i = 1, 2. (3.4) i N i i ⊕ 00

Since Ti is a patrial isometry, then Ti∗Ti = P (Ti) for i = 1, 2. Thus (Ti∗Ti) N ⊥ R and T ∗T ( (T T )) = (T T ) (T ) are closed subspaces. Note that for x i i N 1 2 N 1 2 N i ∈ (T T ) (T ∗T ) we have x = T ∗T x T ∗T ( (T T )). It follows the inclusion N 1 2 ∩ R i i 1 1 ∈ 1 1 N 1 2 (T T ) (T ∗T ) T ∗T ( (T T )) for i = 1, 2. The reverse inclusion holds by N 1 2 ∩ R i i ⊂ i i N 1 2 Lemma 3.6(i). Consequently, G = (T ) (T ∗T ) and G = (T ) (T ∗T ). 1 N 1 ∩ R 2 2 2 N 2 ∩ R 1 1 This means that subspaces G1,G2 consist of all vectors which are in the kernel of one operator and are orthogonal to the kernel of the other one. The result in the following lemma is known. We give it without proof. Lemma 3.9. Consider closed subspaces A, B of a Hilbert space K. Then

(A B)⊥ = A + B . ∩ ⊥ ⊥

We use the lemma above to describe some properties of the subspaces F1,F2. Proposition 3.10. Let T ,T L(H) be commuting quasinormal partial isometries 1 2 ∈ and F ,F be subspaces deﬁned in (3.4). Then F ,F (T ∗) (T ∗). 1 2 1 2 ⊂ N 1 ∩ N 2 Proof. Let F = (T T ) (G G ), where G ,G are subspaces deﬁned in (3.3). 12 N 1 2 2 ⊕ 1 1 2 Denote for convenience: K = (T T ),A = T ∗T (K),B = (T ) for i = 1, 2. Note N 1 2 i i i i N i that G = B A ,G = B A and consequently 1 1 ∩ 2 2 2 ∩ 1 F = K ((B A ) (B A )) = (K (B A )) (K (B A )). 12 2 ∩ 1 ⊕ 1 ∩ 2 2 ∩ 1 ∩ 1 ∩ 2 If we consider K as a whole space, then by Lemma 3.9 we obtain

F = (K B ) + (K A ) (K B ) + (K A ). 12 2 1 ∩ 1 2 428 Zbigniew Burdak

On the other hand, Lemma 3.6(i),(ii) shows that K = B A for i = 1, 2 and i ⊕ i A ,B (T ∗),A ,B (T ∗). Thus 1 2 ⊂ N 2 2 1 ⊂ N 1

F = (A + B ) (A + B ) (T ∗) (T ∗). 12 2 1 ∩ 1 2 ⊂ N 1 ∩ N 2 By the deﬁnition of F , it is a subspace of (T ) (T T ) and it is orthogonal 1 N 1 ⊂ N 1 2 to G . Since G is orthogonal to (T ), then it is also orthogonal to F (T ). 1 2 N 1 1 ⊂ N 1 Consequently, F F . Similarly F F . It follows that 1 ⊂ 12 2 ⊂ 12 F (T ∗) (T ∗) for i = 1, 2. i ⊂ N 1 ∩ N 2

We will now construct the maximal T1,T2 reducing subspace generated by F1,F2. Theorem 3.11. Let T ,T L(H) be a pair of commuting quasinormal partial isome- 1 2 ∈ tries and F1,F2 be subspaces given in (3.4). The subspace

H = T n(F ) T n(F ) T T (F ) + T T (F ) . F 1 2 ⊕ 2 1 ⊕ 1∗ 1 2 2∗ 2 1 n 1 n 1 X≥ X≥ is the minimal subspace reducing T1,T2 and containing F1,F2.

