Probability of a random $n \times n$ matrix over $\mathbb F_2$ being nonsingular

Solution 1:

The probability that a matrix over $\mathbb{F}_2$ has determinant $1$ is

$$\frac{|SL_n(\mathbb{F}_2)|}{|\mathcal{M}_n(\mathbb{F}_2)|} = \frac{\displaystyle\prod_{k=0}^{n-1}(2^n - 2^k)}{2^{n^2}} = \prod_{k=1}^{n} \Big(1 - \frac{1}{2^k} \Big).$$

More generally, the probability that a matrix over $\mathbb{F}_q$ has determinant 1 is

$$\frac{|SL_n(\mathbb{F}_q)|}{|\mathcal{M}_n(\mathbb{F}_q)|} = \frac{\frac{1}{q-1}\displaystyle\prod_{k=0}^{n-1}(q^n - q^k)}{q^{n^2}} = \frac{1}{q-1}\prod_{k=1}^{n} \Big(1 - \frac{1}{q^k} \Big).$$

For an explanation on how to calculate $|SL_n(\mathbb{F}_q)|,$ see this note by Gabe Cunningham.

As this is too long for a comment, I've posted it as an addendum to my original answer. The probability does indeed converge to a positive limit as $n\rightarrow \infty.$ Observe,

$$\begin{align} \log \left(\prod_{n \in \mathbb{Z}_+}(1 - \frac{1}{q^n}) \right) &= \sum_{n \in \mathbb{Z}_+} \log \Big(1 - \frac{1}{2^n} \Big) \\ &= -\sum_{n \in \mathbb{Z}_+} \sum_{k \in \mathbb{Z}_+} \frac{1}{k} \left( \frac{1}{q^n} \right)^k \\ &= -\sum_{k \in \mathbb{Z}_+} \frac{1}{k} \left(\sum_{n \in \mathbb{Z}_+} \Big(\frac{1}{q^{k}} \Big)^n \right) \\ &= -\sum_{k \in \mathbb{Z}_+} \frac{1}{k} \left(\frac{q^{-{k}}}{1 - q^{-{k}}} \right) \\ &= -\sum_{k \in \mathbb{Z}_+} \frac{1}{k} \left(\frac{1}{q^{k} -1} \right) \\ &\gt -q\sum_{k \in \mathbb{Z}_+} \frac{1}{k} \Big(\frac{1}{q} \Big)^k \\ &= \log(1/q^q). \end{align} $$

It follows $\displaystyle\prod_{n \in \mathbb{Z}_+} \Big(1 - \frac{1}{q^n} \Big) > 1/q^q$, so the product converges.