Show right eigenvectors are parallel to a vector that relates to the eigenvalues

I am given a system of first-order linear hyperbolic PDE's that can be written as $n_{\theta} + A\eta_{\xi} = 0, n = (\eta \,\,\,u)^T.$ Thus, A is diagonalizable, $A' = R^{-1}AR$. Multiplying by R from the left the PDE system,the matrix of eigenvectors, one gets two decoupled equations $m^i_{\theta} + \lambda_im^i_{\xi} = 0, \,\,\lambda_i$ being the eigenvalues of $A.$ With initial conditions $n_0 = (\eta_0\,\,\,u_0)^T, \,\,m_0(0,0,\xi) = R^{-1}n_0$ one gets the solution $m^i = m^i_0(\xi - \lambda_i\theta)$ and finally the sought solution $n(\theta,\xi) = Rm(\theta,\xi).$ It turns out that the eigenvectors $r_1,r_2$, i.e. the column vectors of $R,$ are parallel to the vectors $(1\,\,\,\lambda_1)^T$ and $(1 \,\,\,\lambda_2),$ respectively. I tried to write down all the details and still can not see that this holds. Can somebody show this ? Thanks.