Let $A \in M_n(F)$, and $\lambda \in F$ an eigenvalue of A with algebric multiplicity $k$ and multiplicity $r$ as a root in the minimal polynomial, $m_A\;$; Now for some matrix $B$, let's define $V_j=V_j(B):=\mathcal N(B^j)$ (Nullity of $B^j)$.
I already proved that $V_j \subseteq V_{j+1}$, and there is $l \in \Bbb N$ such that for every $j < l, \;V_j \subsetneq V_{j+1},$ and for every $j \ge l, \; V_j = V_{j+1}$, meaning:
$$\text{{0}} \subsetneq V_1 \subsetneq V_2 \subsetneq \dots \subsetneq V_l = V_{l+1} = V_{l+2} = \dots$$
Now I'm trying to prove that if we look at the sequence of spaces: $V_j(A-\lambda I)$, then $r = l$, and $\;dim\;V_l(A-\lambda I)=k$.
Any ideas?