Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

About the isomorphism of 2 infinite-dimensional vector spaces

  1. Apr 27, 2009 #1
    V denotes a vector space
    A, B, C, D denote subspaces of V respectively
    ≈ denotes the isomorphic relationship of the left and right operand
    dim(?) denotes the dimension of "?"

    Find a vector space V and decompositions:
    V = A ⊕ B = C ⊕ D
    with A≈C but B and D are not isomorphic.

    My opinion:
    dim(V)=dim(A)+dim(B)=dim(C)+dim(D) and dim(A)=dim(C), but dim(B)≠dim(D) since V may not be finite-dimensional. It's an idea not an example, would you make a concrete example of V?

    Thanks for any help!
  2. jcsd
  3. Apr 27, 2009 #2
    (Let ~ denote isomorphism, + a direct sum, and <S> the span of the set S. Sorry, but Latex seems to be out of commission.)

    I think this works:

    Let V be the set of all infinite ordered-tuples of real numbers with only finitely many nonzero entries, i.e., the set of all infinite sequences that eventually terminate, such as {3,2,1,0,0,...}. Let ei denote the sequence with a 1 in the ith place and zeros elsewhere. Then B = {e1,e2,...} is a basis for V over R. Let N1 = {e1} and N2 = {e1,e2}. Then V ~ V + <N1> ~ V + <N2> (I think). Clearly, <N1> is not isomorphic to <N2>, since their dimensions differ.
  4. Apr 27, 2009 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    VKint, I think the OP is looking for internal sums decompositions V = A ⊕ B = C ⊕ D.

    But I think this slight variation on your idea works: take A:=<e_2,e_3,...>, B:=<e_1>, C:= <e_3,e_4,...>, D:=<e_1,e_2>.
  5. Apr 29, 2009 #4

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook