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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 $k$?

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    $\begingroup$ I am confused: you say the rank is $k$ when $k\geq n$, but how can the rank be larger than $n$? $\endgroup$ – Carlo Beenakker Nov 3 '15 at 7:05
  • $\begingroup$ It was a mistake. The rank was $n$ in those cases. $\endgroup$ – Jeff Nov 3 '15 at 7:19
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    $\begingroup$ 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. $\endgroup$ – Robert Israel Nov 3 '15 at 7:26
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    $\begingroup$ @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? $\endgroup$ – Jeff Nov 3 '15 at 10:22
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    $\begingroup$ 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.) $\endgroup$ – Gro-Tsen Feb 1 '16 at 14:43
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For a lot more on this, see the work by Rudelson, Vershynin, Tao, and Vu. There are nice notes on Vershynin's web page, or see Terry Tao's blog post.

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What you need is to show that the singular values are away from zero. Knowing that the singular values are invariant under transposition, you can use the following results. They in particular show that with high probability, the smallest singular value is bounded away from 0. They have special use cases for almost square matrices, which might be interesting for you.

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