I have been struggling with a probably easy question but I cannot prove it, so any insights would be really helpful.
If I have a random matrix $X \in GOE(N)$, namely from the Gaussian orthogonal ensemble, we know that it can be written as $X=ULU^{T}$, where $U$ is an orthogonal matrix consisting of the eigenvectors of $X$ and $L$ is diagonal with the eigenvalues as entries. Can I say that this particular $U$ is Haar distributed?
I think that the answer is related to the $GOE$ being translation-invariant and the fact that the Haar measure is the unique measure on $\mathbb{O}(N)$, with this property, but I cannot write a proper proof.
Careful: $U$ is not unique. In particular, you can always replace $U$ by $-U$. So you could arrange e.g. that $U_{11} \ge 0$, in which case it is certainly not Haar distributed. But maybe the question is whether it is possible to do the diagonalization in such a way that $U$ is Haar distributed.
I think the way to proceed is as follows: let $V$ be a Haar distributed orthogonal matrix independent of $X = U L U^T$, and consider $Y = V U L U^T V^T = (VU) L (VU)^T$. Then $Y$ is again in the GOE, and $VU$ is Haar distributed.