Consider a $N\times N$ normalized matrix sample from GOE (the definition see [https://www.lpthe.jussieu.fr/~leticia/TEACHING/Master2019/GOE-cuentas.pdf](https://www.lpthe.jussieu.fr/%7Eleticia/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 $\alpha=2/3$?



The probability upper bound on the ratio of the eigenvalues
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 $N$.


