Consider the following:
1.Suppose $k_i$ is an eigenvalue of $A$ with algebraic multiplicity $n_i$
2.dim $V_i = m_i$ (geometric multiplicity), $V_i$ is the eigenspace corresponding to $k_i$.
3. $m_i < n_i$ (not diagonalizable)
can we say:
If yes, why???
I have learned that: N$(A-k_i*I) =$ N$(A-k_i*I)^2$
proof:
suppose $v$ belongs to N$(A-k_i*I)^2$
then,
$\langle v ,$ N$(A-k_i*I)^2*v \rangle = ($ N$(A-k_i*I)^2*v )^T*v = ($ N$(A-k_i*I)*v )^2 = 0$
so, N$(A-k_i*I)*v = 0$. So $v$ belongs to N$(A-k_i*I)$.
suppose $v$ belongs to N$(A-k_i*I)$
N$(A-k_i*I)^2*v =$ N$(A-k_i*I)*$ N$(A-k_i*I)*v =$ N$(A-k_i*I)*0 = 0$
So $v$ belongs to N$(A-k_i*I)^2$
so N$(A-k_i*I) =$ N$(A-k_i*I)^2$
Consider \begin{bmatrix}1&1\\0&1\end{bmatrix}
Then $(0,1)^T \in $ N$((A-I)^2)$ but $(0,1)^T \notin $ N$(A-I)$. So there is a problem with your proof.
Suppose $0\neq v \in$ N$(A-k_iI)$, then $(A-k_iI)v=0$. Multiply both sides by $(A-k_iI)$, then $(A-k_iI)^2v=(A-k_iI)(A-k_iI)v=(A-k_iI)0=0$ so that $v \in$ N$(A-k_iI)^2$.
So we have N$(A-k_iI) \subseteq$ N$(A-k_iI)^2$ $\ldots$ induce, and you have the series of subsets.
Now from the above we see that dim(N$(A-k_iI)^l)\leq$ dim(N$(A-k_iI)^{l+1})$. If $A$ is of finite order, then for some integer $l$ we must have either dim(N$(A-k_iI)^l)=h$ where $h \times h$ is the order of $A$ (in which case we must have dim(N$(A-k_iI)^l)=$ dim(N$(A-k_iI)^{l+1})$) or dim(N$(A-k_iI)^l)=$ dim(N$(A-k_iI)^{l+1})$. To prove that $l=n_i$ requires some more work - you have to make use of the fact that the generalized eigenspaces of $A$ are invariant under the operator represented by $A$ and that the entire underlying vector space is the direct sum of the generalized eigenspaces...I will leave this to you...