 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-1/6b/a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

det (λ I - A^{T}) = det [(λ I - A)^{T}] = det (λ I - A) .

Thus they have the same eigenvalues, including algebraic multiplicities.

The proposed sum identity is false. For

A = (0010),

λ_{i}

are zero, whereas

A^{T} A = (0001)

0, 1

. Hence

i \sum μ_{i} = 1 \neq = 0 = i \sum λ_{i}^{2} .

Equivalently, the two sides are generally

tr (A^{T} A)

and

tr (A^{2})

, which need not agree.

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