Conjugacy of matrices of order three in $PGL(2,k)$, where $k$ is any field 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.
 A: 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.
A: For $GL_2$, replace the first entry of your matrix with -1 and then the answer is yes.
You can assume that the field is algebraically closed. A matrix $A$ with $A^3=\lambda I$ can be divided by a third root of lambda to get to $A^3=I$. Again by multiplication you can make sure the eigenvalues are the two different third roots of the unit (unless you are in characteristc 3 in which case there is only one third root of the unit, but that case is even easier).
But then the characteristic polynomial is determined and this determines the conjugacy class in $GL_2$.
Now for $PGL_2$. Let $K$ be our field. Assume $A^3=\lambda I$ with some $\lambda\in K$. Then a third root of $\lambda$ must also lie in $K$, as the minimal polynomial of $A$ divides $x^3-\lambda$.
So you can divide $A$ by that third root to obtain $A^3=I$.
By the $GL_2$-case this matrix is conjugate to your standard matrix. 
