For an $m \times n$ matrix $A$, it is possible using the power method to find the eigenvector corresponding to the largest eigenvalue of $AA^T$ by factoring it into a matrix vector product $(AA^T)v = A(A^Tv)$, so the computationally expensive $\mathcal{O}(m^2n)$ operation of computing $AA^T$ need not be done.
But if I have two $m \times n$ matrices $A$ and $B$, and want to find the eigenvector corresponding to the largest eigenvalue of $AA^T \oslash BB^T$ (i.e. the Hadamard or entrywise division between the elements of $AA^T$ and $BB^T$), my suspicion is I can no longer perform a clever factorization and find this eigenvector, without performing the computationally expensive $\mathcal{O}(m^2n)$ calculations of explicitly finding $AA^T$ and $BB^T$. Any directions on whether or not this is true (ideally with a proof/counterexample) would be appreciated.