Let $\lambda\in \sigma(A).$ Then, $e^{\lambda t}\in \sigma(e^{tA})$ where $\sigma(A)$ is set of all eigenvalues of matrix $A$

Question

Let $\lambda\in \sigma(A).$ Then, $e^{\lambda t}\in \sigma(e^{tA})$ where $\sigma(A)$ is set of all eigenvalues of matrix $A$.

MY TRIAL

Let $\lambda\in \sigma(A).$ Then, $\exists:t\neq 0$ s.t. \begin{align}Ax=\lambda x \end{align} \begin{align}\sum^{\infty}_{n=0}\frac{A^n}{n!}=\sum^{\infty}_{n=0} \frac{ \lambda^n}{n!} \end{align} \begin{align}e^{A}=e^{\lambda } \end{align} So, \begin{align}e^{At}=\sum^{\infty}_{n=0}\frac{A^n t^n}{n!}=\sum^{\infty}_{n=0}\frac{\lambda^n t^n}{n!}=e^{\lambda t}\end{align} and we're done!

I'm skeptical about my proof. Can someone please, check for me? If the proof is wrong, alternative proofs will be highly regarded. Thanks

Answer 1

Your idea is right but your proof is muddled (due to mixing up the scalar $t$ with the eigenvector of $A$ corresponding to $\lambda$).

Suppose $v$ is an eigenvector with $Av = \lambda v$. Then $$e^{At}v = \sum_{i=0}^{\infty} \frac{(At)^i}{i!}v = \sum_{i=0}^{\infty} \frac{t^i}{i!} (A^i v) = \sum_{i=0}^{\infty} \frac{(\lambda t)^i}{i!} v = e^{\lambda t}v.$$

Therefore $v$ is an eigenvector of $e^{At}$ with eigenvalue $e^{\lambda t}$. The one additional detail that you may want to check (depending on the level of rigor expected) is that the sums in the above calculation are well-defined (they are).