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 $k$ 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 $k$?


1 Answer 1


There are theorems for the lower bound of the smallest singular value of a matrix. If you want the smallest singular value to be exactly zero, then the probability to that is zero.

BUT - Most likely you are interested in the numerical rank. The bound below is taken from Z. Chen and J.J. Dongarra, "Condition numbers of Gaussian random matrices", SIAM J. Matrix Anal. Appl. 27(3), 2005.

Here is a theorem:

Suppose $G$ is a real $l \times k$ matrix whose entries are i.i.d Gaussian random variables of zero mean and unit variance, and $\beta$ is a positive real number, such that:

$1-\frac{1}{\sqrt{2\pi(l-k+1)}}\left(\frac{e}{(l-k+1)\beta}\right)^{l-k+1} >0$

Then, the smallest singular value of $G$ is at least $\left( \sqrt{l}\beta\right)^{-1}$.

With this theorem, you can set the a lower bound, such as the machine precision, say $10^{-16}$ and this will answer the question.. at least for the case of a rectangular matrix with one dimension is much bigger than the other... it is not very helpful for square matrices and $10^{-16}$ but it is good for your first question.

Another reference that might be helpful is by A. Litvak et. al "Smallest singular value of random matrices and geometry of random polytopes" Advances in Mathematics, 195, 2005.

You can find there bounds for square matrices as well. It might give you a better bound, but the numbers are much more difficult to substitute.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .