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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 $m$.

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  • $\begingroup$ 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. $\endgroup$
    – George
    Commented Oct 23, 2012 at 22:55
  • $\begingroup$ 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). $\endgroup$
    – Robert Israel
    Commented Oct 24, 2012 at 4:35
  • $\begingroup$ 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! $\endgroup$
    – George
    Commented Oct 24, 2012 at 6:34
  • $\begingroup$ Just to add to my previous comment, I found in some slides from Terence Tao the following argument why a gaussian random matrix is non-singular with probability 1: a) The set of singular matrices has positive codimension and thus measure 0. b)the pdf of the gaussian ensemble is absolutely continuous. The claim follows. Part a) makes sense. Part b) not. I guess he is referring to some well-known theorem, that i happen not to know it. I guess what you are saying is that: the distribution that is calculated in Theorem 2 is not absolutely continuous since it singular. Am I correct? $\endgroup$
    – George
    Commented Oct 24, 2012 at 6:43
  • $\begingroup$ Actually regarding the product of the matrices, there is no need for any further explanation. I got it. Thanks! $\endgroup$
    – George
    Commented Oct 24, 2012 at 6:56
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There is a very nice set of notes by Roman Vershynin...

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  • $\begingroup$ Thanks Igor for this reference. Actually I had skimmed through this work but I didnt find something that is relevant to the question that i have. If I am have skipped something, please tell me so! (But it is true that this set of notes is very good.) $\endgroup$
    – George
    Commented Oct 24, 2012 at 6:36

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