proportions of invertible matrices

Let $GL(d,R)$ denote the invertible $d\times d$-matrices over a ring $R$, and$M_d(R)$ the set of all $d\times d$-matrices over $R$. When $R$ is a finite field of order $q$, commonly denoted $GF(q)$ or ${𝔽}_q$, we prefer to write simply $q$. In particular, $q$ is a power of a prime.

Proposition 1.

$$\frac{|GL(d,q)|}{|M_d(q)|}=\prod _{i=1}^d\left(1-\frac{1}{q^i}\right).$$

Proof.

The number of $d\times d$-matrices over a $GF(q)$ is $q^{d^2}$. When a matrixis invertible, its rows form a basis of the vector space $GF{(q)}^d$ and thisleads to the following formula

$$|GL(d,q)|=q^{\left(\genfrac{}{}{0pt}{}{d}{2}\right)}\prod _{i=1}^d(q^i-1).$$

(Refer to order of the general linear group. (http://planetmath.org/OrdersAndStructureOfClassicalGroups))

Now we prove the ratio holds:

$$\prod _{i=1}^d\left(1-\frac{1}{q^i}\right)=\prod _{i=1}^d\frac{q^i-1}{q^i}=\frac{1}{q^{\left(\genfrac{}{}{0pt}{}{d+1}{2}\right)}}\prod _{i=1}^d(q^i-1)=\frac{1}{q^{d^2}}{q}^{\left(\genfrac{}{}{0pt}{}{d}{2}\right)}\prod _{i=1}^d(q^i-1)=\frac{|GL(d,q)|}{|M_d(q)|}.$$

∎

Corollary 2.

As $q\mathrm{\to }\mathrm{\infty }$ with $d$ fixed, the proportion of invertible matricesto all matrices converges to 1. That is:

$$\underset{q\to \mathrm{\infty }}{lim}\frac{|GL(d,q)|}{|M_d(q)|}=1.$$

Corollary 3.

As $d\mathrm{\to }\mathrm{\infty }$ and $q$ is fixed, the proportion of invertible matrices decreases monotonically and converges towards a positive limit. Furthermore,

$$\frac{1}{4}\le 1-\frac{q^2-q+1}{q^2(q-1)}\le \prod _{i=1}^d\left(1-\frac{1}{q^i}\right)\le 1-\frac{1}{q}.$$

Proof.

By direct expansion we find

So setting $a_0=1$ and

$$a_{i+1}=a_i\sum _{j=i+1}^{\mathrm{\infty }}\frac{1}{q^j}=a_i\frac{1}{q^i(q-1)}$$

for all $i\in \mathbb{N}$, we have

$$\prod _{i=1}^{\mathrm{\infty }}\left(1-\frac{1}{q^i}\right)=\sum _{i=0}^{\mathrm{\infty }}{(-1)}^i{a}_i.$$

As $a_i\ge 0$, $a_i\ge a_{i+1}$ for all $i\in \mathbb{N}$ and $a_i\to 0$as $i\to \mathrm{\infty }$, we may use Leibniz’s theorem to conclude the alternatingseries converges. Furthermore, we may estimate the error to the $N$-th term witherror within $\pm a_{N+1}$. Using $N=2$ we have an estimate of $1-1/q$ with error$\pm \frac{1}{q^2(q-1)}$. Since $q\ge 2$ this gives $1/2$ with error $\pm 1/4$. Thuswe have at least $1/4$ chance of choosing an invertible matrix at random.∎

Remark 4.

$q=2$ is the only field size where the proportion of invertible matrices to all matrices is less than $\mathrm{1}\mathrm{/}\mathrm{2}$.

Acknowledgements: due to discussions with Wei Zhou, silverfish and mathcam.