Timeline for Matrix approximation
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Dec 2, 2014 at 22:02 | history | edited | user9072 |
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| Mar 31, 2010 at 9:53 | vote | accept | Danu | ||
| Mar 31, 2010 at 4:05 | vote | accept | Danu | ||
| Mar 31, 2010 at 9:52 | |||||
| Mar 31, 2010 at 4:05 | history | edited | Danu | CC BY-SA 2.5 |
Clarify about element-wise comparison and A being non-negative.
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| Mar 31, 2010 at 4:03 | comment | added | Danu | Yes, I meant element-wise inequality and assume that A is nonnegative. I will clarify these points on the problem statement. | |
| Mar 30, 2010 at 22:32 | comment | added | Tom LaGatta | @Harald: a statement like $A \le B$ for matrices $A$ and $B$ often means that the matrix $A-B$ is non-negative definite. | |
| Mar 30, 2010 at 20:49 | answer | added | Sergei Ivanov | timeline score: 4 | |
| Mar 30, 2010 at 16:50 | comment | added | Sergei Ivanov | Do you assume that the elements of $A$ are nonnegative? If not, it may happen that there is no $\epsilon$ at all. | |
| Mar 30, 2010 at 15:54 | comment | added | Harald Hanche-Olsen |
I suppose $A\leq A(\epsilon I + B)$ means element-wise inequality? Otherwise, I can't make sense of the question. (But then, why didn't you write condition (1) as $B\ge0$?) I guess the main difficulty stems from requirement (3), which seems to give the problem a rather combinatorial flavour. Without that, it looks like a standard linear programming problem.
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| Mar 30, 2010 at 13:21 | history | edited | Danu |
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| Mar 30, 2010 at 13:06 | history | asked | Danu | CC BY-SA 2.5 |