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Math Help - Nilpotent

  1. #1
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    Nilpotent

    A^k=0 for some k>0

    If A and B are nilpotent with AB=BA, then A+B is nilpotent.

    Suppose B^n=0.

    Then (A+B)^{k+n}=\sum_{i=0}^{k+n}\binom{k+n}{i}A^iB^{k+  n-i}

    if 0\leq i<k, then B^{k+n-i}=0 since k+n-i>n

    if k\leq i\leq k+n, then A^i=0.

    Does this prove it?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Nilpotent

    Quote Originally Posted by dwsmith View Post
    A^k=0 for some k>0

    If A and B are nilpotent with AB=BA, then A+B is nilpotent.

    Suppose B^n=0.

    Then (A+B)^{k+n}=\sum_{i=0}^{k+n}\binom{k+n}{i}A^iB^{k+  n-i}

    if 0\leq i<k, then B^{k+n-i}=0 since k+n-i>n

    if k\leq i\leq k+n, then A^i=0.

    Does this prove it?
    Yes.

    Note that (A+B)^{k+n}=\sum_{i=0}^{k+n}\binom{k+n}{i}A^iB^{k+  n-i} because AB=BA.
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