Non-degenerate quadratic form and non-singular matrix Let $(V,Q)$ be a finite-dimensional quadratic space over a field $\mathbb{K}$.
From my definition, $Q$ is non-degenerate if $\operatorname{rad}(V)=\{0\}$.
How can I prove that $Q$ is non-degenerate iff the matrix associated to the scalar product defined by $Q$ (in any basis of $V$) is non-singular?
Surely this is trivial, but I can't conclude...

$$(u,v) = \frac{1}{2}(Q(u+v, u+v) - Q(u) - Q(v)) = u^T B v$$
Letting $u$ vary over all the standard basis vectors, we see that $I^T B v = I B v = Bv = 0$ has a non-trivial solution if and only if $B$ is non-singular.
ie) if $B$ is singular, let $B v = 0$.  Then $u^T B v = 0$ for all $u \in V$.
On the other hand $B$ is non-singular, there is no $v$ satisfying $I^T B v = 0$ and so for any $v$, there is some choice of $i$ such that $(e_i, v) \neq 0$.
