Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2012 / 1 / b / Solution

The annihilator of a vector subspace consists of the linear functionals that vanish on every element of that vector subspace:For the dimension identity, choose a basis $w_1,\dots ,w_k$ of $W$ and use basis extension to obtain $w_1,\dots ,w_n$ for $V$. A linear functional is determined by its values on these $n$ basis vectors. Belonging to the annihilator of a vector subspace sets its first $k$ values to zero and leaves its last $n-k$ values arbitrary. Equivalently the last $n-k$ elements of the dual basis form a basis of $W^0$.