Making matrix upper triangular by finding Jordan Normal Form I must make the following matrix upper triangular:
\begin{bmatrix}
    2    & 1 & 0 & 0 \\
    0 & 2 & 0 & 1 \\
    1    & 0 & 2 & 0 \\
    0 & 0 & 0 & 2
\end{bmatrix}
To do this, I'm calculating the Jordan Normal form outlined here.
There is only one eigenvalue, 2. So I must find three generalized eigenvectors (ugh).
To find the first one, I try to solve $(A-λI)v_2=v_1$, where $v_1$ is the first eigenvector:
\begin{bmatrix}
    0  \\
    0 \\
    1  \\
    0
\end{bmatrix}
So the equation is the following ($v_{21},v_{22},v_{23},v_{24}$ are the four components of $v_2$)
$$\begin{bmatrix}
    0    & 1 & 0 & 0 \\
    0 & 0 & 0 & 1 \\
    1    & 0 & 0 & 0 \\      
    0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
    v_{21}  \\
    v_{22} \\
    v_{23}  \\
    v_{24}
\end{bmatrix}=
\begin{bmatrix}
    0  \\
    0 \\
    1  \\
    0
\end{bmatrix}$$
My problem is that there is a column of all zeros in the first matrix, which means that there are infinite solutions for $v_{33}$? I'm not sure what I'm doing wrong here but any help would be appreciated!
 A: HINT
Since there is only an eigenvalue the Jordan canonical form is:
$$\begin{bmatrix}
    2    & 1 & 0 & 0 \\
    0 & 2 & 1 & 0 \\
    0    & 0 & 2 & 1 \\
    0 & 0 & 0 & 2
\end{bmatrix}$$ 
Since $J$ is obtained by $A$ from a similarity transfomation: $$P^{-1}AP=J$$
to find the generalized eigenvectors you need to solve the following system:
$$AP=PJ$$
More in detail, let ($v_i$ are the generalized eigenvectors, columns of P):
$$P=[v_1,v_2,v_3,v_4]$$
then P has to satisfy the following systems:
$$\begin{cases}Av_1=2v_1 \\ Av_2=v_1+2v_2 \\ Av_3=v_2+2v_3 \\ Av_4=v_3+2v_4\end{cases}$$ 
You have already solved the first now you can find $v_2$ from the second system of equation and so on.
For example the second system $(A-2I)v_2=v_1$ is:
$$\begin{bmatrix}
    0    & 1 & 0 & 0 \\
    0 & 0 & 0 & 1 \\
    1    & 0 & 0 & 0 \\      
    0 & 0 & 0 & 0
\end{bmatrix}
    v_2=
\begin{bmatrix}
    0  \\
    0 \\
    1  \\
    0
\end{bmatrix}$$
thus
$$  v_2=
\begin{bmatrix}
    1  \\
    0 \\
    0  \\
    0
\end{bmatrix}$$
Finally you should obtain:
$$P=\begin{bmatrix}
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    1 & 0 & 0 & 0 \\      
    0 & 0 & 0 & 1
\end{bmatrix}$$
