Is it possible to find the eigenvalues of $AB$ if we know the eigenvalues of $A$, say $\lambda_1, \lambda_2,...,\lambda_n$ and those of $B$ say $\lambda_1, \mu_2,...,\mu_n$ and given that $A$ and $B$ are positive semi/definite symmetric complex valued matrices. Even if not possible can we build a relation of magnitude of the eigenvalues?
Thank you. Related to https://math.stackexchange.com/questions/492697/possible-determinant-inequality-det-leftiaaib-right-1-leq-det-l
$A\left\vert\alpha\right\rangle = \alpha\left\vert\alpha\right\rangle.\quad$ $B\left\vert n\right\rangle = n\left\vert n\right\rangle$
$$ \left\langle m\left\vert AB\right\vert n\right\rangle = n\left\langle m\left\vert \left(A% \sum_{\alpha}\left\vert\alpha\right\rangle\left\langle\alpha\right\vert\right) \right\vert n\right\rangle = n\sum_{\alpha}\alpha\, \left\langle m\right\vert\left.\alpha\right\rangle \left\langle\alpha\right\vert\left.n\right\rangle \equiv C_{mn}\,, \qquad C = AB $$ $C_{mn}$ are matrix elements of $C = AB$ in the base of eigenvectors of $B$. Now, you can expand a $C$ eigenvector $\left\vert\Psi\right\rangle$ in the same base: $$ C\left\vert\Psi\right\rangle = \lambda_{\rm i}\left\vert\Psi\right\rangle \qquad\Longrightarrow\qquad \sum_{m}C_{mn}\Psi_{n} = \lambda_{\rm i}\Psi_{m}\,, \quad \Psi_{m} \equiv \left\langle m\right\vert\left.\Psi\right\rangle $$
