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.*



Number of $k$-rank subspaces of $\mathbb{Z}_2^n$ is odd: easy proof?
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 $k = 2$.


