Finding the matrix exponantial I have to calculate the following:
$e^{xA} ,  A= 
\begin{pmatrix}1 & 1 & 1 \\
1&1&1\\
1&1&1\\
\end{pmatrix}$
I use the following rule:
$e^{xA} = \sum \frac{x^k}{k!}A^k$
Now I'm looking for a matrix such that:
$A=S D S^{-1}$.
So I calculated the eigenvalues and eigenvectors of A, these are:
$\lambda _0=3 , \lambda_{1,2} = 0 $
$\vec{y_0} = \begin{pmatrix}
1\\
1\\
1
\end{pmatrix} $,$\vec{y_1} = \begin{pmatrix}
1\\
-1\\
0
\end{pmatrix} $,$\vec{y_2} = \begin{pmatrix}
1\\
0\\
-1
\end{pmatrix}$
So I have the following 3 matrices:
$ 
S = \begin{pmatrix}
1&1&1\\
1&-1&0\\
1&0&-1
\end{pmatrix}
, D = \begin{pmatrix}
3&0&0\\
0&0&1\\
0&0&0
\end{pmatrix} , S^{-1}=
\begin{pmatrix}
 \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
 \frac{1}{3} & \frac{1}{3} & -\frac{2}{3} \\
 \frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \\
\end{pmatrix}
$
But when I calculate $SDS^{-1}$ I don't get A.
Can someone help me where I have gone wrong. I thought I understood it up until this point, but there must be something wrong with my reasoning because I have checked every thing with mathematica.
I also don't know what to do when I have found the correct S and D.

Hint: for $n\gt0$, $A^n=3^{n-1}A$ This should be enough to compute $e^{xA}$.

You have the wrong Jordan Normal Form for $A$. Since $A$ is Hermitian, it is diagonalizable:
$$
\begin{bmatrix}
1&1&1\\
1&-1&0\\
1&0&-1
\end{bmatrix}
\begin{bmatrix}
3&0&0\\
0&0&0\\
0&0&0
\end{bmatrix}
\begin{bmatrix}
1&1&1\\
1&-1&0\\
1&0&-1
\end{bmatrix}^{-1}
=
\begin{bmatrix}
1&1&1\\
1&1&1\\
1&1&1
\end{bmatrix}
$$
