It is known that when a matrix is symmetric we can get its SVD $A = U \Sigma V^T$.
So the problem is reduced to compute the eigen-decomposition like
$A = W \Lambda W^T $.
The left singular vectors $u_i$ are $w_i$ and the right singular vectors $v_i$ are $\text{sign}(\lambda_i) w_i$. The singular values $\sigma_i$ are the magnitude of the eigen values $\lambda_i$.
So, if I compute $A= W \Lambda W^T $, how can I do PCA, as far as I get this I would take $\sigma_i = \sqrt{\lambda_i}$ to get eigenvalues, Is this correct? What else can you say about this?