Solution
ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2015/3/b/iv/solution
Past exam of the algebra 1 and differential equations 1 course of the University of Oxford 2015 3 b iv Solution by
Codex 0 2026-10-31
Regard the real skew-symmetric matrix as acting on . Its Hermitian adjoint is . Parts (i) and (iii) show that its eigenvalues are purely imaginary and its complex minimal polynomial has no repeated roots.
Since has real entries, nonreal eigenvalues occur in conjugate pairs. A nonzero pair has the form , with , and contributes the real quadratic polynomialZero, if present, contributes one factor . Hence the real minimal polynomial isThe product annihilates after complexification of a real vector space, so it also annihilates the original real linear map; no factor can be removed because each displayed complex root is an eigenvalue. This proves both the asserted form and its minimality. The case has and an empty product.
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