3
$\begingroup$

Is there any theoretical reason that Chol() in R breaks for Hilbert matrix of order greater than 12?

Thanks,

$\endgroup$
1

2 Answers 2

4
$\begingroup$

Both Zen's comment and F.Tusell's answer are correct (+1 to both). Just to add a bit on this: If you look at the associated eigenvalues of Hilbert(13) for a example, the smallest of them is smaller in scale than .Machine$double.eps which is roughly 2.2e-16. Therefore this eigenvalue (and others of even smaller scale) is regarded as a machine zero and leads the decomposition algorithm to treat Hilbert(13+) as non-positive definite matrix and thus not having a Cholesky decomposition. (So actually $H(13-15)$ are "semi-PD" numerically speaking, and for $H(16+)$ you even end up with "negative zeros" eigenvalues).

$\endgroup$
3
$\begingroup$

Hilbert matrices are of notorious poor condition, which probably account for the failure you are seeing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.