Let $A \in M^{n\times n}(\mathbb{R})$ and $\{\varphi_i \in \mathbb{R^n};1 \leq i \leq n \}$ be a basis for $\mathbb{R^n}$ consisting of eigenvectors. The corresponding eigenvalues are $\{\lambda_i\}_{i=2}^{n}$, which means $A\varphi_i=\lambda_i\varphi_i$.
Now I've got to show, that $\exp\left(tA\right)\varphi_i=e^{\lambda_it}\varphi_i$.
I'm asking myself, if this is trivial, because $A\varphi_i=\lambda_i\varphi_i$, we get $A=\lambda_i$ and therefore the assertion $\exp(tA)\varphi_i=\exp(t\lambda_i)\varphi_i=e^{\lambda_it}\varphi_i$. Is this right or is there another way to prove it?