Hyperinvariant Subspaces for Some 2×2 Operator Matrices

doi:10.5666/KMJ.2018.58.3.489

Articles

Kyungpook Mathematical Journal 2018; 58(3): 489-494

Published online September 30, 2018

Hyperinvariant Subspaces for Some 2×2 Operator Matrices

Il Bong Jung^*
Department of Mathematics, Kyungpook National University, Daegu 41566, Korea
e-mail : ibjung@knu.ac.kr

Eungil Ko
Department of Mathematics, Ewha Womans University, Seoul 03760, Korea
e-mail : eiko@ewha.ac.kr

Carl Pearcy
Department of Mathematics, Texas A&M University, College Station, TX 77843, USA
e-mail : cpearcy@math.tamu.edu

Received: December 26, 2017; Accepted: February 20, 2018

Abstract

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Abstract
1. Introduction
2. Preliminary Results
3. Applications
References

The first purpose of this note is to generalize two nice theorems of H. J. Kim concerning hyperinvariant subspaces for certain classes of operators on Hilbert space, proved by him by using the technique of “extremal vectors”. Our generalization (Theorem 1.2) is obtained as a consequence of a new theorem of the present authors, and doesn’t utilize the technique of extremal vectors. The second purpose is to use this theorem to obtain the existence of hyperinvariant subspaces for a class of 2 × 2 operator matrices (Theorem 3.2).

Keywords: invariant subspace, hyperinvariant subspace, extremal vector, transitive algebra.

1. Introduction

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Abstract
1. Introduction
2. Preliminary Results
3. Applications
References

Let ℋ be a separable, infinite dimensional, complex Hilbert space and ℬ(ℋ) the algebra of all bounded operators on ℋ. For T ∈ ℬ(ℋ), we write {T}′ for the commutant of T (i.e., for the algebra of all S ∈ ℬ(ℋ) such that TS = ST). A subspace ℳ ⊂ ℋ is invariant for T in ℬ(ℋ) if T ℳ ⊂ ℳ, and a subspace ℳ is hyperinvariant for T if it is an invariant subspace for all S in {T}′. The question whether every operator in ℬ(ℋ) has a nontrivial invariant subspace, which has been around since von Neumann studied it in the 1930’s, is still an open problem. Moreover the question of whether every operator in ℬ(ℋ) ℂ1_ℋ has a nontrivial hyperinvariant subspace is also open. The results in this note contribute to this circle of ideas.

In two recent papers [5] and [6], H. J. Kim, using the technique of “extremal vectors” introduced by Ansari and Enflo in [1] (for more information about this technique, see the book [3]), proved two nice theorems which we combine as

Theorem 1.1

(H. J. Kim) Let T ∈ ℬ(ℋ ⊕ ℋ) be given matricially as

T=(AC0B),

where A, B, and C are arbitrary operators in ℬ(ℋ) such that A is either a nonzero compact operator or a nonscalar normal operator. Then at least one of T and

T˜=(BD0A),

where D is arbitrary operator in ℬ(ℋ), has a nontrivial hyperinvariant subspace (notation: n.h.s.).

The purpose of this note is first to give a short and simpler proof of a better theorem, and then to use this result to obtain the existence of n.h.s. for a class of 2×2 matrices with operator entries (Theorem 3.2). Our first result is the following.

Theorem 1.2

Let A and B be operators in ℬ(ℋ) such that either A or B has a n.h.s. Then either

for every operator C in ℬ(ℋ), the operator T_C ∈ ℬ(ℋ ⊕ ℋ) given matricially as(1.1)TC=(AC0B),
has a n.h.s., or
for every operator D in ℬ(ℋ), the operator T̃_D given matricially as(1.2)T˜D=(BD0A),
has a n.h.s.

2. Preliminary Results

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Abstract
1. Introduction
2. Preliminary Results
3. Applications
References

Before proving Theorem 1.2, we obtain two needed results. The first one is a new theorem of the present authors, and the second is an elementary proposition about transitive subalgebras of ℬ(ℋ ⊕ ℋ) (i.e., subalgebras with the property that has no nontrivial invariant subspace). The first of these two results generalizes theorems from [4].

