TITLE: Motivation to understand double dual space
QUESTION [52 upvotes]: I am helping my brother with Linear Algebra.  I am not able to motivate him to understand what double dual space is.  Is there a nice way of explaining the concept?  Thanks for your advices, examples and theories.

REPLY [11 votes]: This is an old post and the existing answers are good, but I feel like I can add another perspective. It looks fairly abstract at first, but if you're dealing with double duals, you just got to accept some abstraction. (By the way, I vividly remember solving the exercise in my undergrad linear algebra class showing the natural isomorphism $V \simeq V^{**}$. It felt like I had just warped my mind to the highest level of abstraction possible for human beings. The dual of a dual, woah ...)
Let $V$ be our vector space (over a field $K$, say). Let's assume we already understand the dual $V^*$. Then we will, after a short consideration, accept that the map
$$V^* \times V \rightarrow K$$
$$(l,v) \mapsto l(v)$$
is bilinear. We call it a bilinear form because its values are just scalars.
Now imagine we have any other vector space $W$ (over the same field) with a bilinear form
$$f:W \times V \rightarrow K.$$
We want to know how far this $W$ and $f$ are from the "ideal" form $V^* \times V \rightarrow K$ above. Note that for any such $W$ and $f$ we get a map, let's call it "c" for "comparison",
$$c: W \rightarrow V^*$$
$$w \mapsto f(w, \cdot)$$
(i.e. the image of $w$ is the linear form in $V^*$ which sends a given $v$ to $f(w,v)$). So how do $W$ and $V^*$ "compare"?

Easy exercise: $c$ is injective $\Leftrightarrow$ for all $0\neq w \in W$, there is $v\in V$ such that $f(w,v) \neq 0$
Trickier exercise: $c$ is surjective $\Leftrightarrow$ for all $0\neq v \in V$, there is $w\in W$ such that $f(w,v) \neq 0$.

Note that the "trickier" part needs that $\dim(V) < \infty$.
So the vector space $W$ and the bilinear form $f$ are "ideal" (more technical term: "perfect pairing") if and only if the bilinear form is non-degenerate (which just summarises the criteria on the RHS of the above exercises).
But now look at that non-degeneracy criterion: It is symmetric in the first and second component. In vague terms, $W$ is the dual of $V$ iff that form is non-degenerate; but if that form is non-degenerate, then so is the form that you just get from flipping the components; which means that then $V$ is also the dual of $W$. I.e. $V = V^{**}$.
This whole thinking of duality in terms of non-degenerate bilinear forms, as abstract as it seems on first encounter, becomes incredibly helpful in some applications. E.g. in representation theory it's very neat to translate all the time between self-duality of certain representations and existence of certain $G$-invariant forms on them. Or, a generalisation of this kind of thinking basically leads to adjunction of tensor products and Hom, and all that wonderful general algebra stuff, I recently used that e.g. here to motivate the definition of certain Lie algebra actions.