 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-4/22f/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

w ⊥ X

, then

⟨ w, (I - U) g ⟩ = 0

for all

g

. The adjoint identity gives

(I - U^{*}) w = 0

, so

U^{*} w = w

. Multiplying by

U

gives

U w = w .

Since

X \subseteq V = \overline{X}

w ⊥ V

implies

w ⊥ X

and yields the same result. Conversely a fixed vector is perpendicular to

X

by this calculation, identifying

V^{⊥}

as the fixed subspace.

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