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 & 2 & 0 & -1 \\ 0 & 0 & 0 & 1 \\ 1 & 5 & 2 & -1 \\ 0 & -4 & 0 & 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 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2} $ or $ J= \pmatrix{2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 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 & 2 & 0 & -1 \\ 0 & -2 & 0 & 1 \\ 1 & 5 & 0 & -1 \\ 0 & -4 & 0 & 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 & 2 & 0 & -1 \\ 0 & -2 & 0 & 1 \\ 1 & 5 & 0 & -1 \\ 0 & -4 & 0 & 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 $$ (A-2I)^2 = \begin{bmatrix} 0 & 2 & 0 & -1 \\ 0 & -2 & 0 & 1 \\ 1 & 5 & 0 & -1 \\ 0 & -4 & 0 & 2 \\ \end{bmatrix}^2 $$ The $a_{1,2}$ entry is non-zero. What does that tell you?