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. 

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 $tr(S)=tr(PDP')=tr(DP'P)=tr(D)=\sum \text{eigen values of } S.$
