TITLE: Vector with many non-zero coordinates
QUESTION [6 upvotes]: Given finite field $\mathbb{F}_q$, positive integers $n$ and $k<n$. Given $k$-dimensional subspace $X$ of $\mathbb{F}_q^n$, for which $m=m(q,k,n)$ may we find for sure a vector in $X$ with at least $m$ non-zero coordinates?

REPLY [6 votes]: By Gauss' method, you may find a basis $v_1,\dots,v_k$ of $X$ such that $s(v_1)<s(v_2)<\dots<s(v_k)$, where $s(v)$ is the number of the first nonzero entry in $v$. Using these vectors, you may inductively construct a vector with nonzero entries at positions $s(v_1)$, \dots, $s(v_k)$, so at least $k$ nonzero coordinates are always achievable. Clearly, $m=k$ is the maximal value, in view of the example $\mathbb F_q^k\leqslant \mathbb F_q^n$.