Consider a linear operator $T: \mathrm R^n \rightarrow \mathrm R^n $.
Given: $$ e^T = \sum_{k=0}^\infty \frac{T^k}{k!} $$ Is $e^T$ still a linear operator? If it is, why?
One should note that: Given a vector $v \in \mathrm R^n$, the operator $e^T$ will take its effect by $e^T v$ instead of $e^{Tv}$. And then this question will be solved by answers shown below.
The origin and motivation of this question
I am reading a book about differential equation, and it claims this conclusion where I cannot find any clue.
Though I have finish a course on matrix theory and accept the fact that $e^T$ is a linear operator, I still cannot figure out why the product operator must be applied on $e^T v$ instead of $e^{Tv}$.
And now this question has been resolved: Because the exponential of a vector is not (well) defined by a sum of a serie, so we can only have a linear operator work in $e^T v$ fashion. I think it may revolve some confusions coming from beginners.
$e^T$ is absolutely convergent by assumption (although it's easy to show). Let $F_m = \sum_{k=0}^m T^k/k!$ which is itself a linear operator since it's a finite-degree polynomial in the linear operator $T$. Then:
$$ \begin{aligned} e^T v + e^T w &= \lim_{m \to \infty} F_m v + \lim_{m \to \infty} F_m w & \\ &= \lim_{m \to \infty} (F_m v + F_m w) & \text{(by absolute convergence)} \\ &= \lim_{m \to \infty} F_m (v + w) & \text{(by linearity of $F_m$)} \\ &= e^T (v + w). \end{aligned} $$
Similarly,
$$ \begin{aligned} a (e^T v) &= a \left( \lim_{m \to \infty} F_m v\right) & \\ &= \lim_{m \to \infty} a (F_m v) & \text{(by absolute convergence)} \\ &= \lim_{m \to \infty} F_m (av) & \text{(by linearity of $F_m$)} \\ &= e^T (av). \end{aligned} $$
The text alludes to the lines marked "(by absolute convergence)" when it says "by properties of limits".