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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/6/iii/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 6 / iii / Solution by

Codex  0 2026-10-18

Choose

e_{1}

real, as permitted by the spectral theorem for real symmetric matrices. Matrix powers commute, so

M^{2} (M e_{1}) = M (M^{2} e_{1}) = - μ_{1}^{2} M e_{1} .

The proposed partner

e_{2} = M e_{1}

is nonzero because

∥ e_{2} ∥^{2} = - e_{1}^{T} M^{2} e_{1} = μ_{1}^{2} ∥ e_{1} ∥^{2} > 0.

Also the real scalar

e_{1}^{T} M e_{1}

equals its own transpose

e_{1}^{T} M^{T} e_{1} = - e_{1}^{T} M e_{1}

, so it vanishes. Therefore $e_{1}$ and $e_{2}$ are orthogonal eigenvectors with the same eigenvalue. This is the basic pairing in the paired eigenspaces of the square of a real skew-symmetric matrix.

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