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.

I like to keep things as matrices and column vectors, visual. Since $(A - I)^2 = 0$ but $A - I \neq 0,$ we can take the third column of $R$ (for "Right") as anything we like for which 
$$ (A - I) w \neq 0.  $$
I like 
$$
w =
\left(
\begin{array}{c}
0 \\
0 \\
1
\end{array}
\right)
$$ 
The second column will be $v = (A-I)w,$ which is automatically an eigenvector (WHY??) 
$$
v =
\left(
\begin{array}{c}
1 \\
0 \\
1
\end{array}
\right)
$$ 
For the first column we can take any eigenvector that is independent of $v,$ looking back at $A-I$ we can take
$$
u =
\left(
\begin{array}{c}
1 \\
1 \\
0
\end{array}
\right)
$$ 
So, we have
$$
R =
\left(
\begin{array}{ccc}
1 & 1  & 0 \\
1 & 0 & 0 \\
0 & 1 & 1
\end{array}
\right)
$$
which has determinant $1,$ very helpful, and
$$  R^{-1} A R = J. $$
The direction that is actually useful is $R J R^{-1} = A.$ Useful for finding $e^A$ or $A^{100}$ or any $f(A)$ with $f$ analytic.
$$
R^{-1} =
\left(
\begin{array}{rrr}
0 & 1  & 0 \\
1 & -1 & 0 \\
-1 & 1 & 1
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc}
1 & 1  & 0 \\
1 & 0 & 0 \\
0 & 1 & 1
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 0  & 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 1  & 0 \\
1 & -1 & 0 \\
-1 & 1 & 1
\end{array}
\right) =
\left(
\begin{array}{rrr}
0 & 1  & 1 \\
0 & 1 & 0 \\
-1 & 1 & 2
\end{array}
\right)
$$
