Hoffman and Kunze, Linear algebra , Chapter 3 , Corollary of theorem 17

Question

Theorem $17.$ Let $V$ be a finite-dimensional vector space over the field $F.$ For each vector $\alpha$ in $V$ define $$L_{\alpha}(f)=f(\alpha),\hspace{1cm}f\text{ in } V^{*}.$$

The mapping $\alpha \to L_{\alpha}$ is then an isomorphism of $V$ onto $V^{**}.$

Corollary. Let $V$ be a finite-dimensional vector space over the field $F.$ If $L$ is a linear functional on the dual space $V^{*}$ of $V,$ then there is a unique vector $\alpha$ in $V$ such that $$L(f)=f(\alpha)$$ for every $f$ in $V^{*}.$

The corollary being an obvious consequence is something that I am not able to see. This is probably because I started out this way :

Corollary says $\Bigg(\forall L\in V^{**}\Big(\exists \alpha \in V$ s.t$\big(\forall f \in V^{*} , L(f)=f(\alpha)\big)\Big)\Bigg).$

This means that $\Bigg(\exists L\in V^{**}$ s.t $\Big(\forall \alpha \in V\big(\exists f \in V^{*} $ s.t $ L(f)\ne f(\alpha)\big)\Big)\Bigg)$ should be a false statement.

Going this way, suppose there exists such an $L$ and $f$ such that $L(f)\ne f(\alpha)$ for all $\alpha.$ Integrating this with Theorem $17,$ we see that $L\ne L_{\alpha}$ for any $\alpha.$ But $\{L_{\alpha}\}_{\alpha \in V}$ exhausts $V^{**}$ $\square .$

Is this the contradiction that I should arrive at? Also I highly believe that what I have done is unnecessary complication of the problem, if so, please put forth a simple one.

Answer 1

If $V \ni \alpha \mapsto L_\alpha \in V^{\ast \ast}$ is an isomorphism, in particular it is surjective. So given $L \in V^{\ast\ast}$ there is an unique (because of injectiveness) $\alpha \in V$ such that $L = L_\alpha$. So, for all $f \in V^{\ast}$ you have $$L(f) = L_\alpha(f) = f(\alpha),$$as wanted.

Answer 2

Simpler approach: let $L$ be the linear functional in the corollary. It lies in $V^{**}$ so by the theorem (and definition of isomorphism), there exists a unique $\alpha \in V$ that is mapped to $L$ by the isomorphism.

By the definition of the isomorphism, $L(f)=f(\alpha)$ for any $f \in V^*$. Moreover, uniqueness of this $\alpha$ implies that no other $\alpha' \in V$ can satisfy $L(f) = f(\alpha')$ for all $f \in V^*$ (else the isomorphism would also map $\alpha'$ to $L$, which contradicts injectivity of the isomorphism).