Proof. At the beginning we will show that summands in the deﬁnition of HF are indeed orthogonal. By quasinormality (Ti) (Ti∗). Note that (Ti∗) = H (Ti) N ⊂ N n R N ⊃ H (T ∗) = (T ) for i = 1, 2. Thus T (F ) and T ∗T (F ) are included in N i R i n 1 2 1 2 2 1 n ≥ n (T2∗). On the other hand, n 1 T1 (F2) (T2). Thus n 1 T1 (F2) is orthogo- R ≥ P⊂ N ≥ T n(F ) T T (F ) nal to n 1 2 1 and 2∗P2 1 . Recall that for a quasinormalP partial isometry 2 ≥ T ∗ = T ∗ T (see proof of Lemma 3.6(iii).) For any x, y F2 and n 1 by Propo- P n 1 n 1 ∈ ≥ n sition 3.10 we have that 0 = (T1 − x, T1∗y) = (T1 − x, T1∗T1∗T1y) = (T1 x, T1∗T1y). n Thus n 1 T1 (F2) is orthogonal to T1∗T1(F2). We have shown that the subspace n ≥ n 1 T1 (F2) is orthogonal to the remaining summands. Similarly, the subspace ≥ P T n(F ) T T (F ) T T (F ). Pn 1 2 1 is orthogonal to 1∗ 1 2 and 2∗ 2 1 This ﬁnishes the proof of the≥ orthogonality. P We make a construction of the minimal T1,T2 reducing subspace containing n subspaces F1,F2. Note that n 0 T2 (F1) is T2 invariant and since it is a sub- ≥ space of (T1) (T1∗), it is T1 reducing. Since T2 is a partial isometry then n N n⊂1 N P T ∗T (F ) = T − (F ) for n 2. In case n = 0 Proposition 3.10 shows that 2 2 1 2 1 ≥ T2∗(F1) T2∗( (T1∗) (T2∗)) = 0 . However, T2∗T2(F1) may not be included in ⊂n N ∩ N { } n n 0 T2 (F1). We obtain a similar result for the subspace n 0 T1 (F2). It follows ≥ ≥ that the minimal subspace reducing T1,T2 and containing F1,F2 is not smaller than P P n n HF0 := T1 (F2) + T2 (F1) + T1∗T1(F2) + T2∗T2(F1). n 0 n 0 X≥ X≥

We will show that HF0 reduces T1,T2. Note that HF HF0 and by the pre- n ⊂n vious argumentation, the images of n 0 T1 (F2), n 0 T2 (F1) under operators ≥ ≥ T1,T2,T ∗,T ∗ are subspaces of HF . We need to show a similar result for T ∗T1(F2) 1 2 P P 1 On the decomposition of families of quasinormal operators 429

and T2∗T2(F1). Let us prove it for T1∗T1(F2). Note that T2(T2∗T2(F1)) = T2(F1) HF . 2 ⊂ By T2∗ T2 = T2∗ and Proposition 3.10, we have

T ∗T ∗T (F ) = T ∗(F ) T ∗( (T ∗) (T ∗)) = 0 . 2 2 2 1 2 1 ⊂ 2 N 1 ∩ N 2 { } By the deﬁnition of F it follows that T (F ) (T ) (T ∗) and consequently 1 2 1 ⊂ N 1 ⊂ N 1 T ∗(T ∗T (F )) = 0 . To show that T T ∗T (F ) H , take arbitrary x F and 1 2 2 1 { } 1 2 2 1 ⊂ F ∈ 1 decompose it to x = x T ∗T x (T ) (T ). Since T x = 0, then T T ∗T x = 0 ⊕ 2 2 ∈ N 2 ⊕ R 2∗ 1 1 2 2 T ( x ). For any z G we have z = T ∗T z. Note that that 1 − 0 ∈ 2 1 1 (x , z) = (x ,T ∗T z) = (T x ,T z) = ( T T ∗T x, T z) = 0 0 1 1 1 0 1 − 1 2 2 1 = (T ∗T x, T ∗T z) = (T ∗T x, z) = (T x, T z) = (T x, 0) = 0. − 2 2 1 1 − 2 2 − 2 2 − 2 Since vector z has been chosen arbitrary, x0 is orthogonal to the subspace G2. Thus x0 being in (T ) belongs to F . The equality T T ∗T x = T ( x ) shows the inclusion N 2 2 1 2 2 1 − 0 T T ∗T (F ) T (F ) H . Thus H reduces T ,T . 1 2 2 1 ⊂ 1 2 ⊂ F F0 1 2 Note that we have shown that images of HF0 under operators T1,T2,T1∗,T2∗ are not only subspaces of HF0 but also subspaces of HF . Therefore, since HF is a subspace of HF0 it also reduces T1,T2. By the minimality of HF0 we have that HF = HF0 . By Theorem 3.11, we can ﬁnd the maximal subspace L described in Remark 3.8. Corollary 3.12. A subspace H := H (H H ) ort ni 00 ⊕ F is the maximal T ,T reducing subspace orthogonal to H , where (T ) 1 2 Is N 1|Hort ⊥ (T ). N 2|Hort Proof. It is enough to show that any subspace L reducing T1,T2 and such that (T ) (T ) is orthogonal to H and H . Note that by assumed orthog- N 1|L ⊥ N 2|L 00 F onality of kernels it follows that P (T )P (T )L = 0 and P (T )P (T )L = 0 . On N 1 N 2 { } N 2 N 1 { } the other hand, P (Ti) = I Ti∗Ti. Thus an orthogonal projection on any Ti reducing N − subspace commutes with P (T ). N i The orthogonality of L to H00 follows by

