Let $(V^*)^*=V^{**}$. Define $S:V\to V^{**}$ by $s(v)(\alpha)=\alpha(v)$ for all $v\in V$ and $\alpha\in V^*$.
I need to show that $s(v)\in V^{**}$. And show the S is a linear transformation. ($V$ is a finite dimensional vector space over a field $F$ by the way)
So far for the first proof I have, For a vector $v\in V$, we define $s(v):V^*\to F$ by $s(v)(\alpha)=\alpha(x)$ for every $\alpha \in V^*$. To show $s(v)\in V^{**}$, we need to show $s(v)$ is linear. So let $f_1,f_2 \in V^*$ and let $\alpha \in F$ be arbitrary. Then $$(s(v))(\alpha f_1+f_2) = (\alpha f_1 + f_2)((s(v)) = \alpha f_1(s(v))+ f_2(s(v)) = \alpha s(v)f_1 + s(v) f_2.$$ Hence, $s(v)$ is linear so $s(v)\in V^{**}$ for all $v \in V$.
I think that is right, but for the second part, I am a little confused. I think to show S is linear, we need to show for $v_1,v_2 \in V$ and $\alpha \in F$ that $s(\alpha v_1 +v_2)$ = $\alpha s(v_1)+ s(v_2)$. Not sure what do from here. Thanks for the help!
You are correct. Let us consider the first of these equalities. You must show that for any scalar $\alpha$ and any $v\in V^\ast$ that $s(\alpha v)=\alpha s(v)$. What do this mean though? Since we are working in $V^{\ast\ast}=(V^\ast)^\ast$ this means that for every $\varphi\in V^\ast$ one has that $\alpha(s(v)(\varphi))=s(\alpha v)(\varphi)$. The LHS is just $\alpha (\varphi(v))$ and the RHS is just $\varphi(\alpha v)$. Can you finish?