Say $A \in L(\mathbb{R}^n, \mathbb{R}^n)$, and for all eigenvalues of $A$ (let's call them $\lambda$) we have $Re \: \lambda < -\mu$. I need to prove that $||Ax||_{op} \leq C_0 e^{-\mu x} \; \: \forall x > 0$, where $C_0$ is some positive constant.
The standard restriction:
\begin{equation} ||\exp (Ax)|| \leq \exp^{||A||x} \end{equation}
doesn't seem to work here, because the norm of the matrix can clearly be smaller than its largest eigenvalue. I tried to transform the matrix to the Jordan normal form, but it doesn't seem to help, as even for a matrix $B$ in Jordan normal form the estimate $||Bx|| \leq k x$ doesn't seem to work in a case when $Re \: \lambda < k$ for all eigenvalues $\lambda$ of $B$. Could someone please help me with that?
Apply the standard "restriction" as you call it to $$ \| \exp(B) x\|\, \leq e^{\|B\|} \|x\| $$ for $B = t I + A$. Then the eigenvalues of $B$ will be of the form $(t+\alpha_k) + i \beta_k$ where the eigenvalues of $A$ are $\lambda_k \alpha_k+\beta_k$ with all $\alpha_k$ in $(-\infty,-\mu)$. Then the modulus of the eigenvalue of $B$ is $\sqrt{(t+\alpha_k)^2+\beta_k^2}=|t| \sqrt{1+2\alpha_k t^{-1} + (\alpha_k^2+\beta_k^2)t^{-2}}$Using the fact that $\exp(B)=e^t \exp(A)$ (since $I$ commutes with every matrix) this gives $$ \|\exp(A)x\| \leq e^{\|tI+A\|-t} \|x\| $$ Then take the limit as $t \to \infty$.