Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 1 / b

Write $P=I-B$, $Q=I+B$, so $P^T=Q$ and $Q^T=P$. The factors commute because both are polynomials in $B$. Also $P$ has a matrix inverse since it is the transpose of the invertible $Q$. Thus the Cayley transform of a real skew-symmetric matrix, with its parameter replaced by $-B$, satisfiesHence $C$ is an orthogonal matrix. Its inverse is $Q{P}^{-1}$; this lets us distinguish the symmetric and skew-symmetric Cayley transforms below.

A symmetric matrix that is also an orthogonal matrix satisfies $C=C^{-1}$. Multiplying by the commuting invertible factors gives $(I-B{)}^2=(I+B{)}^2$, hence $4B=0$. Thus there is no nonzero choice of $B$. The only symmetric case is $B=0$, giving $C=I$.

Yes, in even dimension. For example, choose the skew-symmetric matrixHere $B^2=-I$, so $(I+B{)}^{-1}=(I-B)/2$ and $C=(I-B{)}^2/2=-B$. It is an orthogonal skew-symmetric matrix. More generally the skew-symmetry condition $C=-C^{-1}$ gives $(I-B{)}^2=-(I+B{)}^2$, or $B^2=-I$; no real odd-dimensional matrix can satisfy that condition.