Every vector space is the dual of some other vector space

[ak416]

Can someone prove this to me? I know that if you have a finite dimensional vector space V with a dual space V*, then every ordered basis for V* is the dual basis for some basis for V (this follows from a theorem). But if you're just given an arbitrary vector space V. Let's say the Space of R^n over the field R, then how could the elements of that space be linear functions of another space (and thus V is the dual space of another space)? And what would that space be? All i can guess is that V can be thought of as set of constant functions...

The two main theorems I am supplied with in this section are

Thm: Suppose V is a finite dimensional vector space with an ordered basis B = {x_1,...,x_n}. Let f_i (1<= i<=n) be the ith coordinate function wrt B, and let B* = {f_1,...,f_n}. Then B* is an ordered basis for V*, and, for any f element of V*, we have f = Sum i=1 to n[f(x_i)f_i

Thm: Let V be a finite dimensional vector space, and and define U: V->V** by U(x) = x' where x': V* -> F is defined by x'(f) = f(x) (F is a field). Then U is an isomorphism.

So again, I would like a proof of the statement "Every vector space is a dual of some other vector space" and an example using let's say R^n as the dual space. Thanks.

 

[matt grime]

I don't think you can prove it because I dont' think it is true. This is because you are missing the important words 'finite dimensional' from your post.

For a finite dimensional vector space it is correct, and you even write down the proof: V** is naturally isomorphic to V.

 

[ak416]

So what if it's isomorphic? It just tells you that given a space V** and another V (so you already know V** is the double dual of a vector space), there is a one-one mapping between all their members. And obviously because a V** and V exists, there must exist something in between, V* (and therefore V** is the dual of V*). But doesn't this depend on the assumption that V** is the double dual of a vector space?

 

[matt grime]

So you know if V is finite dimensional:
(we'll use = to mean isomorphic)

1. V=V**
2. V** is the dual of the vector space V*, i.e. if we set W=V* then V=W*.What is it about these two points that troubles you? The fact that things are only defined up to isomorphism and not genuine equality? That's all we can define things 'up to', though.

(And the result is false, I'm fairly sure, for infinite dimensional vector spaces: I don't believe l_1 is the dual of any vector space)

 

[ak416]

Well what you have proven is IF V IS THE DOUBLE DUAL OF ANOTHER VECTOR SPACE then V is the dual of another vector space. In your case you used V** as that V, and the other vector space as V.

But how do you prove that any vector space V is the dual of some other vector space without assuming that V is the double dual of some other vector space?

 

[matt grime]

But if V is finite dimensional V is the double dual of a vector space, itself; you know this: you write this as one of the two theorems that you knew. (And as I said, the result is I believe false for infinite dimensional vector spaces; since you are dealing with finite dimensional ones, or your second stated theorem that proves the result is false, there is no problem here). There is no need to speak in caps. Just take a look at what has been written, including what you wrote as the two things you know.

Taking duals is a (contravariant) functor from Vect, the category of finite dimensional vector spaces, to Vect^{op}, it is 'self' inverse. Any finite dimensional vector space V is iso to W* where W=V*. What don't you agree with here? I am not assuming anything; all of these statements are easy to prove.

 

[Hurkyl]

This is just algebra. Forget that we're talking about vector spaces and linear functionals and stuff!

If V is finite dimensional, V = V**.
If V = W, then V* = W*

Your goal: If W is finite dimensional, find a V such that V* = W.

 

[HallsofIvy]

Well what you have proven is IF V IS THE DOUBLE DUAL OF ANOTHER VECTOR SPACE then V is the dual of another vector space. In your case you used V** as that V, and the other vector space as V.

But how do you prove that any vector space V is the dual of some other vector space without assuming that V is the double dual of some other vector space?

You don't have to "assume" it- it is true. If V is any vector space, then V is isomorphic to the dual of V*- that is V is isomorphic to V**.

If what is bothering you is that what can be proved "V is isomorphic to the dual of some other space" rather than "V is the dual of some other space", then I don't believe you can prove "V is the dual of some other space" since the definition of "dual of W" is "the set of all linear functions from the vector space W to the real numbers" and clearly, not every vector space V is a set of linear functionals. The most you can prove is that every vector space V is isomorphic to the dual of some other space.

 

[ak416]

HallsOfIvy: Ya, that's what's troubling me. It seems as if everyone is using "is" and "is isomorphic to" interchangeably, and only able to prove the latter but not the former. My book is careful with these words and it had a true or false question stating whether every vector is space is the dual of another vector space and it came out in the solutions that its true. So are you sure you can't prove the "IS" statement?
By the way, linear functionals are defined as taking a vector space V to any field F (not just the real numbers)

 

[Hurkyl]

I thought that might have been your problem at first, but you said things like:

All i can guess is that V can be thought of as set of constant functions...

so I thought that wasn't the problem. (an isomorphism gives us a way to "think of" things as something else)

So are you sure you can't prove the "IS" statement?

Trivially so -- if the vector space contains something that isn't a function, then it certainly cannot literally be a space of functions.