Question
I want to prove that $V$ is naturally isomorphic to $(V^\ast)^\ast$ without making reference to a basis.
Note: $V$ is finite dimensional space.
Attempt
I have already shown that the map between $V$ and $(V^\ast)^\ast$ is linear and injective. What is left is to show that it is surjective.
We define $\Phi :V\rightarrow (V^\ast)^\ast$ by $\Phi (v)(f)=f(v)$ for all $v\in V$, and for all $f\in V^\ast$.
Then for every $\Psi \in (V^\ast)^\ast$, there is a $v\in V$ such that $\Phi (v)=\Psi$.
I feel there should be more to this. Is there any way I can improve this?