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John Baez
science forum Guru Wannabe
Joined: 01 May 2005
Posts: 220
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 Posted: Tue Jun 27, 2006 5:33 am Post subject:
Dual vector spaces
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Every finite-dimensional vector space is isomorphic to
its dual.
Is it true that no infinite-dimensional vector space is
isomorphic to its dual?
Feel free to pick your favorite field if that helps.
Feel free to use the axiom of choice if that helps, too.
But please let me know if you're using it.
(I emphasize that I'm talking about a purely algebraic
question, not about topological vector spaces. Every
Hilbert space is isomorphic to its *topological*
dual, consisting of *continuous* linear functionals.
That's not what I'm talking about. I'm talking about
the *algebraic* dual of a vector space, consisting of
*all* linear functionals. And by "isomorphism", I just
mean a linear operator with a linear inverse.) |
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David Madore
science forum beginner
Joined: 23 Feb 2005
Posts: 24
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| Posted: Tue Jun 27, 2006 11:35 am Post subject:
Re: Dual vector spaces
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John Baez in litteris <e7qfv1$67o$1@glue.ucr.edu> scripsit:
| Quote: | Is it true that no infinite-dimensional vector space is
isomorphic to its dual?
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Yes. Using the axiom of choice, vector spaces over a field k are
uniquely determined by their dimension over k, and one can show that
if k has cardinal mu (finite or infinite) and E is a k-vector space of
infinite dimension lambda, then the dual E' of E has dimension
mu^lambda: thus, always greater than lambda. [Sketch of proof: the
cardinal of E' is clearly mu^lambda, so if mu^lambda is greater than
mu then E' also has dimension mu^lambda. The remaining case is when
mu^lambda = mu; embed lambda in k by a map h:lambda->k and consider
the functions lambda->k given by 1/(h(x)-a) for a in k: they are
linearly independent, so the dimension of E' is at least mu.]
| Quote: | Feel free to pick your favorite field if that helps.
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It works over any field.
| Quote: | Feel free to use the axiom of choice if that helps, too.
But please let me know if you're using it.
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The axiom of choice is needed. In the absence of the axiom of choice,
it is consistent that the dual of R as a Q-vector space is R.
--
David A. Madore
(david.madore@ens.fr,
http://www.dma.ens.fr/~madore/ ) |
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José Carlos Santos
science forum Guru
Joined: 25 Mar 2005
Posts: 1111
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| Posted: Tue Jun 27, 2006 1:30 pm Post subject:
Re: Dual vector spaces
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On 27-06-2006 6:33, John Baez wrote:
| Quote: | Every finite-dimensional vector space is isomorphic to
its dual.
Is it true that no infinite-dimensional vector space is
isomorphic to its dual?
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It is true for every field F. To be more precise,
dim(V*) = (#F)^(dim(V)).
You'll find a proof of this assertion on page 247 of the second volume
of Jacobson's "Lictures in Abstract Algebra". The proof uses Zorn's
lemma (which is equivalent to the axiom of choice).
Best regards,
Jose Carlos Santos |
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G. A. Edgar
science forum Guru
Joined: 29 Apr 2005
Posts: 470
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| Posted: Tue Jun 27, 2006 2:04 pm Post subject:
Re: Dual vector spaces
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In article <e7qfv1$67o$1@glue.ucr.edu>, John Baez
<baez@math.removethis.ucr.andthis.edu> wrote:
| Quote: | Every finite-dimensional vector space is isomorphic to
its dual.
Is it true that no infinite-dimensional vector space is
isomorphic to its dual?
Feel free to pick your favorite field if that helps.
Feel free to use the axiom of choice if that helps, too.
But please let me know if you're using it.
(I emphasize that I'm talking about a purely algebraic
question, not about topological vector spaces. Every
Hilbert space is isomorphic to its *topological*
dual, consisting of *continuous* linear functionals.
