Degenerate eigenvalues problem for a 4x4 system In summary, my question is whether or not I'm allowed to have the zero vector as my generalised eigenvector. I'm given the system $$x'=\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -2 & 2 & -3 & 1 \\ 2 & -2 & 1 & -3 \end{bmatrix}x$$ I also found two eigenvalues: 0 & -2. For 0, I have an eigenvector $$ \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}$$
and for -2, I have $$ \begin{bmatrix} -1 \\ 0 \\ 2 \\ 0 \end{bmatrix} \ \text{and}\  \begin{bmatrix} 0 \\ -1 \\ 0 \\ 2 \end{bmatrix}$$
I tried finding a third generalised eigenvector using eigenvalue 2, but it just doesn't exist (I get a row of zeros equals 4). What's going on and how do I proceed?
 A: Let's name your matrix A. 
The matrix $(A + 2E) = \begin{pmatrix} 2 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ -2 & 2 & -1 & 1 \\ 2 & -2 & 1 & -1 \end{pmatrix}$ has two linearly independent ordinary eigenvectors with eigenvalue 0: $\begin{pmatrix} -1 \\ 0 \\ 2 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ -1 \\ 0 \\ 2 \end{pmatrix}$, that you have already found. They are A's eigenvectors with eigenvalue -2. There are no other ordinary eigenvectors of A with eigenvalue -2 that are linearly independent with these two, because rk(A + 2E) = 2. 
The matrix $(A + 2E)^2 = \begin{pmatrix} 2 & 2 & 1 & 1 \\ 2 & 2 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ has three linearly independent ordinary eigenvectors with eigenvalue 0: $\begin{pmatrix} -1 \\ 0 \\ 2 \\ 0 \end{pmatrix}$, $\begin{pmatrix} 0 \\ -1 \\ 0 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \end{pmatrix}$. The first two ones are A's ordinary eigenvectors with eigenvalue -2, that you have already found. And the vector $\begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \end{pmatrix}$ is a generalised eigenvector of rank 2 of the matrix A with eigenvalue 2, making it the thing you were looking for.
