Results 1 to 5 of 5

Thread: Nilpotent

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    Nilpotent

    An nxn matrix is said to be nilpotent if $\displaystyle A^k=0$ for some positive $\displaystyle \mathbb{Z}$ k. Show that all eigenvalues of a nilpotent matrix are 0.

    I have proved by math induction that, for $\displaystyle m\geq1$, $\displaystyle \lambda^m$ is an eigenvalue of $\displaystyle A^m$.

    I don't know if that should help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,793
    Thanks
    3035
    Quote Originally Posted by dwsmith View Post
    An nxn matrix is said to be nilpotent if $\displaystyle A^k=0$ for some positive $\displaystyle \mathbb{Z}$ k. Show that all eigenvalues of a nilpotent matrix are 0.

    I have proved by math induction that, for $\displaystyle m\geq1$, $\displaystyle \lambda^m$ is an eigenvalue of $\displaystyle A^m$.

    I don't know if that should help.
    Yes, that certainly does help! I presume that in doing that you also showed that if v is an eigenvector of A corresponding to eigenvalue $\displaystyle \lambda$ then v is also an eigenvector of $\displaystyle A^m$ corresponding to eigenvalue $\displaystyle \lambda^m$.

    In particular, if $\displaystyle \lambda$ an eigenvalue of A, then $\displaystyle \lambda^k$ is an eigenvalue of $\displaystyle A^k$- that is, for some non-zero vector v, $\displaystyle A^k v= \lambda^k v$. But $\displaystyle A^kv= 0$ for any vector so we have $\displaystyle \lambda^k v= 0$, with v non-zero.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    Quote Originally Posted by HallsofIvy View Post
    Yes, that certainly does help! I presume that in doing that you also showed that if v is an eigenvector of A corresponding to eigenvalue $\displaystyle \lambda$ then v is also an eigenvector of $\displaystyle A^m$ corresponding to eigenvalue $\displaystyle \lambda^m$.

    In particular, if $\displaystyle \lambda$ an eigenvalue of A, then $\displaystyle \lambda^k$ is an eigenvalue of $\displaystyle A^k$- that is, for some non-zero vector v, $\displaystyle A^k v= \lambda^k v$. But $\displaystyle A^kv= 0$ for any vector so we have $\displaystyle \lambda^k v= 0$, with v non-zero.
    So that is all it is then?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,793
    Thanks
    3035
    Well, what do you conclude from $\displaystyle \lambda^k v= 0$?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    I should probably conclude lambda is zero.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Nilpotent
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 29th 2011, 04:41 PM
  2. Nilpotent
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Dec 20th 2009, 03:14 AM
  3. nilpotent
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Dec 6th 2009, 06:34 PM
  4. nilpotent
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 24th 2009, 10:49 AM
  5. Nilpotent Matrix
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Mar 29th 2008, 01:26 AM

Search Tags


/mathhelpforum @mathhelpforum