Timeline for Inheriting quasiconvexity from convex function after re-parametrisation of space into the Stiefel manifold
Current License: CC BY-SA 4.0
10 events
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| Jun 10, 2019 at 13:46 | comment | added | Suvrit | I see; I misread it then. IMHO the question could be improved by just stating it shortly as: "Is the map $S \mapsto \log\det(I+SS^*)$ quasi-convex?". It turns out (as a simple numerical counterexample will show) that this map is also not quasi-convex (I tried, however, without the assumption $SS^* \le I$, with that assumption it seems q-c holds for the map mentioned in this comment). | |
| Jun 7, 2019 at 12:11 | comment | added | Jesús Rodríguez | Thanks for your reply, but we are interested in the convexity-properties of $g(\cdot)$. | |
| Jun 7, 2019 at 0:42 | comment | added | Suvrit | The function $f(S) = \det(I + S)$ is not quasi-convex; neither is $\log f$; it is easy to find numerical counterexamples. | |
| Jun 6, 2019 at 16:19 | comment | added | Jesús Rodríguez | Sorry, the function f(S) is concave and equal to log det(). | |
| Jun 6, 2019 at 16:18 | history | edited | Jesús Rodríguez | CC BY-SA 4.0 |
added 15 characters in body
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| Jun 6, 2019 at 14:07 | comment | added | Suvrit | The paper I linked indeed works only for strictly +ve matrices, not for semidefinite ones (because in that case, we no longer have a clean geometry). There seems to be some mismatch btw in your question above, because you say $f$ is convex, however the determinant is not convex on $S$, so I am not sure what you are after. | |
| Jun 6, 2019 at 11:44 | comment | added | Jesús Rodríguez | Thanks for your comment. I have read your paper but failed to see how the reduced rank is entering the problem. I have the feeling we are not studying the same problem. Could you please expand your answer? We have also tried numerically over a long simulation to came up with a counterexample when the quasiconvexity in the euclidean sense fails. It would be of a great help for me if you expand also on this. Thank you for your time. | |
| Jun 3, 2019 at 17:45 | comment | added | Suvrit | The function of interest to you turns out to be geodesically convex on the set of (strictly) positive hermitian matrices. For a proof, please see Corollary 2.11 of our paper: epubs.siam.org/doi/pdf/10.1137/140978168 -- on a different note, this function is not quasi-convex in the Euclidean sense | |
| Jun 3, 2019 at 17:30 | review | First posts | |||
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| Jun 3, 2019 at 17:26 | history | asked | Jesús Rodríguez | CC BY-SA 4.0 |