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Thread: Nilpotent

  1. #1
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    Nilpotent

    $\displaystyle A^k=0$ for some $\displaystyle k>0$

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

    Suppose $\displaystyle B^n=0$.

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

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

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

    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
    $\displaystyle A^k=0$ for some $\displaystyle k>0$

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

    Suppose $\displaystyle B^n=0$.

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

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

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

    Does this prove it?
    Yes.

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