 Solution

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

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

B^{T} B = (I + n m^{T}) (I + m n^{T}) = I + m n^{T} + n m^{T} + n n^{T} .

p = m \times n

is an eigenvector with eigenvalue one. On the plane spanned by the orthonormal pair

m, n

, the matrix is

(1112) .

Its eigenvalues are the roots of

μ^{2} - 3 μ + 1

, namely

μ_{\pm} = \frac{3 \pm 5}{2} .

For either root, an eigenvector is

m + (μ_{\pm} - 1) n

. Therefore

eigenvector m \times n m + \frac{1 + 5}{2} n m + \frac{1 - 5}{2} n eigenvalue 1 \frac{3 + 5}{2} \frac{3 - 5}{2}

These positive eigenvalues are the squared singular values of

B

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