suppose $tr(A^k) = tr(B^k)$ for all $k$=$1,2,....$. why $A$ and $B$ are 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?