Assume that $\mathcal{T}=\{T(t)\}_{t\geq 0}$ be uniformly continuous semigroup of bounded linear operators . It is known that there is a bounded linear operator $A$ such that $T(t)=e^{tA}$. Also the eigenvalues of $e^{tA}$ are the $e^\lambda$ where $\lambda$ is an eigenvalue of $tA$.
I know that if linear operator $T$ has an eigenvalue $\lambda$ satisfying $|\lambda|> 1$, then $||T^n||\geq |\lambda|^n$. .
I think for any uniformly continuous semigroup $\mathcal{T}=\{T(t)\}_{t\geq 0}$
of bounded linear operators on finite dimension space $X$, $$\sup_{t\geq 0}||T(t)||=\infty$$
Because for uniformly continuous semigroup $\{T(t)\}_{t\geq 0}$ there is a bounded linear operator $A$ such that $T(t)=e^{tA}$, and if $\lambda$ is eigenvalue of $A$, then there is $r>0$ such that eigenvalue of $e^{rA}$, $e^{r\lambda}>1$. Hence $||T(nr)||\geq e^{nr\lambda}$ implies that $$\sup_{t\geq 0}||T(t)||=\infty$$ Please help me to know my proof is true or not?