Referring to this Wikipedia page, The definition of a non degenerate bilinear form
is given as a bilinear form $f(x,y)$ such that $v \to (x \to f(x,v))$ is an isomorphism.
We are also told that the below construct is an equivalent definition,
when $V$ is finite dimentional:
$$
f(x,y) = 0 \hspace{2ex} \forall\ y \in V \hspace{2ex} \text{implies}\hspace{2ex} x = 0
$$
In other words the kernel of the mapping $\ \psi:V \to V^{*}$ defined by $\ \psi := v \to
(x \to f(x,v))$ is trivial, that is $\operatorname{Ker}(\psi) = \{0\}$.
My question is how do we prove that the two definitions are equivalent? How does the latter
formulation show that $\psi$ is isomorphic?
Best Answer
Assume that $x\mapsto (y\mapsto f(x,y))$ is an isomorphism. Then it has trivial kernel, meaning $\psi(x)=0$ implies $x=0$. This is equivalent to the statement $f(x,y)=0$ for all $y$ implies $x=0$.
For the converse statement (assuming we are on a finite vectorspace), simply use the fact, that being an isomorphism between two vectorspaces of the same dimension is equivalent to having a trivial kernel.