Timeline for Special considerations when using the Woodbury matrix identity numerically
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| S Sep 19, 2023 at 14:56 | history | suggested | Luca Citi | CC BY-SA 4.0 |
Fixed broken link to reference
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| Sep 19, 2023 at 10:02 | review | Suggested edits | |||
| S Sep 19, 2023 at 14:56 | |||||
| Apr 7, 2020 at 20:36 | comment | added | WasAlbi | Any recent update? I am having troubles with Woodbury decomposition too | |
| Nov 8, 2011 at 23:01 | history | edited | Kiyo | CC BY-SA 3.0 |
added 163 characters in body
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| Nov 8, 2011 at 19:54 | comment | added | Kiyo | What ever ships with numpy... probably LU? | |
| Nov 8, 2011 at 19:46 | comment | added | fedja | By the way, what inversion algorithm are you using? | |
| Nov 8, 2011 at 18:00 | history | edited | Kiyo | CC BY-SA 3.0 |
Wrote the furthur update.
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| Nov 8, 2011 at 13:37 | comment | added | Kiyo | Thanks for the tips! I'll try this later today... stay tuned. | |
| Nov 8, 2011 at 13:36 | history | edited | Kiyo | CC BY-SA 3.0 |
Wrote Update.; added 88 characters in body
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| Nov 8, 2011 at 12:46 | comment | added | fedja | Oops, that should be $A^{−1}[I−UB(I+U^TA^{−1}UB)^{−1}U^TA^{−1}]$, of course :). | |
| Nov 8, 2011 at 8:18 | answer | added | Federico Poloni | timeline score: 1 | |
| Nov 8, 2011 at 4:45 | comment | added | fedja | What happens if you use this formula in a slightly different form: $LHS=A^{-1}[I-UB(I+U^TA^{-1}UB)^{-1}UA^{-1}]$? You'll have to choose between $(U^TA^{-1}U)B$ and $(U^TA^{-1})(UB)$ to see which one works faster but it takes only marginally more time than your original formula anyway and eliminates the inversion of B completely. You still have to invert A, but, since it is diagonal, you can detect any problems there very easily and act accordingly. | |
| Nov 8, 2011 at 3:19 | comment | added | fedja | Well, one obvious thing is that the left hand side may be well-defined and having low condition number without $A$ or $B$ being individually invertible at all, which would be a computational disaster for the right hand side. It is hard to say more without additional information. Anyway, as far as I can see, you just want a quick inversion of the LHS and do not really care whether it comes from this formula or from something else. Maybe it is a good idea to ask if someone knows a good alternative. Do I understand it right that the size of B is less than that of A and that B is symmetric? | |
| Nov 7, 2011 at 23:21 | history | edited | Kiyo | CC BY-SA 3.0 |
Fixed typo.
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| Nov 7, 2011 at 23:08 | history | asked | Kiyo | CC BY-SA 3.0 |