Question: Let $V$ be a finite dimensional vector space over some field $F$. Let $V^{\star}$ be the dual space of $V$. Show $V\cong V^{\star}$.
I know this question has been asked several times, and is quite common, I am just looking for a nice direct proof.
If we let $\{v_1,\dots,v_n\}$ be a basis for $V$. Since $V^{\star}$ is the dual space of $V$, that is $V^{\star}$ is the vector space of linear transformations $T:V\rightarrow F$, so $V\cong F^n$, since $V$ is finite dimensional over $F$ (I suppose that is where this breaks, because if $V$ is not finite dimensional, then we can't say this). So, we would be done if we showed that $V^{*} \cong F^n$.
Define $T:V^{\star}\rightarrow F^n$ by $T(x^{\star})=(x^\star(v_1),\dots,x^{\star}(v_n))$. So, we need to show that this map is an isomorphism.
$T$ is a homomorphism: $T(x^\star+y^\star)=T(x^\star)+T(y^\star)$ (I can't quite think of why I can justify this) and $T(kx^\star)=kT(x^\star),$ since $T$ is linear.
How would I go about showing that $T$ is surjective and injective? I think I can show that $T$ is injective, because if $T(x^\star)=T(y^\star)$ , then $x^\star(v_i)=y^\star(v_i)$, for all $i$ since $x^\star$ and $y^\star$ would agree on the same basis, thus $x^\star=y^\star$.
Any help is greatly appreciated! Thank you.
Let $\varphi, \psi \in V^{*}$ and $\lambda \in F$. Then $$\begin{aligned}T(\varphi+\lambda\psi) &= ((\varphi+\lambda\psi)(v_1), \cdots, (\varphi+\lambda\psi)(v_n)) \\ &=(\varphi(v_1) + \lambda\psi(v_1), \cdots, \varphi(v_n) + \lambda\psi(v_n)) \\ & = (\varphi(v_1), \cdots,\varphi(v_n))+(\lambda\psi(v_1), \cdots,\lambda\psi(v_n)) \\&= T(\varphi) + \lambda T(\psi)\end{aligned} $$
hence $T$ is linear. Given any $a = (a_1, \cdots, a_n) \in F^n$, define $g: V \to \mathbb{R}$ by putting $g(v) = \lambda_1a_1 + \cdots +\lambda_n a_n$, for each $v = \lambda_1v_1 + \cdots \lambda_n v_n \in V$ (of course, the $\lambda_i$ will depend on $v$, but this doesn't matter). Then it's clear that $g(v_i) = a_i$ for all $1 \leq i \leq n$, therefore $T(g) = a$ (the motivation to choose $g$ like this comes from the fact that a linear map is uniquely determined by the values it assumes on a basis). By the arbitrariness of $a$ it follows that $T$ is surjective. Since you already showed it is injective, we conclude that $T$ is a isomorphism, as desired (here I used that a bijective linear map between vector spaces is necessarily an isomorphism, which is easy to prove).