Question: Suppose is a linear operator on a finite-dimensional vector space whose characteristic polynomial splits and let be an eigenvalue of . Suppose that is a basis for consisting of the union of disjoint cycles of generalized eigenvectors. Prove that .

Attempt: Denote as

and . Then the is simply the nullity of the transformation . Then I attempted to use contradiction by assuming that . 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!