 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/8b/c/solution

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

First note that if

A

is normal, then

∥ (A - λ I) v ∥ = ∥ (A^{†} - \overline{λ} I) v ∥

for every

v

. Hence an eigenvector

v

with

A v = λ v

also satisfies

A^{†} v = \overline{λ} v .

Normalize

v

and extend it to an orthonormal basis. Let

U

have these basis vectors as columns. The first column of

U^{†} A U

(λ, 0, \dots, 0)^{T}

. For every

w ⊥ v

⟨ v, A w ⟩ = ⟨ A^{†} v, w ⟩ = ⟨ \overline{λ} v, w ⟩ = 0,

so the first row also has no off-diagonal entries. Therefore

U^{†} A U = (λ 0 0 B) .

Comparing the two block products in the normality identity shows that

B B^{†} = B^{†} B

, so

B

is normal.

The result is trivial in dimension one. Applying the induction hypothesis to

B

and adjoining the eigenvector

v

A

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