quantum mechanics - Von Neumann entropy in terms of the mutual overlap?

I have $N$ pure, but nonorthogonal, states $|\psi_n\rangle$ with density matrix $\rho_n=|\psi_n\rangle\langle\psi_n|$.

Say we call the the total density matrix $\rho=\frac{1}{N}\sum_n \rho_n$.

Are there any formulas to calculate $S_{vn}[\rho]=-\mathrm{Tr}[\rho\log{\rho}]$ solely from the overlaps $\langle\psi_n|\psi_m\rangle$ (and the fact that all individual states are pure)? Gram-Schmidt orthogonalization might be a possibility but perhaps there is an easier way/existing result?

Intuitively it would seem at first that knowledge of the $\mathrm{Tr}[\rho_n\rho_m]=|\langle\psi_n|\psi_m\rangle|^2$ would be sufficient because the mixing between the states is noncoherent, but is this true?

Answer

Let us denote the overlap matrix by $O$, this is, $O_{nm}=\langle\psi_n\vert\psi_m\rangle$. Then, $$ S_{\mathrm{vN}}(\rho)=S_\mathrm{vN}(O)\ . $$ More generally, $O$ has the same non-zero eigenvalues as $\rho$, so any function of the eigenvalues (which is insensitive to zero eigenvalues) can be evaluated on $O$ instead of $\rho$.

This can be seen by defining a matrix $X$ whose columns are the states $\vert\psi_n\rangle$, i.e. $X_{kn}=\langle k\vert\psi_n\rangle$. Then, $\rho=XX^\dagger$, while $O=X^\dagger X$. However, two matrices $AB$ and $BA$ always have the same non-zero eigenvalues, and thus, the non-zero eigenvalues of $\rho$ and $O$ are the same.