the representation of a free group A group $G$ is generated by $\begin{pmatrix}1&n\\0&1\end{pmatrix}$ and $\begin{pmatrix}1&0\\n&1\end{pmatrix}$, then we know $G\cong \mathbb{F}_2$ which is a free group generated by two elements. Now I consider the representation: $G\to GL(2,\mathbb{R})$, it is necessary that the image of $\begin{pmatrix}1&n\\0&1\end{pmatrix}$ is a triangular matrix under conjugation? Thanks in advance.
 A: No, it is not necessary. For instance the matrices 
$$A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}^{100}
$$
and
$$B = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix}^{100}
$$
generate a free group of rank 2 (I just put the exponent $100$ to be safe; any sufficiently large exponent will work). Those matrices have determinant one and they have trace of absolute value $>2$---let me call this a "hyperbolic" matrix---so they are not conjugate to upper triangular matrices with $1$'s on the diagonal. Furthermore, every element of the group they generate is a hyperbolic matrix (except for the identity).
Hyperbolic matrices have distinct eigenvalues and a distinct pair of eigenlines.
The general theorem in this regard is that if $A,B \in SL(2,\mathbb{R})$ are two hyperbolic matrices, and if they do not have a common eigenline, then there exists $M>0$ such that if $m,n \ge M$ then $A^m,B^n$ freely generate a rank 2 free group, every element of which is a hyperbolic matrix (except for the identity).
This is part of the whole "ping-pong" construction, although I see that the Wikipedia page on the ping-pong construction seems to omit mention of hyperbolic matrices.
