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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 ur advice , examples and theories.

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2 Answers 2

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.

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Thanks! I have wanted to know about that for some time, and your explanation was very clear. –  MJD Jul 13 '12 at 21:42
@countinghaus : So what belongs to the double dual space ? They are also functions right ? ie , these functions make the functionals of dual space to take a vector of vector space and spit out field element . right ? –  Theorem Jul 13 '12 at 21:44
@countinghaus : So what kind of functions live in the double dual? –  Theorem Jul 13 '12 at 21:47
can't make a comment because of tex problems in comment box. –  user29743 Jul 13 '12 at 23:05
@Theorem: The double dual space contains linear functions mapping linear functions from the dual space to scalars (members of the field the vector space is defined over). –  celtschk Jul 14 '12 at 10:34

Actually it's quite simple: If you have a vector space, any vector space, you can define linear functions on that space. The set of all those functions is the dual space of the vector space. The important point here is that it doesn't matter what this original vector space is. You have a vector space $V$, you have a corresponding dual $V^*$.

OK, now you have linear functions. Now if you add two linear functions, you get again a linear function. Also if you multiply a linear function with a factor, you get again a linear function. Indeed, you can check that linear functions fulfill all the vector space axioms this way. Or in short, the dual space is a vector space in its own right.

But if $V^*$ is a vector space, then it comes with everything a vector space comes with. But as we have seen in the beginning, one thing every vector space comes with is a dual space, the space of all linear functions on it. Therefore also the dual space $V^*$ has a corresponding dual space, $V^{**}$, which is called double dual space (because "dual space of the dual space" is a bit long).

So we have the dual space, but we also want to know what sort of functions are in that double dual space. Well, such a function takes a vector from $V^*$, that is, a linear function on $V$, and maps that to a scalar (that is, to a member of the field the vector space is based on). Now, if you have a linear function on $V$, you already know a way to get a scalar from that: Just apply it to a vector from $V$. Indeed, it is not hard to show that if you just choose an arbitrary fixed element $v\in V$, then the function $F_v\colon\phi\mapsto\phi(v)$ indeed is a linear function on $V^*$, and thus a member of the double dual $V^{**}$. That way we have not only identified certain members of $V^{**}$ but in addition a natural mapping from $V$ to $V^{**}$, namely $F\colon v\mapsto F_v$. It is not hard to prove that this mapping is linear and injective, so that the functions in $V^{**}$ corresponding to vectors in $V$ form a subspace of $V^{**}$. Indeed, if $V$ is finite dimensional, it's even all of $V^{**}$. That's easy to see if you know that $\dim(V^*)=\dim{V}$ and therefore $\dim(V^{**})=\dim{V^*}=\dim{V}$. On the other hand, since $F$ is injective, $\dim(F(V))=\dim(V)$. However for finite dimensional vector spaces, the only subspace of the same dimension as the full space is the full space itself. However if $V$ is infinite dimensional, $V^{**}$ is larger than $V$. In other words, there are functions in $V^{**}$ which are not of the form $F_v$ with $v\in V$.

Note that since $V^{**}$ again is a vector space, it also has a dual space, which again has a dual space, and so on. So in principle you have an infinite series of duals (although only for infinite vector spaces they are all different).

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