## Disjoint cycles of generalized eigenvectors

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

Attempt: Denote $\displaystyle \gamma$ as
$\displaystyle \{(T-\lambda I)^{p-1}(x_1),...,x_1\} \cup ... \cup \{(T-\lambda I)^{p-1}(x_q),...,x_q\}$ and $\displaystyle E_\lambda = \{x \in V: (T-\lambda I)x = 0\}$. Then the $\displaystyle dim(E_\lambda)$ is simply the nullity of the transformation $\displaystyle (T-\lambda I)$. Then I attempted to use contradiction by assuming that $\displaystyle q > dim(E_\lambda)$. 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!