On the minimal rank in non-reflexive operator spaces over finite fields
  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. 
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