Theorem 2.1

Suppose T = T_C is as in (1.1), where A, B, and C are arbitrary operators in ℬ(ℋ) and there exists X in ℬ(ℋ) satisfying AX = XB. If either

there exists a n.h.s. ℳ for A such that X ℋ ⊂ ℳ, or
there exists a n.h.s.for B such that ker X ⊅ ,

then T has a n.h.s.

Proof

We prove b); the proof of a) is quite similar and left to the reader. Thus we are given a n.h.s. for B and an operator X in ℬ(ℋ) such that AX = XB and ker X ⊅ . With T as in (1.1), denote its commutant by

(2.1){T}′={(LσMσNσPσ):σ∈Σ}.

Since every matrix in (2.1) commutes with T, upon doing the matrix multiplication, for each σ ∈ ∑ we obtain four equations, with the one corresponding to the (2, 1) entry of the product being

(2.2)BNσ=NσA, σ∈Σ.

Multiplication of (2.2) on the right by X gives

BNσX=NσAX=NσXB, σ∈Σ,

so for each σ ∈ ∑, N_σ X commutes with B and therefore , σ ∈ ∑. By hypothesis there exists such that Xy ≠ 0. Finally, let us suppose, to obtain a contradiction, that {T}′ is transitive. It then follows easily from Proposition 2.2 below (which is completely independent of this theorem) that {N_σ Xy : σ ∈ ∑}⁻ = ℋ, which contradicts the fact that for all σ ∈ ∑. Thus {T}′ is not transitive and the proof is complete.

Proposition 2.2

Suppose that

S={(LσMσNσPσ):σ∈Σ}

is a transitive subalgebra of ℬ(ℋ ⊕ ℋ), and let x, y ∈ ℋ with x ≠ 0. Then for every ɛ > 0, there exists L_σ_₁ [respectively, M_σ_₂, N_σ_₃, P_σ_₄ ] such that ||L_σ_₁x − y|| < ɛ [respectively, ||M_σ_₂x − y|| < ɛ, ||N_σ_₃x − y|| < ɛ, ||P_σ_₄x − y|| < ɛ].

Proof

It is well-known (cf., e.g., [8]) that every transitive subalgebra is 1-transitive, meaning that for every x ≠ 0 and y in ℋ and every ɛ > 0, there exists such that ||A_ɛ x − y|| < ɛ. If we apply this fact to the transitive subalgebra and the vectors (x, 0)^t and (y, 0)^t, we obtain an element

(LσMσNσPσ)∈S

such that

ɛ>‖(LσMσNσPσ) (x0)-(y0)‖ =(‖Lσx-y‖2+‖Nσx‖2)1/2≥ ‖Lσx-y‖.

By changing the positions of the vectors x and y in the direct sum ℋ ⊕ ℋ, the other three desired inequalities follow similarly.

Proof of Theorem 1.2

We shall prove i). The proof of ii) is essentially the same. To establish i) there are two cases - that in which A has a n.h.s. and that in which B has a n.h.s. Once again the proofs are virtually indistinguishable, so we shall content ourselves with proving i) under the condition that B has a n.h.s. . Thus, by virtue of Theorem 2.1 b), it is sufficient to exhibit an operator X in ℬ(ℋ) and a nonzero vector v ∈ such that AX = XB and Xv ≠ 0. We may and do suppose that ii) is false, that is, there exists D₀ in ℬ(ℋ) such that T̃_D_₀ as in (1.2) has no n.h.s. We write the commutant of {T̃_D_₀} as

(2.3){T˜D0}′={(Lσ′Mσ′Nσ′Pσ′):σ∈Σ},

and we are given that the algebra {T̃_D_₀}′ is transitive. Since T̃_D_₀ commutes with every matrix in (2.3), we obtain, for each σ ∈ ∑, four equations, one of which is

ANσ′=Nσ′B, σ∈Σ.

Let now v₀ be an arbitrary nonzero vector in . It is an easy consequence of Proposition 2.2 that there exists σ₀ ∈ ∑ such that Nσ0′v0≠0 (take x = v₀, y a unit vector in ℋ, and ɛ = 1/2).

