Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2013 / Algebra / 1 / b

Choose a basis $u_1,\dots ,u_m$ of $U$. Because $v\notin U$, the list $u_1,\dots ,u_m,v$ is linearly independent. Extend it to a basis of $V$. A linear functional is specified uniquely by its values on a basis, so define $f$ to be zero on the $u_j$ and on every added basis vector other than $v$, and set $f(v)=1$.

Then $f$ vanishes on every linear combination of the $u_j$, so $f\in U^0$, whileThis proves that the annihilator of a vector subspace separates every vector outside the subspace from that subspace; no inner product or preferred coordinates are needed.