Results 1 to 4 of 4

Math Help - Algebra, Problems For Fun (29)

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Algebra, Problems For Fun (29)

    Let V be a finite dimensional vector space over \mathbb{C} and T: V \longrightarrow V a linear transformation. Suppose that \text{tr}(T)=0 and T^p = \text{id}_V, for some prime number p. Prove that p \mid \dim V.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Let n=\dim V.

    Expressing T in its Jordan Form J we find that J has zero trace.

    Moreover since \det J=\det T is a pth root of 1, the diagonal entries of J must all be of the form \omega^k,\: 0<k<p,\: \omega^p=1.

    Finally it is an easy exercise to show that if a_1+a_2 \omega+...+ a_p \omega^{p-1}=0=\mbox{ Tr }J then a_1=...=a_p; hence there are as many of each of the elements \{1,\omega,...\omega^{p-1}\} on the main diagonal of J.

    Setting a=a_1=...=a_p, this implies the characteristic polynomial of T is (-1)^n(t^p-1)^a, using the identity (1-t)(\omega-t)...(\omega^{p-1}-t)=t^p-1. Hence n=ap.
    Last edited by Bruno J.; July 3rd 2009 at 03:56 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Proof of the above "easy exercise"

    We have \omega^{p-1}=-(\omega^{p-2}+\hdots+\omega+1)

    Hence if a_1\omega^{p-1}+...+a_{n-1}\omega+a_p=0,\: \: a_j \in \mathbb{N} then (a_2-a_1)\omega^{p-2}+...+(a_{p-1}-a_1)\omega+(a_n-a_1)=0,\: \: a_j \in \mathbb{N}. But 1,\omega,...\omega^{p-2} are linearly independent over \mathbb{Q} hence 0=a_j-a_1 for all 2 \leq j \leq p.
    Last edited by Bruno J.; July 3rd 2009 at 03:57 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Bruno J. View Post
    Let n=\dim V.

    Expressing T in its Jordan Form J we find that J has zero trace.

    Moreover since \det J=\det T is a pth root of 1, the diagonal entries of J must all be of the form \omega^k,\: 0<k<p,\: \omega^p=1.

    Finally it is an easy exercise to show that if a_1+a_2 \omega+...+ a_p \omega^{p-1}=0=\mbox{ Tr }J then a_1=...=a_p; hence there are as many of each of the elements \{1,\omega,...\omega^{p-1}\} on the main diagonal of J.

    Setting a=a_1=...=a_p, this implies the characteristic polynomial of T is (-1)^n(t^p-1)^a, using the identity (1-t)(\omega-t)...(\omega^{p-1}-t)=t^p-1. Hence n=ap.
    a_1= \cdots = a_p is a result of irreducibility of x^{p-1} + \cdots + x + 1 over \mathbb{Q}. to prove that n=ap just note that J is both n \times n and ap \times ap matrix and thus n=ap.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Algebra 2 problems.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 29th 2009, 10:50 AM
  2. Algebra, Problems For Fun (21)
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: June 22nd 2009, 06:57 AM
  3. Algebra, Problems For Fun (22)
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 20th 2009, 11:59 PM
  4. Algebra, Problems For Fun (13)
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: June 5th 2009, 02:53 PM
  5. Need Help With All these Algebra Problems.
    Posted in the Algebra Forum
    Replies: 10
    Last Post: November 26th 2006, 05:42 PM

Search Tags


/mathhelpforum @mathhelpforum