My question is about the following problem from "Introduction to representation theory" by Etingof et al.
Problem 2.15.1 d): Prove that $H$ is diagonalizable on the root subspace $\overline{V}(\lambda)$.
Here $H$ is the "Cartan subalgebra" element of $\mathfrak{sl}_{2}(\mathbb{C})$, such that $[H, E]=2E$, $[H, F]=-2F$, $[E, F]=H$, and $\overline{V}(\lambda)$ is its root subspace corresponding to an eigenvalue $\lambda$, the representation is assumed to be finite-dimensional.
My confusion here is about linear algebra, not representation theory itself. My linear algebra course was quite a while ago, and from it I remember that every finite-dimensional operator can be decomposed into a direct sum of "Jordan normal operators" that act each on its own root subspace. Each root subspace contains precisely one eigenvector and an arbitrary number of generalized eigenvectors; the Jordan's normal operators act on generalized eigenvectors (in a certain basis) as follows: $J|\lambda_{k}\rangle = \lambda |\lambda_{k}\rangle + |\lambda_{k-1}\rangle$, where $k$ enumerates generalized eigenvectors of the given root subspace, corresponding to the eigenvalue $\lambda$, and $|\lambda_{0}\rangle =|\lambda \rangle$ is the eigenvector of the root subspace.
It has been my understanding ever since I learned this that the action of an operator on a root subspace cannot be diagonalized, if the root subspace is not trivial (i.e., does not consist of a single eigenvector). Have I misunderstood this part, is the real statement of Jordan's decomposition theorem that the actions on root subspaces cannot be diagonalized simultaneously, and that on individual root subspaces the action can still be diagonalized? Otherwise the problem's formulation looks odd.
Edit: As another option, the problem may imply that no generalized eigenvectors exist in such representation. However, this is quite a strange way to formulate such statement.