I have a $j$ dimensional vector space $V$ over $\Bbb{F}_2^n$ which contains the vector $\textbf{1}=(1,1,1,...,1)$.
I want to show that there are $2^{j-1} -1$ $(j-1)$-dimensional subspaces of $V$ which also contain the vector $\textbf{1}=(1,1,1,...,1)$.
I've been able to prove that there are $2^{j} -1$ total $(j-1)$-dimensional subspaces of $V$ via the usual method: \begin{equation} \frac{(2^{j}-1)(2^{j}-2)\dots(2^{j}-2^{j-2})}{(2^{j-1}-1)(2^{j-1}-2)\dots(2^{j-1}-2^{j-2})}. \end{equation}
I'm not sure how to do this with the restriction that the subspaces must contain $\textbf{1}$.
Whenever $V$ is a vector space of dimension $m$ and $x\in V$ is a non-zero vector, spanning a $1$-dimensional subspace $L$, there is a bijective correspondence between the set $S_1$ of $\ell$-dimensional subspaces of $V$ containing the vector $x$, and the set $S_2$ of $(\ell-1)$-dimensional subspaces of the $(m-1)$-dimensional quotient space $V/L$. The correspondence maps $U\in S_1$ to $U/L\in S_2$.