Timeline for Finding the smallest eigenvalues of a large, but structured, matrix
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 21, 2016 at 21:39 | comment | added | Jeff | Yes, adding $cI$ and doing inverse iterations works well. | |
| Dec 20, 2016 at 18:49 | comment | added | VF1 | @Jeff have you tried this? I am having the same issue. Is it better Lanczos? It also seems that the Gershgorin bound would be a quickly-computable always-valid value for $c$. | |
| Aug 27, 2012 at 17:08 | vote | accept | Jeff | ||
| Aug 25, 2012 at 1:52 | comment | added | Igor Rivin | You don't need a representation of $M^{-1},$ you just need to iterate $M^{-1}$ on some random starting vector, which is what you do with conjugate gradient, so this requires no storage. | |
| Aug 24, 2012 at 23:52 | comment | added | Jeff | Thanks for the response. So you're proposing I 1) set $M=cI + D - AA'$, 2) solve for $M^{-1}$ with conjugate gradient, and 3) find the eigenvector corresponding to the second largest eigenvalue of $M^{-1}$? I like the use of ridging -- I had forgotten about that technique. But, one concern: Since $M$ isn't sparse, won't conjugate gradient produce a representation of $M^{-1}$ that is too large to fit into memory? | |
| Aug 24, 2012 at 21:49 | history | answered | Igor Rivin | CC BY-SA 3.0 |