Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 6 / b / Solution

Since $C$ is unitary, $C^{-1}=C^{†}$. Hence $B=C^{-1}AC=C^{†}AC$ is Hermitian, and so is its positive integer power:It follows that $B^p$ has real eigenvalues. More explicitly, cancellation of adjacent $C{C}^{-1}$ factors gives the similarity transformationIf the real eigenvalues of $A$ are ${\lambda }_j$, the eigenvalues in question are thereforeFor even $p$ they are nonnegative; for odd $p$ their signs can be either. The similarity argument actually works for every invertible $C$, although the Hermiticity argument uses its unitarity.