Okay, so we know that if lambda is an eigenvector of A, then A(lambda) = (lambda)x, where x is the eigenvector. So in the case above Av3 = A(a1v1 + a2v2), it would be implied that a1v1 + a2v2 would also be an eigenvector belonging to lambda1.
So, then, eigenvectors are unique to the...