0 = PH00 P (T )P (T )L = P (T )P (T )PH00 L = PH00 L. { } N 1 N 2 N 1 N 2

Similarly, 0 = P (T )P (T )PHF L and 0 = P (T )P (T )PHF L. By the def- { } N 1 N 2 { } N 2 N 1 inition of F1, for any x F1 holds P (T )P (T )x = P (T )x = 0. Similarly for F2. ∈ N 2 N 1 N 2 6 Thus PHF L is a proper subspace of HF , reduces T1,T2 and does not contain any vector from F nor F . By the minimality of H , it follows that P L = 0 . 1 2 F HF { } As an easy consequence we can ﬁnd subspaces reducing quasinormal partial isometries to pairs: an isometry – a zero operator and a zero operator – an isometry. Theorem 3.13. Let T ,T L(H) be a pair of commuting quasinormal partial isome- 1 2 ∈ tries such that (T ) (T ). The subspaces N 1 ⊥ N 2 n n H := (T T ∗ ),H := (T T ∗ ) 0,Is N 1 2 Is,0 N 2 1 n 0 n 0 \≥ \≥ 430 Zbigniew Burdak are maximal reducing T ,T such that T = 0,T = 0 and T ,T 1 2 1|H0,Is 2|HIs,0 2|H0,Is 1|HIs,0 are isometries.

Proof. By the inclusion (T ) (T ∗) valid for quasinormal operators, every sub- N i ⊂ N i space of (T ) reduces T . Since H (T ) (T ), then H = T ∗T (H ). N 1 1 0,Is ⊂ N 1 ⊥ N 2 0,Is 2 2 0,Is For any n 1 holds ≥ n 1 n 1 n 0 = T T ∗ − (H ) = T T ∗ − (T ∗T (H )) = T T ∗ (T (H )). { } 1 2 0,Is 1 2 2 2 0,Is 1 2 2 0,Is Also for n = 0 it is obvious that T (H ) T ( (T )) (T ). Consequently, 2 0,Is ⊂ 2 N 1 ⊂ N 1 n T (H ) (T T ∗ ) = H . 2 0,Is ⊂ N 1 2 0,Is n 0 \≥ n Note that H0,Is = n 0 (T1T2∗ ) is the maximal subspace of (T1) invariant for ≥ N N T ∗. Since we have shown that H0,Is is invariant also for T2, it is the maximal subspace 2 T of (T ) that reduce T . Since (T ) is orthogonal to (T ) it follows that H N 1 2 N 1 N 2 0,Is reduces T2 to an isometry.

Denote H = H (H H H H H ). (3.5) G Is ⊕ Is,0 ⊕ 0,Is ⊕ 00 ⊕ F By Lemma 3.6(iii), we have (T T ) (T ∗T ∗). Thus the product of quasinormal N 1 2 ⊂ N 1 2 partial isometries is a quasinormal partial isometry if it is a partial isometry.

Remark 3.14. Subspaces H00,H0,Is,HIs,0 reduce the product T1T2 to a zero operator while HIs deﬁned in (3.2) reduce it to an isometry. Recall that

H = T n(F ) T n(F ) T T (F ) + T T (F ) . F 1 2 ⊕ 2 1 ⊕ 1∗ 1 2 2∗ 2 1 n 1 n 1 X≥ X≥ n n Note that n 1 T1 (F2) (T2), n 1 T2 (F1) (T1). Since ≥ ⊂ N ≥ ⊂ N

P T T T ∗T (F ) = PT T (F ) T T ( (T )) = 0 , 1 2 2 2 1 1 2 1 ⊂ 2 1 N 1 { } then T ∗T (F ) (T T ). Similarly, T ∗T (F ) (T T ). Thus H (T T ). 2 2 1 ⊂ N 1 2 1 1 2 ⊂ N 1 2 F ⊂ N 1 2 Consequently, a product of quasinormal partial isometries restricted to the subspace H H is a quasinormal partial isometry. G In the next paragraph it will be shown that if H = 0 , then the product of G 6 { } quasinormal partial isometries can be, but do not need to be, a quasinormal partial isometry (Examples 4.2 and 4.3). Recall after [5] the following lemma.

