Past exam of the linear algebra course of the University of Oxford / 2021 / 3 / b

For the given nondegenerate bilinear form, define the linear mapBy part (a), $\Phi$ is a linear isomorphism. The paper's bilinear orthogonal complement uses $U$ in the first argument; with this convention,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 $u_1,\dots ,u_k$ of $U$ and extend it to a basis $u_1,\dots ,u_n$ of $V$. In its dual basis $u_1^{\ast },\dots ,u_n^{\ast }$, a linear functional vanishes on $U$ exactly when its first $k$ coefficients are zero. ThereforeConsequentlyThe argument includes $U=0$ and $U=V$, and requires no symmetry assumption on the bilinear form.

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.