Timeline for Descartes rule of signs for a noncommutative polynomial in matrix variables
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 14, 2012 at 15:19 | history | edited | Suvrit | CC BY-SA 3.0 |
put in the assumption.
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| Aug 14, 2012 at 15:05 | history | edited | Suvrit | CC BY-SA 3.0 |
update about current status
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| Aug 14, 2012 at 7:46 | history | edited | Suvrit | CC BY-SA 3.0 |
augmented il titolo
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| Aug 14, 2012 at 7:35 | comment | added | Suvrit | @Mark: Hmm...if all the $C_i$ commute, we diagonalize them, and then we see that there exists a unique diagonal matrix that satisfies the equation. But you are right, just because we have a unique diagonal matrix, does not mean that the solution to the original equation is unique. I'll edit the question to work in these details. Thanks. | |
| Aug 14, 2012 at 1:14 | comment | added | user6976 | @Suvrit: Could you explain why if all $C_i$ pairwise commute, you can diagonalize everything? You cannot assume that $X$ commutes with $C_i$. | |
| Aug 13, 2012 at 23:15 | comment | added | Igor Rivin | Aha, I see, I was careless reading the equation! | |
| Aug 13, 2012 at 21:59 | comment | added | user6976 | @Igor: I think $X$ is supposed to be symmetric too. Then, since $C_i$ are symmetric, the sum is also symmetric. | |
| Aug 13, 2012 at 21:53 | comment | added | Igor Rivin | This may be a stupid question, but while it is obvious that $\mathcal{G}(X)$ is symmetric when $X = I,$ it is not so clear it is ever symmetric otherwise, unless the system is very special. Am I missing something? | |
| Aug 13, 2012 at 21:34 | history | asked | Suvrit | CC BY-SA 3.0 |