Consider an undirected graph $G$ with $N$ vertices and its adjacency matrix $n_{ij}$: $n_{ij} = 1$ if vertices $i$ and $j$ are connected by an edge and $n_{ij} = 0$ otherwise. Consider $A_{ij} \equiv \left(\sum_{k=1}^N n_{ik}\right)\delta_{ij} - n_{ij}$ and its LDL decomposition \begin{equation} A = L D L^T, \end{equation} where $L$ is lower triangular with unit on the diagonal, and $D$ is diagonal.

Suppose now that $G$ is connected: Then it can be shown that $A$ has only one zero eigenvalue, which we denote by $\lambda_1$, and that there is only one zero diagonal entry in $D$, which we denote by $D_{11}$. By diagonalizing $A$ numerically for multiple $G$s, I find heuristically that the following equation \begin{equation} \prod_{i=2}^N \lambda_i = N \prod_{i=2}^N D_i \end{equation} is numerically satistified.

Is this equation true in general? If it is, do you have any idea of how to prove it?

Thanks