Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 44 / 1 / a / Solution

For a unitary matrix, $U{U}^{†}=1$, so $(U^{†}MU{)}^r=U^{†}{M}^{r}U$. Cyclicity of the trace then gives $\mathrm{tr}(U^{†}{M}^{r}U)=\mathrm{tr}(M^r)$ for both powers entering the action. Thus the action is invariant under unitary conjugation.

The spectral theorem for normal operators diagonalizes a finite-dimensional Hermitian matrix by a unitary matrix. Applying the invariance to that diagonal form givesOnly the eigenvalues enter, not the choice of eigenvectors.