Orbit of a vector under the action of a $2\times 2$ matrix Consider the following matrix $A=\pmatrix{0 &1\\
-1&\frac{1}{3}}$ and put  $e_1:=(1,0)^t$. How can I find the elements of the orbit $A^ne_1$, for $n\in\mathbb N$?
It seems to me that $A$ has infinite order in $SL_2(\mathbb R)$ and moreover I don't see any closed formula for the powers of $A$.
 A: First let's note that characteristic polynomial of $A$ is $P_{A}(t) = t^{2}-\frac{1}{3}t+1$, which is irreducible over $\mathbb{R}[t]$ since it's root are $\frac{1}{6}(1 \pm i\sqrt35)$; In particular, it's irreducible over $\mathbb{Q}[t]$.
Since the minimal polynomial of $A$,denoted with $m_{A}$ divide the characteristic polynomial but the second one it's irreducible,
We can now assert that the minimal polynomial of $A$ and the characteristic polynomial coincide.
If $A$ would have finite order,for the Hamilton-Cayley theorem it would follow that $m_{A}(t) \mid t^{n}-1$ for some $n \in \mathbb{N}-\{0\}$
We can notice that $t^{n}-1 = \prod_\limits{d \mid n} \phi_{d}(t)$
(Proving that cyclotomic polynomials have integer coefficients)
Where $\phi_{d}(t)$ are the $d-th$ cyclotomic polynomial, and this factorization lie in $\mathbb{Z}[t]$.
Since we noticed that $P_{A}(t)$ is irreducibile over $\mathbb{Q}[t]$ which is (UFD),
If $P_{A}(t)$ would divide $t^{n}-1$ it would coincide with one of the irreducible factors of $t^{n}-1$, but those lie in $\mathbb{Z}[t]$, while $P_{A}(t)$ don't.

Edit : If orb($e_{1}$) would be finite, even the orbit of $x=Ae_{1}$ would be finite.
But $e_{1},x$ are linearly independent and we have shown that $A^{n}e_{1} = e_{1}$ and $A^{m}x=x$ for some $m,n$. It would follow that $A^{nm} = I_{d}$
Ps : If you're interested in finding the powers of $A$, a way to start could be noticing that the Jordan form of $A$ is indeed, a rotation, which powers are relatively easily computable.
(Questions of this kind or similar have been answered on MSE Powers of Rotation matrix)

