 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/b/solution

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

Adding and subtracting the two coordinates shows that

A (1, 1)^{T} = (c + s) (1, 1)^{T}

and

A (1, - 1)^{T} = (c - s) (1, - 1)^{T}

. The normalized eigenvectors can therefore be chosen as

v_{+} = \frac{1}{2} (1, 1)^{T}, v_{-} = \frac{1}{2} (1, - 1)^{T} .

(1 - 1) /2 = 0

, and their squared norms are one, proving orthonormality. If

s = 0

, every nonzero vector is an eigenvector; this chosen orthonormal basis still works.

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