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Linear Algebra Composition Isomorphism Question

  1. Feb 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Let S : U →V and T : V →W be linear maps.

    Given that dim(U) = 2, dim(V ) = 1, and dim(W) = 2, could T composed of S be an isomorphism?

    2. Relevant equations

    If Dim(v) > dim(W), then T is 1-1
    If Dimv < dim(w), then T is not onto.

    3. The attempt at a solution

    So this seems like a tricky question, but I am having trouble proving whether or not this is an isomorphism. While I know that T composed of S (u) = T(S(u)), S: U→V and T:V→W, this would be an isomorphism if dim(U) = Dim(W). But does the fact that S maps to V, which is one dimension less than U and W, affect that this is an isomorphism?

    I think an example could be helpful in helping me understand this problem.
    Thank you.
     
  2. jcsd
  3. Feb 9, 2014 #2
    That should read "If Dim(V) > Dim(W), then T is not 1-1". I'm sure that's what you meant, but just making sure.

    Anyway, lets suppose that TS is an isomorphism. Which would mean that given w [itex]\in[/itex] W, there is a u [itex]\in[/itex] U. Such that TS(u) = w. Is this a problem?
     
  4. Feb 9, 2014 #3
    That makes sense, kduna, thank you. So Basically this composition bypasses the use of V, and there exists an isomorphism from U to W. Thanks.
     
  5. Feb 9, 2014 #4
    Actually no. w = TS(u) = T(S(u)). So w [itex]\in[/itex] Range(T). But this can be done for any w [itex]\in[/itex] W. This would mean T is surjective, a contradiction since dim(V) < dim(W).

    Therefore TS cannot be an isomorphism.

    Note that since dim(U) = 2 = dim(W), an isomorphism does exist between U and W. But TS is not an isomorphism.
     
  6. Feb 9, 2014 #5
    Ahh, I see. That makes much more sense. Thank you for your help, kduna.
     
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