  Let $U$ and $V$ be vector spaces over a field $\mathbb{K}$, and $\mathcal{S}$
be an $n$-dimensional linear subspace of $\mathcal{L}(U,V)$. The space
$\mathcal{S}$ is called algebraically reflexive whenever it contains every
linear map $g : U \rightarrow V$ such that, for all $x \in U$, there exists $f
\in \mathcal{S}$ with $g(x)=f(x)$. A theorem of Meshulam and \v{S}emrl states
that if $\mathcal{S}$ is not algebraically reflexive then it contains a
non-zero operator $f$ of rank at most $2n-2$, provided that $\mathbb{K}$ has
more than $n+2$ elements. In this article, we prove that the provision on the
cardinality of the underlying field is unnecessary. To do so, we demonstrate
that the above result holds for all finite fields.
