Let $V$ be a finite-dimensional vector space over $\mathbb{R}$. Let $A = (v_{ij})$, with $1 \leq i \leq m$ and $1 \leq j \leq n$, be an $m$ by $n$ array of elements of $V$ (so that for each $i$, $j$ as above, $v_{ij} \in V$). Let us also assume that $m \geq n$.
Consider the following operation. Select from each row (resp. column) of $A$ an element, and then calculate the rank of the $m$ (resp. $n$) chosen elements of $A$, which means the maximal number of these $m$ (resp. $n$) elements which are linearly independent. Define the row-rank (resp. column-rank) of $A$, to be the maximum of all these ranks, as you go over all possible choices of $m$ (resp. $n$) elements of $A$, with one from each row (resp. column) of $A$.
Let $J \subseteq \underline{n}:= \{1,\ldots,n\}$ be a non-empty subset. Define $A^J$ to be the $m$ by $|J|$ subarray of $A$ whose columns are precisely the columns of $A$ indexed by elements of $J$.
Hypothesis: $\operatorname{row-rank}(A^J) \geq |J|$ for any non-empty subset $J \subseteq \underline{n}$.
Question: does it follow from the setup and hypothesis above that $\operatorname{column-rank}(A) = n$?
This post is inspired by Need help to prove the existence of set of linearly independent vectors. Indeed, if one can answer positively my question, then it would imply the statement found in this linked post. Roughly speaking, the elements of the $j$-th column of $A$ belong to the vector space $S_j$. I can explain more why a positive solution of this problem gives a solution to the problem in the linked post if someone is interested.
Remark: the problem extends almost as is to matroids. Instead of assuming that the $v_{ij}$ belong to a vector space, one may instead assume that they belong to a matroid $M$. The rest of the post remains the same.
Proof by induction on $n$; assume it's true for $k=n-1$ (note that $m\geq n$ so $m\geq k$).
Step 1: remove a column (say column $n$). Then row-rank$(A^{\{1,\dots,n-1\}})\geq n-1$ by the hypothesis so by the induction hypothesis has maximal column-rank $\{e_1,\dots, e_{n-1}\}$, with $e_i$ in column $i$. Since row-rank$(A)\geq n$, by matroid properties (e.g., rank) there is an element $v_{i,j}\in A$ such that $v_{i,j}=e_n\notin cl(e_1,\dots,e_{n-1})$. Clearly, if $j=n$ then we are done, so assume $j\neq n$; wlog we may assume $j=n-1$. Now $e_1,\dots, e_n$ are independent with $e_i$ in column $i$ for $i\leq n-1$ and $e_n$ is in column $n-1$ (this column has exactly $2$ $e_i$'s).
Step 2: remove the column $n-1$ from $A$. Then by Step 1 this sub-matrix $B=A^{\{1,\dots,n-2,n\}}$ contains $e_1,\dots, e_{n-2}$ in columns $1,\dots, n-2$ respectively. Additionally, by the hypothesis row-rank$(B)\geq n-1$ so by the induction hypothesis it has maximal column-rank $\{e'_1,\dots, e'_{n-2}, e'_{n-1}\}$, with $e'_i$ in column $i$ (column $n-1$ in $B$ is column $n$ in $A$; note that these $e'_i$'s might be completely different from the $e_i$'s). If $e_{n-1}\notin cl(e'_1,\dots, e'_{n-1})$ or $e_{n}\notin cl(e'_1,\dots, e'_{n-1})$ then we are again done (we can add this element for the missing column), so assume that $e_{n-1},e_n\in cl(e'_1,\dots, e'_{n-1})$.
Step 3: Claim: there is a column $c\leq n-2$ such that $e_c\neq e'_c$ and $\{e'_1,\dots, e'_{n-2}, e'_{n-1}\}\setminus \{e'_c\}\cup \{e_c\}$ is a maximal column-rank for $B$ (i.e., it is possible to replace $e'_c$ with $e_c$ in column $c$).
Proof of claim: $rank(e_1, \dots, e_n)=n>n-1=rank(e'_1,\dots, e'_{n-1})$ so there is $e_c\notin cl(e'_1,\dots, e'_{n-2}, e'_{n-1})$, and by the assumption in Step 2 $c\neq n-1,n$. So $rank(\{e'_1,\dots, e'_{n-2}, e'_{n-1}\}\setminus \{e'_c\}\cup \{e_c\})=n-1$ by standard matroid properties.
Step 4: to conclude the proof we replace $\{e'_1,\dots, e'_{n-2}, e'_{n-1}\}$ in Step 2 with $\{e'_1,\dots, e'_{n-2}, e'_{n-1}\}\setminus \{e'_c\}\cup \{e_c\}$ in Step 3. If it still holds that $e_{n-1},e_n\in cl(e'_1,\dots, e'_{n-1})$ then we can reiterate Step 3 again and again (note that it will have to be a different $c$ each time since the $e_i$'s remain constant while the $e'_i$'s become the $e_i$'s), but at some point this process cannot continue further, since by rank considerations $cl(e_1,\dots, e_{n-2},e'_{n-1})$ cannot contain both $e_{n-1}$ and $e_n$ (i.e., in the extreme case this process stops after we replace all the $e'_i$'s with the $e_i$'s for $i\leq n-1$). At that point (when we cannot continue further) either $e_{n-1}\notin cl(e'_1,\dots, e'_{n-1})$ or $e_{n}\notin cl(e'_1,\dots, e'_{n-1})$, which means that we can add this element from column $n-1$ to form a maximal column-rank for $A$.