# Statistical independence of eigenvectors of real symmetric Gaussian random matrices

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.

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  with a Gaussian distribution (mean zero, variance $1/n$). This "central limit" result does not require that the elements of $M$ have identical distributions.
• @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 17:58
• 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 22, 2018 at 0:59