Measure, Integration & Real Analysis, 2021a

Section 10B Spectrum 305
Because every self-adjoint operator is normal, the following result also holds for
self-adjoint operators.
10.57 orthogonal eigenvectors for normal operators
Eigenvectors of a normal operator corresponding to distinct eigenvalues are
orthogonal.
Proof Suppose α and β are distinct eigenvalues of a normal operator T, with
corresponding eigenvectors f and g. Then 10.56 implies that T ∗ f = α f . Thus
(β − α)〈g, f 〉 = 〈βg, f 〉−〈g, α f 〉 = 〈Tg, f 〉−〈g, T ∗ f 〉 = 0.
Because α ̸= β, the equation above implies that 〈g, f 〉 = 0, as desired.
Isometries and Unitary Operators
10.58 Definition isometry; unitary operator
Suppose T is a bounded operator on a Hilbert space V.
• T is called an isometry if ‖Tf‖ = ‖ f ‖ for every f ∈ V.
• T is called unitary if T ∗ T = TT ∗ = I.
10.59 Example isometries and unitary operators
• Suppose T ∈B(l 2 ) is the right shift defined by
T(a 1 , a 2 , a 3 ,...)=(0, a 1 , a 2 , a 3 ,...).
Then T is an isometry but is not a unitary operator because TT ∗ ̸= I (as is clear
without even computing T ∗ because T is not surjective).
• Suppose T ∈B ( l 2 (Z) ) is the right shift defined by
(Tf)(n) = f (n − 1)
for f : Z → F with ∑ ∞ k=−∞ | f (k)|2 < ∞. Then T is an isometry and is unitary.
• Suppose b 1 , b 2 ,...is a bounded sequence in F. Define T ∈B(l 2 ) by
T(a 1 , a 2 ,...)=(a 1 b 1 , a 2 b 2 ,...).
Then T is an isometry if and only if T is unitary if and only if |b k | = 1 for all
k ∈ Z + .
• More generally, suppose (X, S, μ) is a σ-finite measure space and h ∈L ∞ (μ).
Define M h ∈B ( L 2 (μ) ) by M h f = fh. Then T is an isometry if and only if T
is unitary if and only if μ({x ∈ X : |h(x)| ̸= 1}) =0.
Measure, Integration & Real Analysis, by Sheldon Axler