LU decomposition Consider a $N \times N$ symmetric real matrix $A$: $A_{ij} = (\sum_{k=1}^N n_{ik}) \delta_{ij} - n_{ij}$, where $n_{ij}$ is a real symmetric matrix whose elements are equal to $1$ or $0$. $A$ has one zero eigenvalue $\lambda_1=0$, and suppose that all the other eigenvalues $\lambda_2, ..., \lambda_N$ are strictly positive. So $\det(A) = 0$. I would like to compute $\Delta \equiv \prod_{i=2}^N \lambda_i$, i.e. $\det(A)$ without the zero eigenvalue. 
I am trying to do this is the LU decomposition of $A$. I write $A$ as $A = L U$, where $L$ is a lower triangular matrix with $1$ on the diagonal, and $U$ is an upper triangular matrix. In practice, I do this with a numerical routine from GNU Scientific Library. 
My hope is that by taking the product of the nonzero diagonal elements of $U$, I get $\Delta$. Indeed, I have done some numerical experiments, and I get 
\begin{equation}
\prod_{U_{ii} \neq 0} U_{ii} = \frac{\pm 1}{N} \Delta
\end{equation}
for many values of $N$, which makes me think that this empirical relation is not a coincidence. 
Do you have any ideas of why this should be true? I guess that the $\pm 1$ comes from the sign of a permutation. Can one prove that $\Delta$ is related to $\prod_{U_{ii} \neq 0} U_{ii}$ according to this equation?
Thank you very much for your help
Best,
Michele
 A: Let
$$L=\pmatrix{1&0\cr a&1},U=\pmatrix{c&ac\cr 0&0}.$$
Then
$$LU=\pmatrix{c&ac\cr ac&a^2c}.$$
We get
$$\prod_{U_{ii}\neq 0} U_{ii}=c,\Delta=c+a^2c.$$
A: It can be proved for a special case where the subset in upper right/lower left triangular block consists of the eigenvectors of the upper left block consists of the diagonal elements. Consider such a special matrix:
$$
A1 = \begin{pmatrix}A&u\\ v^T&\alpha\end{pmatrix}, 
$$
where $u$ is the eigenvector of $A$, then $A1$ can be LU decomposed as:
$$
A1 = \begin{pmatrix}A&u\\ v^T&\alpha\end{pmatrix}
={\begin{pmatrix}I_2&0\\ 1/\lambda v^T&1\end{pmatrix}}
\quad
{\begin{pmatrix}A&u\\0&\alpha\end{pmatrix}}
$$ 
Suppose the eigenvalues of $$
{\begin{pmatrix}A&u\\0&\alpha\end{pmatrix}}
$$ are $\lambda$, $\alpha$, and $\beta$.
so,
\begin{align}
\
A1\begin{pmatrix}\frac{1}{\alpha-\beta}u\\1\end{pmatrix}
&=\begin{pmatrix}A&u\\ v^T&\alpha\end{pmatrix}
\begin{pmatrix}\frac{1}{\alpha-\beta}u\\1\end{pmatrix}
=\alpha\begin{pmatrix}\frac{1}{\alpha-\beta}u\\1\end{pmatrix}.
\end{align}
This means $\alpha$ is the eigenvalue of both $A1$ and $U$ from its LU decomposition. You can extend the form to larger matrices as well.
Unfortunately, for a general case, @Michael Renardy already provided one counterexample. 
A: 1) your conjecture is false.
2) The calculation of $\Delta$ has nothing to do with the fact that $A$ is symmetric and with the decomposition $LU$.
3) Prop: $\Delta$ is the sum of $N$ determinants of dimension $N-1$. 
Proof: $\Delta$ is $\pm 1\times$ the coefficient of $x$ in the polynomial $P(x)=\det(A-xI_N)$ (why ?), that is $\Delta=\pm P'(0)$ and it remains to derive a determinant (column by column).  Look at the following instance, where $N=3$.
Let $A=\begin{pmatrix}1&2&3\\2&4&6\\3&-2&1\end{pmatrix}$ be a singular matrix. One obtains $\pm\Delta=\begin{vmatrix}-1&2&3\\0&4&6\\0&-2&1\end{vmatrix}+\begin{vmatrix}1&0&3\\2&-1&6\\3&0&1\end{vmatrix}+\begin{vmatrix}1&2&0\\2&4&0\\3&-2&-1\end{vmatrix}=$
$-\begin{vmatrix}4&6\\-2&1\end{vmatrix}-\begin{vmatrix}1&3\\3&1\end{vmatrix}-\begin{vmatrix}1&2\\2&4\end{vmatrix}=-8$. Note that we calculate the principal $(N-1)-$minors of the matrix $A$ and then the complexity is in $O(N^3)$. 
EDIT: In fact, I am not sure that the complexity of this method is in $O(N^3)$.
Otherwise, one calculates whole polynomial $P$. That can be done, using a random-algorithm, in $O(N^3)$.
EDIT: That follows is a variant in $O(N^3)$. One calculates $V=[1,v_2,\cdots,v_N]^T\in\ker(A)$ and, in $\det(A-xI_N)$, we change the first column $C_1$ with $C_1+v_2C_2+\cdots v_NC_N$. Then $x$ is a factor of the first column and, after, it is easy. Look at the previous instance:
With $V=[1,1,-1]$, one obtains $\det(A-xI_N)=\begin{vmatrix}-x&2&3\\-x&4-x&6\\x&-2&1-x\end{vmatrix}=x\begin{vmatrix}-1&2&3\\-1&4-x&6\\1&-2&1-x\end{vmatrix}$ and $\pm \Delta=\begin{vmatrix}-1&2&3\\-1&4&6\\1&-2&1\end{vmatrix}$.
