Find an invertible matrix such that P^(-1)AP = Jordan normal form

Hi! 

Define A = $$\begin{pmatrix}
3 & 3\\
-3 & -3\\
\end{pmatrix}$$

Find a matrix such that

$$P^{-1}AP = \begin{pmatrix}
\lambda & 1\\
0 & \lambda\\
\end{pmatrix}$$

Where lambda is the unique eigenvalue of A.

What I have is;

the characteristic equation is given by

$$(3-\lambda)(-3-\lambda)+9 = 0\\
\lambda^2+3\lambda-3\lambda-9+9\\
\lambda^2 = 0\\
\lambda = 0\\$$

So we have $$\lambda = 0$$ with $$am(\lambda) = 2$$

To find an eigenvector we have

$$\begin{pmatrix}
3 & 3\\
-3 & -3\\
\end{pmatrix} \begin{pmatrix}
\mu_{11} \\
\mu_{12} \\
\end{pmatrix}
= 0
\begin{pmatrix}
\mu_{11} \\
\mu_{12} \\
\end{pmatrix} \\
3\mu_{11}+3\mu_{12} = 0\\
\mbox{and} \\
-3\mu_{11}-3\mu_{12} = 0\\$$

Pick $$\mu_{11} = 1\\
\mbox{then}\\
\mu_{12} = -1\\
\mbox{and}\\
\mu_1 = \begin{pmatrix}
1 \\
-1
\end{pmatrix}$$

Now normally I'd find a generalized eigenvector using

$$\begin{pmatrix}
3-\lambda & 3\\
-3 & -3-\lambda\\
\end{pmatrix}$$

But in this case this gives

$$\begin{pmatrix}
3-0 & 3\\
-3 & -3-0\\
\end{pmatrix}$$

Which doesn't really lead me anywhere.

How do I find another eigenvector such that I can use

$$P = \begin{pmatrix}
\mu_1 & \mu_2\\
\end{pmatrix}$$

Reinout