Past exam of the linear algebra course of the University of Oxford / 2023 / 1 / b / ii

For a vector subspace $U\subseteq V$, its annihilator of a vector subspace isLet $\dim U=k$ and $\dim V=n$. Extend a basis $u_1,\dots ,u_k$ of $U$ to a basis $u_1,\dots ,u_n$ of $V$, and let ${\epsilon }^1,\dots ,{\epsilon }^n$ be the dual basis. Every linear functional has the unique expression $\phi ={\sum }_{j=1}^n\phi (u_j){\epsilon }^j$. It belongs to $U^0$ precisely when its first $k$ coefficients vanish. ConsequentlyThe listed dual basis elements are linearly independent, so counting them proves the formula, including $U=0$ and $U=V$.