Eigenvalue distribution of a random matrix

Is there any closed form distribution formula for the distribution of the eigenvalues of $$\mathbf{X}^\mathrm{H}\mathbf{X}$$ where the entries of $$\mathbf{X}$$ are independent Gaussian random variables with different variances $$\sigma_{ij}^2$$. Moreover, in the diagonal factorization $$\mathbf{X}^\mathrm{H}\mathbf{X}=\mathbf{U}\mathbf{\Lambda}\mathbf{U}^{\mathrm{H}}$$, the unitary matrix $$\mathbf{U}$$ has Haar distribution? Is it independent of $$\mathbf{\Lambda}$$?

• no, $U$ is not uniformly distributed in the unitary group (no Haar distribution) and eigenvectors and eigenvalues are not independent. There is not much more to say in this completely general case. Feb 28, 2020 at 9:19
• Is there any result for non-identical cases? Feb 28, 2020 at 10:42

More of a comment to indicate how hopeless this is: Consider the edge case of $$\sigma_{ij} = \lambda_i \delta_{ij}.$$ The unitary matrix then is the identity matrix, so is quite far from Haar distributed. The eigenvalues are distributed normally with the covariance matrix $$\mathbf{X},$$ also quite far from the "standard" case. An interesting question is whether there is a less trivial case when there is a discrete set of $$\mathbf{U}.$$
The eigenvalue distribution function is known if $$\sigma^2_{ij}=w_i$$ depends only on the row index, see for example arXiv:1310.2467
$$P(x_1,x_2,\ldots x_n)\propto\prod_{i where $$W={\rm diag}\,(w_1,w_2,\ldots w_n)$$ and $$\beta=1$$ for real matrix elements, $$\beta=2$$ for complex matrix elements.
The measure $$dU$$ is the Haar measure in $${\rm O}(n)$$ for $$\beta=1$$ and in $${\rm U}(n)$$ for $$\beta=2$$. This is the socalled Harish-Chandra-Itzykson-Zuber integral, which has a closed-form expression for $$\beta=2$$ (it’s a ratio of determinants). Note that the integral depends only on the eigenvalues $$x_i$$ of $$XX^H$$. For $$W$$ proportional to the identity matrix we recover the usual Wishart distribution.