Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-331/4/b/vi/solution
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 331 4 b vi Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-05
At the only algebraic normal value , the generator is a skew-Hermitian matrix, so its matrix exponential is a unitary matrix. Both eigenvalues have zero real part; henceThere is no distinguished initial amplitude ratio and no perturbation-energy growth. The relations select the individual orthogonal eigenmodes, but arbitrary nonzero combinations are equally optimal because all singular values are one. If is restricted to its physical positive domain, this part has no applicable normal case at all. This resolves the inconsistency in the original printed parameters rather than assuming an unjustified optimal direction.
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