Partitioning bases of vector spaces [duplicate]

Question

Let $V$ be a $\mathbb{F}_p$-vector space of dimension $d$. Set $W=\bigoplus_{1\leq i\leq n} V$ and let $$S=\{w_i=(v_{i1},\dots,v_{in}): 1\leq i\leq nd\},$$ be a basis for $W$. I am wondering if the following statement holds: $S$ can be partitioned into $n$ sets, $B_1,\dots, B_n$, each of size $d$, such that for any $1\leq \ell\leq n$ the following set $$ \{\pi_\ell(w): w\in B_\ell\} $$ is a basis for $V$, where $\pi_{\ell}(x_1,\dots,x_n)=x_\ell$.

I don't know if the statement holds but I couldn't find a simple counterexample.

Answer 1

Use induction on $n$. For the step of induction, apply the following lemme to $V$ and $\bigoplus_{2\leq i\leq n} V$.

Lemma. If $w_i=(u_i,v_i)$ constitute a basis of $U\oplus V$, then the $w_i$ can be split into two groups such that the $U$-components of the first group and the $V$-components of the second group form bases in the corresponding spaces.

To prove the lemma, express the $w_i$ via some basis in $U$ and some in $V$. In the obtained $(a+b)\times (a+b)$ non-degenerate matrix, you need to find complementary minors of orders $a$ (in the first $a$ rows) and $b$ (in the last $b$ rows) both of which are non-degenerate. The existence of such minors is guaranteed by the general Laplace expansion of the determinant.