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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
    Commented Nov 3, 2015 at 7:05
  • $\begingroup$ It was a mistake. The rank was $n$ in those cases. $\endgroup$
    – Jeff
    Commented Nov 3, 2015 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
    Commented Nov 3, 2015 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
    Commented Nov 3, 2015 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
    Commented Feb 1, 2016 at 14:43

2 Answers 2

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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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  • $\begingroup$ Can you re-up the link (or add a reference)? $\endgroup$
    – ABIM
    Commented Apr 1, 2020 at 9:40
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    $\begingroup$ It now done! I did not know he changed affiliations, I must have missed this. $\endgroup$
    – Jean-Luc Bouchot
    Commented Apr 1, 2020 at 20:53

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