Jordan form exercise What am I doing wrong?

I've been learning how to put matrices into Jordan canonical form and it was going fine until I encountered this $4 \times 4$ matrix:

$A=\begin{bmatrix} 
2 &amp; 2 &amp; 0 &amp; -1 \\ 
0 &amp; 0 &amp; 0 &amp; 1 \\
1 &amp; 5 &amp; 2 &amp; -1 \\
0 &amp; -4 &amp; 0 &amp; 4 \\
\end{bmatrix}
$

Which has as only eigenvalue $\lambda_1=\lambda_2=\lambda_3=\lambda_4=2$ with 2 corresponding eigenvectors, which I will for now call $v_1$ and $v_2$:

$v_1 = \pmatrix{0\\0\\1\\0}, v_2=\pmatrix{-3 \\ 1 \\ 0 \\ 2}
$ 

2 eigenvectors means 2 Jordan blocks so I have 2 possibilities:

$J= \pmatrix{2 &amp; 1 &amp; 0 &amp; 0 \\ 
0 &amp; 2 &amp; 1 &amp; 0 \\
0 &amp; 0 &amp; 2 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 2} $ or $ J= \pmatrix{2 &amp; 1 &amp; 0 &amp; 0 \\
0 &amp; 2 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 2 &amp; 1 \\
0 &amp; 0 &amp; 0 &amp; 2}
$

I consider the first possibility. This gives me the relations:

$Ax_1=2x_1 \\
Ax_2=x_1+2x_2 \\
Ax_3=2x_3+x_2 \\
Ax_4=2x_4 \\
$

where $x_1$ and $x_4$ should be $v_1$ and $v_2$. From the second relation $(A-2I)x_2=x_1$ I see

$\pmatrix{0 &amp; 2 &amp; 0 &amp; -1 \\
0 &amp; -2 &amp; 0 &amp; 1 \\
1 &amp; 5 &amp; 0 &amp; -1 \\
0 &amp; -4 &amp; 0 &amp; 2} \pmatrix{a \\ b \\ c \\ d} =\pmatrix{0 \\ 0 \\ 1 \\ 0} $ 

( $v_2= \pmatrix{ -3 \\ 1 \\ 0 \\ 2}
$ will give an inconsistent system)

Now I get that $x_2 = \pmatrix{-2 \\ 1 \\ 0 \\ 2}
$

From the third relation $(A-2I)x_3=x_2$:

$\pmatrix{0 &amp; 2 &amp; 0 &amp; -1 \\
0 &amp; -2 &amp; 0 &amp; 1 \\
1 &amp; 5 &amp; 0 &amp; -1 \\
0 &amp; -4 &amp; 0 &amp; 2} \pmatrix{e \\ f \\ g \\ h} =\pmatrix{-2 \\ 1 \\ 0 \\ 2} $ 

But this system is inconsistent as well! No matter which vectors I try in which places, when I try to generalize eigenvectors I seem to always end up with some inconsistency.

Is there something staring me in the face that I am overlooking? Or am I doing it completely wrong (even though this method worked fine for me before)?

Sorry for the lengthiness and thank you in advance.
Take a look at
$$
       (A-2I)^2 = \begin{bmatrix} 
0 &amp; 2 &amp; 0 &amp; -1 \\ 
0 &amp; -2 &amp; 0 &amp; 1 \\
1 &amp; 5 &amp; 0 &amp; -1 \\
0 &amp; -4 &amp; 0 &amp; 2 \\
\end{bmatrix}^2 
$$
The $a_{1,2}$ entry is non-zero. What does that tell you?
Look at the image space (column-wise) of $S = ( A - 2I )$, which yields two vectors  $S(v), v = (0,0,2,0) , (-1,1,-1,2)$ [SPECIFICALLY IN THIS ORDER]
Now find $u$ such that $S(u) = v$      say $u = (0,0,0,1)$
Now add any vector $w$ such that $w$ is in $Ker(S)$ and altogether completes a basis for $\Bbb R^4$; clearly $w = (-3,1,0,2)$


You can check that $\mathcal B = \{ (0,0,2,0), (-1,1,-1,2), (0,0,0,1), (-3,1,0,2)\}$does the job for JCF transform.

P.s not experienced in this typefont, but the maths works!:)