Consider a $d$ by $d$ array of vectors $(v_{ij})$, where $1 \leq i,j \leq d$ and each $v_{ij} \in \mathbb{R}^n$, where $n \geq d$. We say that such an array $(v_{ij})$ has maximal row-rank (respectively maximal column-rank), if one can choose from each row (respectively from each column) of this array one vector, such that the $d$ chosen vectors are linearly independent over $\mathbb{R}$.
One question I have is the following. If $(v_{ij})$ has maximal row-rank such that any $k$ by $k$ subarray of $(v_{ij})$ has also maximal row-rank, for all $1 \leq k \leq d$, does it follow that $(v_{ij})$ has maximal column-rank? Or is there perhaps a counterexample?
Edit $1$: someone (T.C.) told me that this generalizes to matroids. Indeed, let $M$ be a matroid of rank $n$. Consider a $d$ by $d$ array $(v_{ij})$ of elements of $M$, for $1 \leq i,j \leq d$ where $1 \leq d \leq n$. We call such an array of maximal row-rank (resp. column-rank), if one can choose an element from each row (resp. column), such that the $d$ chosen elements are linearly independent.
Assume given a $d$ by $d$ array $A=(v_{ij})$ of elements in a given matroid $M$ of rank $n$ (with $1 \leq d \leq n$) such that each $k$ by $k$ subarray of $A$ has maximal row-rank, for $1 \leq k \leq d$, does it follow that $A$ has maximal column-rank?
I'll give it a try, hopefully I didn't miss anything:
By induction, assume it's true for $k=n-1$.
Step 1: remove a row and column (say row & column $n$). Then this sub-matrix has maximal row-rank so has maximal column-rank $\{e_1,\dots, e_{n-1}\}$, with $e_i$ in $v_{r_i,i}$. Since $A$ has maximal row-rank, 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 $n-1$ and a row not in $\{r_1,\dots, r_{n-2}\}$ (e.g., row $n$) from $A$. Then by Step 1 this is an $n-1\times n-1$ sub-matrix $B$ that contains $e_1,\dots, e_{n-2}$ in $v_{r_1,1},\dots v_{r_{n-2},n-2}$ respectively. Additionally, by the assumption it has a maximal row-rank and therefore 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, which is why this question is not so trivial). 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: I claim that 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., that I can 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 $c\neq n-1,n$. 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 do Step 3 again and so forth (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 $e_n$. 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 $A$.