= Solution
Under the <bidual evaluation map>, a vector $y$ lies in $(\operatorname{im}\Phi|_U)^\circ$ exactly when
$$
j(y)(\varphi_x)=\varphi_x(y)=\varphi(x,y)=0\quad\text{for all }x\in U.
$$
That is precisely membership of the <orthogonal complement for a bilinear form>. Therefore
$$
\boxed{U^\perp=(\operatorname{im}\Phi|_U)^\circ.}
$$
The dimension formula in part (ii) gives
$$
\dim U^\perp=n-\operatorname{rank}(\Phi|_U)\geq n-\dim U,
$$
so $\boxed{\dim U+\dim U^\perp\geq n}$. More precisely, the <rank-nullity theorem> gives
$$
\dim U+\dim U^\perp=n+\dim(U\cap\ker\Phi).
$$
This is the <orthogonal complement dimension for a degenerate bilinear form>. The possible excess comes from vectors annihilated by the form against all of $V$.
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