What can you say about the eigenvalues of a real skew-symmetric matrix?
Solution
Let \(A\in M_n(\RR)\) satisfy \(A^t=-A\). Regard \(A\) as a complex matrix. If \(Av=\lambda v\) for \(0\ne v\in\CC^n\), then \[ \lambda\langle v,v\rangle =\langle Av,v\rangle =-\langle v,Av\rangle =-\overline\lambda\langle v,v\rangle. \] Hence \[ \lambda=-\overline\lambda, \] so every complex eigenvalue of \(A\) is purely imaginary (possibly \(0\)).
Because \(A\) has real coefficients, nonzero eigenvalues occur in conjugate pairs \[ i\mu,-i\mu\qquad(\mu\in\RR). \] In particular the only possible real eigenvalue is \(0\). Therefore an odd-dimensional real skew-symmetric matrix is singular: its nonzero eigenvalues pair off, leaving \(0\) as an eigenvalue.
Equivalently, \(iA\) is Hermitian, so the spectral theorem gives a unitary diagonalization of \(A\) over \(\CC\) with purely imaginary diagonal entries.