Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 6 / d / Solution

Use $U^{†}=U^{-1}$ and take the Hermitian conjugate in the reverse product order:The factorization $I-U^{-1}=-U^{-1}(I-U)$ gives $(I-U^{-1}{)}^{-1}=-(I-U{)}^{-1}U$, while $I+U^{-1}=U^{-1}(I+U)$. ConsequentlyThe last equality uses the fact that $I+U$ commutes with $I-U$ and its inverse. Thus $B$ is a Hermitian matrix. The exclusion of the eigenvalue one is essential to define the inverse. Spectrally, an eigenvalue $e^{i\vartheta }\ne 1$ is sent to $-\cot (\vartheta /2)$, a real number. This is the inverse Cayley transform of a unitary matrix without eigenvalue one.