The probability upper bound on the ratio of the eigenvalues

Question

Consider a $N\times N$ normalized matrix sample from GOE (the definition see https://www.lpthe.jussieu.fr/~leticia/TEACHING/Master2019/GOE-cuentas.pdf). If we apply the following result of the edge of the spectrum,

If we denote the $k$ largest eigenvalues by $\lambda_N,\lambda_{n-1},··· ,\lambda_{N-k+1}, $ then for Gaussian ensembles the joint distribution function of rescaled eigenvalues has the limit: $$ \lim_{N\to\infty}P(N^{2/3}(\lambda_N-2)\le s_1,\dots, N^{2/3}(\lambda_{N-k+1}-2)\le s_k)=F_{\beta, k}(s_1,\dots, s_k) $$ where $F_{\beta, k}(s_1,\dots, s_k)$ is the Tracy-Widom distribution.

then we will get the following results by continuous mapping theorem: $$\lambda_N-\lambda_{N-k+1}=O_P(N^{-2/3})$$

Now, if we ordering all eigenvalues by $|\sigma_N|\ge |\sigma_{N-1}|\ge \dots \ge |\sigma_1|$.

I would like have the similar result that for every $\epsilon>0$, there exists constants $C>0,\alpha>0$ so that $$ P\left(N^{\alpha}\left(\frac{|\sigma_N|}{|\sigma_{N-k+1}|}-1\right)\le C\right)\ge 1-\epsilon $$

I am not if we can take $\alpha=2/3$?

Answer 1

The Tracy-Widom distribution says that the level spacing near the edge of the spectrum at $\pm 2$ is of order $N^{-2/3}$, hence we may define $$|\sigma_N|\equiv 2+N^{-2/3}\delta,\;\;|\sigma_{N-k+1}|\equiv 2+N^{-2/3}\delta',$$ with $\delta,\delta'$ of order $N^0$. Now consider $$\Delta=N^{2/3}\left(\frac{|\sigma_N|}{|\sigma_{N-k+1}|}-1\right)=\frac{\delta-\delta'}{2+N^{-2/3}\delta'}=\tfrac{1}{2}(\delta-\delta')+{\cal O}(N^{-2/3}).$$ Normalisation requires that $\lim_{C\rightarrow\infty}P(\Delta\leq C)=1$, so for each $\epsilon>0$ we can find a constant $C>0$ so that $$P(\Delta\leq C)\geq 1-\epsilon,$$ which is the desired inequality. The constant $C$ will depend on $\epsilon$, but it will be independent of $N$ for large $N$.