Eigenvalues of $AB$ from eigenvalues of $A$ and $B$ 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:
$$
C\left\vert\Psi\right\rangle = \lambda_{\rm i}\left\vert\Psi\right\rangle
\qquad\Longrightarrow\qquad
\sum_{m}C_{mn}\Psi_{n} = \lambda_{\rm i}\Psi_{m}\,,
\quad
\Psi_{m}
\equiv
\left\langle m\right\vert\left.\Psi\right\rangle
$$
$A\left\vert\alpha\right\rangle = \alpha\left\vert\alpha\right\rangle.\quad$
$B\left\vert n\right\rangle = n\left\vert n\right\rangle$

$$
\left\langle m\left\vert AB\right\vert n\right\rangle
=
n\left\langle m\left\vert \left(A%
\sum_{\alpha}\left\vert\alpha\right\rangle\left\langle\alpha\right\vert\right)
\right\vert n\right\rangle
=
n\sum_{\alpha}\alpha\,
\left\langle m\right\vert\left.\alpha\right\rangle 
\left\langle\alpha\right\vert\left.n\right\rangle
\equiv
C_{mn}\,,
\qquad
C = AB
$$
$C_{mn}$ 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:
$$
C\left\vert\Psi\right\rangle = \lambda_{\rm i}\left\vert\Psi\right\rangle
\qquad\Longrightarrow\qquad
\sum_{m}C_{mn}\Psi_{n} = \lambda_{\rm i}\Psi_{m}\,,
\quad
\Psi_{m}
\equiv
\left\langle m\right\vert\left.\Psi\right\rangle
$$