Jordan form, number of blocks. [closed]

To answer your first question, each block looks like

$$ \left( \begin{array}{ccccc} \lambda & 1 & & & \\ & \lambda & 1 & & \\ & & \ddots & \ddots & \\ & & & \lambda & 1 \\ & & & & \lambda \end{array} \right) $$

The columns of a matrix correspond to the basis vectors. You can see here that the first column gives an eigenvector, and the rest do not. The columns here tell you that

$$ \begin {align*} Ae_1 &= \lambda e_1 \\ Ae_2 &= e_1 + \lambda e_2 \\ &\vdots \\ Ae_n &= e_{n-1} + \lambda e_n \end {align*} $$

So each Jordan block gives you exactly 1 eigenvector.

To answer the second question, one can assume that $\lambda =0$ (since we are always considering $A-\lambda I$). Let $J$ be the usual Jordan block of size $n$ and eigenvalue zero and note that $\dim\ker J^k = k$. To see this, note that $J^k$ consists of a diagonal of ones of length $n-k$. In particular, it follows that

$$\dim \ker J^k - \dim \ker J^{k-1} =\begin{cases} 1 & \text{ if $1\leqslant k\leqslant n$ } \\ 0 & \text{ if $k>n$. } \end{cases} $$

Hence, if you let $b_k(J) = \dim \ker J^k - \dim \ker J^{k-1}$, this is counting the number of blocks of size at least $k$ in $J$, and it follows that $b_k(J) - b_{k+1}(J)$ counts the number of blocks of size exactly $k$ in $J$: by the formula above it is zero except when $k=n$.

Since conjugation preserves $b_k(A) = \dim \ker A^k$, and since dimensions add up when putting matrices in blocks, it follows that $b_k(A-\lambda)-b_{k+1}(A-\lambda)$ counts the number of blocks of size at least $n$ in $A$ corresponding to the eigenvalue $\lambda$.

(Note that we do not need to care about other eigenvalues in the formula since $J(\lambda)^k$ is always full rank unless $\lambda=0$.)