Number of $k$-rank subspaces of $\mathbb{Z}_2^n$ is odd: easy proof? Let $B(n, k)$ be the number of $k$-rank subspaces of $\mathbb{Z}_2^n$. One can establish $B(n, k) = {n \choose k}_2 = \frac{F(n)}{F(k)F(n - k)}$, where $F(x) = \displaystyle\prod_{i = 1}^x (2^i - 1)$. This expression clearly implies that $B(n, k)$ is odd for all $k, n \in \mathbb{N}_0$ such that $k \leq n$.
One can equally easy establish a recurrence $B(n, k) = B(n - 1, k) + 2^{n - k}B(n - 1, k - 1)$, which again provides a straightforward inductive proof for the fact that $B(n, k)$ is odd.
Question: is there an elementary, simple, non-enumerative, non-inductive argument for the fact that $B(n, k)$ is odd for all $0 \leq k \leq n$? E.g. a concise explicit involution on $k$-rank subspaces with a single fixed point would qualify as such.
 A: Consider the involution which complements the ''free'' entries in the reduced row echelon form of the matrix representing a subspace. That is, we leave the pivots as well as things that are necessarily zero unchanged while changing everything else. This involution has a single fixed point, e.g.
$$\begin{bmatrix} 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{bmatrix}$$
for $n = 5$ and $k = 2$.
An example of an obit is
$$\begin{bmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1\end{bmatrix} \leftrightarrow \begin{bmatrix} 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0\end{bmatrix}$$
again for $n = 5$ and $k = 2$.
A: $\DeclareMathOperator\Un{Un}\DeclareMathOperator\Gr{Gr}$Let $\Un_n(K)$ be the group of upper triangular matrices over the field $K$ with diagonal $1$.
It acts on the $d$-Grassmannian $\Gr_{n,d}(K)$, with, for $0\le d\le n$, a single fixed point, namely the subspace $K^d\times\{0\}^{n-d}$.
Now suppose that $K$ is a finite field of characteristic $2$. Then $\Un_n(K)$ is a 2-group. It acts on $\Gr_{n,d}(K)$ with a single fixed point. So other orbits have even size. Hence $|\Gr_{n,d}(K)|$ is odd whenever $0\le d\le n$.
