Assume that $\Vert e^{tA}\Vert\to 0$ as $t\to \infty$. I want to prove that \begin{align}\sigma(A)\subset \{\lambda\in \Bbb{C}:\,\Bbb{Re}(\lambda)<0\}\end{align} where $\sigma(A)$ is set of all eigenvalues of matrix $A\in M_{n\times n}(\Bbb{R})$
MY TRIAL
Let $\lambda\in \sigma(A).$ Then, then wwe can find $x\neq 0$ s.t. \begin{align}Ax=\lambda x \end{align}
Since $\Vert e^{tA}\Vert\to 0$ as $t\to \infty,$ then $\forall\,\epsilon>0,\,\exists\, n_0\in\Bbb{N}$ s.t. $\forall\,n\geq n_0$ \begin{align}\Vert e^{tA}\Vert< \epsilon\end{align} but I don't know how to put these statements together in bringing up the right proof. Can someone show me how? It seems that the answer provided in the following post gave a related solution A real matrix whose eigenvalues have all negative real parts but I don't know how to employ it
Let $x \in \mathbb{R}^n$ such that $\| x\| =1$ and $Ax = \lambda x$. Then \begin{align*} \\exp(tA)x = \left( \sum_{i=0}^\infty t^nA^n/n! \right)x = \sum_{i=0}^\infty t^nA^n x/n! = \left( \sum_{i=0}^\infty t^n\lambda^n/n! \right)x = \exp(t \lambda) x. \end{align*} Thus $\exp(t \lambda)$ is an eigenvalue of $\exp(tA)$ and $\| \exp(t A)\| \ge |\exp(t \lambda)|$.
Now, what happens with $\exp(t \lambda)$ if $\Re(\lambda) \ge 0$ and we let $t \to \infty$? Can you get a contradiction with $\lim_{t \to \infty} \| \exp(tA) \| = 0$?
Also, this only works if $\lambda$ is real, what happens if $\lambda$ is exactly complex? Can you show that $\exp(t \lambda)$ is again an eigenvalue of $\exp(tA)$?