Eigenvalue problem - Right hand matrix is singular

Question

I am constructing an eigenvalue problem of the form

$$[R]{c} = \lambda [F]{c}$$

The matrices are populated by the results of some integrals

$$ I_{i,j} = \int f(x,y,i,j) dxdy \quad for \quad i=1,..,N \quad j=1,...,M $$

All the numbers are coming out wrong the eigenvalues are nonsensical and do not converge as the matrices get larger, they just get larger in turn, and I am trying to troubleshoot. I noticed that $[F]$ always is singular. I added some "salt" ($1e-10$) so that the program did not rebel on me but I am thinkning that this might indicate some deeper issue about my problem formulation/computation, although I am not really sure what.

So my question is: Does the fact that $[F]$ is singular, point to any such problems and if yes how should I go about correcting it? Furthermore any advice on where to focus the troubleshooting?

Cheers

Answer 1

I don't see how I can mark a comment as the correct answer but after much pain and tears, using the pseudo-inverse matrix as suggested by Jean Marie yielded the best results. Not completely there yet, the resulting eigenvalues are always what I am looking for divided by 2 for some reason, but it's getting there.

Answer 2

If $F$ is rank one (as in the example you gave) the generalized eigenproblem

$$Rc=\lambda Fc \tag{1}$$

has indeed (according to you terms) a "deeper" issue.

Indeed, if $F$ is rank one, we can write it under the form :

$$F=C \mathbb{U}^T \ \tag{2}$$

with $C$ any column of $F$, (e.g. $2.4, 9.8,22.2,39.4$ in the example you have given) and $ \mathbb{U}$ the column vector of $\mathbb{R^4}$ with null entries.

Thus (1) becomes $Rc=\lambda C(\mathbb{U}^T c)$ ; as parentheses enclose in fact a number, one gets $Rc=\mu C$ for a certain $\mu$ ; otherwise said (provided $R$ is invertible):

$$c=\mu R^{-1}C \ \tag{3}$$

giving a a unique family of (generalized) eigenvectors ($\mu$ has no constraint on it).

Having an eigenvector, it is of course easy to get the corresponding eigenvalue.

Remark : in fact, of course, this reasoning works as well in nD.