 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ib/paper-2/17f/i/solution

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

α^{*} α = α α^{*}

. Moving an operator across the inner product gives

∥ αv ∥^{2} = ⟨ v, α^{*} αv ⟩ = ⟨ v, α α^{*} v ⟩ = ∥ α^{*} v ∥^{2} .

One of these norms is zero exactly when the other is zero. Therefore

ker α = ker α^{*} .

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