Exponential of matrices and bounded operators Let $A$ be a complex $n \times n$ matrix, such that the function $t\mapsto e^{tA}x$ is bounded on $\mathbb{R}$ and nonzero, for some vector $x\in \mathbb{C}$. How can we prove that $\inf_{t\in \mathbb{R}}|e^{tA}x|>0$ or in other words $|e^{tA}x|\geq c>0$ for all $t\in \mathbb{R}$.
I can only prove this in the case $A=a \in \mathbb{C}$, because $|e^{ta}x|=e^{Re(a)t}|x|$, and the boundedness implies that $Re(a)=0$, so $|e^{ta}x|=|x|=c>0$ because $t\mapsto e^{ta}x$ is supposed to be nonzero.
I also want to know if we have this property in the case $A$ is a bounded operator on an infinite dimensional Banach space $X$.
 A: $f:=t\rightarrow ||e^{tA}x||\in\mathbb{R}^{+*}$ is a continuous function. Then $\inf_tf(t)=0$ iff $\liminf_{t\rightarrow +\infty}f(t)=0$ or $\liminf_{t\rightarrow -\infty}f(t)=0$. For instance, assume that $\liminf_{t\rightarrow +\infty}f(t)=0$. Let $(E_i)_{i\leq q}$ be the generalized eigenspaces of $A$ associated to the distinct eigenvalues $(\lambda_i)_{i\leq q}$ of $A$. We may write $x=\sum_{k=1}^px_i$ where $x_i\in E_i\setminus \{0\}$ and $p\leq q$. We may assume that, there is $r\leq p$ s.t.for every $j\leq r$, $Re(\lambda_j)=\sup\{Re(\lambda_i)|i\leq p\}=\tau$, with $\tau<0$. Then $\liminf_{t\rightarrow +\infty}f(t)$ is "driven" by $||\sum_{i\leq r}e^{tA}x_i||$ in the following sense:  $||f(t)-\sum_{i\leq r}e^{tA}x_i||=o(e^{\tau t})$ when $t$ tends to $+\infty$. In the same way, $\limsup_{t\rightarrow -\infty}||f(t)||=\limsup_{t\rightarrow -\infty}||\sum_{i\leq r}e^{tA}x_i||\geq \lim_{t\rightarrow -\infty}e^{(\tau +\epsilon)t}$ for every $\epsilon\in ]0,-\tau[$ (cf. Jordan form of $A$). Finally, $\limsup_{t\rightarrow -\infty}||f(t)||=+\infty$, a contradiction.
