Unless there is something special going on, it's always the largest eigenvalue.

The eigenvectors form a basis set, so any vector [B]x[/B] can be represented as

[B]x[/B] = a*[B]e[sub]1[/sub][/B] + b*[B]e[sub]2[/sub][/B]

From which we see that

[B]A[/B][sup]n[/sup][B]x[/B] = a*lambda[sub]1[/sub][sup]n[/sup]*[B]e[sub]1[/sub][/B] + b*lambda[sub]2[/sub][sup]n[/sup]*[B]e[sub]2[/sub][/B]

So the larger eigenvalue will always dominate this process unless there is something special forcing the coefficient to zero. The actual iteration adds a constant at every stage, the vector [B]b[/B] in

[B]x[/B][sub]n+1[/sub] = [B]Ax[/B][sub]n[/sub]+[B]b[/B]

so even if [B]x[/B][sub]0[/sub] is aligned with the smaller eigenvector, [B]x[/B][sub]1[/sub] will have a component of the larger eigenvector unless [B]b[/B] is also a eigenvector for the smaller eigenvalue. (This problem actually picks up the component in [B]x[/B][sub]1[/sub] because [B]x[/B][sub]0[/sub] is the zero vector).

You can concoct special scenarios with lamda=-1 so that the multiplication and addition cancel each other, but that's about all that can get in the way of the largest eigenvalue dominating the process.