 Cayley transform of a real skew-symmetric matrix

For a real skew-symmetric matrix $A$, $I-A$ is invertible: $(I-A)x=0$ would imply $\Vert x{\Vert }^2=x^{T}Ax=0$. Its Cayley transform is an orthogonal matrix, since transposition gives $Q^T=(I+A{)}^{-1}(I-A)$ and the commuting factors cancel in $Q^{T}Q$. Also $\det (I+A)=\det ((I+A{)}^T)=\det (I-A)$, so $\det Q=1$ in every dimension. If $Av=\lambda v$, then $Qv=(1+\lambda )(1-\lambda {)}^{-1}v$. For $\lambda =it$, the transformed eigenvalue is $e^{2i\arctan t}$. In odd dimension $\det A=\det (-A)=-\det A$, so $A$ has a nonzero kernel and $Q$ fixes that direction.