= Solution
{c}
\b[No.] On the <real vector space> $V=\mathbb R^2$, use the <symmetric bilinear form>
$$
((x,y),(s,t))=xs-yt.
$$
Its <matrix of a bilinear form> is $\operatorname{diag}(1,-1)$, whose <determinant> is $-1\ne0$, so it is a <nondegenerate bilinear form>. Let
$$
U=\operatorname{span}\{(1,1)\}.
$$
A <vector> $(s,t)$ belongs to the <bilinear orthogonal complement> $U^\perp$ exactly when $s-t=0$. Hence
$$
\boxed{U^\perp=U,\qquad U\cap U^\perp=U\ne0,\qquad U+U^\perp=U\ne V.}
$$
In particular, the claimed <direct sum> fails. The <isotropic vector> $(1,1)$ has zero self-pairing, although it pairs nontrivially with other <vectors> of $V$.
The <dimension> identity in part (i) counts both <vector subspaces> but does not force their <intersection of vector subspaces> to be zero. For a <symmetric bilinear form>, $U\cap U^\perp$ is the <radical of a bilinear form> restricted to $U$. Thus the precise criterion is
$$
\boxed{V=U\oplus U^\perp\ \Longleftrightarrow\
\text{the restriction of the form to }U\text{ is nondegenerate}.}
$$
Indeed, trivial <intersection of vector subspaces> and part (i) give a <sum of vector subspaces> of <dimension> $\dim V$. A <positive-definite bilinear form> would guarantee this for every $U$; symmetry alone does not.
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