Suppose $V$ is a finite-dimensional vector space. We say that a symmetric 2-tensor $q$ on $V$ is non-degenerate if the linear map $\hat{q}:V\to V^*$ defined by $\hat{q}(v)(w)=q(v,w)$ is an isomorphism. Now I would like to prove that the following two statements are equivalent:
(a) $q$ is non-degenerate.
(b) For each nonzero $u\in V$, $\exists w\in V$ s.t. $q(u,w)\neq 0$.
I have shown (a) implies (b), and it's quite straightforward. It remains to show (b) leads to (a), and I must prove that $\hat{q}$ is an isomorphism. This can be done by showing that $\hat{q}$ is injective and surjective. Actually, all we need to do is justify the injectivity of $\hat{q}$ since $\hat{q}$ is a linear map between vector spaces of equal dimension. But somehow I can't work it out. Does anyone have an idea? Thanks a lot.
Define $\hat q : V \to V^*$ as $\hat q (u) = \left(v \mapsto q(u,v)\right)$. This is a linear map between spaces of the same finite-dimension. We have: $$ \ker \hat q = \{ u \in V \mid \hat q (u) = 0 \} = \{ u \in V \mid \forall v \in V, q(u,v) = 0\}. $$ Hence, it follows that $\ker \hat q = \{0\} \iff (b)$. Therefore:
\begin{align} (a) = q \text{ is non degenerate } & \iff \hat q \text{ is an isomorphim }\\ & \iff \hat q \text{ is injective} \\ &\iff \ker \hat q = \{0\} \\ &\iff (b). \end{align}