Let $V$ be a finite-dimensional vector space over a field $k$, and let $U \subset V$ be a $k$-linear subspace. Set $r = \dim_k U$ and $n = \dim_k V$. Then there exists an ordered basis $(v_1,\ldots,v_n)$ of $V$ such that $(v_1,\ldots,v_r)$ is an ordered basis of $U$. Equivalently,
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\begin{align*}
U = \operatorname{span}_k\{v_1,\ldots,v_r\}.
\end{align*}
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When $r = 0$, the assertion means that the empty initial segment is the ordered basis of the zero subspace $U = \{0\}$.