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Jordan Conanical Form

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Let T: V [tex]\rightarrow[/tex] W be a linear trans. Prove:
    a) N(T) = N(-T)
    b) N(Tk) = N((-T)k)
    c) If V = W and [tex]\lambda[/tex] is an eigenvalue of T, then for any positive integer k:
    N((T - [tex]\lambda[/tex]Iv)k) = N(([tex]\lambda[/tex]Iv - T)k)

    2. Relevant equations



    3. The attempt at a solution
    Im not sure how to start on a. I know if I can get started on that one, I can handle the rest. I like this:

    -T(v) = -0
    -T(v) = 0
    -T(v) = T(v)

    therefore N(T) = N(-T) ?

    Im not sure if that's even remotely on the right track, but this question is in the first Jordan Conanical form section, and Im not sure how that concept can be applie to this very first question. I can see it needing to be applied in the other two, but not this one.
     
  2. jcsd
  3. May 18, 2010 #2

    Mark44

    Staff: Mentor

    How does this follow from what appears before it. You're trying to prove that the nullspace of T is the same as the nullspace of -T.

    Start by assuming that v is in null(T). Can you show that v is also in null(T)?
     
  4. May 18, 2010 #3
    OK so if v is in N(T) then T(v) = 0

    Since -T(v) is equal to T(-v), we need to show that -v is also in N(T)?

    I think Im thinking what Landau was thinking in this thread:
    https://www.physicsforums.com/showthread.php?t=403507
    "It looks correct. Basically, you're proving that for any linear map T, T and -T have the same kernel. This is true because Tv=0 iff -Tv=0"

    T(v) = 0 iff -T(v) = 0, since we are only talking about v in N, then T(v) must equal 0 then -T(v) = 0 this N(T) = N(-T), no?
     
    Last edited: May 18, 2010
  5. May 18, 2010 #4

    Mark44

    Staff: Mentor

    Yes, you need to show that -v is also in N(T).

    Do you know why -T(v) = T(-v)?
     
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