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Let $\mathbb{F}_p$ be a field with $p$ elements and $\mathbb{M}_n(\mathbb{F}_p)$ is the set of all $n \times n$ matrices over the field $\mathbb{F}_p$.

Now, we know that $|\mathbb{M}_n(\mathbb{F}_p)| = p^{n^2}$. The number of matrix with rank $0$ is $1$, namely the null matrix of order $n$. Number of matrices with rank $n$ is $$\prod_{i=0}^{n-1} (p^n-p^i),$$ namley $GL(n, \mathbb{F}_p)$.

How many elements with rank $r~ (0 \leq r \leq n)$ in $\mathbb{M}_n(\mathbb{F}_p)$?

Note: If we denote the number of elements with rank $i~ (0 \leq i \leq n)$ is $R_i$ in $\mathbb{M}_n(\mathbb{F}_p)$ then $$\sum_{i=0}^{n} R_i= p^{n^2}.$$

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    $\begingroup$ the null matrix has rank 0. The matrices with rank 1 are those shuch that the columns are all multiple of a fixed vector. There are plenty of them. $\endgroup$
    – user126154
    Feb 16, 2017 at 13:14

2 Answers 2

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Let us first count the subspaces of dimension $r$ of the underlying vector space $V = F^{n}$. I understand this to be the spans of the rows, say (you may prefer columns) of a matrix of rank $r$.

There are $$ s_{r} = \frac{(p^{n}-1) \cdots (p^{n} - p^{r-1})}{(p^{r}-1) \cdots (p^{r} - p^{r-1})} $$ such subspaces.

The number of matrices of rank $r$ is the number of surjective linear maps from $V$ to one of these subspaces (think the rows or the columns of such a matrix).

For this, I refer to this answer.

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One cal build a recursive formula for computing the number of matrices $m\times n$, with coefficient in $\mathbb F_p$, of a given rank $r$. Let $k(m,n,r)$ be such number.

Let $M$ be a matrix of rank $r$ and let $N$ be the matrix $m\times(n-1)$ obtained by deleting the last column $v$ of $M$. Then $N$ has rank either $r$ or $r-1$. If $N$ has rank $r$, then $v$ belongs to the $r$-dimensional space of $\mathbb F_p^m$ generated by the columns of $N$. If $N$ has rank $r-1$, then $v$ belongs to the complement of the $(r-1)$-dimensional space generated by the columns of $N$.

It follows that $$k(m,n,r)=k(m,n-1,r) p^r + k(m,n-1,r-1)(p^m-p^{r-1})$$

The same reasoning apply to the rows. This gives a recursive formula for $k(n,n,r)$ in terms of $k(n-1,n-1,r), k(n-1,n-1,r-1), k(n-1,n-1,r-2)$

I don't know if from that we can deduce an exact formula.

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