Let $a_i$ be independent vectors in some inner product space and consider the matrix $M_{ij} = \langle a_i|a_j\rangle$, where $y\mapsto\langle x|y\rangle$ is linear and $x\mapsto \langle x|y\rangle$ is anti-linear. Then how would one diagonalize $M$? Indeed if $a_i$ are orthogonal, then $M$ is already diagonalized. My gut tells me that it has something to do with the Gram-Schmidt decomposition, but I'm not quite sure of the details
EDIT. Just to clarify, I understand $M$ is a Hermitian matrix, so it is unitarily diagonalizable, but this is implicit. I'm wondering if there is a formula for the eigenvalues/eigenvectors.
Denote the matrix $A = [a_1, a_2, ..., a_n]$. Then we can write $M =A^H A$. The singular value decomposition of $A$ is $A = U \Sigma V^H$, where the columns of $V$ store a basis of eigenvectors for $M$.
Your question amounts to saying: given $A$, is there a formula for the columns of $V$?
In general, no.