For GOE matrix $A$, we have the following limiting distribution for eigenvalues of $A$ by $\lambda\_N\ge \lambda\_{N-1}\ge \dots \ge \lambda\_1$:



> 
> In this [paper][1], 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:
> $$
> \lim\_{N\to\infty}P(N^{2/3}(\lambda\_N-2)\le s\_1,\dots, N^{2/3}(\lambda\_{N-k+1}-2)\le s\_k)=F\_{\beta, k}(s\_1,\dots, s\_k)
> $$
> 
> 
> 


Now, if we can order the eigenvalues of $A$ by $|\sigma\_N|\ge |\sigma\_{N-1}|\ge \dots \ge|\sigma\_{1}|$, can we still have the similar results as in [The ratio of spectral edge of the GOE matrix](https://mathoverflow.net/questions/434446/the-ratio-of-spectral-edge-of-the-goe-matrix)?



> 
> *[For my notation $\sigma$: For example, the eigenvalues could be the case $1.98>1.1>...>-1.99>-2$ (where I take $\lambda\_N=1.98$, $\lambda\_{N-1}=1.1$ and $\lambda\_2=-1.99$, $\lambda\_1=-2$). Also, we order the eigenvalues by the absolute value: it becomes $|\lambda\_1|>|\lambda\_2|>|\lambda\_N|>....$. So the order changed. I denoted the new notation for $\sigma\_N=\lambda\_1, \sigma\_{N-1}=\lambda\_2, \sigma\_{N-3}=\lambda\_N$.]*
> 
> 
> 


That is
$$
\frac{|\sigma\_{2}|}{|\sigma\_{1}|}=1+O(N^{-2/3})
$$




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To prove the result in the above question, we need the joint limiting distribution also hold for $|\sigma\_{1}|$ and $|\sigma\_{2}|$. Do we still have $$|\sigma\_{2}|-|\sigma\_{1}|=O(N^{-2/3})?$$



Can we still have the order of ratio result of the two smallest eigenvalues?
The $\sigma\_i$'s are a permutation of the $\lambda\_i\in(-2,2)$, ordered by absolute value in the order $|\sigma\_N|\geq|\sigma\_{N-1}|\geq \cdots\geq|\sigma\_1|$. So $\sigma\_1$ and $\sigma\_2$ are the eigenvalues closest to zero (in the bulk of the spectrum), while $\sigma\_N$ and $\sigma\_{N-1}$ are the eigenvalues closest to $\pm 2$ (near the spectral edge). One has $|\sigma\_1|,|\sigma\_2|={\cal O}(N^{-1})$ in view of the Wigner semicircle, while $|\sigma\_N|,|\sigma\_{N-1}|=2+{\cal O}(N^{-2/3})$, in view of the Tracy-Widom law, hence


$$|\sigma\_2|-|\sigma\_1|={\cal O}(N^{-1}),$$
$$|\sigma\_N|-|\sigma\_{N-1}|={\cal O}(N^{-2/3}).$$


(The OP has this the other way around, perhaps a typo?)


