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.

From the Jordan form of a linear map $T$ you can read the dimensions of the regular and generalized eigenspaces. Each Jordan block for the eigenvalue $\lambda$ looks like 
$$ \left( \begin{array}{cccc} \lambda & 1 & \ldots & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{array} \right). $$
If the block is $k \times k$, it comes with exactly one (linearly independent) eigenvector $e_1$ and $k - 1$ generalized eigenvectors $e_2, ..., e_k$. 
If in the Jordan block decomposition of $J$, there are exactly $q$ blocks of this form, each of size $k_i \times k_i$, then there are exactly $q$ (linearly independent) eigenvectors with corresponding eigenvalue $\lambda$, so you have in fact equality.
Try also to see how to read the dimension of $\ker(T - \lambda I)^2$ only from the number of Jordan blocks and their size.
