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Consider the subgroup $H \le \operatorname{GL}(3,3)$ generated by the two matrices $$ A = \begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \quad\mbox{ and }\quad B = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} $$ If $D \in H$ has order $6$, then I want to show that $D^3$ is diagonalisable as $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$ i.e. the eigenvalue $-1$ has geometric multiplicity $2$. I know that as $D^3$ is involutory that $\pm 1$ are the only eigenvalues, also see this post. Also of course $x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x-1)(x^2+x+1)(x+1)(x^2-x+1)$, so this polynomial is divided by the minimal polynomial of $D^6$, but I do not see if this might be helpful?

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StefanH
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  • @Bernard But not when $D^6 = I$ and $D^i \ne I$ for $i =1,2,3,4,5$; I added that to the question (sorry forgot that). For the other comment: That might be possible too, but what is essential is that both eigenvalues occur, and not just $I$ or $-I$ (but I conjecture that $-1$ has to occur twice) – StefanH Jan 31 '16 at 20:30
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    I rewrote the title to coincide with the clarified question. – Lee Mosher Jan 31 '16 at 20:32
  • OK, it's clearer now. – Bernard Jan 31 '16 at 20:34
  • Also, over $\Bbb{F}_3$ we have the factorization $$x^6-1=(x^2-1)^3=(x-1)^3(x+1)^3.$$ This may come in handy. In other words, the quadratic polynomials in your factorization are not irreducible modulo three. – Jyrki Lahtonen Jan 31 '16 at 20:40
  • It looks like (read, I haven't checked, but this is an educated guess) $H$ is the group of monomial matrices with determinant $+1$. That has order 24. Its diagonal matrices form a normal subgroup $N$ of order four, and $H/N\simeq S_3$. Wonder how that will help? – Jyrki Lahtonen Jan 31 '16 at 21:16
  • Well, if an element has order six, the $S_3$ part must be a 3-cycle. This implies that $D^3$ is itself diagonal, not just diagonalizable. That settles the question I think :-) – Jyrki Lahtonen Jan 31 '16 at 21:18
  • @JyrkiLahtonen What do you mean by "$\mathcal S_3$ part", $H$ is not a semi-direct product where we can decompose as $N \rtimes H/N$? And why does this forces that $-1$ occurs twice on the diagonal? – StefanH Jan 31 '16 at 21:22
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    You're right. It is not the semi-direct product. At least it isn't clear to me why it would be. But it is an index two subgroup of the semidirect product $G=C_2^3\rtimes S_3$, Where that Cartesian power of $C_2$ consists of the diagonal matrices. $H$ is a subgroup of $G$ such that all the matrices in $H$ have determinant $+1$. I added another answer. Turns out I didn't need to figure out the order of $H$. – Jyrki Lahtonen Jan 31 '16 at 21:32
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    So we have a homomorphism from $H$ to $S_3$ that simply replaces all the $-1$ entries with $1$s. And (with a bit of hindsight) what I called the $S_3$ part really is the image under that homomorphism. – Jyrki Lahtonen Jan 31 '16 at 21:35

2 Answers2

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This answer is to an earlier version of the question, when the subgroup was not specified.

Collecting some of my comments to an answer. The claim is false as stated. The order six matrix $$ D=\pmatrix{-1&1&0\cr0&-1&0\cr0&0&-1\cr} $$ raised to third power gives $-I$ and thus has $-1$ as an eigenvalue of multiplicity three.

Similarly another order six matrix $$ D=\pmatrix{1&1&0\cr0&1&0\cr0&0&-1\cr} $$ has as the cube the diagonal matrix with $1$ occuring twice and $-1$ once.

These examples show that

  • it is not necessarily the case that both $+1$ and $-1$ occur as eigenvalues (one of the formulations of the question from the comments)
  • $D^3$ may have $-1$ as an eigenvalue with multiplicity one only.

Given this I'm still wondering what the question really was?

Observe that any Jordan block of size two or three can be written in the form $$ J=rI+N, $$ where $N^3=0$. Because we are in characteristic three and $I$ and $N$ commute, the binomial formula gives $$ J^3=r^3I+3r^2N+3rN^2+N^3=r^3I. $$ This helped me in constructing those examples.

Jyrki Lahtonen
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  • Thanks for your effort! Mhh, I edited the question to just consider a special subgroup instead of the whole of $\operatorname{GL}(3,3)$, now I am unsure, would it be better to open a new question so that your answer could stand as it is? – StefanH Jan 31 '16 at 21:01
  • Probably not. I marked the answer as belonging to an earlier version. – Jyrki Lahtonen Jan 31 '16 at 21:09
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This answer deals with the specified group $H$.

The matrices $A$ and $B$ are both monomial matrices, i.e. they have a single non-zero entry on each row and column. Furthermore, they both have determinant one. Because monomial matrices form a group, and $SL_3(\Bbb{F}_3)$ is a group, we see that these two properties are shared by all the matrices in $H$.

Let's say that a matrix $D\in H$ belongs to the permutation $\sigma\in S_3$, if the non-zero entry of column $j$ appears on row $\sigma(j)$. It is easy to see that if $D$ belongs to the permutation $\sigma$, then $D^k$ belongs to the permutation $\sigma^k$. Therefore $\ell=\operatorname{ord}(\sigma)$ is the lowest power of $D$ that is diagonal. Because the non-zero elements of $\Bbb{F}_3$ are $\pm1$, we see that the order of $D$ is then either $\ell$ or $2\ell$.

We were given that $D$ has order $6$. Because $\ell\in\{1,2,3\}$ we must have $\ell=3$. And because $D$ had order $6$, we know that $D^3$ is a diagonal matrix with entries $\pm1$, not all $=1$. Because $\det D^3=1$, we can deduce that the number of entries $=-1$ in $D^3$ is exactly two. Q.E.D.

Jyrki Lahtonen
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  • Thank you! Let me add that your conclusions about the order could be drawn as we have a homomorphism on $\mathcal S_3$ as you wrote in the comment. – StefanH Jan 31 '16 at 21:47