# Minimal perturbation of a Wigner matrix needed to produce an orthogonal top eigenvector

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.

• Any directional or strategical suggestion is welcome. Sep 13, 2020 at 18:51
• 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 19:34
• 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 19:48
• 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 19:53
• 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 19:57