Using inner product $(x,My)$ in Lanczos is equivalent to using $M^{-1}$ preconditioner in its preconditioned version. As far as I know, preconditioned version of Lanczos is used for generalized eigen-problems: $A x = \lambda M x$

Using inner product $(x,My)$ in Lanczos is equivalent to using $M^{-1}$ preconditioner in its preconditioned version. As far as I know, preconditioned version of Lanczos is used for generalized eigen-problems: $A x = \lambda M x$

Preconditioned Lanczos operates with two black-boxes: $Ax$ multiplier and $M^{-1}x$ multiplier.