Timeline for Maximize sum of largest eigenvalues
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Oct 28, 2011 at 11:22 | vote | accept | Kap | ||
| Oct 25, 2011 at 13:39 | history | edited | Kap | CC BY-SA 3.0 |
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| Oct 25, 2011 at 11:59 | vote | accept | Kap | ||
| Oct 25, 2011 at 11:59 | |||||
| Oct 24, 2011 at 23:16 | answer | added | Noah Stein | timeline score: 3 | |
| Oct 24, 2011 at 22:06 | history | edited | Kap | CC BY-SA 3.0 |
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| Oct 24, 2011 at 21:26 | history | edited | Kap | CC BY-SA 3.0 |
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| Oct 24, 2011 at 21:14 | comment | added | Kap | However, for it to be true, I think the vectors $v_j$ should be such that the rank 1 matrices $v_jv_j^T$ should span the space of $n$ dimensional matrices. | |
| Oct 24, 2011 at 21:01 | comment | added | Kap | My first intuitive guess is that this is true because the function is convex (linear) over the independent dimensions of the matrix (the eigenvalues), and thus the optima is always attained at the outermost extreme points of our region, which are obtained by the intersection of the planes and the cone. | |
| Oct 24, 2011 at 20:44 | history | edited | Kap | CC BY-SA 3.0 |
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| Oct 24, 2011 at 20:39 | comment | added | Kap | Now I have added a more general statement of the problem, where some weigts are also included. I have also explained a bit better. So the objective function is still convex in the elements of $X$, so if the solution is full rank then it must be at the vertices of the polyhedron. But if it is on the boundary, then convexity cannot be directly exploited since the boundary is not convex. However, it turns out the solution is still defined uniquely from the hyperplanes, as I explain in the post. Is this behavior well known? | |
| Oct 24, 2011 at 20:33 | history | edited | j.c. | CC BY-SA 3.0 |
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| Oct 24, 2011 at 20:23 | history | edited | Kap | CC BY-SA 3.0 |
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| Oct 24, 2011 at 20:16 | history | edited | Kap | CC BY-SA 3.0 |
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| Oct 24, 2011 at 14:35 | comment | added | Noah Stein | Could you clarify a bit what your question is? Let's say the feasible reason is compact so an optimum exists. Then there will be one at an extreme point $X$ of the feasible region. Of course, if $X$ is not extreme for the polyhedron defined by dropping the semidefiniteness constraint, then $X$ cannot be positive definite, so it lies on the boundary of the positive semidefinite cone. Also, why do you say the intersections between some of the hyperplanes and the semidefinite cone are unique? | |
| Oct 23, 2011 at 22:26 | comment | added | Kap | Yes you can assume there are more than $k$ independent constraint vectors. | |
| Oct 23, 2011 at 21:06 | comment | added | cardinal | Do you know anything about how many constraint vectors $v_j$ there are? For example, more than $k$? Less than or equal to $k$? Something else? | |
| Oct 23, 2011 at 9:46 | comment | added | Kap | I edited my post now, so you see that $k$ is $1 \leq k < n$. | |
| Oct 23, 2011 at 9:45 | history | edited | Kap | CC BY-SA 3.0 |
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| Oct 23, 2011 at 9:18 | comment | added | Suvrit | what is $k$...? | |
| Oct 21, 2011 at 9:53 | history | asked | Kap | CC BY-SA 3.0 |