Prove that there is a natural isomorphism between $V$ and $(V^*)^*$ Let me just start by saying I'm very very new to this material. I have very little idea what's going on. I've read Wikipedia and a few other sources but this is still very hard for me, so I would much appreciate if someone could help me solve this question, slowly and patiently.
We are given $V$ a vector space over field $F$, $\mathrm{dim}(V)$ is a finite number. Show that there is an isomorphism $i: V \longrightarrow (V^{*})^{*}$, where $V^*$ is the dual space of $V$.
Could someone please help me with this?
 A: The isomorphism you are looking for is given by $\Phi:V\to V^{**}$ by $v\mapsto(\lambda\mapsto\lambda(v))$, i.e. you associate to an element $v$ of $V$ the element of the dual of the dual mapping the element $\lambda$ of $V^*$ (a mapping $\lambda:V\to k$) to $\lambda(v)$.
I know it can look very confusing at first, but you'll get used to it after reading it a couple of times and using it in various exercises and proofs.
A: For every vector $v \in V$, you have to produce some $i(v) \in V^{**}$. And this one must be a linear map
$$
i(v) : V^* \longrightarrow \mathbb{K} \ .
$$
So, you have to produce an element of $\mathbb{K}$, for every $\omega \in V^*$. That is
$$
\omega \mapsto i(v)(\omega ) \ .
$$
Now, having as all data a linear form $\omega \in V^*$ and a vector $v\in V$, how would you obtain, in the more natural way, an element of $\mathbb{K}$?
A: The easiest way I know to see this, for finite dimensional vector spaces, is to simply observe that for such, $V \cong V^*$.  This assertion follows as follows:  let $\{e_i\}$, $1 \le i \le \dim V$, be any basis for $V$.  Define elements $\theta_j \in V^*$, $1 \le j \le \dim V$, via $\theta_j(e_i) = \delta_{ij}$.  It is easy to see that the $\theta_j$ form a basis for $V^*$, since for any $\theta \in V^*$ and $v \in V$ with $v = \sum \alpha_i e_i$, $\alpha_i \in F$, $\theta(v) = \sum_i \alpha_i \theta(e_i) = \sum_i \alpha_i \sum_j \theta(e_j) \theta_j(e_i)= \sum_j \theta(e_j) \theta_j(\sum_i \alpha_i e_i) = \sum_j \theta(e_j)\theta_j(v)$, whence $\theta = \sum_j \theta(e_j) \theta_j$, so the $\theta_j$ span $V^*$; they are also linearly independent, since $\sum_j \beta_j \theta_j = 0$ yields $0 = \sum_j \beta_j \theta_j(e_i) = \beta_i$ for all $i$.  These remarks show $\dim V^* = \dim V$, since each has a basis of $\dim V$ elements.  Now apply what we have just proved to $V^*$, resulting in $V^{**} \cong V^*$.  But since $V^* \cong V$, we have $V^{**} \cong V$.  QED.
The preceding remarks show $V^{**} \cong V$, but do not provide an explicit isomorphism.  However, taking $f_k \in V^{**}$ dual to the $\theta_j$, i.e. $f_k(\theta_j) = \delta_{kj}$, and then mapping $f_k \leftrightarrow e_k$, probably yields as "natural" an isomorphism as any.
Hope this helps.  Happy New Year,
and as always, 
Fiat Lux!!!
