Prove that $q \leq \dim(E_\lambda)$ given that $J$ has $q$ Jordan blocks associated with $\lambda$ Let $T: V \rightarrow V$ be linear, $V$ is a finite dimensional vector space, and the characteristic polynomial of $T$ splits.  Also let $\lambda$ be an eigenvalue of $T$ and $B$ be a Jordan Canonical basis for $V$ with respect to $T$.  Suppose that $J=[T]_B$ has $q$ Jordan blocks associated with $\lambda$.  

Prove that $q \leq \dim(E_\lambda)$.

I'm having trouble even starting off this proof.  I know that $K_\lambda$, the generalized eigenspace corresponding to $\lambda$, has an ordered basis of a union of disjoint cycles of generalized eigenvectors.  But I'm not sure how this even relates to $E_\lambda$, since for the $K_\lambda$ and $E_\lambda$ to be equal it must be diagonalizable. 
Or is this something to do with the initial vectors of the generalized eigenvectors of $T$ corresponding to $\lambda$ and the fact that the union of those generalized eigenvectors is disjoint.
Thanks a lot in advance.  I really appreciate any help on this particular problem.

Every Jordan block for $~\lambda$ has an eigenspace for$~\lambda$ of dimension$~1$ spanned by the first vector of$~B$ associated to this block. Since the $q$ such Jordan blocks form a direct sum, the sum of their eigenspaces for$~\lambda$ is also direct, and it is contained in the eigenspace$~E_\lambda$; this gives you $q\leq\dim(E_\lambda)$. In fact $E_\lambda$ is the sum of the eigenspaces of those Jordan blocks, so you even have $q=\dim(E_\lambda)$.
