= Solution
{c}
For the given <nondegenerate bilinear form>, define the <linear map>
$$
\Phi:V\longrightarrow V',\qquad \Phi(v)(u)=(u,v).
$$
By part (a), $\Phi$ is a <linear isomorphism>. The paper's <bilinear orthogonal complement> uses $U$ in the first argument; with this convention,
$$
U^\perp=\Phi^{-1}(U^0),\qquad
U^0=\{f\in V':f(u)=0\text{ for every }u\in U\}.
$$
Here $U^0$ is the <annihilator of a vector subspace> $U$. Since $\Phi$ is a <linear isomorphism>, its <restriction> gives a <linear isomorphism> $U^\perp\to U^0$.
For completeness, choose a <basis> $u_1,\ldots,u_k$ of $U$ and extend it to a <basis> $u_1,\ldots,u_n$ of $V$. In its <dual basis> $u_1^*,\ldots,u_n^*$, a <linear functional> vanishes on $U$ exactly when its first $k$ <coefficients> are zero. Therefore
$$
U^0=\operatorname{span}\{u_{k+1}^*,\ldots,u_n^*\},\qquad \dim U^0=n-k.
$$
Consequently
$$
\boxed{\dim U^\perp=\dim U^0=\dim V-\dim U,\qquad
\dim U+\dim U^\perp=\dim V.}
$$
The argument includes $U=0$ and $U=V$, and requires no symmetry assumption on the <bilinear form>.
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