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Local Spectral Property of 2 × 2 Operator Matrices

. In this paper we study the local spectral properties of 2 × 2 operator matrices. In particular, we show that every 2 × 2 operator matrix with three scalar entries has the single valued extension property. Moreover, we consider the spectral properties of such operator matrices. Finally, we show that some of such operator matrices are decomposable.


Introduction
Let H denote a separable, infinite dimensional, complex Hilbert space, and write B(H) for the algebra of all bounded, linear operators on H. Let σ(T), σ p (T), σ ap (T), and σ su (T) denote the spectrum, the point spectrum, the approximate point spectrum, and the surjective spectrum of T, respectively.
The invariant subspace problem is the question whether every operator in B(H) has a nontrivial invariant subspace. This problem has been around since the early 1930's when von Neumann raised it, and despite much effort by many mathematicians, it still remains one of intractable problems (see [3] for more details).
The purpose of this note is to study the local spectral properties and the invariant subspace problem for 2 × 2 operator matrices.
Every operator T in B(H ⊕ H) is associated with a unique 2 × 2 operator matrix where the T ij ∈ B(H), i, j = 1, 2, and the action of T on an arbitrary vector (x, y) t ∈ H ⊕ H is given by multiplying the matrix (T i j ) by the column vector above, yielding T(x ⊕ y) t = T 11 x + T 12 y ⊕ T 21 x + T 22 y .
Thus the operators we consider will be given as 2 × 2 operator matrices, as above. The following problem might be considered by those mathematicians interested in invariant subspaces. We write, as usual, C1 H for the set of all scalar multiples of the identity operator 1 H on H.
Let α, γ, and δ be three different complex scalars, and let B C1 H be arbitrary in B(H). Does the operator have a nontrivial invariant or hyperinvariant subspace?
This question may look trivial, but we have thus far been unable to solve it positively. However, for the operators if R has nice properties and σ(S) has nonempty interior in C, then it can be shown that S has a nontrivial invariant subspace. Proposition 1.1 (see below) gives some positive answer. Before that, we recall some basic concepts.
An operator T ∈ L(H) has the single valued extension property (i.e., SVEP) at λ 0 ∈ C if for every open neighborhood U of λ 0 the only analytic function f : U −→ H which satisfies the equation is the constant function f ≡ 0 on U. The operator T is said to have the single valued extension property if T has the single valued extension property at every λ ∈ C. For an operator T ∈ L(H) and for a vector x ∈ H, the local resolvent set ρ T (x) of T at x is defined as the union of every open subset G of C on which there is an analytic function f :
Any of the converse implications does not hold, in general (see [9] for more details).
If R has the Bishop's property (β) (or the SVEP), then S has the Bishop's property (β) (or the SVEP). In addition, if σ(S) has nonempty interior in C, then S has a nontrivial invariant subspace.
Proof. Since S 2 = R ⊕ R and R has the Bishop's property (β) (or the SVEP), so does S 2 . Hence S has the Bishop's property (β) (or the SVEP) from [9]. In addition, if σ(S) has nonempty interior in C, the proof follows from [9].

Main results
In this section we study the local spectral properties of 2 × 2 operator matrices. In particular, we show that every 2×2 operator matrix with three scalar entries has the single valued extension property. Moreover, we consider the spectral properties of such operator matrices. Finally, we show that some of such operator matrices are decomposable. The assumption in Theorem 2.1 that α, γ, δ are different scalars is made to avoid the same cases with the matrix in Proposition 1.1 by translations and rotations of the matrix in Theorem 2.1 by scalar multiples of 1 H . We begin with the following theorem for our program.
Theorem 2.1. Suppose that α, γ, and δ are three different scalars and R is any operator in B(H). Let If R has the single valued extension property, so does T.
Proof. If γ = 0, then it is well known that T has the single valued extension property. Hence it suffices to show that T has the single valued extension property when γ 0. Let G be a domain in C and let f = f 1 ⊕ f 2 be analytic H ⊕ H-valued function defined on G, where f 1 : G → H and f 2 : G → H are analytic functions such that . Choose an open disk G 0 of G such that G 0 ⊂ G\{α, δ}. Consider an analytic map sending µ onto z. Then (4). Thus T has thethe single valued extension property.
We note from Theorem 2.1 that every operator similar to T (in Theorem 2.1) has the single valued extension property. As some applications of Theorem 2.1, we get the following corollary.

Corollary 2.2.
Suppose that α, γ, and δ are three different scalars and R is any operator ∈ B(H). Let Then the following statements hold.
(i) If R is hyponormal, then T has the single valued extension property and σ(T) = ∪ x∈H⊕H σ T (x).
(ii) If R is normal, then T and T * have the single valued extension property.
Proof. (i) Since R has the single valued extension property, the proof follows from Theorem 2.1 and [2].
(ii) It follows from Theorem 2.1 that T has the single valued extension property. Since T * is unitarily has the single valued extension property from Theorem 2.1, T * has the single valued extension property.
We next study some spectral properties of of 2 × 2 operator matrices with three scalar entries.
Proof. Since C\σ(T) = {z ∈ C : T − z1 H⊕H is invertible.} and we know from [6] that T − z1 H⊕H is invertible if and only if the operator Thus To complete the proof, it suffices to find σ ap (T). z ∈ σ ap (T) if and only if there exist a sequence {x n }, Then we get that lim n→∞ (α − z)x 1 n + Rx 2 n = 0 and lim n→∞ γx 1 n − (δ − z)x 2 n = 0. Since Since x 2 n 0 from (5) and x n = 1, (z − α)(z − δ) ∈ σ ap (γR). Hence Thus So we complete the proof.

