Timeline for Bounding the eigenvalues of $B A B^T$ with the eigenvalues of $A$
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2023 at 19:09 | vote | accept | jay | ||
| Nov 6, 2018 at 2:37 | comment | added | Brendan McKay | Take $A=I$ and recall that any psd matrix can be written as $B^TB$. | |
| Nov 6, 2018 at 1:38 | history | edited | jay | CC BY-SA 4.0 |
edited title
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| S Nov 6, 2018 at 1:37 | history | suggested | David G. Stork | CC BY-SA 4.0 |
Simple MathJax
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| Nov 5, 2018 at 22:21 | review | Suggested edits | |||
| S Nov 6, 2018 at 1:37 | |||||
| Nov 5, 2018 at 21:04 | answer | added | Suvrit | timeline score: 7 | |
| Nov 5, 2018 at 20:57 | comment | added | Alex M. | @JayStanley: "Bounding the eigenvalues of $BAB^T$ with $A$" does not seem fixed to me; it simply makes no sense in English. | |
| Nov 5, 2018 at 20:26 | comment | added | jay | @AlexM. I fixed the title. PSD means positive semi definite. | |
| Nov 5, 2018 at 20:20 | history | edited | jay | CC BY-SA 4.0 |
edited title
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| Nov 5, 2018 at 19:51 | comment | added | Christian Remling | In your second paragraph, you forgot that $B^tx$ need not have the same norm as $x$. But of course you can make trivial observations along these lines, for example if $B^t x$, $\|x\|=1$, is an eigenvector of the min ev $\lambda$ of $A$, then the smallest ev of $BAB^t$ is $\ge \lambda\|B^tx\|^2$. | |
| S Nov 5, 2018 at 17:16 | history | suggested | Amir Sagiv | CC BY-SA 4.0 |
title + English + formatting + tag
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| Nov 5, 2018 at 16:33 | comment | added | Nik Weaver | @AlexM. surely "positive semidefinite". | |
| Nov 5, 2018 at 16:26 | comment | added | Alex M. | Your title doesn't make sense. Also, what is PSD? | |
| Nov 5, 2018 at 16:20 | review | Suggested edits | |||
| S Nov 5, 2018 at 17:16 | |||||
| Nov 5, 2018 at 15:20 | review | First posts | |||
| Nov 5, 2018 at 16:20 | |||||
| Nov 5, 2018 at 15:15 | history | asked | jay | CC BY-SA 4.0 |