Results 1 to 6 of 6

Math Help - [SOLVED] Linear Algebra, Nilpotent

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    51

    [SOLVED] Linear Algebra, Nilpotent

    Let V be a finite dimensional vector space over \mathbb C. Let T be a linear transformation
    on V all of whose eigenvalues are zero. Show that T^{n} = 0 for some n, i.e.
    that T is nilpotent.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by JJMC89 View Post
    Let V be a finite dimensional vector space over \mathbb C. Let T be a linear transformation
    on V all of whose eigenvalues are zero. Show that T^{n} = 0 for some n, i.e.
    that T is nilpotent.
    Do you know the Cayley-Hamilton theorem?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by JJMC89 View Post
    Let V be a finite dimensional vector space over \mathbb C. Let T be a linear transformation
    on V all of whose eigenvalues are zero. Show that T^{n} = 0 for some n, i.e.
    that T is nilpotent.
    http://www.mathhelpforum.com/math-he...nilpotent.html
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    This is not the same.

    In your topic, you wanted to show that if A is nilpotent, then all of its eigenvalues are 0.

    Here he wants to prove that if all of A's eigenvalues are 0, then it is nilpotent.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2010
    Posts
    51

    Thumbs up

    Quote Originally Posted by Defunkt View Post
    Do you know the Cayley-Hamilton theorem?
    Yes.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,583
    Thanks
    1418
    Well, then you have it, don't you? If all eigenvalues are 0, then the characteristic equation is \lambda^n= 0 and every matrix satisfies its own characteristic equation.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Independence, Nilpotent Matrices
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: November 30th 2011, 02:21 PM
  2. linear operator nilpotent
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 20th 2011, 07:21 PM
  3. Nilpotent Linear Transformation
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: February 4th 2011, 07:42 AM
  4. Replies: 7
    Last Post: August 30th 2009, 10:03 AM
  5. [SOLVED] linear algebra textbooks
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: May 4th 2008, 05:57 AM

Search Tags


/mathhelpforum @mathhelpforum