= Finite-dimensional subspace is an intersection of kernels
{title2=$U=\bigcap_{f\in U^0}\ker f$}
For a <vector subspace> $U$ of a finite-dimensional <vector space> $V$, its <annihilator of a vector subspace> $U^0\subseteq V^*$ determines $U$ exactly: $U=\bigcap_{f\in U^0}\ker f$. Indeed, a <basis> of $U$ can be extended to one of $V$. If $v\notin U$, include $v$ among the added basis vectors and define a <linear functional> taking value one at $v$ and zero on all the other basis vectors. This functional vanishes on $U$ but not at $v$, excluding every outside vector from the intersection.
It suffices to intersect the <kernel of a linear map>[kernels] of any <basis> $g_1,\ldots,g_k$ of $U^0$: every other annihilating functional is their <linear combination>. The <dual basis> for a basis of $V$ extending a basis of $U$ shows $k=\dim V-\dim U$. If $U=V$, the empty intersection is $V$. This translates a subspace description by spanning vectors into an exact description by independent linear equations.
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