Let $L$ be a Laplacian matrix of a connected graph $G = (V, E)$ and $D$ its degree matrix. Then, for a vector $z \in \mathbb{R}^{|V|}$ and a constant $\rho \in \mathbb{R}$, I have the following:
\begin{equation} (L + \rho D)x = \rho Dz. \end{equation}
This paper (page 264) says that, assuming that ${1}^\top_{|V|} Dz = 0$, the above equation has a unique solution. I don't understand why it is so, since it seems to me that $L + \rho D$ isn't always full rank for $\rho \leq 0$. I may be missing something here. Any help will be appreciated. Thanks!
I think I understand better the problem more now. First of all, the matrix $L+ \rho D$ only has a non-trivial null space when $\rho=0$ or $\rho=−\lambda_i(G)$, where $\lambda_i(G)$ is the $i$-th Generalized Eigenvector of $L$ with respect to $D$. Then, the authors of 1 showed that $\rho >−\lambda_i(G)$, leaving $\rho=0$ to be analysed (remember that $L+ \rho D$ is non singular for $\rho>0$). Now, we are just left to show the uniqueness when $\rho=0$. Maybe we'll use $1^{\top}_{|V|}Dz=0$ here.