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. 

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
$$
\operatorname{Tr}A=\sum_kv_k'Av_k=\sum_kv_k'\lambda_kv_k=\sum_k\lambda_kv_k'v_k=\sum_k\lambda_k.
$$
