Timeline for Product of Positive Matrices
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Sep 1, 2018 at 0:31 | history | edited | darij grinberg | CC BY-SA 4.0 |
corrections
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| Aug 2, 2015 at 12:38 | history | edited | darij grinberg | CC BY-SA 3.0 |
LaTeX broken by automatic transformations?
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| Mar 23, 2010 at 13:08 | comment | added | darij grinberg | Ehm. Of course, I am talking about the case when $u$ is nonsquare only. | |
| Mar 23, 2010 at 13:08 | comment | added | user2734 | I was so concentrated on the real case that I forgot about the square root problem... @MM, thanks for pointing this out, and @DG, thanks for the remedy | |
| Mar 23, 2010 at 12:46 | comment | added | darij grinberg | The buggy LaTeX should be $k\left[X\right]/\left(X^2-u\right)$. | |
| Mar 23, 2010 at 12:45 | comment | added | darij grinberg | Square roots aren't particularly evil. It's easy to show that if $u$ is a nonnegative element of an ordered field $k$, then $k\left[X\right]/\sqrt(X^2-u\right)$ can be ordered as well. Much harder is the spectral theorem, as it requires adjoining the root of an $n$-th degree equation. | |
| Mar 23, 2010 at 12:35 | comment | added | Mark Meckes | The Kronecker product proof appears to work more generally, though, since it doesn't require the existence of square roots. | |
| Mar 23, 2010 at 11:49 | comment | added | darij grinberg | @unknown: Thanks. I knew that there was such a proof, but forgot how to do it. | |
| Mar 23, 2010 at 11:47 | comment | added | Mark Meckes | Of course if you use the spectral theorem, you get the trace result (properly rewritten as $\mathrm{Tr}(AB)\ge 0$) for complex Hermitian matrices too. This appears in Horn and Johnson's book "Matrix Analysis" as "Fejer's Theorem". | |
| Mar 23, 2010 at 11:47 | comment | added | user2734 | I think that the second result can also be proved from $\mathrm{Tr(AB)}=\mathrm{Tr}(BA)$, without using the Kronecker product. In detail, $\mathrm{Tr}(UDU^TVEV^T)$ equals $\mathrm{Tr}(DU^TVEV^TU)$ equals $\mathrm{Tr}(\sqrt{D}U^TV\sqrt{E}\sqrt{E}V^TU\sqrt{D})$. The matrix inside is a Gram matrix, and hence symmetric non-negative definite, and so has a non-negative trace. | |
| Mar 23, 2010 at 10:50 | history | edited | darij grinberg | CC BY-SA 2.5 |
added 307 characters in body
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| Mar 23, 2010 at 10:26 | history | answered | darij grinberg | CC BY-SA 2.5 |