Corollary 2.5.
Suppose that α, γ, and δ are three different scalars and R ∈ B(H) has closed range where γ is nonzero. Let If R is quasinilpotent, then T has a nontrivial hyperinvariant subspace.
consists of two different points from Theorem 2.4. So the proof follows from [12].
For an operator T in B(H), we write σ e (T), σ le (T), and σ re (T) for the essential, left essential, and right essential spectra of T, respectively. Recall that z ∈ σ le (T) if and only if there exists a sequence {x n } of unit vectors in H such that {x n } converges weakly to zero and lim n→∞ (T − z)x n = 0.
for j = le, re, e.
We next consider the following lemma, which is very useful for our program. be an operator matrix where ran e(C) is closed. Then R has the following matrix representation; which maps from H ⊕ ker (C) ⊕ ker (C) ⊥ to ran e(C) ⊥ ⊕ ran e(C) ⊕ H where C 1 = C| ker (C) ⊥ , A 1 = P ran e(C) ⊥ A| H , A 2 = P ran e(C) A| H , B 1 denotes a mapping B from ker (C) into H, B 2 denotes a mapping B from N(C) ⊥ into K , P R(C) ⊥ denotes the projection of H onto ran e(C) ⊥ , and P ran e(C) denotes the projection of H onto ran e(C).
where ran e(C) is closed, we denote M by the matrix representation as (6) in Lemma 2.7 for every Z ∈ L(H). Since ran e(C) is closed, C 1 = C| N(C) ⊥ : ker(C) ⊥ → ran e(C) is invertible. Let λ ∈ C be given. Using the representation (6) in Lemma 2.7, we write M − λ as follows; As some applications of Lemma 2.7, we get the following theorem.
If B has closed range, then both T and T * are decomposable, and have the Bishop's property (β), the Dunford's property (C), and the single valued extension property.
Proof. Since T is decomposable if and only if both T and T * have the Bishop's property (β), it suffices to show that T and T * have the Bishop's property (β). If γ = 0, it is well known that T has the Bishop's property (β). Let γ 0. Since B ha closed range, B 1 = B| ker(B) ⊥ : ker(B) ⊥ → ran e(B) is invertible. Hence for any z ∈ C S − z has the following matrix representation from Lemma 2.7; for every compact subset K of D. Then we get that Since B 1 is invertible, lim n→∞ 3 n (z) K = 0 from (8). Since α 1 and δ 1 have the Bishop's property (β), we obtain from (8) that lim n→∞ j n (z) K = 0 for j = 1, 2. Since  Suppose that one of the following statements holds; (i) there exists a constant c > 0 such that for any x ∈ H there exists a y ∈ H such that Bx = By and y ≤ c Bx .
(ii) B is semi-Fredholm. Then both T and T * are decomposable, and have the Bishop's property (β), the Dunford's property (C), and the single valued extension property.
Proof. (i) By [11], B has closed range. Hence the proof follows from Theorem 2.8.
(ii) Since B has closed range, the proof follows from Theorem 2.8.
Recall that a subspace M of H is said to be spectral maximal for T ∈ B(H) if for every T-invariant subspace N of H the inclusion σ(T| N ) ⊆ σ(T| M ) implies N ⊆ M.
where B has closed range. Then the following statements hold. Proof. (i) Since both T and T * have the Bishop's property (β) by Theorem 2.8, the proof follows from [9].
(ii) Since T is decomposable from Theorem 2.8, the proof follows from [4].
Recall that A and B in B(H) are norm equivalent if there exist two positive real numbers k 1 and k 2 such that k 1 Ax ≤ Bx ≤ k 2 Ax for all x ∈ H. and S * are decomposable, and have the Bishop's property (β), the Dunford's property (C), and the single valued extension property.
Proof. Assume A and B are norm equivalent. Since B has closed range, so does A from [11]. Hence it follows from Theorem 2.8 that the proof is completed.
Recall that T ∈ B(H) is said to be a Weyl operator if it is a Fredholm operator having index 0. An operator T ∈ B(H) is called a Browder operator if it has both finite ascent and finite decent. Let σ w (T) and σ b (T) denote the Weyl spectrum and the Browder spectrum of T, respectively, defined as  Proof. (i) Since both T and T * have the single valued extension property from Theorem 2.8, we get from [2] that σ ap (T * ) = σ(T) = σ su (T) = ∪ x∈H σ T (x) and σ ap (T) = σ(T * ) = σ su (T * ) = ∪ x∈H σ T * (x).