Lemma 3.15. Let T1,T2 be partial isometries (possibly not commuting). Product T1T2 is a partial isometry if and only if T1∗T1T2T2∗ = T2T2∗T1∗T1. The subspace H H H H H reduce T ,T such that the product Is ⊕ Is,0 ⊕ 0,Is ⊕ 00 ⊕ F 1 2 T1T2 is a partial isometry. However, it is not the maximal such subspace. On the decomposition of families of quasinormal operators 431

Proposition 3.16. Let T1,T2 be a pair of commuting quasinormal partial isometries such that H = HG, where HG is given in (3.5). Let Hp be the maximal subspace of (T T ) reducing T ,T . Then H is the maximal subspace reducing T ,T where the N 1 2 1 2 p 1 2 product T T is a quasinormal partial isometry. 1 2|Hp Proof. Let H be the maximal subspace of (T T ) reducing T ,T . Obviously p N 1 2 1 2 (T T ) being a zero operator is a quasinormal partial isometry. Let L H H 1 2 |Hp ⊂ p be any non zero subspace reducing T1,T2. Since we have assumed H = HG, then L can not reduce T ,T to a pair of isometries. Consequently, (T T ) L = 0 . 1 2 N 1 2 ∩ 6 { } Since L is orthogonal to H , we can choose x (T T ) L such that T ∗x or T ∗x p ∈ N 1 2 ∩ 1 2 is not in (T T ). Assume that T T T ∗x = 0. It follows that T ∗T T T ∗x = 0. On N 1 2 1 2 2 6 1 1 2 2 6 the other hand, x (T T ) (T ∗T ). Hence T T ∗T ∗T x = T T ∗T ∗T x = 0. By ∈ N 2 1 ⊂ N 2 1 2 2 1 1 2 1 2 1 Lemma 3.15 the product is not a partial isometry.

We can formulate the decomposition theorem for pairs of commuting quasinormal partial isometries.

Theorem 3.17. Let H be a Hilbert space and T ,T L(H) be a pair of commuting 1 2 ∈ quasinormal partial isometries. There is a decomposition

H = H H H H , J ⊕ F ⊕ p ⊕ n where HJ ,HF ,Hp,Hn are the maximal subspaces reducing operators T1,T2 such that:

(i) T ,T are jointly quasinormal, 1|HJ 2|HJ (ii) T ,T are completely non q-compatible, 1|HF 2|HF (iii) T ,T are q-compatible, completely non jointly quasinormal and the product 1|Hp 2|Hp T1T2 is a partial isometry, (iv) ( (T ) H ) ( (T ) H ) and there is no non trivial T ,T reducing subspace N 1 ∩ n ⊥ N 2 ∩ n 1 2 of Hn, where the product T1T2 is a partial isometry.

Proof. Define H = H H H H , where H defined in (3.2) reduces J Is ⊕ 0,Is ⊕ Is,0 ⊕ 00 Is T1,T2 to a pair of isometries, HIs,0,H0,Is are given by Theorem 3.13 and H00 = (T ) (T ). By Theorem 3.2, every jointly quasinormal pair has a multiple injective N 1 ∩N 2 canonical decomposition. Note that an injective partial isometry is just an isometry. On the other hand, HIs,H0,Is,HIs,0,H00 are the maximal subspaces reducing T1,T2 to suitably: a pair of isometries, T1 to a zero operator and T2 to an isometry, T1 to an isometry and T2 to a zero operator, a pair of zero operators. Consequently, the maximality of HJ follows from the maximality of their summands. Define HF ,Hp,HG suitably by Theorem 3.11, Proposition 3.16, formula (3.5) and H = H H . From these results follows also that restrictions of T ,T to the n G p 1 2 subspaces H ,H ,H have suitable properties and H H = H H H . F p n J F ⊕ p ⊕ n

It may be surprising that in the subspace where the product of quasinormal partial isometries is not a partial isometry we have the orthogonality of kernels. 432 Zbigniew Burdak

4. EXAMPLES

Each subspace in the decomposition in Theorem 3.17 can be non trivial. In this paragraph we give examples of non jointly quasinormal pairs. We use the fact that in case of partial isometries the inclusion (T ) (T ∗) is equivalent to quasinormality. N ⊂ N The ﬁrst example concerns the non q-compatible case.

Example 4.1. Let H = n 1 Hn, where Hn = Ce Cf for every n = 1, 2,... ≥ ⊕ and e, f are orthonormal vectors. Denote the canonical basis in H by ei = L (0, 0,... 0, e, 0,... ) and fi = (0, 0,... 0, f, 0,... ) with non zero value on the i-th coor- dinate. Deﬁne:

T1(ei) = ei+1,T1(fi) = 0, for i = 1, 2,..., T (√2/2e + √2/2f ) = 0,T (√2/2e √2/2f ) = f , 2 1 1 2 1 − 1 2 T2(ei) = 0,T2(fi) = fi+1, for i = 2, 3,...

