 Finite-dimensional subspace is an intersection of kernels

For a vector subspace $U$ of a finite-dimensional vector space $V$, its annihilator of a vector subspace $U^0\subseteq V^{\ast }$ 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 kernels of any basis $g_1,\dots ,g_k$ 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.