TITLE: A free subgroup of GL(2,Z)?
QUESTION [7 upvotes]: Is the subgroup of $GL(2,\mathbb Z)$ generated by the matrices 
$$ \left( \begin{array}{cc}
1 & 1  \\\
1 & 0 \end{array} \right) \ \ \text{and} \ \
\left( \begin{array}{cc}
2 & 1  \\\
1 & 0 \end{array} \right)
$$
free of exponential growth? More generally, how does one find all the relations between two matrices? 
I am sure this is well known, so any relevant references will be appreciated. 
My motivation comes from dynamical systems where these matrices specify two automorphisms of the 2-torus; I am interested in studying the orbits of their joint action.

REPLY [13 votes]: I interpret the question "how does one find all the relations between the matrices" as "find a set of defining relations for the group generated by the two matrices".
To do that we need a presentation of $GL(2,\mathbb{Z})$. I found one in the paper:
T. Brady, Automatic structures on Aut$F_2)$, Arch. Math. 63, 97-102 (1994).
$\langle p,s,u \mid p^2=s^2=(sp)^4=(upsp)^2=(ups)^3=(us)^2=1 \rangle$
where $p= \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$, $s= \left(\begin{array}{cc}-1&0\\0&1\end{array}\right)$, $u= \left(\begin{array}{cc}1&1\\0&1\end{array}\right)$.
Denoting your two matrices by
$a= \left(\begin{array}{cc}1&1\\1&0\end{array}\right)$,
$b= \left(\begin{array}{cc}2&1\\1&0\end{array}\right)$ 
we have $a=up$, $b=u^2p$.
Putting $H = \langle a,b \rangle$ and using coset enumeration in Magma, it turns out that $H={\rm GL}(2,\mathbb{Z})$. So your two matrices actually generate all of GL$(2,\mathbb{Z})$.
In fact, denoting $a^{-1}$ and $b^{-1}$ by $A$ and $B$, we have 
$p=aBa$, $u=a^2Ba$, $s=abaBAbabA$.
Using the modified coset enumeration algorithm, we can compute a presentation of $H$, which came out as the not particularly enlightening
$H = \langle a,b \mid (aBa)^2, (AbaBA)^2, aBabABAbAbABAbAbABAbAbAB, (abABabaB)^3 \rangle$.