 Solution

No. On the real vector space $V={\mathbb{R}}^2$, use the symmetric bilinear formIts matrix of a bilinear form is $\mathrm{diag}(1,-1)$, whose determinant is $-1\ne 0$, so it is a nondegenerate bilinear form. LetA vector $(s,t)$ belongs to the bilinear orthogonal complement $U^{\perp }$ exactly when $s-t=0$. HenceIn 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 isIndeed, 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.