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

Proceed by induction on the dimension. For positive dimension, the fundamental theorem of algebra gives a root of the characteristic polynomial, hence a nonzero eigenvector $v$. Normalize it to have norm one. Its span is an invariant subspace, and part (ii) makes its orthogonal complement invariant too:The restriction of $T$ to the complement remains skew-adjoint, since the same inner product identity holds there. The induction hypothesis supplies an orthonormal eigenbasis for that smaller space. Adjoining $v$ gives an orthonormal eigenbasis for $V$.

The zero-dimensional base case is immediate. Thus $T$ is diagonalizable, indeed unitarily diagonalizable. This proves the finite-dimensional case of the spectral theorem for a skew-adjoint operator without assuming that distinct eigenvalues have multiplicity one.