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. 



Partitioning bases of vector spaces
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](https://en.wikipedia.org/wiki/Laplace_expansion#Laplace_expansion_of_a_determinant_by_complementary_minors) of the determinant.


