Timeline for Nonlinear least squares and regularization
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 28, 2018 at 3:00 | history | tweeted | twitter.com/StackSciComp/status/1078485826537963520 | ||
| Dec 14, 2018 at 0:02 | comment | added | Brian Borchers | @vibe Yes, sorry for the confusion. | |
| Dec 13, 2018 at 14:56 | comment | added | vibe | @Brian: do you mean the nullspace of $J$, not $f$? | |
| Dec 13, 2018 at 14:55 | comment | added | vibe | In my application, $C$ is symmetric and banded, but it seems lapack doesn't have an LDL routine for those matrices :/ | |
| Dec 13, 2018 at 4:21 | comment | added | Brian Borchers | Keep in mind that if there is a nonzero vector in the nullspace of $C$ and also in the null space of $f$, then your problem won't be effectively regularized. | |
| Dec 12, 2018 at 22:13 | comment | added | Wolfgang Bangerth | That's what I would do: compute the $LDL^T$ decomposition. If you want to, you can then write $C=(D^{1/2}L^T)^T(D^{1/2}L^T)$, which is a product of a triangular matrix with itself. | |
| Dec 12, 2018 at 18:24 | comment | added | Richard Zhang | The Cholesky factorization does exist, but square-root based routines can run into stability issues. Instead, you can also just compute the LDL decomposition and round all negative elements in D to zero. | |
| Dec 12, 2018 at 18:16 | comment | added | vibe | I didn't consider that, thank you | |
| Dec 12, 2018 at 17:49 | comment | added | Brian Borchers | Another option is to use the symmetric square root of your positive semidefinite matrix. | |
| Dec 12, 2018 at 17:11 | comment | added | vibe | So for my problem, the choice is between Option 2 and 3 - I cannot use Option 1 since the Cholesky factorization doesn't exist for my $C$. | |
| Dec 12, 2018 at 17:06 | history | edited | vibe | CC BY-SA 4.0 |
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| Dec 12, 2018 at 17:03 | comment | added | vibe | It is positive semi-definite (eigenvalues are either 0 or positive) | |
| Dec 12, 2018 at 16:43 | comment | added | Richard Zhang | When you say that $C$ may not be positive definite, are you saying that it might be indefinite, or just positive semidefinite? (The former is much, much harder than the latter.) | |
| Dec 12, 2018 at 16:25 | history | edited | vibe | CC BY-SA 4.0 |
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| Dec 12, 2018 at 16:08 | history | edited | vibe | CC BY-SA 4.0 |
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| Dec 12, 2018 at 4:34 | history | asked | vibe | CC BY-SA 4.0 |