Let $\mathbf{A}$ be a matrix of size $N\times N$ whose elements $A\_{ij}$ (with $1\leq i,j\leq N$) are I.I.D following some distribution.


If we set set $\langle A\_{ij}\rangle=0$ and $\langle {A\_{ij}}^2\rangle=\frac{1}{N}$ (where $\langle \cdot\rangle$ denotes the average over the distribution) then we know that for $N\to \infty $ the eigenvalues of $\mathbf{A}$ will be distributed **uniformly** in the unit circle on the complex plane (which could be called the disk law really).


I know that the expectation value of the determinant of $\mathbf{A}$ should be zero ([Expected determinant of a random NxN matrix](https://mathoverflow.net/questions/13008/expected-determinant-of-a-random-nxn-matrix)) meaning that **at least one** of the eigenvalues is zero.


However my issue in understanding is mainly heuristic: even as $N$ grows to infinity, the number of eigenvalues is still **countable** and the points to cover are **uncountable**. Even after generating $N$ random points on the unit circle, no matter how big $N$ is, we still have an uncountable number of points to cover. Therefore if the eigenvalues were truly random, the probability of having an eigenvalue $\lambda\_i=0$ should be null. Therefore the determinant of $\mathbf{A}$ should not be zero.


Where is the flaw in this heuristic argument?


Edit: Speaking about the expectation value of the determinant was not a good idea. What I ask is that even if $N\to \infty$, det$(A)$ will always be non zero?



Why is the determinant of a large random matrix equal to zero? (Heuristics)
The determinant ${\rm det}\,A$ is a polynomial $P(\{a\_{nm}\})$ in the $N^2$ elements of the matrix, which we can consider as a point in $\mathbb{R}^{N^2}$. The probability distribution of the random matrix gives you some measure in this space. Your question for the probability that ${\rm det}\,A=0$ amounts to the question what is the measure of the set of points with vanishing polynomial $P=0$. This set has measure zero for the Lebesgue measure and for any continuous deformation of that measure.


So if your matrix elements are taken from a discrete set, the probability of zero determinant can be nonzero, as in the case considered in [On the probability that a random $\pm 1$-matrix is singular](https://www.ams.org/journals/jams/1995-08-01/S0894-0347-1995-1260107-2/S0894-0347-1995-1260107-2.pdf). But for a continuous, say Gaussian, distribution, the probability will be zero.


