Suppose that $V, W$ are $n$ dimensional vector spaces on a field $K$ and $b:V \times W \to K$ is a bilinear form. And let $\phi$ and $\psi$ be the maps defined by the folloing:
$$\phi : V \to W^* ; v \mapsto (w \mapsto b(v, w)),$$ $$\psi : W \to V^* ; w \mapsto (v \mapsto b(v, w)).$$
Then, why does the map $\psi$ is a isomorphism if $\phi$ is a isomorphism?
I know that $$\phi \ {\rm is \ a \ isomorphism} \Leftrightarrow [\forall v \in V, (\forall w \in W,b(v,w) = 0) \Rightarrow v = 0].$$ (That is, the right hand side means $\phi$ is injective.)
Let $w \in W$ be given such that $b(v,w)=0$ for all $v \in V$.
Suppose towards contradiction that $w \neq 0$. Let $w' \in W^{*}$ such that $w$ is not in the kernel of $w'$. One such $w'$ is to send the subspace generated by $cw$ to $c$, and everything else to 0.
Observe that $\exists v'\in V$ such that $\phi(v')=w$ since $\phi$ is an isomorphism, and hence surjective. Then, $0\neq w'(w)=ϕ(v')(w)=b(v',w)=0$. This is a contradiction. Thus $\psi$ is injective.
So by dimension counting, $\psi$ is an isomorphism.