Motivation to understand double dual space

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 ij. 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 vV. 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 vΦv 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 vΦv is injective, but not necessarily surjective, in this case.

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

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