Obviously, T1,T2 are partial isometries. We leave to the reader to check that T1T2 = 0 = T T and (T ) (T ∗), for j = 1, 2. It follows that T ,T are commuting 2 1 N j ⊂ N j 1 2 quasinormal operators. From (3.3) and (3.4) follows that:

G = f : i = 2, 3,... , 1 { i } G = e : i = 2, 3,... , 2 { i } F = f ,F = √2/2e + √2/2f . 1 { 1} 2 { 1 1}

Thus H = HF .

The next example concerns the q-compatible case, where the product is quasinormal – the case of the Hp subspace in the decomposition.

Example 4.2. Let H = i∞= Cei be a Hilbert space generated by orthonormal −∞ vectors ei i∞= . Deﬁne operators: { } −∞ L T (e ) = 0 for i 1,T (e ) = e for i 0, 1 i ≤ − 1 i i+1 ≥

T2(ei) = ei 1 for i 0,T2(ei) = 0 for i 1. − ≤ ≥ The operators are partial isometries. By a simple calculation we can check that (T ) (T ∗) for j = 1, 2. Since T T = 0 = T T , the operators commute and N j ⊂ N j 1 2 2 1 their product is a partial isometry. Moreover G = (T ) and consequently F = 0 j N j j { } for j = 1, 2. By formulas in Theorem 3.13, also H = H = 0 . Eventually, 0,Is Is,0 { } H = Hp.

The last example concerns the q-compatible pair, where the product is not quasinormal – the case of the Hn subspace in the decomposition. On the decomposition of families of quasinormal operators 433

Example 4.3. Consider a Hilbert space generated by an orthonormal basis e ∞ , { i}i=0 f , g ∞ . Deﬁne operators: { i i}i=1 T (g ) = 0,T (e ) = e ,T (f ) = f for i 1,T (e ) = 1/√2e + 1/√2f , 1 i 1 i i+1 1 i i+1 ≥ 1 0 1 1 T (f ) = 0,T (e ) = e ,T (g ) = g for i 1,T (e ) = 1/√2e + 1/√2g . 2 i 2 i i+1 2 i i+1 ≥ 2 0 1 1 One can check that the operators commute and

∞ ∞ (T2) = Cfi, (T1) = Cgi. N N i=1 i=1 M M Thus the kernels are orthogonal. On the other hand, n (T2∗ gn, e0) = (T2∗g1, e0) = (g1,T2e0) = (g1, 1/√2e1 + 1/√2g1) = 1/√2.

n Note that for every x (T1), there is n such that T2∗ x is not orthogonal to ∈ N n e0. Since e0 is orthogonal to (T1) then T2∗ x / (T1). It follows that H0,Is = n N ∈ N n 0 (T1T2∗ ) = 0 . Similarly, HIs,0 = 0 . Thus H = HG. ≥ N { } { } Check that T T e = 1/√2e = 1/√2. We will show that the product T T T k 1 2 0k k 2k 1 2 is not a partial isometry, if we check that e (T T ). Decompose e = x y 0 ⊥ N 1 2 0 ⊕ ∈ (T T ) (T T ). Note that e and x are orthogonal to both kernels (T ), (T ). R 1∗ 2∗ ⊕N 1 2 0 N 1 N 2 Thus y = e x is orthogonal to both kernels, precisely y is orthogonal to f and g for 0 − i i i 1. Since T T e = e = e and the product T T is a contraction, then ≥ k 1 2 kk k k+2k k kk 1 2 e (T T ) for k 1. Consequently, y is orthogonal to every vector in the basis k ⊥ N 1 2 ≥ except e0. On the other hand, y (T1T2), while e0 is not in (T1T2). Therefore y = 0. ∈ N N

5. APPLICATION TO PAIRS OF QUASINORMAL OPERATORS

The subspaces in the decomposition Theorem 3.17 have been described by geometrical properties of kernels. The kernel of any operator is equal to the kernel of a partial isometry in the polar decomposition of this operator. Thus the decomposition of a pair of quasinormal partial isometries may be used to ﬁnd the decomposition of a pair of arbitrary quasinormal operators. We will generalize Theorem 3.17 to a pair of quasinormal operators. Theorem 5.1. Let H be a Hilbert space and T ,T L(H) be a pair of commuting 1 2 ∈ quasinormal operators. There is a decomposition

H = H H H , J ⊕ 0 ⊕ n where HJ ,H0,Hn are the maximal subspaces reducing T1,T2 such that: (i) T ,T are jointly quasinormal, 1|HJ 2|HJ (ii) H (T T ) and T ,T are completely non jointly quasinormal, 0 ⊂ N 1 2 1|H0 2|H0 (iii) Hn reduces T1,T2 to a completely non jointly quasinormal pair and the product T1T2 can not be a zero operator on any nontrivial subspace of Hn reducing T1,T2. 434 Zbigniew Burdak

