 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-1/3/c/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

A v = λ v

, then

A^{2} v = λ^{2} v

. Alternatively, direct matrix multiplication gives

B = (1 sin 2 θ sin 2 θ 1) .

Thus the normalized eigenvectors

v_{+}, v_{-}

from part (b) have respective eigenvalues

(c + s)^{2} = 1 + sin 2 θ, (c - s)^{2} = 1 - sin 2 θ .

sin 2 θ = 0

, then

B = I

, and the whole two-dimensional space is its eigenspace. This degeneracy includes cases where the two eigenvalues of

A

were distinct but opposite.

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