$$k=\frac{f^{T}Be}{f^{T}e}\\ A=-B^{-1}L$$
e is the eigenvector of A and f is the eigenvector of L.
L is a Laplacian matrix of an undirected graph (real symmetric, singular & positive semidefinite). A is also singular.
B is also a real symmetric matrix. $B=((M^{T}GM-H)^{-1}+D)$
where D, G, H are diagonal matrices and M is an incidence matrix such that $M^{T}M = X,$
where X is a Laplacian matrix
$$e^{T}e=f^{T}f=1$$
Is there any peculiar property for k?
Why do I observe $k$ to be almost constant, for a particular $B$ and all possible $L$ matrices?
For example, for a 4 node graph, when I change L and keep B constant, k is lying within values 5.5 and 7.5 for all possible L matrices. What is the explanation for that?
$f^T Be = f^T (-LA^{-1}e) = f^T (-L \frac{1}{\lambda_{A}}e) = - \frac{\lambda_L}{\lambda_A}f^Te$.
Thus the ratio is equal to $-\frac{\lambda_L}{\lambda_A}$.
Without doing more working, I would guess that if $B$ is fixed, this essentially fixes the ratio.
(Workings beneath)
$A^{-1} = -L^{-1}B$
$Ae = \lambda e$, so $\frac{1}{\lambda_{A}}e = A^{-1} e$
$Lf = \lambda_L f$ so $f^T L = \lambda_L f^T$