Why is the determinant of a large random matrix equal to zero? (Heuristics) [closed]

Question

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) 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?

Answer 1

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. But for a continuous, say Gaussian, distribution, the probability will be zero.