Timeline for Largest symmetric matrix given rank
Current License: CC BY-SA 3.0
3 events
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| Oct 18, 2019 at 13:52 | comment | added | Aditya Guha Roy | Not sure if this would help or not, but one can show (it is simply by using Cauchy Schwarz inequality with the non zero eigenvalues) that, if A is a n by n matrix then rank of A is at least ( (trace(A))^2 ) / trace(A^2), and for symmetric matrices the mentioned ratio of traces can be written in a compact way. | |
| Jun 28, 2015 at 12:03 | answer | added | Dustin G. Mixon | timeline score: 4 | |
| Jun 28, 2015 at 10:26 | history | asked | Turbo | CC BY-SA 3.0 |