TITLE: Conjugacy of matrices of order three in $PGL(2,k)$, where $k$ is any field
QUESTION [9 upvotes]: We know that every matrix of order three in $PGL(2,\mathbb Z)$ is conjugate to the following matrix 
$$
\left(
\begin{array}[cc]
 &1 & -1 \\
 1&0
\end{array}
\right)
$$
I want to know if the same result holds in $PGL(2,k)$, where $k$ is any field, under some/no constraints.

REPLY [9 votes]: The following argument does not need any assumption on the field $\Bbbk$. 
Let $[A]\in PGL(2,\Bbbk)$ be of order three. Then $A$ is invertible and not a multiple of $E=\binom{1\;0}{0\;1}$, so there exists a vector $v$ such that $v$ and $Av$ form a basis. With respect to this basis, $A$ is of cyclic normal form
$$\begin{pmatrix}0&b\\1&a\end{pmatrix}\quad\text{with}\quad
\begin{pmatrix}0&b\\1&a\end{pmatrix}^3=\begin{pmatrix}ab&b(a^2+b)\\a^2+b&ab+a(a^2+b)\end{pmatrix}\;.$$
So $[A]$ is of order three if and only if $b=-a^2$. Conjugating (in $PGL(2,\Bbbk)$) gives
$$\begin{pmatrix}0&-a\\1&0\end{pmatrix}\begin{pmatrix}0&-a^2\\1&a\end{pmatrix}\begin{pmatrix}0&-a\\1&0\end{pmatrix}=\begin{pmatrix}a^2&-a^2\\a^2&0\end{pmatrix}\;,$$
which represents the matrix in the question.