1

We know that \begin{align*} \frac{|GL_n(\mathbb{F}_q)|}{|\mathcal{M}_N(\mathbb{F}_q)|} &= \frac{\prod \limits_{k=0}^{n-1} (q^n - q^k)}{q^{n^2}} \\ &= \prod_{k=1}^n \left( 1-q^{-k} \right) \\ &= \left(\frac{1}{q}, \frac{1}{q} \right)_n, \end{align*} where the last expression denotes the q-Pockhammer symbol.

How can we generalize this to $m\times n$ matrices. I mean what is the ratio of matrices which have full rank to total number of matrices.

Thanks.

Shweta Aggrawal
  • 5,659
  • There is a much simpler question here: "How can I count the number of full rank matrices?". Have you tried to do this? You can WLOG assume $n > m$ and work it out. – xxxxxxxxx Feb 28 '21 at 06:28

2 Answers2

2

For an $m \times n$ matrix of rank $m$ (where $m \le n$), there are $q^n-1$ possibilities for the first row, then $q^n - q^1$ for the second, etc., or $\prod_{j=0}^{m-1} (q^n - q^j)$ in all.

Robert Israel
  • 470,583
0

I don't think it will be too hard to write down the numerator and denominator, because the matrices with full rank will have $n$ linearly independent columns, so the first column can be any nonzero vector, there are $q^2$ linear combinations of the first two columns, etc... and that gives $(q^m-1)q^m-q)\dots(q^m-q^{n-1})$.

Meanwhile the denominator is just $q^{mn}$.

  • Depending on what the first and second column are, there may not be $q^{2}$ linear combinations of the first two columns (for example, the first column may be zero, or the second column could be a multiple of the first); you also don't seem to count the possibilities for the second column? – xxxxxxxxx Feb 28 '21 at 06:33
  • Actually, the product that you give makes very little sense. Does the second power go down by 2 each time? What happens if $n > m$? – xxxxxxxxx Feb 28 '21 at 06:35
  • Oh, that should have been a $q$. Typo. @MorganRodgers Might you have figured that out? The first column can't be zero. That's the reason for the $-1$! –  Feb 28 '21 at 06:40
  • @MorganRodgers I thought we were counting full rank matrices. –  Feb 28 '21 at 06:43
  • Full row rank? Full column rank? I suppose it is not clear. It makes no difference in the count though, just the reasoning. If we just mean rank is $\mathrm{max}(m,n)$ then it could have a zero column (as long as there are more columns than rows); that is what I meant to say. – xxxxxxxxx Feb 28 '21 at 06:47
  • Ok. I see your point. @MorganRodgers And thanks for the input. What I did will work, I think, if $m\ge n$. Heck, we would get zero, otherwise. So there's some ambiguity. –  Feb 28 '21 at 06:54