Prove that the dual -space of the dual-space of V is isomorphic to V without using bases

Question

Given a vector space $V$ the dual space $V^*$ is the space of all linear operators from $V$ to $\mathbb{C}$. $V^*$ is itself a vector space and I know how to prove $V \cong (V^*)^*$ by using a standard basis for $V$ and the corresponding dual basis for $V^*$. However, I was wondering whether or not it would be possible to prove this equivalence without making reference to a standard basis for $V$. Does anyone know how to do this?

Answer 1

If $V$ is finite dimensional, there is a natural isomorphism \begin{align*} V&\longrightarrow V^{**}\\ x&\mapsto x^{**} \end{align*} where $x^{**}: V^*\to \mathbb{F}$ is defined by $x^{**}(f):=f(x)$.

As the dimensions of $V$ and $V^{**}$ are the same, it suffices to show that the above natural map is injective in order to conclude that it is an isomorphism. If $x^{**}=0$, i.e. $f(x)=0$ for all $f\in V^*$, then $x$ must be 0 for otherwise, there exists $f$ such that $f(x)=1$ and $f$ is 0 on some subspace complement to $\text{span}\{x\}$.