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What is known about the statistical independence of the eigenvectors of a real symmetric matrix with independent Gaussian entries with zero mean, and finite variance? The matrix elements are not assumed to have same variance.

I see some results for Wigner matrices in literature, where the entries are i.i.d. standard Gaussian (except diagonal) - though even in this case, whether the eigenvectors are in fact statistically independent is not entirely clear to me (though I suspect that to be the case for Wigner matrices).

So, does there exist results regarding statistical independence of eigenvectors of random real symmetric matrices with non-identical, but statistically independent Gaussian entries? Any references for this in literature would be helpful.

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A precise answer exists for the Gaussian Orthogonal Ensemble (all variances the same): then the eigenvectors are the columns of an orthogonal matrix which is uniformly distributed with the Haar measure; they are therefore not independent --- they cannot be because they must be orthogonal to one another.

In the limit $n\rightarrow\infty$ of large $n\times n$ matrices $M$, any finite subset of elements of the eigenvectors does become statistically independent [1] with a Gaussian distribution (mean zero, variance $1/n$). This "central limit" result does not require that the elements of $M$ have identical distributions.

[1] How Many Entries of a Typical Orthogonal Matrix Can Be Approximated by Independent Normals? (2006).

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  • $\begingroup$ 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? $\endgroup$ – Brendan McKay Jul 21 '18 at 12:56
  • $\begingroup$ @BrendanMcKay --- yes, Gram-Schmidt works provided that you start from independent vectors with Gaussian elements. $\endgroup$ – Carlo Beenakker Jul 21 '18 at 13:13
  • $\begingroup$ @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? $\endgroup$ – user125930 Jul 21 '18 at 17:58
  • $\begingroup$ I added a reference. $\endgroup$ – Carlo Beenakker Jul 21 '18 at 18:13
  • $\begingroup$ 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! $\endgroup$ – user125930 Jul 22 '18 at 0:59

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