Algebra, Problems For Fun (29)

**NonCommAlg** · July 2nd 2009, 10:41 AM

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 $p \mid \dim V.$

**Bruno J.** · July 2nd 2009, 10:54 PM

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 $p$ th 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 $n=ap$ .

**Bruno J.** · July 3rd 2009, 09:50 AM

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 $2 \leq j \leq p$ .

**NonCommAlg** · July 3rd 2009, 05:22 PM

Originally Posted by Bruno J.

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 $p$ th 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 $n=ap.$

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