Any unitary matrix $U$ can be diagonalized by another unitary matrix $V$,
$$U=VDV^\dagger,$$
where $D={\rm diag}(z\_1,z\_2,...,z\_N)$ is diagonal.


If $U$ is taken at random uniformly with respect to Haar measure, then $V$ and $D$ are independent and $D$ has the Weyl distribution, $P(D)\propto \prod\_{j<k}|z\_k-z\_j|^2$. I would like to know what is the space of all $V$'s. Which unitary matrices are eigenvectors of unitary matrices? What is their distribution?


On the one hand I would guess that $V$ is also uniformly distributed in the unitary group, but on the other hand this seems paradoxical. Because integration over $U$ can be decomposed as integration over $D$ and $V$ and then integration over $V$ would be the same as integration over $U$ again?


I have consulted many references about this subject, but they tend to focus on the eigenvalues.



Eigenvectors of random unitary matrices
The invariance of the Haar measure implies that the probability to draw the matrix $U$ from the unitary group is unchanged if you replace $U$ by $U\_0 U U\_0^\dagger$, with $U\_0$ an arbitrary unitary matrix. Since this conjugation changes the unitary matrix of eigenvectors from $V$ into $U\_0V$, it means that $V$ and $U\_0V$ are equally probably in the unitary group, which is another way to say that the matrix of eigenvectors is uniformly distributed in the unitary group.


If you wish to carry out the integration over $U$ by first integrating over the eigenvalues $D$ and then over the eigenvectors $V$, you will need to first specify a parameterization which uniquely fixes the eigenvectors. The uniformity of the eigenvector distribution holds for any such parameterization.


