Timeline for Is adding an outer product to a matrix guaranteed to not decrease the eigenvalues? [closed]
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2016 at 11:41 | history | closed |
Christian Remling Wolfgang Stefan Waldmann Stefan Kohl♦ Alexey Ustinov |
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| Oct 14, 2016 at 4:14 | review | Close votes | |||
| Oct 14, 2016 at 11:41 | |||||
| Oct 14, 2016 at 3:59 | comment | added | Christian Remling | Each eigenvalue increases; this is immediate from the min-max principle: en.wikipedia.org/wiki/Min-max_theorem. Questions of this type are not really suited for this site. Please ask at MSE instead. | |
| Oct 14, 2016 at 3:40 | comment | added | Noam D. Elkies | In that case, it was already answered (for the version where you subtract $vv^T$ and reduce the eigenvalues) in an earlier MO question, with the additional information that each eigenvalue increases to at most the next-largest one, attributed to Cauchy: mathoverflow.net/questions/193527/… | |
| Oct 14, 2016 at 2:43 | history | edited | Christopher Johnson | CC BY-SA 3.0 |
added 3 characters in body
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| Oct 14, 2016 at 2:43 | comment | added | Christopher Johnson | Yes, thank you for the clarification, I have edited the question. A is assumed symmetric. | |
| Oct 14, 2016 at 2:19 | comment | added | Noam D. Elkies | Are you assuming that $A$ is symmetric? In that case it should be true because $vv^T$ is positive semidefinite, but in general the eigenvalues need not even be real numbers. | |
| Oct 14, 2016 at 1:59 | review | First posts | |||
| Oct 14, 2016 at 2:26 | |||||
| Oct 14, 2016 at 1:54 | history | asked | Christopher Johnson | CC BY-SA 3.0 |