Cayley transform of a real skew-symmetric matrix ($Q=(I+A)(I-A{)}^{-1}$)

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.

For a real skew-symmetric matrix $B$, its Cayley transform of a real skew-symmetric matrix with the opposite sign convention has inverse $C^{-1}=(I+B)(I-B{)}^{-1}$. Since $C$ is an orthogonal matrix, symmetry means $C=C^{-1}$, or $(I-B{)}^2=(I+B{)}^2$. This forces $B=0$. Skew-symmetry means $C=-C^{-1}$, or $(I-B{)}^2=-(I+B{)}^2$, which is equivalent to $B^2=-I$. In that case $C=-B$. Such examples exist in even dimension by taking two-dimensional right-angle rotation matrices in diagonal blocks; in odd real dimension the determinant equation $(\det B{)}^2=(-1{)}^n$ rules them out.