Trace of symmetric matrix equals sum eigenvalues 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. 
 A: 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 $tr(S)=tr(PDP')=tr(DP'P)=tr(D)=\sum \text{eigen values of } S.$
A: 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
$$
\operatorname{Tr}A=\sum_kv_k'Av_k=\sum_kv_k'\lambda_kv_k=\sum_k\lambda_kv_k'v_k=\sum_k\lambda_k.
$$
A: Every symmetric real matrix $M$ is diagonalizable (spectral theorem). So, $M$ is similar to a diagonal matrix $D$ and therefore\begin{align}\operatorname{tr}M&=\operatorname{tr}D\\&=\sum\text{ eigenvalues of }\\&=\sum\text{ eigenvalues of }M.\end{align}
A: As the coefficient of $\lambda^{n-1}$ in characteristic polynomial of $A$ is
$(-1)^{n-1}\cdot\operatorname{Tr}(A)$
and $A$ is diagonalizable, the claim follows by Vieta.
