Generators for $PSL(2,\mathbb{Z})$ with a specific property Does there exist generators $S$ and $T$ for the modular group $\Gamma=PSL(2,\mathbb{Z})$ with the following property:
$$S+S^{-1}+T+T^{-1}=0$$
Here is a candidate:
$$S=\left[\begin{array}{cc} -1 & 0 \\ 1 & -1 \\ \end{array}\right], \,T=\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}\right]$$
Just not sure if these two generate $\Gamma$.
 A: Yes, your choice for $S$ and $T$ generate the modular group $\Gamma=PSL_2(\mathbb{Z})$.  Other choices would also work.
The standard choice of generators is 
$$ \Gamma = \left \langle \left[ \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right], \left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right] \right \rangle.$$  
You already have $T=\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right]$.   Notice that $TST=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] 
= -  \left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right]$.  Since we can obtain the standard generators for $\Gamma$ as products of $S$, $T$, and their inverses, we conclude that $S$ and $T$ generate $\Gamma$.

Just to clarify for anyone not familiar with all the matrix groups, $PSL_2(\mathbb{Z})=SL_2(\mathbb{Z})/\left\{1,-1\right\}$ is the group of 2x2 matrices over the integers with determinant 1, modulo the subgroup of scalar transformations, which in this case is just $\{1,-1\}$.  That's why it doesn't matter if I obtain $\left[ \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right]$ or $\left[ \begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix} \right]$; they represent the same element in $PSL_2(\mathbb{Z})$.
