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Rank proof

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Given a linear operator T, show that if rank(T^2)=rank(T), then the range and null space are disjoint.

    So I know that I can form a the same basis for range(T^2) and range(T), but I'm not sure where to go from there.
     
  2. jcsd
  3. Oct 8, 2009 #2

    HallsofIvy

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    I recommend you go back and reread the problem. The range and nullspace of any linear operator are subspaces- they both include the 0 vector and so are never disjoint. Perhaps the problem is to show that the only vector in both the range and null space is the 0 vector?

    If so try a proof by contradiction. Let v be a non-zero vector in both range and null space of T. Since v is in the range of T there exist u such that v= Tu. Now, what is T2u?
     
  4. Oct 8, 2009 #3
    You're right, I meant disjoint other than 0. I get that T2u=Tv=0, but what does that say about the rank?
     
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