 Solution

Since both real matrices are antisymmetric, $A^{†}=-A$ and $B^{†}=-B$. Thus $U^{†}=-A+iB$. Expand the two unitary identities:Subtracting and adding give $\boxed{AB=BA,A^2+B^2=-I}$. Because $A$ has simple eigenvalues, each of its complex eigenspaces is one-dimensional. Commutation makes each one invariant under $B$: if $Av=av$, then $A(Bv)=a(Bv)$. Hence $Bv=bv$ for some scalar $b$. The complete eigenbasis of $A$ is therefore a common eigenbasis, which also diagonalizes $U$.

The skew-Hermitian decomposition of a unitary matrix is explicit:If $Uv=uv$, unitarity gives $U^{†}v=u^{-1}v=\overline{u}v$. ConsequentlyEquivalently, for $u=e^{i\theta }$ the paired eigenvalues are $\boxed{a=i\sin \theta ,b=-i\cos \theta }$. The hypotheses can only occur in even dimension: $U^T=-U$ and $U$ is invertible, whereas an odd-dimensional skew-symmetric matrix has zero determinant.