Is there a good way to show that a subgroup is Zariski-dense? Let $G$ be a semisimple algebraic group, and let $\Gamma$ be a finitely generated subgroup.  Given the generators of $\Gamma$, is there a good way to determine if $\Gamma$ is Zariski-dense in $G$?
For example, let $\Gamma \subseteq \mathrm{PSL}_2$ where $ \displaystyle
\Gamma = \left\langle \gamma_1 , \gamma_2  \right\rangle$ and $\gamma_1 = \begin{pmatrix} 1 & i \newline 0 & 1 \end{pmatrix}$ and $\gamma_2 = \begin{pmatrix} 1 & 0 \newline
 1 & 1 
 \end{pmatrix}$.
 A: *

*What do you mean by "good"? In practice, a theorem of either Lubotzky or Weigel (depending on who you ask) states that if a congruence quotient is surjective for SOME prime (bigger than 3, say), then the subgroup is Zariski-dense, so generally checking for a couple of primes gives a certificate.

*In $PSL_2,$ Zariski-dense is equivalent to non-elementary, so your group is Zariski-dense.
EDIT To answer the OP's comment: in this paper: Zariski density and genericity
I Rivin - International Mathematics Research Notices, 2010, I show that two "random" matrices generate a Zariski dense subgroup. R. Aoun in this paper: Transience of algebraic varieties in linear groups and application to generic Zariski density
R Aoun - Arxiv preprint arXiv:1103.0944, 2011 - arxiv.org
Generalized it (at the cost of losing the effectiveness). Aoun also showed (in a different paper) that a random subgroup is free (which, in higher rank) means that it is not a lattice. In rank one a corresponding result (not yet written up fully) was shown by Elena Fuchs, myself, and Peter Sarnak: there you show that generically the Hausdorff dimension of the limit set is small, so again you don't get a lattice generically (of course lattices are Zariski dense by the Borel Density Theorem).
You might also look at these slides: 
