# Probability of a submatrix to be full rank in a N x N Random Matrix of rank m.

Consider a random matrix $\mathbf{A} \in \mathbb{C}^{N \times N}$ of rank $m$ with $m < N$ that follows the Wishart distribution ( http://en.wikipedia.org/wiki/Wishart_distribution ).

I have a feeling that any submatrix that has $m$ columns is going to have rank $m$ with probability 1. Might be obvious to some of you in this forum but I would really like your help.

I found this work ( http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1628747 ) that gives the density function of a Wishart matrix of rank $m$, ($m < N$) but after that I dont know how to show that any submatrix will also have rank $m$.

If the rank of the matris is $1$ then obviously any 1-column vector has rank 1 (a.s.). But, lets say that the rank is $2$. If I take any two column vectors, I believe that the probability that the first is going to be a linear combination of the second has measure 0. Is this obvious?

Actually, I am not sure whether the Wishart distribution makes any difference to the problem. Probably in the case of a random matrix distributed according a continuous density function (lets say gaussian random matrix) similar statements should hold.

Thank you very much for any references, ideas, suggestions.

George

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Your matrix $A = X^T X$ where $X$ is a random $m \times N$ matrix with a continuous distribution having a density. An $m \times m$ submatrix of $A$ is $Q^T A P = (XQ)^T XP$ where $P$ and $Q$ are $N \times m$ matrices each consisting of $m$ columns of the $N \times N$ identity matrix. $XP$ and $XQ$ are $m \times m$ submatrices of $X$. With probability $1$, any $m \times m$ submatrix of $X$ has rank $m$ (its determinant is a non-constant polynomial in the matrix entries, and since the distribution of $X$ has a density the value of this polynomial is almost surely nonzero). So with probability $1$, $(XQ)^T XP$ has rank $m$.
Thank you Robert for your quick reply. I have a question on the last part of your argument. You say that since XP and XQ are rank m, then also their multiplication has rank m with probability 1. Is this obvious? Also, after reading your argument, why cant I say from the beginning that any $m \times m$ submatrix of $\mathbf{A}$ has rank $m$? For any submatrix, all the determinants are non-constant polynomials of the entries of the submatrix and since the distribution of the submatrix has a density, the value of the polynomials will be almost surely nonzero. – George Oct 23 '12 at 22:55
The point is that it's a polynomial in the entries of $X$ (which has a continuous distribution with a density), not just a polynomial in the entries of $A$ (which has a singular distribution and not a density). – Robert Israel Oct 24 '12 at 4:35
Thanks once more for your reply. Yet in the paper that I cited above, the authors derive the complex singular Wishart density (Theorem 2). What is the problem with that density and I cannot use it as you are using the gaussian density? Is there any big theorem somewhere that I could help me to clarify this issue? Also regarding the product of $\mathbf{(XQ)^T}$ and $\mathbf{XP}$, I still dont see why it should be full rank. Thank you! – George Oct 24 '12 at 6:34