Timeline for How can we do LDU decomposition modulo $p$?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 10, 2016 at 12:21 | vote | accept | Matt Groff | ||
| Jun 10, 2016 at 9:58 | answer | added | J. M.'s missing motivation♦ | timeline score: 2 | |
| Jun 7, 2016 at 15:03 | review | Close votes | |||
| Jun 9, 2016 at 14:56 | |||||
| Jun 5, 2016 at 16:14 | comment | added | Daniel Lichtblau | @J.M Yes, that's what I meant by "except when it must". | |
| Jun 5, 2016 at 15:57 | comment | added | J. M.'s missing motivation♦ |
@Daniel, yep, here's a random example where it pivots: LUDecomposition[{{4, 4, 4, 3}, {0, 0, 4, 1}, {3, 3, 4, 2}, {4, 1, 0, 4}}, Modulus -> 5] But that would be because the leading $2\times 2$ block is singular, which is exactly when pivoting is necessary.
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| Jun 5, 2016 at 15:44 | comment | added | Daniel Lichtblau | @J.M. I'm not sure it pivots in the modular case, except when it must. But I don't recall for certain (I think I added that option, and it would have been around 20 years ago). | |
| Jun 5, 2016 at 2:57 | comment | added | J. M.'s missing motivation♦ |
Note that LUDecomposition[] pivots, so you have an extra permutation matrix to contend with. If you need the version without pivoting, you'll have to write your own.
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| Jun 5, 2016 at 2:56 | history | edited | J. M.'s missing motivation♦ |
edited tags
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| Jun 4, 2016 at 22:59 | comment | added | Matt Groff | @DanielLichtblau: Thanks! That Modulus trick makes it easy! | |
| Jun 4, 2016 at 22:46 | comment | added | Daniel Lichtblau |
Do LUDecomposition[mat,Modulus->p] and separate out the diagonal from the upper part.
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| Jun 4, 2016 at 22:15 | history | asked | Matt Groff | CC BY-SA 3.0 |