- #1

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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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- Thread starter Markjdb
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- #1

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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?

- #2

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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.

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

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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- #4

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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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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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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.

I might be wrong, but it seems to me that in this vector space, any two subspaces without finite bases should be isomorphic

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