Timeline for Eigenvalues of sum of a non-symmetric matrix and its transpose $(A+A^T)$
Current License: CC BY-SA 4.0
12 events
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Dec 27, 2021 at 17:03 | history | edited | LSpice | CC BY-SA 4.0 |
Capitalise title; `<<` -> `\ll` while this is on the front page
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Jun 9, 2021 at 20:50 | history | edited | Denis Serre | CC BY-SA 4.0 |
edited title
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Jun 8, 2020 at 0:13 | comment | added | valle | Related: math.stackexchange.com/questions/3709480/… | |
Jan 24, 2011 at 17:18 | vote | accept | Abhishek Kumar | ||
Jan 21, 2011 at 0:26 | history | edited | Abhishek Kumar | CC BY-SA 2.5 |
added 401 characters in body
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Jan 21, 2011 at 0:21 | comment | added | Abhishek Kumar | There are some relations on eigenvalues as pointed out by Denis Serre. I am also looking for a relation between eigenvectors of $M$ and $(M+M^T)/2$. Can we write the eigenvectors of $M+M^T$ in terms of eigenvectors of $M$? or any relation between the subspaces the eigenvectors span? | |
Jan 20, 2011 at 10:53 | comment | added | BS. | See my answer to this related question : mathoverflow.net/questions/31238/a-signature-inequality Any symmetric matrix with positive trace is of the form $AB+BA$ for symmetric positive definite $A,B$ (and any $M$ in your question is an $AB$). I realize that this doesn't answer your question, but methods in Ballantine's paper might help. | |
Jan 20, 2011 at 7:03 | answer | added | Denis Serre | timeline score: 13 | |
Jan 20, 2011 at 6:22 | comment | added | drbobmeister | If I am not mistaken, the requirement that all eigenvalues of $M$ are positive implies $det(M)$ is positive and hence $M$ has maximal rank, since the rows must be linearly independent. So what gives? How can $M$ have low rank? Or am I missing something? | |
Jan 20, 2011 at 5:45 | answer | added | Aaron Meyerowitz | timeline score: 0 | |
Jan 20, 2011 at 4:53 | answer | added | Igor Rivin | timeline score: 1 | |
Jan 20, 2011 at 4:43 | history | asked | Abhishek Kumar | CC BY-SA 2.5 |