Timeline for Signature of a quadratic form
Current License: CC BY-SA 3.0
11 events
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| Apr 27, 2017 at 18:47 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Apr 27, 2017 at 17:59 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Apr 27, 2017 at 17:35 | comment | added | Noam D. Elkies | G-S doesn't require square roots either unless you insist that your basis be orthonormal, not just orthogonal. | |
| Apr 27, 2017 at 17:12 | comment | added | Will Jagy | @Noam, right. I asked Denis for commentary. Maybe I should have said that, starting with a symmetric matrix of integers, this method or Denis's method never requires square roots, all numbers used in all the matrices are rational. | |
| Apr 27, 2017 at 17:08 | comment | added | Noam D. Elkies | Gram-Schmidt doesn't find eigenvalues either. | |
| Apr 27, 2017 at 16:45 | comment | added | Will Jagy | @Noam I don't think so. We don't ever find out the eigenvalues this way. I think it is the same as Denis Serre's answer, which he calls Gauss reduction. That way, you gradually produce $D$ and the rows of some $Q$ with $Q^T DQ = A.$ This method, we gradually produce $P$ and get $P^T AP = D.$ So, the real difference is just $P = Q^{-1},$ unless different choices led to different diagonal $D$ | |
| Apr 27, 2017 at 4:18 | comment | added | Noam D. Elkies | Is this basically Gram-Schmidt? (Gram-Schmidt does find an equivalent quadratic form that's diagonal.) | |
| Apr 27, 2017 at 2:37 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Apr 27, 2017 at 2:15 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Apr 27, 2017 at 2:12 | comment | added | Igor Rivin | Very cool! This is clearly superior over integer or rationals, a little harder to tell with floating point! | |
| Apr 27, 2017 at 1:49 | history | answered | Will Jagy | CC BY-SA 3.0 |