Solution

The spectral theorem for compact Hermitian operators says that the nonzero spectrum of a compact Hermitian

T

consists of real eigenvalues of finite multiplicity, with zero as the only possible accumulation point. Eigenvectors for distinct eigenvalues are orthogonal, and an orthonormal basis of

ℓ^{2}

can be chosen from eigenvectors together with a basis of

ker T

. Thus

T x = \sum_{n \geq 1} λ_{n} ⟨ x, e_{n} ⟩ e_{n},

(271)

where the nonzero eigenvalues are repeated according to multiplicity and

λ_{n} \to 0

if there are infinitely many.

Define

T_{N} x = \sum_{n = 1}^{N} λ_{n} ⟨ x, e_{n} ⟩ e_{n} .

(272)

Each

T_{N}

is finite-rank and Hermitian, and

∥ T - T_{N} ∥ = sup_{n > N} ∣ λ_{n} ∣ ⟶ 0.

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Hence every compact Hermitian operator is a norm limit of finite-rank Hermitian operators, by finite-rank truncation of a compact Hermitian operator.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 22I / b / Solution

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