Motivation to understand double dual space

If $V$ is a finite dimensional vector space over, say, $\mathbb{R}$, the dual of $V$ is the set of linear maps to $\mathbb{R}$. This is a vector space because it makes sense to add functions $(\phi + \psi)(v) = \phi(v) + \psi(v)$ and multiply them by scalars $(\lambda\phi)(v) = \lambda(\phi(v))$ and these two operations satisfy all the usual axioms.

If $V$ has dimension $n$, then the dual of $V$, which is often written $V^\vee$ or $V^*$, also has dimension $n$. Proof: pick a basis for $V$, say $e_1, \ldots, e_n$. Then for each $i$ there is a unique linear function $\phi_i$ such that $\phi_i(e_i) = 1$ and $\phi_i(e_j) = 0$ whenever $i \neq j$. It’s a good exercise to see that these maps $\phi_i$ are linearly independent and span $V^*$.

So given a basis for $V$ we have a way to get a basis for $V^*$. It’s true that $V$ and $V^*$ are isomorphic, but the isomorphism depends on the choice of basis (check this by seeing what happens if you change the basis).

Now let’s talk about the double dual, $V^{**}$. First, what does it mean? Well, it means what it says. After all, $V^*$ is a vector space, so it makes sense to take its dual. An element of $V^{**}$ is a function that eats elements of $V^*$, i.e. a function that eats functions that eat elements of $V$. This can be a little hard to grasp the first few times you see it. I will use capital Greek letters for elements of $V^{**}$.

Now, here is the trippy thing. Let $v \in V$. I am going to build an element $\Phi_v$ of $V^{**}$. An element of $V^{**}$ should be a function that eats functions that eat vectors in $V$ and returns a number. So we are going to set
$$\Phi_v(f) = f(v).$$

You should check that the association $v \mapsto \Phi_v$ is linear (so $\Phi_{\lambda v} = \lambda\Phi_v$ and $\Phi_{v + w} = \Phi_v + \Phi_w$) and is an isomorphism (one-to-one and onto)! This isomorphism didn’t depend on choosing a basis, so there’s a sense in which $V$ and $V^{**}$ have more in common than $V$ and $V^*$ do.

In fancier language, $V$ and $V^*$ are isomorphic, but not naturally isomorphic (you have to make a choice of basis); $V$ and $V^{**}$ are naturally isomorphic.

Final remark: someone will surely have already said this by the time I’ve edited and submitted this post, but when $V$ is infinite dimensional, it’s not always true anymore that $V = V^{**}$. The map $v \mapsto \Phi_v$ is injective, but not necessarily surjective, in this case.