Timeline for LU factorization for $I+A$ (A skew-symmetric)
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 19, 2014 at 16:42 | comment | added | Denis Serre | The question is more interesting if you modify a bit the definition of $LU$. Usually, the diagonal of $L$ is the unit, and that of $U$ is free. But since you estimate $\|L\|+\|U\|$, it is more natural to require that the diagonal of $L$ equals that of $U$. | |
| Nov 18, 2014 at 19:29 | answer | added | loup blanc | timeline score: 4 | |
| S Jul 23, 2014 at 12:33 | history | bounty ended | CommunityBot | ||
| S Jul 23, 2014 at 12:33 | history | notice removed | CommunityBot | ||
| Jul 15, 2014 at 15:36 | comment | added | Nathaniel Johnston | For what it's worth, it seems like the smallest possible $C$ in the $3\times 3$ case is also approximately $2.1846$, based on parametrizing all such $3\times 3$ matrices and constructing a mesh over them. So we might (naively) guess that this $C$ works in all dimensions. | |
| Jul 15, 2014 at 12:03 | comment | added | Winfried | That is true. However, most of the worst-case examples I know are not uniformly positive. Nevertheless, I would be also happy with a counterexample for the estimate. | |
| Jul 15, 2014 at 11:58 | comment | added | Federico Poloni | I would be very careful in assuming this is true based on numerical examples only. The worst-case bounds for the growth factor in LU factorization of general (non-antisymmetric) matrices are quite hard to reach in practice. | |
| S Jul 15, 2014 at 10:57 | history | bounty started | Winfried | ||
| S Jul 15, 2014 at 10:57 | history | notice added | Winfried | Draw attention | |
| Jul 15, 2014 at 7:57 | comment | added | Winfried | I should have written $C\lesssim 1$, I just wanted to say that the constant is small. However, I don't really care about the size of the constant. In my computations $L$ is normalized to have one's at the diagonal, but your normalization would also be fine. | |
| Jul 14, 2014 at 20:35 | comment | added | Neil Strickland | Are you assuming that $L$ has ones on the diagonal? The claim will be false without some kind of normalisation like this, but that particular choice of normalisation feels quite arbitrary in this context. It would seem more natural to set things up so that if the chosen factorisation of $I+A$ is $LU$, then the chosen factorisation of $I-A$ is $U^TL^T$. | |
| Jul 14, 2014 at 13:42 | comment | added | Nathaniel Johnston | $C \leq 1$ isn't possible, even for $2\times 2$ matrices. Let $a = 0.379$ and consider $A = \begin{bmatrix}0 & a\\ -a & 0\end{bmatrix}$. Then $\|L\| + \|U\| = 2.4984$ but $1 + \|A\|^2 = 1.1436$. In fact, a direct calculation shows that for $2 \times 2$ matrices the smallest possible $C$ is approximately $2.1846$. | |
| Jul 14, 2014 at 13:17 | comment | added | Piotr Migdal | @NathanielJohnston You are right, I deleted my remarks. | |
| Jul 14, 2014 at 12:59 | comment | added | Winfried | I observed $C\leq 1$. | |
| Jul 14, 2014 at 12:46 | history | edited | Winfried | CC BY-SA 3.0 |
added 18 characters in body
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| Jul 11, 2014 at 11:16 | history | asked | Winfried | CC BY-SA 3.0 |