I'm working on the following problem:

Let $\langle,\rangle$ be a symmetric bilinear form on a finite dimensional vector space $V$. Assume that the bilinear form is nondegenerate; that is $\langle x,y\rangle=0$ for all $y\in V$ if and only if $x$ is the zero vector in $V$. Let $W$ be a subspace of $V$ and $W^{\perp}=\{v\in V:\langle w,v\rangle=0\text{ for all }w\in W\}$. Prove that if $W\subseteq W^{\perp}$, then $$\dim(W)\leq\frac{\dim(V)}{2}.$$

I'm studying for a qualifying exam and this is the first time I've encountered bilinear forms. After doing a bit of reading, I understand that a symmetric, nondegenerate bilinear form is almost an inner product, as the only thing missing is positivity. (I'm assuming that's why the angle bracket notation is used here, as that doesn't seem to be common for bilinear forms.) This seemingly minor exclusion appears to make a large difference though because in an inner-product space, the intersection of a space with its orthogonal complement is the empty set, but here we can have $W\subseteq W^{\perp}$.

All this exposition is to say that I'm not really sure how to approach this. First, I don't even think this is true in general. If we take $W^{\perp}=W=V=\{0,v\}$, then $\dim(W)=\dim(V)=1$, right? But maybe this is the only counterexample? So if we assume that $\dim(V)\neq1$ and suppose that $W\subseteq W^{\perp}$, then $\langle a,b\rangle=0$ for all $a,b\in W$. And maybe from here we can argue that if the dimension of $W$ were more than half the dimension of $V$, then $V$ would have to be the trivial space, for which this statement is vacuously true?


Indeed, for a symmetric bilinear form it is possible that $W \subseteq W^{\perp}$. However, what still remains true is that

$$ \dim W + \dim W^{\perp} = \dim V. $$

To see why, consider the map $\psi \colon V \rightarrow V^{*}$ given by $\psi(v)(w) = \left< v, w \right>. $ The fact that $\left< \cdot, \cdot \right>$ is non-degenerate precisely says that $\psi$ is injective and since $V$ is finite dimensional and $\dim V = \dim V^{*}$, this means that $\psi$ is an isomorphism.

Now, recall that given any subspace $U \subseteq V^{*}$, we have $$ \dim U + \dim U_0 = \dim V^{*} $$ where $$ U_0 = \{ v \in V \, | \, \varphi(v) = 0 \,\,\forall \varphi \in U \}. $$

Using the isomorphism $\psi$, we have $\psi(W)_{0} = W^{\perp}$ and so

$$ \dim V = \dim V^{*} = \dim \psi(W) + \dim \psi(W)_{0} = \dim W + \dim W^{\perp}. $$

Returning to your question, if $W \subseteq W^{\perp}$ then $$ 2\dim W \leq \dim W + \dim W^{\perp} = \dim V \implies \dim W \leq \frac{\dim V}{2}. $$

  • $\begingroup$ Excellent, thank you for the thorough explanation! $\endgroup$ – Atsina Jan 4 '20 at 22:45

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