Problem with generalized eigenvectors in a 3x3 matrix. I have this matrix:
$$
  A=  \begin{pmatrix}
    0 & 1 & 1 \\
    0 & 1 & 0 \\
    -1 & 1 & 2 \\
    \end{pmatrix}
$$
I have founded the eigenvalues:
$$\lambda_{1,2,3}=1$$
So
$$\lambda=1$$$$\mu=3$$
I'm expecting to have one eigenvector plus two generalized eigenvectors. But, proceding I have some troubles:
$$
  A=  \begin{pmatrix}
    -1 & 1 & 1 \\
    0 & 0 & 0 \\
    -1 & 1 & 1 \\
    \end{pmatrix}
   \begin{pmatrix}
    x \\
     y\\
     z\\
    \end{pmatrix}
 = \begin{pmatrix}
    0 \\
    0 \\
    0 \\
    \end{pmatrix}
$$
Which cleary brings to two equal equations:
$$-x+y+z=0$$
I don't know how I should proceed. I can find a solution by trying some values but I don't like this method. 
Which is the best and secure method to solve this problem? Thank you very much.

Since $A-I=\begin{pmatrix}-1&1&1 \\0&0&0\\ -1&1&1\end{pmatrix}$, the eigenspace is defined by the single equation $x=y+z$, and it has dimension $2$, being isomorphic to $\mathbf R^2$ by the isomorphism:
\begin{align}\mathbf R^2&\longrightarrow \ker(A-I)\\\begin{pmatrix} y\\z\end{pmatrix}&\longmapsto \begin{pmatrix}y+z\\ y\\z\end{pmatrix}\end{align}
You can take as a basis of the eigenspace, say the image of the canonical basis of $\mathbf R^2$, i.e.
$$v_1=\begin{pmatrix}1 \\1\\0\end{pmatrix},\quad v_2=\begin{pmatrix}1 \\0\\1\end{pmatrix}.$$
To obtain a Jordan basis, you need to complete it with a vector $v_3$ such that $(A-I)v_3=v_2$ In such a basis the matrix of the endomorphism associated to $A$ will be
$$J=\begin{pmatrix}1&0&0 \\0&1&1\\ 0&0&1\end{pmatrix} \quad\text{since }\enspace Au_1=u_1,\;Au_2=u_2,\;Au_3=u_3+u_2.$$
Now to solve the vector equation $(A-I)v_3=v_2$, we have to solve the linear system
$$\begin{cases}1=-x+y+z, \\0=0,\\1=-x+y+z,\end{cases}$$
which has an obvious solution: $x=0,\;y=1,\;z=0$, in other words $u_3=e_2$.
As a conclusion, $A$ has a Jprdan normal form in basis
$$\mathcal B=(e_1+e_2,\, e_1+e_3,\, e_2).$$
