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I need to show that if $mathbf{S}$ is symmetric, then it's trace sums to the sum of the eigenvalues. But I don't know how to show this. Can anybody give me a hint?

P.S. Shame on my google skills, buy I really can't find any pages on this specific issue. Not with the assumption that $mathbf{S}$ is symmetric, and no proofs.

If $S$ is a symmetric matrix then $S$ has a spectral decomposition as $S=PDP'$ where $D$ is the diagonal matrix consisting the eigenvalues of $S$ and $P$ is orthogonal. Then $tr(S)=tr(PDP')=tr(DP'P)=tr(D)=sum text{eigen values of } S.$

The trace can be expressed as
$$
operatorname{Tr}A=sum_ke_k'Ae_k,
$$
where $e_k$ are any orthonormal vectors. Since we assume that the matrix $A$ is symmetric, its eigenvectors are orthogonal. Suppose that $v_k$ are eigenvectors of $A$ with their corresponding eigenvalues $lambda_k$. Then
$$
operatorname{Tr}A=sum_kv_k'Av_k=sum_kv_k'lambda_kv_k=sum_klambda_kv_k'v_k=sum_klambda_k.
$$

Every symmetric real matrix $M$ is diagonalizable (spectral theorem). So, $M$ is similar to a diagonal matrix $D$ and thereforebegin{align}operatorname{tr}M&=operatorname{tr}D\&=sumtext{ eigenvalues of }\&=sumtext{ eigenvalues of }M.end{align}

As the coefficient of $lambda^{n-1}$ in characteristic polynomial of $A$ is
$(-1)^{n-1}cdotoperatorname{Tr}(A)$
and $A$ is diagonalizable, the claim follows by Vieta.