 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-3/9g/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For every

x

∥ αx ∥^{2} = ⟨ x, α^{*} αx ⟩, ∥ α^{*} x ∥^{2} = ⟨ x, α α^{*} x ⟩ .

α

is normal, these quantities are equal. Conversely, equality of the norms for all

x

gives

⟨ x, (α^{*} α - α α^{*}) x ⟩ = 0.

The operator in parentheses is self-adjoint, so polarization makes it zero. Therefore

α α^{*} = α^{*} α ⟺ ∥ αx ∥ = ∥ α^{*} x ∥ for all x .

The same equivalence holds over a real inner product space: for a self-adjoint operator

T

4 ⟨ T x, y ⟩ = ⟨ T (x + y), x + y ⟩ - ⟨ T (x - y), x - y ⟩

shows that a vanishing quadratic form forces

T = 0

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