Jordan form of a specific 2x2 matrix I'm trying to follow an algorithm for finding the Jordan form for the matrix:
$$     \begin{pmatrix}
        0 & -1 \\
        4 & 4 \\
       \end{pmatrix}    $$
Its eigenvalue is 2 with algebraic multyplicity of 2.
$$
A-2I = 
        \begin{pmatrix}
        -2 & -1 \\
        4 & 2 \\
        \end{pmatrix}
$$
therefore $\ker(A-2I) = sp\begin{pmatrix} 1\\-2 \end{pmatrix}$, and $\dim(\ker(A-2I))=1$. Next:
$$
(A-2I)^2 = 
        \begin{pmatrix}
        0 & 0 \\
        0 & 0 \\
        \end{pmatrix}
$$
so $\dim(\ker(A-2I)^n)=2$ for $n \geq 2$. The next step is to find a vector $u_1$ in $ker(A-2I)^2 \setminus \ker(A-2I)$. Problem is that for every $u_1$ (unless I made some dumb arithmetic error along the way, which I believe I haven't) I get that $(A-2I)u_1 = \alpha\begin{pmatrix} 1\\-2 \end{pmatrix}$ for some $\alpha\in\mathbb R$. e.g. $$(A-2I)e_1 = \begin{pmatrix} -2\\4 \end{pmatrix}$$
Okay, so I got a 1x1 Jordan block, and I'm left to find another. That means I ought to find some $m_2$ for which
$$ m_2 = max \left\{ m : \ker(A-2I)^m \setminus \ker(A-2I)^{m-1} \cup sp\begin{pmatrix} 1\\-2 \end{pmatrix}\right\}$$
but since the spaces spanned by my eigenvector and $(A-2I)u_1$ are the same, the only m that can work is 0, that is if I interpret $\ker(A-2I)^-1$ as an empty set or something.
How should I interpret it? And is that the correct choice for $m$?
 A: There is a theorem in Friedberg's linear algebra book that makes computing the Jordan form easy and algorithmic. We will construct a "dot diagram" for each eigenvalue that will tell us the size of the jordan blocks corresponding to that eigenvalue.
Let $\lambda$ be an eigenvalue of a linear transformation $T$. Compute the numbers $r_i$ defined as follows:
\begin{align}
r_1 &= dim(V) - rank(T-\lambda I)\\
r_i &= rank((T-\lambda I)^{i-1}) - rank((T-\lambda I)^{i})
\end{align}
Now create a left-justified figure with rows and columns of dots. In the $i^{th}$ row, put $r_i$ dots. Once done, each column corresponds to a Jordan block! The number of dots in that column tells you the size of that block.
In this problem the only eigenvalue is 2. So we create the dot diagram corresponding to 2: $r_1 = 1$ and $r_2 = 1$. ($r_3 = 0$ and you can stop once you get zero). So we create a dot diagram with 1 dot in the first row and 1 dot in the second row. 
\begin{matrix}
\cdot \\ 
\cdot
\end{matrix}
This has one column with 2 dots, so there is a Jordan block of size 2.
A: Simply $A$ has characteristics polynomial $(x-2)^2$ and $A-2I\neq O$, so $A$ has minimal polynomial $(x-2)^2$. So, if we denote $N=A-2I$, then $N$ is a nilpotent matrix of index 2. So, $N$ has canonical form $$\pmatrix{ 0&1\\0&0}.$$ Hence Jordan canonical form of $A$ is $$\pmatrix{2&1\\0&2}.$$
