Timeline for Computing signature
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 20, 2012 at 16:44 | history | edited | Ian Agol | CC BY-SA 3.0 |
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| Sep 20, 2012 at 14:28 | history | edited | Ian Agol | CC BY-SA 3.0 |
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| Sep 20, 2012 at 5:33 | comment | added | Denis Serre | OK, I see. But the vocabulary (diagonalization) suggested that you were going to compute the eigenvalue. Everything is OK now. | |
| Sep 20, 2012 at 0:58 | comment | added | Igor Rivin | This is a nice way of doing it, but I wonder what the complexity actually is, since one is likely to get horrible rational numbers. The usual way of dealing with this is to do chinese remaindering, but I am not 100% sure that this works here... | |
| Sep 19, 2012 at 21:41 | comment | added | Ian Agol | @ Suvrit and Denis: you're right, I shouldn't have used the term "orthogonal transformation", which I've elminated from the description. I guess I meant "orthogonalization" or something, but as David points out, it's just rational operations. | |
| Sep 19, 2012 at 21:39 | history | edited | Ian Agol | CC BY-SA 3.0 |
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| Sep 19, 2012 at 21:03 | comment | added | Suvrit | Thanks David, that was the clarification I sought; The name Gram-Schmidt invokes the idea of orthogonal transformations, whereas Agol is invoking the (constructive) a general congruence transform. | |
| Sep 19, 2012 at 20:37 | comment | added | David E Speyer | But Denis, we're acting on this matrix by $A \mapsto S A S^T$, not $A \mapsto S A S^{-1}$. With the $A \mapsto S A S^T$ transformation, you can get to a diagonal matrix with only rational operations, exactly as Agol says. (I'm not sure whether or not this what you call Gram-Schmidt, although I always thought of it as Gram-Schmidt.) | |
| Sep 19, 2012 at 20:17 | comment | added | Denis Serre | I confirm Suvrit's comment. Gram-Schmidt is not a mean for diagonalization. It is used to perform the QR factorization, which is a piece of a step of the QR algorithm. More generally, Abel's Theorem states that solving a general polynomial equation $P(x)=0$ is impossible in finitely many operations. Because you can associate a symmetric matrix with an arbitrary polynomial with real roots, you cannot diagonalize a symmetric real matrix by finitely many operations. | |
| Sep 19, 2012 at 19:51 | comment | added | Suvrit | Maybe I am missing something, but you cannot just diagonalize a matrix merely by orthogonal transformations (at best one gets a tridiagonal form) --- | |
| Sep 19, 2012 at 19:02 | history | answered | Ian Agol | CC BY-SA 3.0 |