Can every isomorphism between a vector space and its dual be written as a non-degenerate bilinear form? If I have understood correctly, non-degeneracy of a bilinear form:
$$\omega:V\times V\rightarrow \Bbb F \tag{1},$$
in which $\Bbb F$ the underlying field of $V$, is a strong enough condition to conclude that $\omega$ provides an isomorphism between $V$ and $V^*$: non-degeneracy $\Rightarrow$ trivial kernel $\Rightarrow$ injectivity $\Rightarrow$ surjectivity if dimesions are the same.
My question is does the converse hold, can all isomorphisms between a vector space and its dual be induced by some non-degenerate bilinear form on $V$?
 A: If $f:V\to V^*$ is an isomorphism, and we define $\omega:V\times V\to \Bbb{F}$ as $\omega(v_1,v_2):= [f(v_1)][v_2]$, then $\omega$ is bilinear, and note that from the definition, we have that $\omega(v_1,\cdot)= f(v_1)$. In other words, the mapping $v_1\mapsto \omega(v_1,\cdot)$ is equal to $f$, which by assumption is an isomorphism of $V$ onto $V^*$.
If $V$ is infinite-dimensional, there are no isomorphisms $V\to V^*$, so the statement above becomes vacuous.

Edit:
The previous version of my answer made it seem like there was a distinction to be made (because it wasn't obvious to me in the case of non-symmetric/skew-symmetric bilinear forms). But, thanks to @Marc van Leeuven for pointing out that the distinction between left/right non-degeneracy is unnecessary.
For the sake of completeness, here's the proof of the equivalence. Let $\omega_l:V\to V^*$ be the left map, $x\mapsto \omega(x,\cdot)$, and let $\omega:V\to V^*$ be the right map $y\mapsto \omega(\cdot, y)$. Also, let $\iota:V\to V^{**}$ be the canonical map into the double dual. One can easily verify by unwinding definitions that
\begin{align}
\omega_r^*\circ \iota=\omega_l\quad\text{and} \quad\omega_l^*\circ\iota=\omega_r
\end{align}
where the stars on the maps indicate the dual/transposed mapping.
Now, make the assumption that $V$ is finite-dimensional. Then, $\iota$ is an isomorphism.
So, if $\omega_r$ is assumed an isomorphism, then so is $\omega_r^*$, and thus (by first equation) $\omega_l$ being the composition of two isomorphisms is as well. Similarly, $\omega_l$ being an isomorphism implies by the second equation that $\omega_r$ is.
One can also rephrase the argument using matrices since we're in finite dimensions, but I'd rather not :)
