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Homework Help: Linear transformation from V to V proof

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Let V be a vector space over a field F and let L(V) be the vector space of linear transformations from V to V. Suppose that T is in L(V). Do not assume that V is finite-dimensional.
    a) Prove that T^2 = -T iff T(x) = -x for all x in R(T).
    b) Suppose that T^2 = -T. Prove that the intersection of N(T) and R(T) = {0}.


    2. Relevant equations

    3. The attempt at a solution
    The T^2 is throwing me off slightly. Does it just mean take the square of the original linear transformation T?
    So if T^2 = -T,
    T^2 (x) = -T(x)
    Also rank(T^2) = rank (-T)
    That's all I have so far. How can I approach this if I can't assume that V is finite-dimensional?
    If I could get a few tips on how to start this question, it would be very helpful :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 9, 2008 #2

    Dick

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    You haven't gotten very far. Can you show if T(x)=-x, then T^2(x)=-T(x)? That's the easy direction. For the 'only if', if x is in R(T) then there is a y in V such that x=T(y). Act with T on both sides of that and use T^2=-T.
     
  4. Nov 9, 2008 #3
    Hmm....I might be completely missing the point here...

    Show that if T(x) = -x, then T^2(x) = -T(x)
    If T(x) = -x,
    T^2(x) = (-x)^2 = x^2
    -T(x) = -(-x) = x
    T^2(x) does not equal -T(x)
     
  5. Nov 9, 2008 #4

    Dick

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    You are missing the point. T^2(x) means T(T(x)). You can't really 'square' a vector.
     
  6. Nov 9, 2008 #5
    Thanks. I got it.
    Just to make sure, for part b), would N(T) = 0 since it is a transformation from V to V? Also would R(T) = x?
     
  7. Nov 9, 2008 #6

    Dick

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    Why would you think N(T)=0 (null space, right?)? And why would R(T) (range, right?) be x, whatever that is? R(T) is a subspace. I think you'd better look up those definitions again.
     
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