Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 3 / b / iv / Solution

Regard the real skew-symmetric matrix as acting on ${\mathbb{C}}^n$. Its Hermitian adjoint is $A^{\ast }=A^T=-A$. Parts (i) and (iii) show that its eigenvalues are purely imaginary and its complex minimal polynomial has no repeated roots.

Since $A$ has real entries, nonreal eigenvalues occur in conjugate pairs. A nonzero pair has the form $i\alpha ,-i\alpha$, with $\alpha >0$, and contributes the real quadratic polynomialZero, if present, contributes one factor $x$. Hence the real minimal polynomial isThe product annihilates $A$ 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 $A=0$ has $\epsilon =1$ and an empty product.