Following these two questions: Can we still have the order of ratio result of the two smallest eigenvalues? and The ratio of spectral edge of the GOE matrix.

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: $$\lambda_N-\lambda_{N-k+1}=O_P(N^{-2/3})$$

(See the related questions: Can we apply the continuous mapping theorem for the limiting joint distribution of the Tracy-Widom law?, Does there exist a constant $c>0$ such that $$ P(N^{2/3}(\lambda_N-\lambda_{N-1})\ge c)\ge 1-\epsilon? $$, Can we get that $ P(N^{2/3}(\lambda_N-\lambda_{N-1})\le c)\ge 1-\epsilon$?)

Now, if we ordering all eigenvalues after taking the absolute value by $|\sigma_N|\ge |\sigma_{N-1}|\ge \dots \ge |\sigma_1|$. Can we have a similar result about the joint limiting distribution of $|\sigma_N|\ge |\sigma_{N-1}|\ge \dots \ge |\sigma_1|$? Moreover, 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 $\alpha=2/3$?

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 $\alpha=2/3$?