Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 5 / iv

Multiplying the defining equation on the right by $I+A$ gives $(I+U)A=I-U$. Consequently the only possible answer isThe factors commute because they are rational functions of $U$. Its conjugate transpose, using the unitarity $U^{†}=U^{-1}$, isThus $A$ is Skew-Hermitian. Moreover $I+A=2(I+U{)}^{-1}$ and $I-A=2U(I+U{)}^{-1}$, so substitution recovers $U$ and establishes existence as well as uniqueness. The unitary matrix $U=-I$ is a counterexample to invertibility of $I+U$; more generally the obstruction is an eigenvalue $-1$.