# Jordan form, number of blocks.

Suppose I have an eigenvalue $\lambda$, now I want to determine the number of Jordan blocks corresponding to that eigenvalue, as well as size of each block. I know that:
1) number of blocks is equal to the number of eigenvectors
2) there are $\dim \ker (A - \lambda I)^k - \dim \ker (A - \lambda I)^{k-1}$ blocks of size at least $k$
Why exactly are those statements true. Could someone provide me with a proof or at least some resoning behind this?

$$\left( \begin{array}{ccccc} \lambda & 1 & & & \\ & \lambda & 1 & & \\ & & \ddots & \ddots & \\ & & & \lambda & 1 \\ & & & & \lambda \end{array} \right)$$
\begin {align*} Ae_1 &= \lambda e_1 \\ Ae_2 &= e_1 + \lambda e_2 \\ &\vdots \\ Ae_n &= e_{n-1} + \lambda e_n \end {align*}