Find $T$ and $T^{-1}$ from Matrix to solve differential equation We have the differential equation $y'= \begin{pmatrix}
0 & 1 & 0 \\
-4 & 4 & 0 \\
-2 & 1 & 2
\end{pmatrix} \cdot y$  
I found out that the eigenvalues are 2 and the jordan normal form is $\begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}$
For my next step I need to find $T$ such that $\begin{pmatrix}
0 & 1 & 0 \\
-4 & 4 & 0 \\
-2 & 1 & 2
\end{pmatrix}$ = $T \cdot \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix} \cdot T^{-1}$
How can I find this $T$ (Change of basis)? I got the problem that the eigenvector is$ \begin{pmatrix}
x_1 \\
2x_1 \\
x_3
\end{pmatrix}$

The characteristic polynomial is
\begin{align}
     \mbox{det}\begin{pmatrix}
\lambda & -1 & 0 \\
4 & \lambda-4 & 0 \\
2 & -1 & \lambda-2
\end{pmatrix} &= \lambda(\lambda-4)(\lambda-2)+4(\lambda-2) \\
         &= (\lambda-2)(\lambda^2-4\lambda+4) =(\lambda-2)^3
\end{align}
This affords a lot of simplification. The solution of the vector ODE is
$$
             y = e^{tA}y_0,\;\;\; \mbox{where } y_0 =y(0).
$$
The solution may be expressed as a truncated power series:
$$
       y = e^{t(A-2I)}e^{2t}=e^{2t}\left(I+t(A-2I)+\frac{t^2}{2}(A-2I)^2\right)y_0.
$$
