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?



Vector with many non-zero coordinates
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 $\mathbb F\_q^k\leqslant \mathbb F\_q^n$.


