Diagonalizable matrices A and B with $\mathrm{Tr}(A^k)=\mathrm{Tr}(B^k)$ have the same characteristic polynomial?
If $ \operatorname{Tr}(M^k) = \operatorname{Tr}(N^k)\;\forall 1\leq k \leq n$, then how do we show the $M$ and $N$ have the same eigenvalues?