Vector space isomorphism

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I came across this problem today and haven't been able to figure it out...

Give an example of a vector space V which isomorphic to a proper subspace W, i.e. V != W.

It seems to me that V can't have a finite basis, but can't think of any examples regardless...any thoughts?
 

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  • #2
Polynomials over a field

Let V=C[x,y]. A basis for this space is [tex] $ B= \{ x^i y^j \mid i,j=0,1,2,...\} $ [/tex]. It is well known that there is a bijection [tex] $ f: Z_{+} \times Z_{+} \rightarrow Z_{+} $ [/tex]. Therefore, if we let [tex] $ t_{i,j} =x^iy^j \ \forall i, j $ [/tex], then we have a bijective map from B to the set [tex] $ B' = \{T^k \mid k \in Z_{+}\} $ given by $ F(t_{i,j}) = T^{f(i,j)} $ [/tex]. Clearly F linearly extends from the basis to all of V and is an isomorphism onto C[T]. You then may trivially send C[T] to C[x] via the isomorphism T -> x.
 
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  • #3
Hurkyl
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It seems to me that V can't have a finite basis, but can't think of any examples regardless...any thoughts?
I assume you've considered infinite dimensional vector spaces; where did you run into difficulty showing that one might be isomorphic to a proper subspace?

Analogy might help -- can you think of any other infinitary structure that is isomorphic to a proper substructure? What about the simplest kind of structure: that of simply being a set?
 
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Even easier, consider the vector space of infinite sequences of real numbers (or equivalently, countably infinite tuples)

I might be wrong, but it seems to me that in this vector space, any two subspaces without finite bases should be isomorphic
 
  • #5
mathwonk
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any bijection between bases yields and isomorphism between the spaces, so just find a basis and a bijection with a proper subset.

e.g. if the basis is the natural numbers, the usual bijection (x-->x+1)with those > 1 induces the famous "shift operator" on the space of (finite) sequences.
 
  • #6
morphism
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Even easier, consider the vector space of infinite sequences of real numbers (or equivalently, countably infinite tuples)

I might be wrong, but it seems to me that in this vector space, any two subspaces without finite bases should be isomorphic
The sequences with only finitely many nonzero terms form a subspace that admits a countably infinite basis, whereas the space of absolutely summable sequences (l_1) has as its dimension the cardinality of the continuum. So these two spaces aren't isomorphic. But on the other hand, all the l_p spaces (for 1<=p<infinity) are isomorphic as vector spaces, and l_p is a proper subspace of l_q when p<q.
 
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