Rank-robustness of the parallel complexity of linear algebra problems
I think the following procedure computes a basis of the column span of an $n \times n$-matrix of rank at most $\log n$ in $\mathrm{AC}_1$.

If you have a matrix of size $n \times 2 \log n$, you can find a basis of its column span in $\mathrm{AC}_0$ by running over all subsets in parallel and checking in parallel whether a nontrivial linear combination vanishes (say over $\mathbb{F}_2$). From all sets of maximal rank you select the lexicographically smallest. 

Now given an $n \times n$-matrix of rank at most $\log n$, we can find a basis of its column span in $\mathrm{AC}_1$ as follows: Divide the columns into two halfves, compute a basis recursively. From these two bases we can compute a common basis using the algorithm from above.