Consider the following $n\times n$ random matrix $V\_{n}$ where the $(p,q)$ entry is given by
$$
V\_{n}(p,q):= \frac{1}{\sqrt{n}}\exp(2\pi i(p-1) x\_{q})
$$
where $x\_{1},x\_{2},\ldots,x\_{n}$ are iid random variables with uniform distribution on $[0,1]$.


It is not difficult to prove that $V\_{n}^{\*}V\_{n}$ has the same eigenvalues as $X\_{n}$ where
$$
X\_{n}(p,q)=\frac{\sin(n(x\_p-x\_q)/2)}{n\sin((x\_p-x\_q)/2)}.
$$
This matrix is positive definite and invertible with probability one. The minimum eigenvalue, $\lambda\_{1}(n)$, goes to zero as $n\to\infty$. I'm interested in the rate at which this eigenvalue goes to zero. Simulations suggest that 
$$
\mathbb{E}(\lambda\_1(n))\sim \exp(-\alpha n),
$$
the expected value decays exponentially. Using the Cauchy interlacing theorem I can only get the upper bound $O(\frac{1}{n^2}).$ 


Any idea of what can work here? 


Thanks!


--Gabriel



A Question on Random Matrices
It is actually more like $e^{-\sqrt n}$. Let's look at the norm of the inverse matrix. The entries are $\pm\prod\_{i:i\ne j}\frac 1{z\_j-z\_i}\sigma\_m(z\_1,\dots,z\_{j-1},z\_{j+1},\dots,z\_n)$ where $z\_k=e^{2\pi i x\_k}$ is a random point on the unit circle and $\sigma\_m$ is the $m$-th symmetric sum. Since $\log |Z-z\_j|$ has zero mean and finite variance, you expect the first factor to be $e^{O(\sqrt n)}$ most of the time. The size of second factor is essentially the size of the random polynomial $\prod\_i(z-z\_i)$ on the unit circle. The typical value at one point is $e^{O(\sqrt n)}$ and we need about $n$ points to read the true maximum (plus we have $n$ rows to serve), so my educated guess (which I can try to convert into a proof if this subexponential dependence is of any value for you) would be $e^{-\alpha \sqrt{n\log n}}$ with some $\alpha$ (with high probability; the expectation may be a bit larger because there is a chance that the rare large values will still dominate). 


I hope it makes sense but I'm in quite a hurry right now, so accept my apologies if I said some nonsense somewhere.


