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Vector space isomorphism

  1. Jun 9, 2008 #1
    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?
  2. jcsd
  3. Jun 9, 2008 #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.
    Last edited: Jun 9, 2008
  4. Jun 9, 2008 #3


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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?

    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?
    Last edited: Jun 9, 2008
  5. Jun 9, 2008 #4
    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
  6. Jun 10, 2008 #5


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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.
  7. Jun 10, 2008 #6


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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.
    Last edited by a moderator: Jun 10, 2008
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