Ytukyg

For simplicity, assume $n$ and $w$ are even.

Let $mathbf{M} in mathbb{F}_2^{n times m}$ be an arbitrary matrix with $operatorname{rank} mathbf{M} = w$. What is the probability that we can write
$$
mathbf{M} = begin{bmatrix} mathbf{M}_1 \ mathbf{M}_2 end{bmatrix}text,
$$
where $mathbf{M}_1, mathbf{M}_2 in mathbb{F}_2^{(n / 2) times m}$ and $operatorname{rank} mathbf{M}_1 = operatorname{rank} mathbf{M}_2 = w / 2$?

What follows is what I have worked on so far.

There are (see 1)

$$
C := binom{n}{w}_2 prod_{i = 0}^{w - 1} left(2^m - 2^iright)
$$

choices for $mathbf{M}$, where $binom{n}{w}_2$ denotes the $2$-binomial coefficient.

Given some $mathbf{M}_2$ with $operatorname{rank} mathbf{M}_2 = w / 2$, the number of ways we can "extend" it (as per problem statement) by an $mathbf{M}_1$ and get a rank-$w$ matrix is (see Introduction in 2)

$$
C_1 := 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^{w / 2 + i}right)text.
$$

Analogously to $mathbf{M}$, there are

$$
C_2 := binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^iright)
$$

choices for $mathbf{M}_2$.

Therefore, the wanted probability is

$$
frac{C_1 C_2}{C} = 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2^2 binom{n}{w}_2^{-1}text.
$$

(The products cancel out.) Any hopes to verify this? Are you aware of related work that might help?