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Nilpotent Operator

  1. Apr 3, 2006 #1
    Suppose V is finite dimensional and N:V->V is a linear transformation such that N^e=0. Is it possible to show that e[tex]\leq[/tex]dim V? Is it even true?
     
  2. jcsd
  3. Apr 3, 2006 #2

    matt grime

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    Yes, it is very easy to show this, or rather the correct statement when e is the minimal positive integer such that N^e=0, so try it.
     
  4. Apr 3, 2006 #3
    I got it. It's part of a much larger proof. Thanks.
     
  5. Feb 6, 2011 #4
    Yes it's true. If V were represented via its "basis" as R^n, (n=dim V) then, N can be represented as a nilpotent matrix. We knew that with appropriate representation, N is a strictly upper triangular (diagonal and lower triangular elements are all zeroes) matrix. And now we just have to show that a strictly upper triangular matrix the property that N^n is always zero. (Sorry, I use N (the operator, as the matrix).
     
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