Can't find the Jordan form of this 3x3 I have the matrix 
$$\begin{pmatrix}
2 & 2 & -1 \\
-1 & -1 & 1 \\
-1 & -2 & 2
\end{pmatrix}$$
and need to find its Jordan canonical form.  I can find that the only eigenvalue is 1 with algebraic multiplicity 3, and I can find two independent eigenvectors, 
$$\begin{pmatrix}1\\0\\1\end{pmatrix},\qquad \begin{pmatrix}0\\1\\2\end{pmatrix}$$
I am told that to find a generalized eigenvector I need to solve 
$$A\vec{x} = \vec\xi$$
where $\vec\xi$ is an eigenvector.  However, when I try to solve in this case, 
$$\begin{pmatrix}
1 & 2 & -1 \\
-1 & -2 & 1 \\
-1 & -2 & 1
\end{pmatrix}\vec{x} = \begin{pmatrix}1\\0\\1\end{pmatrix}$$
I find that there is no solution, and likewise for the other eigenvector.
I've also seen instruction that to find the generalized eigenvector you can solve $(A-\lambda I)^2\vec x=\vec0$ which in this case is
$$\begin{pmatrix}
1 & 2 & -1 \\
-1 & -2 & 1 \\
-1 & -2 & 1
\end{pmatrix}^2\vec{x} = \vec{0}$$
But the square of this matrix is the zero matrix.  I thought perhaps that means I can select any independent vector I want, but when I select $\begin{pmatrix}0\\0\\1\end{pmatrix}$ I find that the resulting transformation matrix does not produce a matrix in Jordan form.  That is to say, with 
$$T = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 2 & 1 \end{pmatrix}$$
If I compute $J=T^{-1}AT$ I get 
$$J = \begin{pmatrix} 1 & 0 & -1 \\
0 & 1 & 1 \\
0 & 0 & 1 \end{pmatrix}$$
Since that's not in Jordan form, I'm not sure what I'm doing wrong.
 A: Since the only eigenvalue is $1$ and it has multiplicity $3$, you know that the Jordan normal form is one of a short list:
$$\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\hspace{1cm}
\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\hspace{0.5cm}\text{or}\hspace{0.5cm}
\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}.$$
Call these $A_1,A_2$, and $A_3$. As you say, we can determine the size of the Jordan blocks by looking at $\dim\ker(A-I)$, where $I$ is the $3\times 3$ identity matrix. Notice that
$$A-I=\begin{pmatrix}1 & 2 & -1 \\ -1 & 2 & 1 \\ -1 & -2 & 1 \end{pmatrix}$$
from which, by row operations for example, we see $\dim\ker(A-I)=2$. This means that $A$ has two Jordan blocks, so we have ruled out $A_1$. Next, compute $\dim\ker(A-I)^2$. Actually, $(A-I)^2$ is the $0$ matrix, so $\dim\ker(A-I)^2=3$. This tells us that the biggest Jordan block has rank $2$, so the correct answer is $A_2$.

As confirmation, Mathematica returns the Jordan normal form
$$J=\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 1 \\
 0 & 0 & 1 \\
\end{array}
\right)$$
with the conjugating matrix as
$$T=\left(
\begin{array}{ccc}
 0 & -1 & 0 \\
 1 & 1 & 0 \\
 2 & 1 & 1 \\
\end{array}
\right)$$
That is, $T^{-1}AT=J$.

A: Once you have found that $(A-\lambda I)^2=0$, you can pick any vector $x_1$ that is not en eigenvector, e.g. as you suggested $x_1=(0,0,1)^T$. After that, you construct the corresponding eigenvector as $x_2=(A-\lambda I)x_1=(-1,1,1)^T$. Next, you pick an eigenvector that is linearly independent with $x_1$ and $x_2$ (independent with $x_2$ is enough), e.g. $x_3=(1,0,1)^T$. Now make $T=[x_1\ x_2\ x_3]$.
As to your first approach, it is not guaranteed that you can solve $(A-\lambda I)x=\xi$ for any eigenvector $\xi$, it is only possible to solve the system for some $\xi$. It means that you need to solve
$$
(A-\lambda I)x=t\begin{bmatrix}1\\0\\1\end{bmatrix}+s\begin{bmatrix}0\\1\\2\end{bmatrix}
$$
for some $t,s$. Since the last two equations are dependent, you get easily that the RHS must be the same, i.e. $s=t+2s$ etc.
