If $T$ is an endomorphism, there exists a basis, according to which $T$ will be a block-diagonal matrix. Because if $V$ is the domain of $T$, $V$ will be the direct sum of the generalized eigenspaces, so we can choose a basis in each generalized eigenspace and extends to a basis of $V$, according to which $T$ will indeed be a block-diagonal matrix.
Every block of this matrix can be written as the sum of a multiple of the identity matrix (the multiple is an eigenvalue and the size of the identity matrix conforms with the algebraic multiplicity of the eigenvalue) and a matrix $N$, which is nilpotent with some index $s$.
Now the claim is:
"The order of nilpotency of $N$ equals the multiplicity of the according eigenvalue in the minimal polynomial of the endomorphism $T$".
(This claim has been made in the beginning of a chapter of Jordan-block matrices, but I don't think this is the Jordan-block matrix already). I hope my question is clear and someone can help me out.
The point is that is enough to consider the case when $V$ is the generalized eigenspace relative to a single eigenvalue $\alpha$.
$N$ can represented as a Jordan matrix, a blockwise diagonal matrix, with blocks that are Jordan blocks relative to the eigenvalue $0$.
A Jordan block relative to the eigenvalue zero is nilpotent. Then it corresponds to a linear mapping that maps $e_1 \mapsto e_2 \mapsto \dots \mapsto e_n \mapsto 0$, where the $e_i$ are suitably chosen independent vectors. It is not difficult to see that this has nilpotence index $n$.
Now if the Jordan blocks of $N$ have size $n_1 \le n_2 \le \dots \le n_k$, clearly $N^{n_k} = 0$, and $N^{n_k - 1} \ne 0$, so the nilpotence index is $n_k$. But then $T = \alpha I + N$ has minimal polynomial $(x-\alpha)^{n_k}$, because it is a root of it, but not of $(x-\alpha)^{n_k-1}$.