True/False question about simmetrical bilinear forms

Question

Let $V$ be a finite dimensional vector space of dimension $n$ equipped with a symmetrical bilinear form. If for every subspace $W \not= {0}$ of $V$, $\dim W^{\perp} > \dim V - \dim W$ then the bilinear form $\phi$ is $\phi = 0$. Is that true? Any thoughts? It was an exercise in one of the past papers of my professor, but I can't get my head around it.

Answer 1

I think I solved it. (I realise it is pretty dumb to answer your own question, but whatever; it may work even to check if my reasoning is right). Now, let's call $n $ the dimension of $V$. Let $B = {v_{1},...,v_{n}}$ be a base of $V$. If $B$ is a base, then every vector $v\neq0$ (in case $v=0$ the claim is trivial) can be written as a linear combination of the elements of the base. $$v = {a_{1}v_{1} +...+ a_{n}v_{n}}$$ Now let $W=span(v)$. This subspace has obviously dimension 1. Our hypothesis says that $\dim W^{\perp} > \dim V - \dim W$ so $\dim W^{\perp}=n$. But then $W^{\perp}=V$. So the claim is true.