Singular values of random uniform matrix Suppose  $X \in \mathbb{R}^{N \times M}$ with elements sampled i.i.d. from $\mathcal{U}(-\sigma, \sigma)$.
I would like to find the marginal distribution of the unordered singular values of $X$.
The case for $X \sim \mathcal{N}(0, \sigma)$ has been extensively studied, but i can't seem to find any work on the uniform distribution case.
Or equivalently the distribution of the eigenvalues of $X^TX$. The distribution of the $X^TX$ seems to follow some gaussian with mean zero and a small gaussian in the tail as observed in some simulations. This leads to the eigenvalues not being distributed as gaussian i.i.d.

 A: Rescale $Y=(\sqrt{3}/\sigma)X$, so that the matrix elements of $Y$ have uniform distribution with zero mean and unit variance.
In the limit $N,M\rightarrow\infty$ at fixed $M/N=r\in(0,1]$ the $M$ eigenvalues $\lambda_n$ of $N^{-1}Y^{\rm T}Y$ have the Marcenko-Pastur distribution,
$$\rho(\lambda)=\frac{1}{2\pi \lambda r}\sqrt{(\lambda_+-\lambda)(\lambda-\lambda_-)},\;\;\lambda_-<\lambda<\lambda_+,\;\;\lambda_\pm=(1\pm\sqrt r)^2.$$
For a proof, see for example these notes.
The plot below is a test for $M=10^3$, $N=10^4$: the histogram shows the eigenvalue distribution for a randomly generated matrix with uniformly i.i.d. matrix elements, the curve is the Marcenko-Pastur distribution, which as you can see agrees very nicely.
(No idea why the histogram in the OP is so different.)


The OP has asked in the comment for non-asymptotic results. I don't think there are exact closed-form expressions for any $N,M$, but the large-$N$,$M$ asymptotics is already reached quite accurately for moderately large values of $N,M$. In the plot below I show the case $M=10,N=20$, which is already quite close to the Marcenko-Pastur limit.

And even $M=5,N=10$ is not too bad...

