Measure, Integration & Real Analysis, 2021a

Section 10D Spectral Theorem for Compact Operators 329
The next result is one of the major highlights of the theory of compact operators
on Hilbert spaces. The result as stated below applies to both real and complex Hilbert
spaces. In the case of a real Hilbert space, the result below can be combined with
10.103(a) to produce the following result: A compact operator on a real Hilbert
space is self-adjoint if and only if there is an orthonormal basis of the Hilbert space
consisting of eigenvectors of the operator.
10.106 Spectral Theorem for self-adjoint compact operators
Suppose T is a self-adjoint compact operator on a Hilbert space V. Then
(a) there is an orthonormal basis of V consisting of eigenvectors of T;
(b) there is a countable set Ω, an orthonormal family {e k } k∈Ω in V, and a family
{α k } k∈Ω in R \{0} such that
for every f ∈ V.
Tf = ∑ α k 〈 f , e k 〉e k
k∈Ω
Proof Let U denote the span of all the eigenvectors of T. Then U is an invariant
subspace for T. Hence U ⊥ is also an invariant subspace for T and T| U ⊥ is a selfadjoint
operator on U ⊥ (by 10.102). However, T| U ⊥ has no eigenvalues, because
all the eigenvectors of T are in U. Because all self-adjoint compact operators on a
nonzero Hilbert space have an eigenvalue (by 10.99), this implies that U ⊥ = {0}.
Hence U = V (by 8.42).
For each eigenvalue α of T, there is an orthonormal basis of null(T − αI) consisting
of eigenvectors corresponding to the eigenvalue α. The union (over all eigenvalues
α of T) of all these orthonormal bases is an orthonormal family in V because eigenvectors
corresponding to distinct eigenvalues are orthogonal (see 10.57). The previous
paragraph tells us that the closure of the span of this orthonormal family is V (here
we are using the set itself as the index set). Hence we have an orthonormal basis of
V consisting of eigenvectors of T, completing the proof of (a).
By part (a) of this result, there is an orthonormal basis {e k } k∈Γ of V and a family
{α k } k∈Γ in R such that Te k = α k e k for each k ∈ Γ (even if F = C, the eigenvalues
of T are in R by 10.49). Thus if f ∈ V, then
( )
Tf = T ∑ f , e k 〉e k
k∈Γ〈
= ∑ 〈 f , e k 〉Te k = ∑ α k 〈 f , e k 〉e k .
k∈Γ
k∈Γ
Letting Ω = {k ∈ Γ : α k ̸= 0}, we can rewrite the equation above as
Tf = ∑ α k 〈 f , e k 〉e k
k∈Ω
for every f ∈ V. The set Ω is countable because T has only countably many
eigenvalues (by 10.93) and each nonzero eigenvalue can appear only finitely many
times in the sum above (by 10.82), completing the proof of (b).
Measure, Integration & Real Analysis, by Sheldon Axler