Sum of entries of powers of symmetric matrix related to eigenvalues?

Question

I wish to prove:

For any symmetric matrix $A$ with nonnegative entries, the grand sum (sum of all entries) of $A^n$ is equal to the sum of the $n$th powers of the eigenvalues of $A$.

Proof sketch:

Diagonalize the matrix. We know we can because it is symmetric and all entries are positive or zero. Say $S$ is the diagonalization matrix (i.e. the combination of all the eigenvectors), so that $B = S^{-1} A S$ is a diagonal matrix with the eigenvalues of $A$ on the diagonal. Then $$A^n = ({S B S}^{-1})^n = {S B^n S}^{-1}$$ so $$grandsum(A^n)=grandsum({S B^n S}^{-1}) \stackrel{?}{=} grandsum(B^n) = trace(B^n) = \sum_{a \in eigvals(A)}{a^n}$$

The only thing I'm missing to complete the proof is to show that the grand sum of ${S B^n S}^{-1}$ is the grand sum of $B^n$. My thought is that the diagonalization shouldn't change the magnitude of $A$ but I don't know how to express this.

Answer 1

This is wrong already for $n=1$, where, for instance, $\pmatrix{1&1\\1&1}$ has eigenvalues $0$ and $2$ but the sum of the entries is $4$.

Answer 2

The claim is false. For starters, check it for $n=1$ and almost any matrix $A$, say $A=\begin{bmatrix}1&2\\2&1\end{bmatrix}$ (or any non-diagonal symmetric matrix).

For a matrix $A$, the sum of its eigenvalues is equal to its trace $\operatorname{tr}(A)=a_{11}+a_{22}+\cdots+a_{nn}$. As long as you have nonzero elements outside of the main diagonal, i.e. if your matrix is not diagonal, the claim doesn't hold true.