If $I+A$ were a <singular matrix>, a nonzero <vector> $v$ would satisfy $(I+A)v=0$, so $Av=-v$. This would make $-1$ an <eigenvalue> of the <anti-Hermitian matrix>, contrary to part (a). Thus $\boxed{I+A\text{ is invertible}}$. The same argument shows $I-A$ is an <invertible matrix>, which will also be used below.

Solution

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 5 / b / Solution

 Solution (source code)