Timeline for Minimum off-diagonal elements of a matrix with fixed eigenvalues
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| S Aug 11, 2018 at 11:06 | history | suggested | Rodrigo de Azevedo | CC BY-SA 4.0 |
Minor improvements to increase readability
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| Aug 11, 2018 at 9:37 | review | Suggested edits | |||
| S Aug 11, 2018 at 11:06 | |||||
| May 23, 2013 at 4:14 | vote | accept | mermeladeK | ||
| May 23, 2013 at 4:14 | history | bounty ended | mermeladeK | ||
| May 18, 2013 at 3:38 | answer | added | Dustin G. Mixon | timeline score: 8 | |
| May 18, 2013 at 2:31 | history | bounty started | mermeladeK | ||
| May 17, 2013 at 20:45 | history | edited | mermeladeK |
Added 2 more tags
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| May 17, 2013 at 19:08 | answer | added | mermeladeK | timeline score: 2 | |
| May 17, 2013 at 18:38 | comment | added | mermeladeK | I sort of forgot about this question. :) Yes, the eigenvalues are prescribed. Ahhh.. it seems nobody has an answer. | |
| Dec 17, 2012 at 17:37 | history | edited | Felix Goldberg |
edited tags
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| Dec 17, 2012 at 17:37 | history | edited | Felix Goldberg |
edited tags
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| Apr 25, 2012 at 19:27 | comment | added | fedja | As far as I understand it, lambda's are prescribed, not free to choose. | |
| Apr 25, 2012 at 17:27 | comment | added | Suvrit | Perhaps I am misunderstanding something... Given your description, the minimum absolute value among all the off-diagonals can be zero. For example, the identity matrix satisfies your hypothesis. Also, since the matrix that you have is a correlation matrix (positive semidefinite, ones on diagonal), we know that the off-diagonals must be bounded in absolute value by 1. | |
| Apr 25, 2012 at 13:37 | comment | added | Felix Goldberg | I think you should more tags like "linear algebra" to this question. | |
| Apr 25, 2012 at 9:40 | history | asked | mermeladeK | CC BY-SA 3.0 |