My problem: The coefficients of the characteristic polynomial are expressed as $\chi_A(\lambda) = \lambda^n + D_1\lambda^{n-1}+...+D_n$. Prove that $Tr(A^m) =\sum_{} \lambda^m_i$, where $\lambda_i$ are the eigenvalues of $A$ (a matrix) and from this prove the reccurent formula:
$mD_m + D_{m-1}Tr(A) + D_{m-2}Tr(A^2)+...+Tr(A^m)$ = 0
My solution: Every diagonalisable matrix $A$ can be expressed as the product od the similar diagonal matrix $D$ with the eigenvalues on its diagonal and the bases containing the eigenvectors. Traces of those matrixes are same: $Tr(A) = Tr(D) = \sum_{i=1}^na_{ii} = \sum_{i=i}^n \lambda_i$, so the same follows for $Tr(A^m) = Tr(D^m) = \sum_{i=i}^n \lambda^m_i$, for $1\le m \le n$.
I think I would prove that reccurent formula with Viet's relation, not sure how. Is there any proof of $Tr(A) = Tr(D)$?
If you want to prove the recurrence formula, you would rather applies the Hamilton theorem which says that every matrix satisfies its own characteristic equation. As such, in the characteristic equation, substitute $\lambda$ by $A$. and then apply the equality on the trace operator given that: the trace of the sum of matrices is equal to sum of traces of each matrix.
Best