# Algebra, Problems For Fun (29)

• Jul 2nd 2009, 10:41 AM
NonCommAlg
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 $\displaystyle p \mid \dim V.$
• Jul 2nd 2009, 10:54 PM
Bruno J.
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 $\displaystyle n=ap$.
• Jul 3rd 2009, 09:50 AM
Bruno J.
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 $\displaystyle 2 \leq j \leq p$.
• Jul 3rd 2009, 05:22 PM
NonCommAlg
Quote:

Originally Posted by Bruno J.
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 $\displaystyle n=ap.$