Proof. First note some property we will use in the proof. Let K H be any subspace. ⊂ The maximal subspace of K reducing T1,T2 is an orthogonal complement of the minimal T1,T2 reducing subspace containing K⊥. Thus the maximal subspace of K reducing T1,T2 is the following

⊥ L reducing T ,T : K⊥ L . { 1 2 ⊂ } \ We will construct the subspace H . The commutants of T ∗T and T are equal. J i i | i| Thus T ,T are jointly quasinormal, when T commutes with T for i, j = 1, 2. 1 2 i | j| Consequently, the subspace reduces T1,T2 to a jointly quasinormal pair if and only if it is a T ,T reducing subspace of ( T T T T ) ( T T T T ). By the 1 2 N | 1| 2 − 2| 1| ∩ N | 2| 1 − 1| 2| previous argumentation, the subspace

⊥ H = L reducing T ,T :( ( T T T T ) ( T T T T ))⊥ L J 1 2 N | 1| 2 − 2| 1| ∩ N | 2| 1 − 1| 2| ⊂ \ is the maximal T ,T reducing subspace of ( T T T T ) ( T T T T ). 1 2 N | 1| 2 − 2| 1| ∩ N | 2| 1 − 1| 2| Thus HJ is the maximal subspace reducing T1,T2 to a jointly quasinormal pair. We will construct the H0 subspace. The subspace

⊥ L = L reduce T ,T : (T T )⊥ L max 1 2 N 1 2 ⊂ \ is the maximal subspace of (T T ) reducing T ,T . By Theorem 3.2 applied to N 1 2 1 2 the pair T ,T we obtain H = H H H H . Note that H is 1|HJ 2|HJ J 11 ⊕ 10 ⊕ 01 ⊕ 00 11 orthogonal to (T T ) and H ,H ,H (T T ). Since H ,H ,H reduce N 1 2 01 10 00 ⊂ N 1 2 01 10 00 T ,T , it follows they are also subspaces of L . Consequently, H = L (H 1 2 max 0 max 10 ⊕ H H ) is the maximal subspace of (T T ) reducing T ,T to a completely non 01 ⊕ 00 N 1 2 1 2 jointly quasinormal pair. The subspace H := H (H H ) have required properties since it is the n J ⊕ 0 orthogonal complement of H H . J ⊕ 0 Theorem 5.1 has been proved independently to Theorem 3.17. However, for fur- ther decompositions we will use some of the previous results on quasinormal partial isometries. Remark 5.2. Let T ,T L(H) be commuting quasinormal operators and L H 1 2 ∈ ⊂ be a subspace reducing T ,T such that T = 0 or T = 0. Restrictions T ,T 1 2 1|L 2|L 1|L 2|L are jointly quasinormal since suitable products are equal to 0. Therefore, to ﬁnd a subspace reducing T1,T2 to a pair where at least one of operators is a zero operator, it is enough to check only those subspaces where they are jointly quasinormal. The next result generalize some formulas to pairs of quasinormal operators. Theorem 5.3. Let T ,T L(H) be a pair of commuting quasinormal operators, 1 2 ∈ where T = W T ,T = W T are the polar decompositions. Denote by H the 1 1| 1| 2 2| 2| J maximal subspace reducing T1,T2 to a jointly quasinormal pair and by H0,Is,HIs,0 the maximal subspaces reducing W1,W2 to pairs: a zero operator – an isometry, an isometry – a zero operator. Then H H ,H H are the maximal subspaces reducing 0,Is ∩ J Is,0 ∩ J On the decomposition of families of quasinormal operators 435

T1,T2 to pairs: a zero operator – an injective operator, an injective operator – a zero operator.

Proof. Let L be any subspace reducing T1 to a zero operator and T2 to an injective operator. By Remark 3.3 and properties of the polar decomposition, the subspace L reduces W to a zero operator and W to an isometry. Thus L H . By Remark 5.2, 1 2 ⊂ 0,Is we have the inclusion L H . Thus every subspace reducing T to a zero operator ⊂ J 1 and T to an injective operator is a subspace of H H . 2 0,Is ∩ J By Remark 3.3, the subspace H reduces W ,W and consequently H H J 1 2 0,Is ∩ J reduces W1,W2, since it is an intersection of such subspaces. Denote K = Span T n(H H ): n 0 . Since K H , it follows that T T n(H H ) = {| 2| 0,Is ∩ J ≥ } ⊂ J 1| 2| 0,Is ∩ J T nT (H H ) = 0 . Therefore K reduces T to a zero operator. Since H H | 2| 1 0,Is∩ J { } 1 0,Is∩ J reduces W2 it follows that

