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Nilpotent operators, matrix

  1. May 7, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that any square triangular matrix with each diagonal entry equal to zero is nilpotent

    3. The attempt at a solution
    Drawing out the matrix and multiplying seems a little tedious. Perhaps there is a better way?
    Is there another way to do this without assuming that the eigenvalues of a nilpotent operator are all 0?

    Thanks for your help!
  2. jcsd
  3. May 7, 2007 #2


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    This might not be the easiest way, but trying a few examples shows that as you raise the matrix to higher powers, the zeros creep up towards the top-right corner one space at a time. This suggests trying to prove a stronger result, that if Aij=0 for i>j-k and Bij=0 for i>j-l, then (AB)ij=0 for i>j-k-l (or something like that).
    Last edited: May 7, 2007
  4. May 8, 2007 #3

    matt grime

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    Since it is triangular with zero diagonal you know the eigenvalues are all 0, thus you know the min poly is x^n for some n. Why are you 'assuming' that the eigenvalues of a nilpotent operator are all 0? It is clearly true (over a field), and isn't important for this question, really (you state you don't want to assume nilpotent implies all e-values 0, but we actually need all e-values 0 implies nilpotent).
    Last edited: May 8, 2007
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