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Let $M$ be an $m \times n$ tall ($m > n$) matrix over $\mbox{GF}(2)$. Let $M^{'}$ be an $r \times n$ submatrix of $M$ (where $r \leq m$) whose rows are randomly selected from $M$. Is it possible that the $M^{'}$ is full row-rank?

If it is, why?

If it is not, how about the case where $r < m$?

I have run some experiments on Matlab. In experiments, I first randomly generated $10^{6}$ matrices over $\mbox{GF}(2)$. Each matrix has $m = 288$ rows and $n = 216$ columns. Turn out they all have the rank of $216$. Then, I randomly generated $10^{6}$ matrices over GF(2). Each matrix has $m = 84$ rows and $n = 216$ columns. Turn out they all have the rank of $84$. So I am very confused now.

Rodrigo de Azevedo
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S. Y.
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1 Answers1

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Note that a matrix can only have full row-rank if the number of rows it has is greater than or equal to the number of columns. So, $M'$ can only have full row-rank if $r \leq n$.

That being said: because $M$ has full column-rank (according to your title), there is necessarily at least one $n \times n$ submatrix of $M$ that is invertible (full row and column rank), and every submatrix taken by choosing rows from the invertible submatrix will have full row rank.

As I note here, if both matrices are randomly selected, then the probability that either fails to have full row-rank is smaller than $10^{-20}$, which in practice means that both matrices will "almost always" have full rank.

Ben Grossmann
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  • Thanks for answering. Do you mean I have to find out which $n \times n$ submatrices are invertible first, and then select $M{'}$ from these $n \times n$ submatrices? – S. Y. Oct 19 '19 at 07:37
  • That wasn't quite what I was saying, since I'm really not sure what you're trying to do with your matrix. Putting that aside for the moment: what you describe is not only way to get full-rank $r \times n$ submatrices, but it is true that every full-rank $r \times n$ submatrix can be attained in this way. – Ben Grossmann Oct 19 '19 at 07:50
  • For sufficiently large matrices, it is notable that there is a very high probability that your randomly selected $m \times n$ matrix will have full-rank, and that your randomly selected $r \times n$ submatrix will also have full rank. Are you interested in finding out what these probabilities are? – Ben Grossmann Oct 19 '19 at 07:54
  • Yes. I have been reading a paper involving this problem. The authors said that $M$ is a full column-rank matrix, and they claimed that the $M'$ they arbitrarily selected from $M$ is a full row-rank submatrix. That confused me.

    Is it because $M$ is over GF(2)?

    If you are interested in this paper (maybe not), click [https://www-users.cs.umn.edu/~tianhe/Papers/WEBee.pdf] and read section 4.

     – S. Y. Oct 19 '19 at 08:07
  • I'll take a look. I suspect that the authors mean that $M'$ is very likely to be a full row-rank submatrix. Notably, even if $M$ is a matrix over an infinite field like $\Bbb R$, it is possible to obtain an $m \times n$ matrix $M$ that has full row-rank for which most $r \times n$ matrices fail to have full row-rank. – Ben Grossmann Oct 19 '19 at 08:11
  • Now that I have read the section you're talking about, it seems that they are just selecting $M'$ randomly, and then discuss what happens if $M'$ has full row-rank (page 5, right column, second paragraph: "Given $M'$ is now a full row-rank matrix..."). Presumably, one could in practice select a random submatrix $M'$ and then check the rank. If the low-probability event that $M'$ fails to have full rank occurs, we simply choose another random submatrix. – Ben Grossmann Oct 19 '19 at 08:23
  • I have asked my own question about your post. Hopefully we get a useful answer here. If I find the time and energy I might answer it myself. – Ben Grossmann Oct 19 '19 at 08:40
  • I got an answer to my question. As I suspected, the probability that either matrix fails to have full row-rank is small enough that it can be considered negligible in practice. – Ben Grossmann Oct 19 '19 at 09:01
  • Thank you! Your answer is great! – S. Y. Oct 19 '19 at 13:55
  • @ShuMinYao You're welcome. If you're satisfied with my answer, please click the $\checkmark$ to the left of my post to "accept" the answer and mark your post as resolved. – Ben Grossmann Oct 20 '19 at 09:20