= Solution
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:
$$
V=\operatorname{span}\{v\}\oplus\operatorname{span}\{v\}^{\perp}.
$$
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 \b[$T$ 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.
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