The Jordan Normalform $J$ to $A$ is a Matrix, on whose Diagonal are Jordanblocks $J(\lambda,s)$, where $s$ is the size of the block to eigenvalue $\lambda$.
For $i=1,...,r$ and $j=1,...,n_i$ let $k_{i,j}$ be the multiplicity of $J(\lambda_i, j)$ in $J$. (The number of times it occurs in $J$).
We then have $k_{i,1} + 2k_{i,2} +...+n_ik_{i,n_i} = n_i$ since the multiplicity of $\lambda_i$ is $n_i$.
We now use the following to calculate all of the values $k_{i,j},j=n_i,n_i-1,...,2$:
$rg\,J(\lambda,m)^l=\begin{cases} \mathrm{max}\{0,m-l\} \text{ for} \lambda =0\\ m\text{ for } \lambda\ne 0\end{cases}$
We then obtain
$rg(J-\lambda_iE)^{n_i} = n-n_i\\ rg(J-\lambda_iE)^{n_i-1} = n-n_i+k_{i,n_i}\\ rg(J-\lambda_iE)^{n_i-2} = n-n_i+2k_{i,n_i}+k_{i,n_{i-1}}\\ ...\\ rg(J-\lambda_iE)^{1} = n-n_i+(n_i-1)k_{i,n_i}+...+k_{i,2}$
This was an excerpt of the book "Lineare Algebra" by Siegfried Bosch, 5.ed, p. 234-235. I can't understand how the rank formulars were obtained.
The key formula for the rank formulas is the first, for $rg\,J(\lambda,m)^l$. Its holds because when $\lambda$ is non-zero then $l$-th power of the $J(\lambda,m)$ is an upper triangular matrix with $\lambda^l$ on its main diagonal, so its rank is $m$; when $\lambda$ is zero then $J(\lambda,m)^l=\|a_{i,j}\|$ is a $0$-$1$ matrix which non-zero elements are exactly $a_{i,i+l}$, so its rank is $\mathrm{max}\{0,m-l\}$. The next rank equalities follow from this formula. The sums of $k$’s in each of them count the number of powers of $J(\lambda_i, m)-\lambda_i I$ which are still non-zero, each of them multiplied by a rank of the respective matrix. For instance, sum $2k_{i,n_i}+k_{i,n_{i-1}}$ in the third sum formula means that we have $k_{i,n_{i}}$ non-zero powers of blocks $J(\lambda_i, n_i)-\lambda_i I$ each with rank $2$ and $k_{i,n_{i-1}}$ non-zero powers of blocks $J(\lambda_i, n_{i-1})-\lambda_i I$ each with rank $1$.