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.

**Answer**

If V is a finite dimensional vector space over, say, R, the dual of V is the set of linear maps to R. This is a vector space because it makes sense to add functions (ϕ+ψ)(v)=ϕ(v)+ψ(v) and multiply them by scalars (λϕ)(v)=λ(ϕ(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∨ or V∗, also has dimension n. Proof: pick a basis for V, say e1,…,en. Then for each i there is a unique linear function ϕi such that ϕi(ei)=1 and ϕi(ej)=0 whenever i≠j. It’s a good exercise to see that these maps ϕi 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∈V. I am going to build an element Φv 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

Φv(f)=f(v).

You should check that the association v↦Φv is linear (so Φλv=λΦv and Φv+w=Φv+Φw) 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↦Φv is injective, but not necessarily surjective, in this case.

**Attribution***Source : Link , Question Author : Theorem , Answer Author : Community*