Results 1 to 4 of 4

Thread: Algebra, Problems For Fun (29)

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Algebra, Problems For Fun (29)

    Let $\displaystyle V$ be a finite dimensional vector space over $\displaystyle \mathbb{C}$ and $\displaystyle T: V \longrightarrow V$ a linear transformation. Suppose that $\displaystyle \text{tr}(T)=0$ and $\displaystyle T^p = \text{id}_V,$ for some prime number $\displaystyle p.$ Prove that $\displaystyle p \mid \dim V.$
    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 $\displaystyle n=\dim V$.

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

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

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

    Setting $\displaystyle a=a_1=...=a_p$, this implies the characteristic polynomial of $\displaystyle T$ is $\displaystyle (-1)^n(t^p-1)^a$, using the identity $\displaystyle (1-t)(\omega-t)...(\omega^{p-1}-t)=t^p-1$. Hence $\displaystyle n=ap$.
    Last edited by Bruno J.; Jul 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 $\displaystyle \omega^{p-1}=-(\omega^{p-2}+\hdots+\omega+1)$

    Hence if $\displaystyle a_1\omega^{p-1}+...+a_{n-1}\omega+a_p=0,\: \: a_j \in \mathbb{N}$ then $\displaystyle (a_2-a_1)\omega^{p-2}+...+(a_{p-1}-a_1)\omega+(a_n-a_1)=0,\: \: a_j \in \mathbb{N}$. But $\displaystyle 1,\omega,...\omega^{p-2}$ are linearly independent over $\displaystyle \mathbb{Q}$ hence $\displaystyle 0=a_j-a_1$ for all $\displaystyle 2 \leq j \leq p$.
    Last edited by Bruno J.; Jul 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 $\displaystyle n=\dim V$.

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

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

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

    Setting $\displaystyle a=a_1=...=a_p$, this implies the characteristic polynomial of $\displaystyle T$ is $\displaystyle (-1)^n(t^p-1)^a$, using the identity $\displaystyle (1-t)(\omega-t)...(\omega^{p-1}-t)=t^p-1$. Hence $\displaystyle n=ap$.
    $\displaystyle a_1= \cdots = a_p$ is a result of irreducibility of $\displaystyle x^{p-1} + \cdots + x + 1$ over $\displaystyle \mathbb{Q}.$ to prove that $\displaystyle n=ap$ just note that $\displaystyle J$ is both $\displaystyle n \times n$ and $\displaystyle ap \times ap$ matrix and thus $\displaystyle n=ap.$
    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: Jul 29th 2009, 10:50 AM
  2. Algebra, Problems For Fun (21)
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Jun 22nd 2009, 06:57 AM
  3. Algebra, Problems For Fun (22)
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jun 20th 2009, 11:59 PM
  4. Algebra, Problems For Fun (13)
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Jun 5th 2009, 02:53 PM
  5. Need Help With All these Algebra Problems.
    Posted in the Algebra Forum
    Replies: 10
    Last Post: Nov 26th 2006, 05:42 PM

Search Tags


/mathhelpforum @mathhelpforum