Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 3.
HYPONORMAL AND SUBNORMAL THEORY
(b) Similar to (a).
(c) From (a) and (b).
(d) Observe that
A is pure hyponormal =⇒ A − λ is pure hyponormal
=⇒ ker (A − λ) = {0} (by Proposition 3.1.5)
=⇒ A − λ is not onto since λ ∈ σ(A)
=⇒ index (A − λ) = dim (ker (A − λ)) − dim ( ran(A − λ) ⊥)
= −dim ( ran(A − λ) ⊥) ≤ −1.
Write F denotes the set of Fredholm operators. We here give a direct proof
showing that Weyl’s theorem holds for hyponormal operators.
Proposition 3.1.8. If A ∈ B(H) is hyponormal then
σ(A)\ω(A) = π 00 (A),
where π 00 (A) = the set of isolated eigenvalues of finite multiplicity.
Proof. (⇐) If λ ∈ π 00 (A) then ker (A − λ) reduces A. So
A = λI ⊕ B,
where I is the identity on a finite dimensional space, B is hyponormal and λ /∈ σ(B).
So λ /∈ ω(A).
(⇒) Suppose λ ∈ σ(A)\ω(A), and so A−λ not invertible, Fredholm with index (A−
λ) = 0. We may assume λ = 0. Since A ∈ F and index A = 0, it follows that 0 is an
eigenvalue of finite multiplicity.
It remains to show that 0 ∈ iso σ(A). Observe that
ker (A) ⊆ ker (A ∗ ) = (ranA) ⊥ and 0 = index(A) = dim (ker (A)) − dim ( ranA) ⊥) ,
so that ker(A) = (ranA) ⊥ . So
where B is invertible.
σ(A).
A = 0 ⊕ B,
Since σ(A) = {0} ∪ σ(B), 0 must be an isolated point of
Corollary 3.1.9. If A ∈ B(H) is a pure hyponormal then
∥A∥ ≤ ∥A + K∥ for every compact operator K.
Proof. Since A is pure, π 0 (A) = ∅. So σ(A) = ω(A) = ∩ K∈K(H)
σ(A + K). Thus for
every compact operator K, σ(A) ⊆ σ(A + K). Therefore, ∥A∥ = r(A) ≤ r(A + K) ≤
∥A + K∥.
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