Timeline for The probability upper bound on the ratio of the eigenvalues
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 6, 2023 at 6:07 | comment | added | Carlo Beenakker | Tracy-Widom applies to both edges of the spectrum, the upper edge near 2, and the lower edge near -2. | |
| Mar 6, 2023 at 4:45 | comment | added | Hermi | Sorry for one more question. So Tracy-Widom law is the joint limiting distribution of the $k$-largest eigenvalues. It seems that we also need this law for $k$ bottom eigenvalues? Dose this one still hold? | |
| Mar 3, 2023 at 11:53 | comment | added | Carlo Beenakker | Yes, certainly. | |
| Mar 3, 2023 at 5:38 | comment | added | Hermi | Thanks. So we can also find a constant $C'>0$ so that $P(\Delta\ge C')\ge 1-\epsilon$? | |
| Mar 3, 2023 at 5:37 | vote | accept | Hermi | ||
| Mar 1, 2023 at 18:16 | comment | added | Carlo Beenakker | no one has considered that, but it's obvious what happens: the levels near $-2$ are statistically independent from the levels near $+2$; so when you take the absolute value you just superimpose two independent Tracy-Widom distributed sequences; this means there will be no level repulsion, that is the main difference, but for the estimate you are seeking that does not matter. | |
| Mar 1, 2023 at 17:15 | comment | added | Hermi | Thanks! Can I ask if there is some relevant reference about the Tracy-Widom law for the absolute value of eigenvalue? | |
| Mar 1, 2023 at 17:03 | comment | added | Carlo Beenakker | Tracy-Widom applies to the eigenvalues near $+2$ and near $-2$, these have a spacing that is of order $N^{-2/3}$ or smaller; taking the absolute value will give you quantities near $+2$, still with a spacing of order $N^{-2/3}$ or smaller; for fixed $k$ this applies to any $|\sigma_{N-k+1}|$ at large $N$, so $|\sigma_{N-k+1}|=2+{\cal O}(N^{-2/3})$. | |
| Mar 1, 2023 at 16:50 | comment | added | Hermi | Thanks. But I am a little bit confused about if we can still apply the Tracy-Widom law of $k$ largest eigenvalue after taking the absolute value. We know that the largest eigenvalue is near 2. But why $|\sigma_N|$ is still near 2? Also, can you explain why we have $|\sigma_{N-k+1}=2+O(N^{-2/3})$? Thanks! | |
| Mar 1, 2023 at 9:28 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |