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?


$$
\begin{aligned}
\min\limits\_{A^+} \ & \|{A^+}^Tw\|\_1\\
\text{s.t.} \ & A^+A= I
\end{aligned}
$$



Is there a closed-form solution for this problem?
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
$\|B^\top w + v\|\_1$.


