Let $V$ be a k-dimensional subspace of $(\mathbb{F}\_2)^n$, such that vector $\vec{j}=(1,1,...,1) \in V$.

Standard linear algebra shows that it is possible to find a $(k-1)$-dimensional space $W$ such that $\langle \vec{j},W\rangle=V$. However, this choice is not unique. 

Is there any "canonical" choice for $W$, i.e. one that does not depend on making certain positions special, like there is the orthogonal complement in the $\mathbb{R}$-case?

In case it matters, my parameters are $n=2058$, $k=52$ and all vectors in $V$ have even weight (so orthogonal complement is pointless).

Edit: clarification of what I mean with "without making certain positions special". Consider the action of $S\_n$ as a permutation group on the positions of the vector space, and let $G\le S\_n$ be the stabilizer of $V$ in that action. Then $G$ should also stabilize $W$, as in the orthogonal complement case. (Note that $G$ automatically stabilizes $\vec{j}$.) I'm unsure if such spaces exist in general, so a feasible algorithm to find one is also appreciated.