Calculation of the Jordan canonical form Given the matrix $ F= \begin{bmatrix}
  3 & -1 & 0 \\
  1 & 1 & -2 \\
  0 & 0 & 2
 \end{bmatrix}$ calculate the Jordan canonical form such that $F = T F_j T^{-1}$.
The characteristic polynomial is $ (\lambda -2)^3= 0$ so the eigenvalue is $\lambda = 2$
The eigenvectors $v$ are given by $(F - \lambda I)v = 0$ so  $ \begin{bmatrix}
  1 & -1 & 0 \\
  1 & -1 & -2 \\
  0 & 0 & 0
 \end{bmatrix}v = 0 $ so the kernel is $ < \begin{bmatrix}
  1 \\
  1 \\
  0
 \end{bmatrix}>$
Now I need to calculate the kernel of $(F - \lambda I)^2$
$(F - \lambda I)^2=\begin{bmatrix}
  1 & -1 & 0 \\
  1 & -1 & -2 \\
  0 & 0 & 0
 \end{bmatrix}*\begin{bmatrix}
  1 & -1 & 0 \\
  1 & -1 & -2 \\
  0 & 0 & 0
 \end{bmatrix}=\begin{bmatrix}
  0 & 0 & 2 \\
  0 & 0 & 2 \\
  0 & 0 & 0
 \end{bmatrix}$
$(F - \lambda I)^2 v_2 = 0$ so the kernel is $<\begin{bmatrix}
  1  \\
  0  \\
  0 
 \end{bmatrix} ><\begin{bmatrix}
  0  \\
  1  \\
  0 
 \end{bmatrix} >$
So far my calculations should be right but now come the problems.
The matrix $T $ should be $\begin{bmatrix}
  1 & 0 & x_1 \\
  0 & 1 & x_2 \\
  0 & 0 & x_3
 \end{bmatrix}$ where the third column is:
$(F - \lambda I) \begin{bmatrix}
  1 \\
  0 \\
  0 
 \end{bmatrix} = \begin{bmatrix}
  1 & -1 & 0 \\
  1 & -1 & -2 \\
  0 & 0 & 0
 \end{bmatrix} \begin{bmatrix}
  1 \\
  0 \\
  0 
 \end{bmatrix} = \begin{bmatrix}
  1 \\
  1 \\
  0 
 \end{bmatrix}$
But this value makes the matrix $T$ not invertible. Where is my mistake?
 A: The first step is ok. You obtain $v_1= (1,1,0)^T$ such that $(F-\lambda I) v_1 =0$.
For the second step, you have to find a vector $v_2$ such that $$ 
(F-\lambda I) v_2 = v_1.$$
The solution is $v_2= (1,0,0)^T$.
For the third step, you have to find a vector $v_3$ such that $$ 
(F-\lambda I) v_3 = v_2.$$
The solution is $v_3= (1,0,1/2)^T$. The transformation matrix is then
$$ T= \begin{pmatrix} v_1 & v_2 & v_3 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\
 1 & 0 & 0 \\
 0 & 0 & \frac{1}{2} \end{pmatrix}.$$
And $$F= T \begin{pmatrix}2 & 1 & 0 \\
 0 & 2 & 1 \\
 0 & 0 & 2 \end{pmatrix} T^{-1}.$$
A: If we don't care about finding the change of basis matrix $T$, then we can arrive at the Jordan form quickly.
First, note that the characteristic polynomial of $F$ is
$$
\chi_F(t)=(t-2)^3
$$
This implies that $2$ is the only eigenvalue of $F$. 
Next, note that
$$
\dim\DeclareMathOperator{null}{null}\null(F-2\,I)=1
$$
That is, the eigenvalue $2$ has geometric multiplicity one. This means that the Jordan form of $F$ has one Jordan block. Hence the Jordan form of $F$ is
$$
\begin{bmatrix}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{bmatrix}
$$
Of course, the change of basis matrix can be computed using @Fabian's answer.
