About surjectivity of $V\cong (V^*)^*$

Question

Let $V$ be a finite dimensional vector space. We know that $V\cong V^{**}$ naturally . For example, See $V$ is isomorphic to $V^{\ast\ast}$, the double dual space of $V$. My my question is that how can we show that the map $E:V\to V^{**}$ by $v\mapsto E_v (\phi )=\phi (v)$ is surjective directly. Indeed, I need a formula for that, i.e., if $\psi \in V^{**}$, then I need explicitely a $v\in V$ such that $E_v=\psi$.

Answer 1

Let $\beta=[v_1,\ldots,v_n]$ be an ordered basis for $V$. For each $i$, let $\delta_i\in V^*$ be the functional defined by $\delta_i(v_j)=\delta_{ij}$.

Given $\psi\in V^{**}$, let $v=\psi(\delta_1)v_1+\cdots+\psi(\delta_n)v_n$. I claim that $E_v=\psi$.

To verify this, it is enough to show that $E_v(\delta_i)=\psi(\delta_i)$, since $[\delta_1,\ldots,\delta_n]$ is a basis for $V^*$. Indeed, we have $$E_v(\delta_i) = \delta_i(v)= \delta_i(\psi(\delta_1)v_1+\cdots+\psi(\delta_n)v_n) = \psi(\delta_i),$$ as required.