Is there a method to solve for $(\lambda, \vec{v})$ eigenpair in the following system? $$L\vec{v} = \lambda M \vec{v}$$ Here $\lambda \in R$, $\vec{v} \in R^n$, $L \in R^{n \times n}$, and $M \in R^{n \times n}$ is a diagonal matrix.
For more information this problem arises when trying to find the balanced cut of a graph in which each vertex has weight given by $M_{ii}$.
If $\lambda M$ is invertible (which is easy to check for diagonal matrices) then this equation is equivalent to $(\lambda M)^{-1} L v = v$, so you are simply looking for eigenvectors of $(\lambda M)^{-1} L$ of eigenvalue $1$.
In general, you're looking for (presumably nonzero) elements of the kernel of $L - \lambda M$, which you can do by hand for any explicit choices of $L$, $\lambda$, and $M$.