Does the concatenation of linearly independent vectors give a set of linearly independent vectors?

Question

Suppose you have a set of $m$ linearly independent vectors $u_1,\dots,u_m$, as well as a larger set $S$ of $n\times m$ vectors $v_1,\dots,v_{nm}$ such that any $n$-subset of $S$ is linearly independent. Now consider an arbitrary partitioning of $S$ into $m$ sets of size $n$ each. Say this is implicit in the relabelling $v^{i}_{j}$ ($i \in[n], j \in [m]$). Does it hold that the concatenations $(u_i | v^{i}_{j})$ of column vectors (apologies for the clumsy notation) give a set of $n\times m$ linearly independent vectors?

Answer 1

No, for dimension reasons alone sometimes: if the $u_i$ span an $m$-dimensional vector space and $S$ spans an $n$-dimensional vector space, then the concatenations live in an $(m+n)$-dimensional vector space, so it's impossible for $mn$ vectors therein to be linearly independent if $m\ge3$ or $n\ge3$.

For a concrete example with $(m,n)=(3,2)$, take $\{u_1,u_2,u_3\} = \bigl\{(1,0,0),\, (0,1,0),\, (0,0,1) \bigr\}$ and $\{v_j^i\} = \bigl\{ (1,0),\, (1,1),\, (1,2),\, (1,3),\, (1,4),\, (1,5) \bigr\}$. (To be honest, I'm not sure about the indexing of your concatenations.)