most recent 30 from http://mathoverflow.net 2013-05-24T03:59:43Z http://mathoverflow.net/feeds/question/110467 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110467/probability-of-a-submatrix-to-be-full-rank-in-a-n-x-n-random-matrix-of-rank-m George 2012-10-23T20:32:10Z 2012-10-24T03:05:16Z

<p>Consider a random matrix $\mathbf{A} \in \mathbb{C}^{N \times N}$ of rank $m$ with $m &lt; N$ that follows the Wishart distribution ( <a href="http://en.wikipedia.org/wiki/Wishart_distribution" rel="nofollow">http://en.wikipedia.org/wiki/Wishart_distribution</a> ).</p> <p>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. </p> <p>I found this work ( <a href="http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&amp;arnumber=1628747" rel="nofollow">http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&amp;arnumber=1628747</a> ) that gives the density function of a Wishart matrix of rank $m$, ($m &lt; N$) but after that I dont know how to show that any submatrix will also have rank $m$.</p> <p>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? </p> <p>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. </p> <p>Thank you very much for any references, ideas, suggestions.</p> <p>George</p>

http://mathoverflow.net/questions/110467/probability-of-a-submatrix-to-be-full-rank-in-a-n-x-n-random-matrix-of-rank-m/110474#110474 Robert Israel 2012-10-23T20:58:22Z 2012-10-23T20:58:22Z

<p>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 $m$.</p>

http://mathoverflow.net/questions/110467/probability-of-a-submatrix-to-be-full-rank-in-a-n-x-n-random-matrix-of-rank-m/110503#110503 Igor Rivin 2012-10-24T02:50:55Z 2012-10-24T02:50:55Z

<p>There is a <a href="http://www-personal.umich.edu/~romanv/papers/non-asymptotic-rmt-plain.pdf" rel="nofollow">very nice set of notes by Roman Vershynin...</a></p>