The proof is completed by setting v = v₀ and X=Nσ0′.

Recall finally that for T ∈ ℬ(ℋ), then Lat(T) is by definition the lattice of all invariant subspaces of T and AlgLat (T) is the algebra of all operators A in ℬ(ℋ) such that Lat(T) ⊂ Lat(A). An operator T in ℬ(ℋ) is said to be reflexive if , the smallest unital subalgebra of ℬ(ℋ) that contains T and is closed in the weak operator topology on ℬ(ℋ).

Corollary 2.3

Suppose that A is an arbitrary nonreflexive contraction in ℬ(ℋ) such that the spectrum of A contains the unit circle. Let also B, C, and D be arbitrary operators in ℬ(ℋ). Then either i) or ii) of Theorem 1.2 is valid.

Proof

One knows from [2, Corollary 7.3] that every such operator A has a n.h.s.

3. Applications

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Abstract
1. Introduction
2. Preliminary Results
3. Applications
References

In this section we show that in certain situations Theorem 1.2 can be applied to yield a n.h.s. for the operator T_C in (1.1).

Proposition 3.1

Suppose A ∈ ℬ(ℋ) has a n.h.s., and B is an arbitrary operator in ℬ(ℋ) such that B commutes with A. Then the operator Q ∈ ℬ(ℋ ⊕ ℋ) given matricially by

Q=(A1H0B)

has a n.h.s.

Proof

We know from Theorem 1.2 that either Q or

Q˜=(B1H0A)

has a n.h.s., so it suffices to prove that Q and Q̃ are similar, which we now do. It is well-known that the operator S ∈ ℬ(ℋ ⊕ ℋ) given by

S=(1H0X1H),

where X is arbitrary in ℬ(ℋ), is invertible, and

S-1=(1H0-X1H).

Therefore we calculate S⁻¹QS, where X is yet to be chosen. Thus

S-1QS=(1H0-X1H) (A1H0B) (1H0X1H) =(A+X1H-XA+(-X+B)X-X+B) =(B1H0A)=Q˜

if X is chosen to be X = B − A.

The following shows that Theorem 1.2 can be useful in obtaining a n.h.s. for certain classes of operators.

Theorem 3.2

Suppose A ∈ ℒ(ℋ) has a n.h.s., and B and C are arbitrary operators in ℒ(ℋ) that commute with A, with C invertible. Then

TC=(AC0B)

has a n.h.s.

Proof

We know from Theorem 3.1 that

Q=(A1H0B)

has a n.h.s., and the calculation

(C-1001H) (AC0B) (C001H)=(C-1AC1H0B)=(A1H0B)

shows that T_C and Q are similar.

Remark 3.3

It would be interesting to show that Theorem 3.2 remains valid without the commutativity assumptions made there.

References

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Abstract
1. Introduction
2. Preliminary Results
3. Applications
References

Ansari, S, and Enflo, P (1998). Extremal vectors and invariant subspaces. Trans Amer Math Soc. 350, 539-558.
Chevreau, B, Exner, G, and Pearcy, C (1989). On the structure of contraction operators III. Michigan Math J. 36, 29-62.
Chalendar, I, and Partington, J (2011). Modern approaches to the invariant-subspace problem. Cambridge Tracts in Math. Cambridge: Cambridge Univ. Press
Douglas, R, and Pearcy, C (1972). Hyperinvariant subspaces and transitive algebras. Michigan Math J. 19, 1-12.
Kim, HJ (2011). Hyperinvariant subspaces for operators having a normal part. Oper Matrices. 5, 487-494.
Kim, HJ (2012). Hyperinvariant subspaces for operators having a compact part. J Math Anal Appl. 386, 110-114.
Lomonosov, V (1973). Invariant subspaces of the family of operators that commute with a completely continuous operator. Funktsional Anal i Prilozen. 7, 55-56.
Radjavi, H, and Rosenthal, P (1973). Invariant subspaces. New York, NY: Springer-Verlag