That's not what I'm talking about. I'm talking about
the *algebraic* dual of a vector space, consisting of
*all* linear functionals. And by "isomorphism", I just
mean a linear operator with a linear inverse.)
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Is it true, using AC, that if a vector space has infinite
Hamel dimension m, then the dual has Hamel dimension 2^m ?
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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Daniel Asimov
science forum beginner
Joined: 03 May 2005
Posts: 13
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| Posted: Thu Jun 29, 2006 3:15 am Post subject:
Re: Dual vector spaces
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Gerald Edgar ( http://www.math.ohio-state.edu/~edgar/) asks:
<<
Is it true, using AC, that if a vector space has infinite Hamel
dimension m, then the dual has Hamel dimension 2^m ?
(A Hamel basis for R over Q is familar, but I had to look up "Hamel
dimension". According to Wikipedia, it is just the ordinary dimension
of a vector space V, aka dim(V).)
According to the posts of Madore and Santos, the Axiom of Choice
implies dim(V*) = (#(F))^(dim(V)), where V is a vector space over the
field F.
As long as dim(V) > 0, and #(F) >= beta > 2^(dim(V)) for some cardinal
beta, we will have dim(V*) = beta^(dim(V)) >= beta > 2^(dim(V)).
So the answer to the question is No if we can find a field F with #(F)
| Quote: | = beta > 2^(dim(V)).
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For this, just take pick any beta > 2^(dim(V)) and let F = the field of
rational functions in a set of variables of cardinality beta, say over
Z/2Z.
Or am I missing something here?
--Dan Asimov |
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Robert E. Beaudoin
science forum beginner
Joined: 24 May 2005
Posts: 10
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| Posted: Thu Jul 20, 2006 12:04 pm Post subject:
Re: Dual vector spaces
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David Madore wrote:
| Quote: | John Baez in litteris <e7qfv1$67o$1@glue.ucr.edu> scripsit:
Is it true that no infinite-dimensional vector space is
isomorphic to its dual?
Yes. Using the axiom of choice, vector spaces over a field k are
uniquely determined by their dimension over k, and one can show that
if k has cardinal mu (finite or infinite) and E is a k-vector space of
infinite dimension lambda, then the dual E' of E has dimension
mu^lambda: thus, always greater than lambda. [Sketch of proof: the
cardinal of E' is clearly mu^lambda, so if mu^lambda is greater than
mu then E' also has dimension mu^lambda. The remaining case is when
mu^lambda = mu; embed lambda in k by a map h:lambda->k and consider
the functions lambda->k given by 1/(h(x)-a) for a in k: they are
linearly independent, so the dimension of E' is at least mu.]
Feel free to pick your favorite field if that helps.
It works over any field.
Feel free to use the axiom of choice if that helps, too.
But please let me know if you're using it.
The axiom of choice is needed. In the absence of the axiom of choice,
it is consistent that the dual of R as a Q-vector space is R.
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Could you provide a reference (or better still a proof sketch) for this
consistency result? It's plausible enough, but my off the cuff attempts
to find a model have failed so far.
Robert E. Beaudoin |
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G. A. Edgar
science forum Guru
Joined: 29 Apr 2005
Posts: 470
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| Posted: Thu Jul 20, 2006 2:00 pm Post subject:
Re: Dual vector spaces
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| Quote: | David Madore wrote:
The axiom of choice is needed. In the absence of the axiom of choice,
it is consistent that the dual of R as a Q-vector space is R.
|
Robert E. Beaudoin <rbeaudoin@comcast.net> wrote:
| Quote: |
Could you provide a reference (or better still a proof sketch) for this
consistency result? It's plausible enough, but my off the cuff attempts
to find a model have failed so far.
Robert E. Beaudoin
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I thought I knew an answer, but when I wrote it out I find that it is
something else...
Take Solovay's model where every subset of R has the property of
Baire. It follows that every additive homomorphism R -> R is
continuous. Therefore, every additive homomorphism R -> Q is
zero. So in THAT model, the dual of R (as a Q-vector space) is 0.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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