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
Best Answer
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 $$\prod_{U_{ii}\neq 0} U_{ii}=c,\Delta=c+a^2c.$$