Timeline for Minimal perturbation of a Wigner matrix needed to produce an orthogonal top eigenvector
Current License: CC BY-SA 4.0
8 events
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Sep 13, 2020 at 21:04 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
more informative title
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Sep 13, 2020 at 21:00 | history | edited | YCor |
edited tags
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Sep 13, 2020 at 19:57 | comment | added | Daniel Li | Oh oh ok. But spacing between first and second eigenvalue is $N^{-1/6}$, right? This means before reaching that level, changing k entries contributes roughly $\sqrt{k}/N$. This is where we get $N^{5/3}$? | |
Sep 13, 2020 at 19:53 | comment | added | Carlo Beenakker | the shift of an eigenvalue by the eigenvalue spacing creates independent eigenvectors, which would then become orthogonal for the reason mentioned in the question. | |
Sep 13, 2020 at 19:48 | comment | added | Daniel Li | Thank you but how shifting largest eigenvalue by $\delta_N$ gives orthogonality? Ps. this is not supposed to be trivial because we are asked to genuinely think about it. | |
Sep 13, 2020 at 19:34 | comment | added | Carlo Beenakker | this seems quite nontrivial to me; the top eigenvector refers to the largest eigenvalue, which has a Tracy-Widom distribution with a level spacing $\delta_N\propto N^{-2/3}$ -- larger than in the bulk of the spectrum, where the spacing scales $\propto N^{-1}$. The question would seem to ask for the minimal rank $k$ of a perturbation that shifts the largest eigenvalue by $\delta_N$, to obtain an orthogonal eigenvector. | |
Sep 13, 2020 at 18:51 | comment | added | Daniel Li | Any directional or strategical suggestion is welcome. | |
Sep 13, 2020 at 18:21 | history | asked | Daniel Li | CC BY-SA 4.0 |