Non-degenerate quadratic form and non-singular matrix

Question

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...

Answer 1

$$(u,v) = \frac{1}{2}(Q(u+v, u+v) - Q(u) - Q(v)) = u^T B v$$

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 $(e_i, v) \neq 0$.

Answer 2

One can make above answer characteristic free by taking $B(u,v)=Q(u+v)-Q(u)-Q(v)$. Note that $\mathrm rad(V)=0$ is condition 2) below.

If $B$ is a bilinear form over a vector space $V$, then note that following are equivalent: 1) The matrix $\{B(e_i,e_j)\}$ is non-singular. 2) $u\mapsto B(~,u)$ defines an isomorphism $V\rightarrow V^*$ 3) For $u\in V$, $B(u,v)=0$ for all $v\in V$ implies that $u=0$.

Proof: 2) implies 3) is straightforward. For 1) implies 2): In this case $\{B(~,e_j)\}$ will form a basis: $\sum a_jB(~,e_j)=0$ implies $\sum a_jB(e_i,e_j)$ for every $i$. Thus $(a_1,\cdots,a_n)$ is in the kernel of matrix $\{B(e_i,e_j)\}$ hence it is zero. Thus we will get 2. Now to show 3) implies 1): we will show that rows of $\{B(e_i,e_j)\}$ are linearly independent. Consider $\sum\alpha_i(B(e_i,e_1),\cdots,B(e_i,e_n))=0$. This will imply $B(\sum_i\alpha_ie_i,e_j)=0$ for every $j$. Hence by 3) $\sum_i\alpha_ie_i=0$, this $\alpha_i=0$ for each $i$.