Timeline for Statistical independence of eigenvectors of real symmetric Gaussian random matrices
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 22, 2018 at 20:03 | comment | added | Carlo Beenakker | Heuristically, the uniformity of the eigenvectors in the orthogonal group appears once the random matrix $M$ does not have a preferential basis. So it will fail if you choose a banded random matrix (all elements zero except those close to the diagonal in a certain basis), but the independence of the matrix elements does not matter in the large-$n$ limit. | |
| Jul 22, 2018 at 6:49 | vote | accept | user125930 | ||
| Jul 22, 2018 at 6:49 | comment | added | user125930 | Very interesting. Thanks for the explanations. I will mark the question as answered. I wonder if there any similar results for real symmetric random Gaussian matrices when the elements are just pairwise uncorrelated and not necessarily independent? | |
| Jul 22, 2018 at 6:05 | comment | added | Carlo Beenakker | you cannot apply the Gaussian approximation to the full eigenvector, only to a subset of elements (the cited reference says a subset of order $\sqrt n$ elements); so for $n\rightarrow\infty$ this does not constrain the number of degrees of freedom. | |
| Jul 22, 2018 at 0:59 | comment | added | user125930 | The "central limit" result for real symmetric $M$ with non-identical, independent Gaussian entries seems intuitive, except it also seems to lead to an issue: In the limit as $n \rightarrow \infty$, the matrix $M$ still has $\frac{n(n+1)}{2}$ degrees of freedom (the variances) when the variables are not identical. If each element of each eigenvector is $\mathcal{N}(0, 1/\sqrt{n})$, then the eigen decomposition of $M$ would have only $n$ degrees of freedom, right? How can we reconcile with this? Can you explain? Also the reference seems to deal with the case of iid normal entries of $M$. Thanks! | |
| Jul 21, 2018 at 18:13 | comment | added | Carlo Beenakker | I added a reference. | |
| Jul 21, 2018 at 18:12 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
added 192 characters in body
|
| Jul 21, 2018 at 17:58 | comment | added | user125930 | @CarloBeenakker -- Thanks for the answer. So for finite $n$ with non-identical variances, the eigenvectors cannot be statistically independent since they need to be orthogonal due to the real symmetry of the random matrix. The large limit case with $n \rightarrow \infty$ is still quite interesting - is there a reference for this CLT independence result in literature, or is it trivial to prove? | |
| Jul 21, 2018 at 13:13 | comment | added | Carlo Beenakker | @BrendanMcKay --- yes, Gram-Schmidt works provided that you start from independent vectors with Gaussian elements. | |
| Jul 21, 2018 at 12:56 | comment | added | Brendan McKay | As you say, orthogonality rules out independence. But it could still be that each pair of eigenvectors looks like two independent vectors modified by projecting one onto the orthogonal complement of the other. Does that follow? Most generally, does the full set of eigenvectors look like the result of applying the Gram-Schmidt process to a set of independent vectors? | |
| Jul 21, 2018 at 11:25 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
added 122 characters in body
|
| Jul 21, 2018 at 10:17 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |