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The instructor proposed a the following statement in the passing and suggested that we think about it (although it is not required):

For any $N \times N$ Wigner matrix, we replace $k$ entries with resampled copy (i.e. the new entries have exact same distribution as the original but are resampled independently). Suppose that $k/N^{5/3}\to\infty$. Denote top eigenvector of original and resampled matrix as $v,v'$. Show that $\mathbb{E}|\langle v,v' \rangle| \to_N0$.

Attempt: the intuition is of course that because of resampling too many entries, the two eigenvectors become less and less correlated. And in high dimensions, two independent vectors distributed uniformly on a ball tend to be orthogonal. But I struggle to explain the $N^{5/3}$ threshold or to formulate a rigorous proof.

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  • $\begingroup$ Any directional or strategical suggestion is welcome. $\endgroup$
    – Daniel Li
    Commented Sep 13, 2020 at 18:51
  • $\begingroup$ 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. $\endgroup$
    – Carlo Beenakker
    Commented Sep 13, 2020 at 19:34
  • $\begingroup$ 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. $\endgroup$
    – Daniel Li
    Commented Sep 13, 2020 at 19:48
  • $\begingroup$ the shift of an eigenvalue by the eigenvalue spacing creates independent eigenvectors, which would then become orthogonal for the reason mentioned in the question. $\endgroup$
    – Carlo Beenakker
    Commented Sep 13, 2020 at 19:53
  • $\begingroup$ 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 $N^{5/3}$? $\endgroup$
    – Daniel Li
    Commented Sep 13, 2020 at 19:57

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