Use natural identification of V with its double dual to prove that $U^{\bot \bot}=U$

Question

I meet this question when doing my homework: Given$U\subset V$,prove that $U^{\bot\bot}=U$,Hint: you may use the natural identification of $V$ with its double dual $V^{**}$,i.e,let $v\in V$ acts on $f\in V^*$ by $v(f):=f(v)$

I know how to prove $U^{\bot\bot}=U$ with addition theorem of basis, but I don't know how to prove it with notion of double dual space.

Answer 1

Presumably, $V$ is finite dimensional and $U$ is a subspace.

Hint: Let $\alpha :V \to V^{**}$ denote the natural identification. The restriction of $\alpha$ to $U$ (i.e. $\alpha|_{U}$) defines an injective map whose image is a subset of $U^{\perp\perp}$. By considering the dimension of the spaces of interest, conclude that the image is in fact equal to all of $U^{\perp\perp}$.