Given the constant matrix
A = \begin{bmatrix}4&-1&0\\3&1&-1\\1&0&1\end{bmatrix}
Find the fundamental matrix.
After finding the eigenvalue $\lambda$=2 with multiplicity 3, then trying to find the eigenvector corresponding to the eigenvalue, I end up with matrix \begin{bmatrix}2&-1&0\\3&-1&-1\\1&0&-1\end{bmatrix}, then after trying to find the corresponding eigenvector I end up with u=(1,2z,z), which I obtained from letting x=1 and subtracting the first and second rows with appropriate multiplication of 3$R_1$ and -2$R_2$ and adding the rows. Also, the reduced matrix has the form
\begin{bmatrix}1&0&-1\\0&1&-2\\0&0&0\end{bmatrix}
Then the first eigenvector u is u=(1,2,1). Nonetheless, since the eigenvalue has muliplicity 2 shouldn't there be two more eigenvectors in terms of some variable t?