The equation Ax= 0 has a non-trivial solution if and only if A is not one-to-one. That is the same as saying that its determant is 0 and that it has 0 as an eigenvalue. The standard way to find an eigenvalue, [itex]\lambda[/itex] for matrix A is to solve the equation [itex]det(A- \lambda I)= 0[/itex]. If A is an n by n matrix, that will be a polynomial equation of degree n and so has n solutions (not necessarily all distinct, not necessarily real).
IF [itex]\lambda[/itex] really is an eigenvalue, then [itex]Ax= \lambda x[/itex] or [itex]
Ax- \lambda x= (A- \lambda I)x= 0[/itex] has, by definition of "eigenvalue", a non-trivial solution. That is, some of the equations you get by looking at individual components will be dependent. Note that x= 0 always will be a solution, just not the only one.
Perhaps if you posted a specific example, we could point out errors. The most obvious one, if you "keep getting x=0", is that what you think is an eigenvalue really isn't!