T T n(H H ) = T n+1W (H H ) T n+1(H H ). 2| 2| 0,Is ∩ J | 2| 2 0,Is ∩ J ⊂ | 2| 0,Is ∩ J n n+1 Similarly, T ∗ T (H H ) T (H H ). Therefore K reduces T to an 2 | 2| 0,Is ∩ J ⊂ | 2| 0,Is ∩ J 2 injective operator. By the ﬁrst part of the proof K H H . The reverse inclusion ⊂ 0,Is ∩ J follows by the deﬁnition of K. Thus K = H H reduces T ,T . 0,Is ∩ J 1 2

It can be shown that commutativity of jointly quasinormal operators is inherited on partial isometries in their polar decompositions. Unfortunately, it is not true for arbitrary pairs of quasinormal operators. Another problem is that a subspace reducing a partial isometry in the polar decomposition of T does not need to reduce T. Note the following.

Lemma 5.4. Let T = W T ,T = W T be the polar decompositions of commuting 1 1| 1| 2 2| 2| quasinormal operators. Denote by L a subspace of (T T ). N 1 2 (i) If L reduces T ,T , then L (W W ) and L reduces W ,W . 1 2 ⊂ N 1 2 1 2 (ii) If L reduces W ,W and (T ) L for i = 1, 2, then L reduces T ,T . 1 2 N i ⊂ 1 2 Proof. To show (i) take arbitrary x L such that x = x T y, where x (W ) = ∈ 0⊕| 2| 0 ∈ N 2 ( T ) and y is orthogonal to ( T ). Since L reduces T ,T , by Remark 3.3 it N | 2| N | 2| 1 2 reduces also W ,W . It follows that W x = T y L and consequently y = T ∗W x 1 2 2 2 ∈ 2 2 ∈ L (T T ) = (W T ). On the other hand, W W x = W T y = 0. Since x has ⊂ N 1 2 N 1 2 1 2 1 2 been taken arbitrary in a dense set in L, we have the thesis. To show (ii) it is enough to prove that L is T , T invariant. By (W ) = | 1| | 2| N 1 ( T ), it follows that W is an isometry on ( T ). Thus T = W ∗W T = W ∗T N | 1| 1 R | 1| | 1| 1 1| 1| 1 1 and consequently

T (L) = W ∗T (L) W ∗T ( (T T )) W ∗( (T )) W ∗(L) L. | 1| 1 1 ⊂ 1 1 N 1 2 ⊂ 1 N 2 ⊂ 1 ⊂ It follows that L is T invariant. Similarly, L is T invariant which ﬁnishes the | 1| | 2| proof. 436 Zbigniew Burdak

As a consequence of Lemma 5.2, we obtain the following corollary. Corollary 5.5. Denote by T = W T ,T = W T the polar decompositions of com- 1 1| 1| 2 2| 2| muting quasinormal operators. If L (T T ) reduces T ,T , then L (W W ) ⊂ N 1 2 1 2 ⊂ N 1 2 and W1,W2 commute on L.

Proof. By the ﬁrst part of the Lemma 5.4 and commutativity of T1,T2, we have W W = 0 = W W . 1 2|L 2 1|L As it was shown above, with appropriate assumptions subspaces reducing W1,W2 reduce also T1,T2. If necessary, some of subspaces in the decomposition of W1,W2 can be treated as generators of subspaces reducing T1,T2. By Corollary 5.5, commutativity of T1,T2 implies commutativity of W1,W2 on the subspace H0 in Theorem 5.1. By Theorem 3.17 applied to restrictions of operators to the subspace H0, we obtain the decomposition H = H H . Since H is a subspace of (W W ) it follows that 0 F ⊕ p 0 N 1 2 H = 0 . If necessary, we can extend H to the minimal subspace reducing T ,T . n { } F 1 2 In this way we obtain the maximal subspace reducing T1,T2 to a completely non q-compatible pair. For any operator T = W T we have (T ) = ( T ). For the product we do | | N N | | not have the similar property, not always (T T ) = ( T T ). On some of the N 1 2 N | 1|| 2| subspaces in Theorem 3.17 the above equality never holds. Remark 5.6. Consider a pair of commuting quasinormal operators T = W T , 1 1| 1| T = W T , where W ,W are partial isometries. Assume that H = 0 , where 2 2| 2| 1 2 F 6 { } HF is the subspace deﬁned in Theorem 3.11. Since T1,T2 are not jointly quasinormal on H , there is x H such that T T x = T T x or T T x = T T x. On F ∈ F 2| 1| 6 | 1| 2 1| 2| 6 | 2| 1 the other hand, by (T ) = ( T ), it follows that (T T ) = ( T T ). Since N 1 N | 1| N 1 2 N | 1| 2 H (T T ), then T T x = 0 and 0 = T T x = W T T x. Consequently, F ⊂ N 1 2 | 1| 2 6 2| 1| 2| 2|| 1| T2 T1 x = 0 and (T1T2) = ( T2 T1 ). The similar result can be obtained if |H ||= |0 .6 N 6 N | || | p 6 { } As a corollary we obtain the following proposition.

