# Rank of a fat random matrix

Let $\mathbf{R} \in \mathbb{C}^{~n \times k}$ with $n \leq k$ be a random matrix, whose entries are i.i.d zero mean random variables with circularly symmetric Normal distribution. Two questions:

(1) What is the rank of $\mathbf{R}$? I guess for $k \to \infty$, the matrix would become a full-rank matrix. Is it true?

(2) Moreover, I simulated such matrix for many times, and for all values of $k \geq n$, the rank was $n$ even in $k=n$. Is there any closed-form expression to show the probability of not having a full-rank matrix as a function of $n$ and $k$?

• I am confused: you say the rank is $k$ when $k\geq n$, but how can the rank be larger than $n$? – Carlo Beenakker Nov 3 '15 at 7:05
• It was a mistake. The rank was $n$ in those cases. – Jeff Nov 3 '15 at 7:19
• For any absolutely continuous distribution of random variables $X_1, \ldots, X_m$,, any nonconstant polynomial in the $X_j$ is a.s. nonzero. Apply that to the determinant of an $n \times n$ submatrix. – Robert Israel Nov 3 '15 at 7:26
• @Robert, I see. So it means that the rank of any $n \times m$ random matrix with i.i.d. entries taken from an absolutely continuous distribution, is $\min (n,m)$, right? – Jeff Nov 3 '15 at 10:22
• To put Robert Israel's answer differently, non-full-rank matrices are a (singular) algebraic subvariety of $\mathbb{C}^{n\times k}$ that is not the full space, so it has codimension at least $1$ and Lebesgue measure zero: with probability $1$ a random matrix has full rank, you don't need to take a limit. (Over finite fields, of course, things would be different.) – Gro-Tsen Feb 1 '16 at 14:43