How is it possible to calculate the state transition matrix of the following LTI-System:
$$ \frac{d\mathbf{x}}{dt}=\mathbf{A}\mathbf{x}+\pmatrix {1 \\ -1}u$$ $$y=\pmatrix {2 & -2}\mathbf{x}$$ $$s_1=2,s_2=-1$$ $$\mathbf{p}_{r1}=\pmatrix {1 \\ 1},\mathbf{p}_{l1}=\pmatrix {0 \\ 1} $$ $s_1$ and $s_2$ are the known eigenvalues of $\mathbf {A} $, $\mathbf{p}_{r1}$ is the right eigenvector and $\mathbf{p}_{l1}$ the left eigenvector, both belonging to the $s_1$.
You have to reconstruct the (right) eigenbasis of $A$ first. Since two eigenvectors are given, there is enough information to find the second right eigenvector. Let $v$ denote this eigenvector, then $Ap_{r2}=-p_{r2}$.
Multiplying this from the left by $p_{l1}^T$, we get $$ -p_{l1}^Tp_{r2} = p_{l1}^TAp_{r2} = 2p_{l1}^Tp_{r2}, $$ which proves $p_{l1}^Tp_{r2}=0$. Hence $p_{r2}=\pmatrix{1\\0}$ is such a vector. Now you can proceed with standard techniques.