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



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


