Timeline for Lower Bound on the Cost of Solving Linear System
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 21, 2012 at 23:42 | vote | accept | CommunityBot | ||
| Jan 20, 2011 at 4:57 | comment | added | Igor Rivin | Also, you should remember that this is for solving/multiplying under the assumption that arithmetic operation is unit cost. This is not true if your are trying to solve equations over $\mathbb{Q}$ or $\mathbb{Z}.$ | |
| Jan 20, 2011 at 1:58 | comment | added | user11000 | Thanks David and Igor. Yes I am aware that solving a system and multiplication of two matrices have the same order of cost. lehigh.edu/~gi02/m242/08linstras.pdf. Anyway I will look if someone else answers this question or provides other useful insights before accepting the answer. | |
| Jan 19, 2011 at 23:28 | comment | added | Igor Rivin | The wikipedia article is very good: en.wikipedia.org/wiki/… | |
| Jan 19, 2011 at 23:23 | history | answered | David Harris | CC BY-SA 2.5 |