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Homework Help: Linear Algebra, Linear Maps

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Suppose that T is a linear map from V to F, where F is either R or C. Prove that if u is an element of V and u is not an element of null(T), then

    V = null(T) (direct sum) {au : a is in F}.

    2. Relevant information
    null(T) is a subspace of V
    For all u in V, u is not in null(T)
    For all n in V, n is in null(T)
    T(n) = 0, T(u) not= 0

    3. The attempt at a solution
    I think I should let U = {au : a is in F} and show that it's a subspace of V. Then I can show that each element of V can be written uniquely as a sum of u + n. Should I do this by showing that (u, n) is a basis for V?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 20, 2007 #2

    Dick

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    What's the dimension of null(T) in terms of the dimension of V?
     
  4. Sep 20, 2007 #3
    if V were finite dimensional then I could say, dim{null(T)} = dim(V) - dim{range(T)}.

    But nothing given in the problem statement will let me assume V is finite.
     
  5. Sep 20, 2007 #4

    Dick

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    Ok, I think I was thinking about this wrong. Suppose x=a*u+n and x=a'*u+n' where n and n' are in the null space. Then T(x)=a*T(u)=a'*T(u). Since T(u) is nonzero, a=a'. Right? So x=a*u+n and x=a*u+n'. Can n and n' be different?
     
  6. Sep 20, 2007 #5
    n and n' could definitely be different, but I don't think it matters much since they both get mapped to zero.

    Is the result of a = a' is enough to prove uniqueness for a direct sum?
     
  7. Sep 20, 2007 #6

    Dick

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    It does matter in V. But if x=a*u+n and x=a*u+n' (after you've shown a=a') then n=x-a*u and n'=x-a*u. Conclusion?
     
  8. Sep 20, 2007 #7
    Gosh, I must be getting sleepy to overlook the importance of n being unique.

    So, I can show that each element of V can be written uniquely as a sum of u + n.

    Should I also prove U = {au : a is in F} is a subspace of V
     
  9. Sep 20, 2007 #8

    Dick

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    Even being sleepy, I think you could, right? I think it could actually be considered as 'obvious' and not deserving of proof.
     
  10. Sep 20, 2007 #9
    :zzz:

    I should be able to stay awake long enough to write down my solution.

    Thanks for the help!
     
  11. Sep 20, 2007 #10

    Dick

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    Very, very welcome.
     
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