You've pretty much got everything backwards here. Ax= 0 for x= 0 for any A. What is true is that A is invertible if and only if there is NO non-trivial solution. That is if x= 0 is the only x such that Ax= 0.
Saying that $\displaystyle \lambda$ is an eigenvalue of A means there exist non-zero x such that $\displaystyle Ax= \lambda x$. Use the fact that A is invertible to multiply both sides by $\displaystyle A^{-1}$.
So there exists x = 0 where Ax = 0
0 = lamda. 0 = 0
I'm not quite sure this proves that lamda is not equal to zero though?!