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



Connection between eigenvalues of A and its LDL decomposition
More generally, the following appears to be true. Suppose $A$ is a real $N \times N$ symmetric matrix with rank $N-1$ and $A b = 0$ where $b \ne 0$.
Let $A = LDL^T$ be the $LDL^T$ decomposition of $A$ (assuming no pivoting needed), 
and $d\_1, \ldots, d\_N$ the diagonal entries of $D$, where $d\_N = 0$. If $c\_1$ is the coefficient of $\lambda$ in the characteristic polynomial $\det(A - \lambda I)$of $A$, then 
$$c\_1 = d\_1 \ldots d\_{N-1} \sum\_{i=1}^N (b\_i/b\_N)^2 $$ 


EDIT: Here is a proof.


Let $L$ be a lower triangular matrix with diagonal entries $1$, $D$ a diagonal matrix with diagonal entries $d\_1, \ldots, d\_N$ where $d\_N = 0$, and $A = L D L^T$.
Then $b = (L^T)^{-1} e\_N$ is in the null space of $A$, and $\sum\_{i=1}^N b\_i^2 = b^T b = (L^{-1} (L^T)^{-1} )\_{NN} = \left((L^T L)^{-1}\right)\_{NN}$. By Cramer's rule that is $\det(C\_{NN})$, where $C\_{NN}$ is the upper left $(N-1) \times (N-1)$ submatrix of $L^T L$ (note that $\det(L^T L) = (\det L)^2 = 1$). On the other hand, $b\_N = 1$ (because $(L^T)^{-1}$ is upper triangular with $1$ on the diagonal).


Now by Sylvester's determinant theorem the characteristic polynomial of $A = L D L^T$ is also the characteristic polynomial of $D L^T L$. Since the last row of $D L^T L$ is $0$, we expand along the last row and find that the coefficient of $\lambda$ in the characteristic polynomial is the determinant of the upper left $(N-1) \times (N-1)$ submatrix of $D L^T L$. This submatrix is the product of the upper left $(N-1) \times (N-1)$ submatrices of $D$ and $L^T L$, so its determinant is
the product of the determinants of those, namely $d\_1 \ldots d\_{N-1} \det(C\_{NN})$.


