 Solution

The annihilator of a vector subspace isChoose a basis $e_1,\dots ,e_r$ of $U$ and extend it to a basis $e_1,\dots ,e_d$ of $V$. In the dual basis, precisely $f_{r+1},\dots ,f_d$ span $U^{\circ }$. ThereforeThe canonical restriction linear transformationhas kernel of a linear map $U^{\circ }$. It is surjective: extend any linear functional on $U$ by assigning arbitrary values, for instance zero, to the additional basis vectors. The First isomorphism theorem for vector spaces gives the explicit linear isomorphismIt is natural because the displayed map itself makes no basis choice; a basis was used only to prove surjectivity. In particular, $U^{\prime }$ is canonically a quotient vector space of $V^{\prime }$, rather than canonically a subspace of $V^{\prime }$.