Generalized eigenvector space with $\lambda_1=\lambda_2=\lambda_3=2$ Let  be $$M:=\begin{pmatrix}
1 & 1 & 1 \\
-1 & 3 & 1 \\
0 & 0 & 2
\end{pmatrix}$$
The characteristic polynomial would then be $\chi_M(\lambda)=\lambda^3-6\lambda^2+12\lambda-8$, and his zeros points are $\lambda_1=\lambda_2=\lambda_3=2$. Now we have the matrix:
$$N:=\lambda_1I_3-M=\begin{pmatrix}
1 & -1 & -1 \\
1 & -1 & -1 \\
0 & 0 & 0
\end{pmatrix}$$
From there we calculate that there are two eigenvectors: $v_1=\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $v_2=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$. If we want to calculate the generalized eigenvectors we need to find a vector $v_3$ such that $N^2v_3=0$, but we get that $N^2=0$, a nullmatrix 3x3. How can we calculate the third vector? As I know there should be a minimum of $\dim{M}=3$ generalized eigenvectors.

Let $$v_3=\begin{pmatrix}-1\\0\\0\end{pmatrix}.$$ Then $$(M-2I)v_3 = \begin{pmatrix}-1&1&1\\-1&1&1\\0&0&0\end{pmatrix}\begin{pmatrix}-1\\0\\0\end{pmatrix} = \begin{bmatrix}1\\1\\0\end{bmatrix}=v_1, $$
so 
$$(M-2I)^2v_3 = (M-2I)(M-2I)v_3=(M-2I)v_1=0, $$
and hence $v_3$ is a generalized eigenvector of $M$.
