Timeline for Why can ill-conditioned linear systems be solved precisely?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 3, 2017 at 11:17 | comment | added | percusse | Just give another example from Moler's blog blogs.mathworks.com/cleve/2015/02/16/… | |
| Sep 30, 2017 at 9:06 | vote | accept | Zoltan Csati | ||
| Sep 29, 2017 at 20:03 | history | edited | GoHokies | CC BY-SA 3.0 |
added the full title of the paper in case the arxiv link gets broken for some reason
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| Sep 29, 2017 at 19:31 | answer | added | Kirill | timeline score: 1 | |
| Sep 29, 2017 at 19:28 | answer | added | Brian Borchers | timeline score: 8 | |
| Sep 29, 2017 at 18:16 | comment | added | rchilton1980 | To be more specific, my thought experiment here is something like A = diag([1 1 1 1 1 eps]), b = [b1 b2 b3 b4 b5 0]. It is contrived, but I think it is suffcient to justify the original claim: "sometimes ill conditioned A's can be solved accurately for particular choices of b" | |
| Sep 29, 2017 at 18:11 | comment | added | rchilton1980 | That's a fair counterpoint - perhaps an artifact of picking infinite condition number (an exactly zero eigenvalue). However I think you can replace that zero eigenvalue with machine epsilon and my point still stands. (That is, the sytem has very large condition number, the system is nonsingular with a a unique solution, which we can compute very accurately provided b has no component along that tiny eigenpair). | |
| Sep 29, 2017 at 17:57 | comment | added | Zoltan Csati | @rchilton1980 "yet Ax=b may still possess a solution" But that solution is not unique. And the examples I'm referring to possess a unique solution. | |
| Sep 29, 2017 at 17:37 | comment | added | rchilton1980 | My intuition is that solvability/accuracy of an Ax=b system is tied to the forcing vector b, not just the matrix A. Perhaps if b doesn't "probe" or "excite" the ill conditioned modes of A, then accurate solution remains possible. As a limiting example, A can be exactly singular (infinite condition number), yet Ax=b may still possess a solution, that can be computed accurately, if the forcing data b resides in the range of A. I admit this is pretty hand-wavy, which is why I only comment instead of answer. | |
| Sep 29, 2017 at 14:08 | history | tweeted | twitter.com/StackSciComp/status/913767406371254272 | ||
| Sep 29, 2017 at 13:29 | history | edited | GoHokies | CC BY-SA 3.0 |
improved wording of title
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| Sep 29, 2017 at 12:59 | history | asked | Zoltan Csati | CC BY-SA 3.0 |