TITLE: Misconception on Jordan Canonical Form
QUESTION [5 upvotes]: Say we have matrix 
$$M= \begin{pmatrix} 
2 & 0 & 1 & -3\\
0 & 2 & 4 & 8\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 3
\end{pmatrix}\DeclareMathOperator{\Id}{Id}$$
It follows that 
$$\chi_{M}=(x-2)^{3}(x-3)$$
I found that $\ker(M-2\Id)=\{(1,0,0,0)^{T},(0,1,0,0)^{T}\}$, so dimension 2.
Similarly, we get dimension of $\ker(M-3\Id)=1$.
I want to focus on $ker(M-2\Id)$:
Looking at $ker(M-2Id)^{2}$, we get a basis of $\{(1,0,0,0)^{T},(0,1,0,0)^{T}, (0,0,1,0)^{T}\}$, so dimension 3. 
In our notes, we have written down:
Find a vector $v \in \ker(M-2\Id)^{2}$, such that $v \notin \ker(M-2\Id)$. It follows that $Mv \in \ker(M-2\Id)$ ( First question, should this not be $(M-2\Id)v \in \ker(M-2\Id)?).$ How does this help us in terms of the Jordan Form? I'm not sure how invariant subspaces fit into all of this either, other than the fact 
$\ker(M-2\Id)\subset \ker(M-2\Id)^{2}$.
An intuitive explanation would be of great assistance.

REPLY [2 votes]: $\DeclareMathOperator{\Id}{Id}$In Your case You already found a vector $v\in ker (M-2\Id)^2\backslash ker(M-2\Id)$, namely $v=(0,0,1,0)^T$ and  $(M-2\Id)v=(1,4,0,0)^T$. Furthermore $M(1,4,0,0)^T=2(1,4,0,0)^T$ and $Mv=2v+(1,4,0,0)^T$. This corresponds to the Jordan block $$\begin{pmatrix}2&1\\0&2\end{pmatrix}.$$
The subspace $\operatorname{span}\{(0,0,1,0)^T,(1,4,0,0)^T\}$ is thus an invariant subspace.