Skip to main content
8 events
when toggle format what by license comment
Sep 13, 2020 at 21:04 history edited Carlo Beenakker CC BY-SA 4.0
more informative title
Sep 13, 2020 at 21:00 history edited YCor
edited tags
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