Why induced linear transformations of bilinear function are duals of each other?

Question

Suppose we have a bilinear function $b:V\times W \to R$. We also define $V^*=Hom(V,R)$ and $L^* = \circ L.$ Question asks to prove that if $K:V \to W^*,L:W \to V^*$ are induced linear transformations of $b$, then we must have $L = K^*$. I got the fact that for all $x \in V, y \in W$ $$b(x,y) = L(y)(x) = K(x)(y)$$ I am a little confused by the fact that $K^*:W^{**} \to V^{*}$ (which I understand), but $L:W \to V^*$, so how can they be equal then? Perhaps it is another way of saying two things are equal (which apparently I don't know) or some confusion with terminology. I'd love to hear the explicit explanation with (perhaps) examples.

Answer 1

This answer does not assume finite-dimensionality of any vector space. Furthermore, the base field can be any field (not just $\mathbb{R}$).

Let $\phi:W\to W^{**}$ denote the canonical injection $$w\mapsto \big(f\mapsto f(w)\big)$$ for all $w\in W$ and $f\in W^*$. We shall prove that $K^*\circ \phi=L$.

For $v\in V$ and $w\in W$, we see from the definition that $$\big(L(w)\big)(v)=b(v,w)\,.$$ On the other hand, $$\begin{align}\big((K^*\circ \phi)(w)\big)(v)&=\Big(K^*\big(\phi(w)\big)\Big)(v)=\Big(\big(\phi(w)\big)\circ K\Big)(v) \\&=\big(\phi(w)\big)\big(K(v)\big)=\big(K(v)\big)(w)\\&=b(v,w)\,.\end{align}$$ Thus, $\big(L(w)\big)(v)=\big((K^*\circ\phi)(w)\big)(v)$ for all $v\in V$ and $w\in W$. This shows that $L=K^*\circ \phi$.

Now, if $W$ is a finite-dimensional vector space, then $\phi$ is a canonical isomorphism. Hence, we can naturally identify $W^{**}$ with $W$ (and with this identification, $\phi$ is the identity map). Therefore, $L=K^*$ in this case.