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