Consider $A$, a random binary matrix of zeros and ones in $\mathbb{R}^{{M\times N}}$, and $M>N$. We assume that $P(a\_{i,j}=0)=P(a\_{i,j}=1)=0.5$ (although I appreciate any advice on the case of non-even probabilities). Are there any results that provide a lower bound (and maybe an interesting upper bound) on the eigenvalues of $A^TA$? The reference ring is certainly $\mathbb{R}$. 


Thanks in advance!



Bounds on the eigenvalues of a random binary matrix
The distribution of eigenvalues will follow the [Marchenko-Pastur distribution](http://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution), scaled appropriately.


Since the mean entry will be equal to $0.5$, then the largest eigenvalue will be (approximately) equal to $0.25MN$, and the corresponding eigenvector will be the vector of all $1$'s (again, approximately). The rest of the eigenvalues will follow Marchenko-Pastur. In this case the random variables in the matrix have variance $\sigma^2=0.25$, which means that the next largest eigenvalue will be $0.25(1+\sqrt{N/M})^2M$ and the smallest will be
$0.25(1-\sqrt{N/M})^2M$. In other words, the range of eigenvalues other than the first will be
$$\bigl[0.25(\sqrt{M}-\sqrt{N})^2,0.25(\sqrt{M}+\sqrt{N})^2\bigr].$$


You can similarly compute the spectrum for other probablitites besides $0.5$


