Question: Suppose T is a linear operator on a finite-dimensional vector space whose characteristic polynomial splits and let \lambda be an eigenvalue of T. Suppose that \gamma is a basis for K_\lambda consisting of the union of q disjoint cycles of generalized eigenvectors. Prove that q \leq dim(E_\lambda).

Attempt: Denote \gamma as
\{(T-\lambda I)^{p-1}(x_1),...,x_1\} \cup ... \cup \{(T-\lambda I)^{p-1}(x_q),...,x_q\} and E_\lambda = \{x \in V: (T-\lambda I)x = 0\}. Then the dim(E_\lambda) is simply the nullity of the transformation (T-\lambda I). Then I attempted to use contradiction by assuming that q > dim(E_\lambda). How does this generate a contradiction? Is it because of using a theorem based on of the multiplicities of eigenvalues of a characteristic-splitting transformation? Thanks!