Proposition 5.7. Let T1,T2 be commuting quasinormal operators and their partial isometries in the polar decompositions W1,W2 also commute. Then the following con- ditions are equivalent:

(i) T1,T2 have a multiple injective canonical decomposition, (ii) W1,W2 have a multiple isometric canonical decomposition, (iii) (T T ) = ( T T ) and the product W W is a partial isometry. N 1 2 N | 1|| 2| 1 2 Proof. We have (i) (ii) by Remark 3.3. ⇒ We will show implication (iii) (i). Using the notation of Theorem 3.17 applied ⇒ to W ,W we have H = 0 and by Remark 5.6 also H = H = 0 . We need 1 2 n { } F p { } to show that subspaces H00,H0,Is,HIs,0,HIs reduce T1,T2. By Lemma 5.4(ii), the subspace H H H reduces T ,T . Consequently, H reduces T ,T . By 0,Is ⊕ Is,0 ⊕ 00 1 2 Is 1 2 the inclusion (T ) (T ∗) every subspace of (T ) reduces T for i = 1, 2. Since N i ⊂ N i N i i each of the operators T1,T2 is zero on two of the subspaces H00,H0,Is,HIs,0, then it is reduced by these two subspaces and consequently by all three subspaces. The On the decomposition of families of quasinormal operators 437 equality (T ) = (W ) for i = 1, 2 shows that the obtained decomposition of T ,T N i N i 1 2 is a multiple injective canonical decomposition. We will show implication (ii) (iii). Consider the decomposition of W ,W given ⇒ 1 2 by Theorem 3.17. It follows immediately that product W1W2 is a partial isometry. We need to show that (T T ) = ( T T ). Similarly like in the proof of the implication N 1 2 N | 1|| 2| (iii) (i) the subspaces in the decomposition reduce also T ,T . Since (iii) trivially ⇒ 1 2 holds on H and H , we can assume for convenience that H = H H and 00 Is 0,Is ⊕ Is,0 consequently that H = (T T ). An arbitrary x can be decomposed to x = x +x N 1 2 1 2 ∈ H H . Since 0,Is ⊕ Is,0

T x = T W ∗W x + T x = T ∗W x H (T ) = ( T ), | 2| | 2| 2 2 1 | 2| 2 2 2 1 ∈ 0,Is ⊂ N 1 N | 1| then T T x = 0. | 1|| 2|

REFERENCES

[1] H. Bercovici, R.G. Douglas, C. Foias, Canonical models for bi-isometries, Operator Theory: Advances and Applications 218 (2012), 177–205.

[2] A. Brown, On a class of operators, Proc. Amer. Math. Soc. 4 (1953).

[3] Z. Burdak, On decomposition of pairs of commuting isometries, Ann. Polon. Math. 84 (2004), 121–135.

[4] X. Catepillán, M. Ptak, W. Szymański, Multiple Canonical decompositions of families of operators and a model of quasinormal families, Proc. Amer. Math. Soc. 121 (1994), 1165–1172.

[5] P.R. Halmos, L.J. Wallen, Powers of partial isometries, J. Math. Mech. 19 (1970), 657–663.

[6] K. Horak, V. Müller, On the structure of commuting isometries, Commentatores Math- ematicae Univ. Carolinae 28 (1987), 165–171.

[7] A. Lubin, Weighted shifts and commuting normal extentions, J. Aust. Math. Soc. (Ser. A) 27 (1979) 1, 17–26.

[8] D. Popovici, A Wold-type decomposition for commuting isometric pairs, Proc. Amer. Math. Soc. 132 (2004), 2303–2314.

[9] M. Slocinski, On the Wold-type decomposition of a pair of commuting isometries, Ann. Polon. Math. 37 (1980), 255–262.

[10] L. Suciu, Some invariant subspaces for A-contraction and applications, Extracta Math- ematicae 21 (2006), 221–247

[11] H. Wold, A study in the analysis of stationary time series, Uppsala 1938. 438 Zbigniew Burdak

Zbigniew Burdak [email protected]

University of Agriculture Department of Applied Mathematics ul. Balicka 253c, 30-198 Krakow, Poland

Received: July 17, 2012. Revised: February 7, 2013. Accepted: February 18, 2013.