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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

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  • $\begingroup$ Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix. $\endgroup$ – Jean Marie Nov 22 '18 at 11:25
  • $\begingroup$ I would try at first to convert your issue into the eigenvalue problem $F^+Rc=\lambda c$ where $F^+$ is the pseudo-inverse of $F$. $\endgroup$ – Jean Marie Nov 22 '18 at 11:29
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    $\begingroup$ I am trying to find the lowest value that would cause buckling on a composite plate. This would be the lowest eigenvalue, and the program should converge on that solution as the matrices get larger. Here is a sample $R$ matrix: Here an F matrix: $\endgroup$ – Tristan Greenwood Nov 24 '18 at 17:25
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    $\begingroup$ $$R = \begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\\ \end{array} $$ $\endgroup$ – Tristan Greenwood Nov 24 '18 at 17:39
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    $\begingroup$ Typical (singular) F matrix $$ F = \begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\\ \end{array} $$ $\endgroup$ – Tristan Greenwood Nov 24 '18 at 17:54
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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.

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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.

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  • $\begingroup$ $ \mu = \lambda (U^{T}c)$ ? $\endgroup$ – Tristan Greenwood Nov 26 '18 at 11:47
  • $\begingroup$ Yes, exactly... $\endgroup$ – Jean Marie Nov 26 '18 at 18:18
  • $\begingroup$ Has my answer been satisfying for you ? $\endgroup$ – Jean Marie Dec 19 '18 at 21:19
  • $\begingroup$ The final answer has not been much of a help to be honest, but the suggestion of the pseudoinverse has helped a lot, mainly since I was not aware of its existence sadly. After a lot more tweaking on different parts of the code it yielded the best results. Thank you very much for taking all this time to help $\endgroup$ – Tristan Greenwood Jan 9 at 10:55
  • $\begingroup$ I understand. Nevertheless, there was another side in my contribution : the remark that your matrix has identical columns (= rank one). $\endgroup$ – Jean Marie Jan 9 at 11:12

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