Let $A,B \in \mathbb{C}^{n \times n}$ be related by $B=PAP^{*}$, where $P \in \mathbb{C}^{n \times n}$ is a unitary matrix. Furthermore, $A$ has an eigenbasis $f_{j}$, and corresponding eigenvalues $\lambda_{j}$ ordered such that $\lambda_{1} \leq \lambda_{2} \leq \cdots \leq \lambda_{n}$.
What is the simplest way to show that $A$ and $B$ have the same eigenvalues? Is proof of Schur's theorem sufficient for this task?
Using the characteristic polynomial:
$$\chi_B(x)=\det(B-xI)=det(PAP^*-xI)=\det P\det(A-xI)\det P^*=\chi_A(x)$$ and I leave for you some justifications to do.