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?
$$ \begin{aligned} \min\limits_{A^+} \ & \|{A^+}^Tw\|_1\\ \text{s.t.} \ & A^+A= I \end{aligned} $$
I don't think that there is an explicit solution, but you can translate this problem into a linear program.
You can parameterize $A^+$ by means of the cokernel of $A$. Let $$ A^+ = (A^TA)^{-1}A^T + XB $$ where $X$ is an $n\times (m-n)$ matrix and $B$ is an $(m-n) \times m$ that contains a basis of the cokernel of $A$ as its rows. Let $y\in\mathbb{R}^m$ and $$ y= ((A^TA)^{-1}A^T + XB)^Tw $$ Introduce $z\in\mathbb{R}^m.$ Then the linear program is $$ \min \sum_{i=1}^m z_i $$ such that \begin{eqnarray} y - B^TX^Tw & = & A(A^TA)^{-1}w \\ y_i & \leq & z_i \\ -y_i & \leq & z_i \end{eqnarray} (The variables are $X$, $y$ and $z$)