We have $V$ a $\mathbb{R}$-vector space of finite dimension $n$ with non-degenerate bilinear form $\phi:V\times V\rightarrow \mathbb{R}$ and subspace $U$ of $V$ with $U\subset U^{\perp}$. How do I show that the dimension of $U$ is at most $n/2$?

Edit $U^{\perp}=\{v\in V:\phi(v,u)=0\space \forall u\in U\}$

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    $\begingroup$ Hmm, I thought $U\cap U^{\perp} = \{0\}$ for any set $U$? $\endgroup$ – MPW Jan 13 '16 at 13:24
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    $\begingroup$ Presumably, $U^{\perp}$ is defined here in terms of the bilinear form, and not the usual dot product. @MPW $\endgroup$ – Thomas Andrews Jan 13 '16 at 13:30
  • $\begingroup$ @ThomasAndrews You are right $\endgroup$ – Lucas George Jan 13 '16 at 13:32
  • $\begingroup$ Is your bilinear form symmetric - that is - if $\phi(v,w)=\phi(w,v)$? If not, you are going to need to define $U^{\perp}$ specifically. It would also help to be specific about what you mean by "degenerate." I'm taking it to mean that $\phi$ isn't zero everywhere, but I don't know. $\endgroup$ – Thomas Andrews Jan 13 '16 at 13:34
  • $\begingroup$ @ThomasAndrews Not (necessarily) symmetric, see edit $\endgroup$ – Lucas George Jan 13 '16 at 13:36

Non-degenerate means in this case that the linear map

$$V \rightarrow V^\ast; \quad v \mapsto \phi(v, \cdot)$$

is an isomorphism. Under this isomprphism, the subspace $U^\perp$ is isomprphic to what we call the anihilator of $U$ in $V^\ast$, i.e. the subspace $U' := \lbrace v^\ast \in V^\ast \, |\, v^\ast(u) = 0 \quad \forall u \in U \rbrace$ of $V^\ast$.

Now choose a basis $(u_1, \dots, u_k, w_{k+1}, \dots, w_n)$ of $V$ such that $(u_1, \dots, u_k)$ is a basis of $U$. Let $(u_1^\ast,\dots, u_k^\ast, w_{k+1}^\ast, \dots, w_n^\ast)$ be its dual basis in $V^\ast$. Then $(w_{k+1}^\ast, \dots, w_n^\ast)$ is a basis of $U'$. The assumption $U \subset U^\perp \cong U'$ then implies that $$k = \dim{U} \leq \dim{U^\perp} =\dim{U'} = n-k$$ which is equivalent to $k \leq \frac{n}{2}$.

  • $\begingroup$ Is this answer unclear, Lucas? $\endgroup$ – m.s Jan 26 '16 at 19:02

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