Let $A\in\mathbb{R}^{m\times n}$ be a full column rank matrix. Then there exists a left inverse $A^+$ of $A$. Let $w\in \mathbb{R}^n$ be a vector. Is there a closed-form solution for the following problem?
$$ \begin{aligned} \min\limits_{A^+} \ & \|{A^+}^Tw\|_1\\ \text{s.t.} \ & A^+A= I \end{aligned} $$
If $B$ is one left inverse of $A$, then $B+X$ is a left inverse of $A$ (where $X$ is $n \times m$) iff $X A = 0$, i.e. the restriction of $X$ to $\text{Ran}(A)$ is $0$. Of course if $A$ is surjective, there is no choice: $X$ must be $0$, so let's suppose it is not. We may also assume $w \ne 0$.
We can choose $X$ so that $X^T w$ is any member of $\text{Ran}(A)^\perp = \text{Ker}(A^\top)$. So the question is reduced to: find $v \in \text{Ker}(A^\top)$ to minimize $\|B^\top w + v\|_1$.
