Proof of $\ker (A - \lambda_1 I )^N \cap \ker (A - \lambda_2 I)^N = 0 $ Let $V $ be an $N$-dimensional vector space. Let $A$ be a nonzero linear operator on $V $. Let $\lambda_{1,2}$ be two different eigenvalues. 
It is known that eigenvectors corresponding to different eigenvalues are linearly independent. 
But for an eigenvalue, there are generalized eigenvectors besides eigenvectors. How are the generalized eigenvectors corresponding to different eigenvalues related? 
It is suspected that 
$$ \ker (A - \lambda_1 I )^N \cap \ker (A - \lambda_2 I)^N = 0. $$ 
How to prove it? 
Here I take the $N$th power of $A - \lambda I $ because we know there is the ascending chain condition for $\ker (A - \lambda I )^n$. 
 A: The kernel of any polynomial in $A$ is $A$-stable, in particular that holds for the two kernels in your question; one can therefore consider the restrictions of (the action of) $A$ to each of those two kernels. For the first restriction $\lambda_1$ is the only eigenvalue, for the second $\lambda_2$ is the only eigenvalue. (More precisely the minimal polynomials of the restrictions are powers of $X-\lambda_1$ respectively of $X-\lambda_2$.)
Now the intersection of the kernels is again an $A$-stable subspace, but now the restriction of $A$ the the subspace cannot have any eigenvalues (they would have to equal both $\lambda_1$ and $\lambda_2$), so it can only be that the intersection has dimension $0$. (Or you can argue that the minimal polynomial of the restriction divides both a power of $X-\lambda_1$ and a power of $X-\lambda_2$, so that minimal polynomial is $1$ and the subspace acted upon of dimension$~0$. This latter argument does not depend on working over an algebraically closed field, as the first argument that uses "no eigenvalues implies $\dim=0$".)
A: Suppose $v \in  \ker(A - \lambda_1 I)^N \cap \ker(A - \lambda_2 I)^N$ is a nonzero vector. There exists $k$ such that $v ' = (A - \lambda_1 I)^k v$  is not zero but $(A - \lambda_1 I)^{k + 1} v = 0$. This implies that $v'$ is an eigenvector with eigenvalue $\lambda_1$. In particular, $$(A - \lambda_2 I)^{N} v' = (\lambda_1 - \lambda_2)^N v' \neq 0$$ in contradiction to $$(A - \lambda_2 I)^{N} v' = (A - \lambda_1 I)^k (A - \lambda_2 I)^{N} v = 0.$$ Note that we used here that $(A - \lambda_1 I)^k$ and $(A - \lambda_2 I)^{N}$ commute.
