Timeline for The intuition behind the Hilbert projective metric and the Perron Frobenius Theorem
Current License: CC BY-SA 3.0
12 events
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| Sep 29, 2019 at 16:17 | answer | added | M. Dus | timeline score: 2 | |
| Nov 18, 2015 at 14:16 | comment | added | Suvrit | It is a shorthand for how you would implement this iteration. Choose $x_0$, and then iterate $x' = Ax_k$, $x_{k+1} = x' / (1^Tx')$ as the iteration. That's all I meant. | |
| Nov 18, 2015 at 13:15 | comment | added | user11178 | @Suvrit I Can't understand the notation (a) $x←Ax$, (b)$ x←x/xT1.$ Can you explain? | |
| Nov 18, 2015 at 13:14 | vote | accept | CommunityBot | ||
| Nov 15, 2015 at 19:51 | answer | added | Vaughn Climenhaga | timeline score: 7 | |
| Nov 15, 2015 at 4:04 | comment | added | Suvrit | (of course, in (b) above you could also use the usual $x \gets x / \|x\|$; the point is to be able to analyze iterations of the form $x \gets Ax/\alpha(x)$, for which Hilbert's metric proves very handy. | |
| Nov 15, 2015 at 3:50 | comment | added | Suvrit | One reason why a projective metric is the right one is because the actual power iteration for computing the Perron eigenvector is (a) $x \gets Ax$, (b) $x \gets x / x^T1$. Now it is clear that if we have a projective metric, then a contraction / nonexpansion based analysis of the nonlinear map corresponding to (a) and (b) will become simpler... | |
| Nov 14, 2015 at 18:30 | history | edited | user11178 | CC BY-SA 3.0 |
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| Nov 14, 2015 at 18:22 | history | edited | user11178 | CC BY-SA 3.0 |
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| Nov 14, 2015 at 18:13 | history | edited | user11178 | CC BY-SA 3.0 |
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| Nov 14, 2015 at 18:07 | history | edited | user11178 | CC BY-SA 3.0 |
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| Nov 14, 2015 at 17:51 | history | asked | user11178 | CC BY-SA 3.0 |