Timeline for How to maximize the determinant of a matrix of the form VDV^H
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| May 17, 2013 at 18:05 | answer | added | Suvrit | timeline score: 1 | |
| May 17, 2013 at 17:50 | history | edited | Suvrit | CC BY-SA 3.0 |
added some tags, fixed TeX and some grammar plus typos
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| May 17, 2013 at 10:44 | comment | added | Carlo Beenakker | thank you, Federico, I stand corrected; the correct formula is $${\rm Det}A=\sum_{S}|{\rm Det}V_{S}|^2\prod_{k\in S}D_{kk}$$ where $S$ is a subset of $M$ indices out of $1,2,...2M$ and $V_S$ is an $M\times M$ matrix constructed from $V$ by deleting the $M$ columns that are not in $S$. it would seem that to maximize this is in general not trivial. | |
| May 17, 2013 at 7:49 | comment | added | Federico Poloni | Not sure about that: $V^HV$ is a rank-$M$ $2M\times 2M$ matrix, so its determinant would be zero. | |
| May 17, 2013 at 7:28 | comment | added | Carlo Beenakker | since $Det A = (Det V^H V) \prod_n D_n$, this is trivial | |
| May 17, 2013 at 6:56 | comment | added | user34079 | btw: ^H means hermitian transpose | |
| May 17, 2013 at 6:55 | history | asked | user34079 | CC BY-SA 3.0 |