I know that when you add a constant to the diagonal of a standard eigenvalue problem, it has the effect of adding that same constant to the eigenvalues:
$Av=\lambda v$ and $A'=A+cI$
$A'v = (A+cI)v = Av+cIv = \lambda v + c v = (\lambda+c)v$
So if you want to know the eigenvalues of the original matrix $A$, you can simply deduct $c$ of the eigenvalues of $A'$. However, I wonder if a similar relationship can be found when you add a constant to the right matrix $B$ in a generalized eigenvalue problem $Av=Bv\lambda$. This way I could calculate the eigenvalues of $Av'=(B+cI)\lambda' v'$ and then use that relationship $\lambda = f(\lambda')$ to calculate the original eigenvalues $\lambda$. However, I can't seem to figure out a relationship like it exists for the standard formulation. Am I looking in the wrong direction?
PS: I am interested in doing this because I am solving such a generalized problem with $A$ and $B$ Hermitian but $B$ unfortunately is not positive definite, so I want to add a constant to make it positive definite to obtain real eigenvalues. Afterwards, I could then use this relationship to find the original eigenvalues.