Convergence of the exponential matrix. 
Hello. I find myself studying the exponential matrix and I am stuck on a statement that is given in the image.
Why does the fact that the series converges uniformly on any bounded subset of $gl(n,R)$ implies that the series converges uniformly on all $gl(n,R)$?
 A: When I read your question I’ve thought the following:
Take a matrix $A$ and let $ c:=|A|$. Then the sequence $a_t:=\sum_{k=1}^t\frac{A^k}{k!}$ is contained in $cl(B(0, r))$, where $r:= \sum_{k=1}^t\frac{c^k}{k!}=e^c $.
However the closure of the ball of radius $r$ is always compact in the space $M_n(\mathbb{R})$, so there exists a subsequence $\{a_{t_s}\}_s$ that is convergent to a certain matrix, that we call $e^A$ (in another way you can simply say that the sequence is convergent showing that is Cauchy by comparing it with the series $\sum_{k=0}^t\frac{c^k}{k!}$ of $e^c$). By definition of $\{a_t\}$, also $\{a_t\}$ is convergent and it converges to $e^A$.
Moreover you can prove that $e\colon M_n(\mathbb{R})\to M_n(\mathbb{R})$ satisfies $e^Ae^B=e^{A+B}$, so $e^{-A}$ is the inverse of $e^A$, that means $e^A\in GL(n, \mathbb{R})$. This shows also that the exponential map $e$ is a morphism from the group $(M_n(\mathbb{R}),+)$ to the group $(GL(n, \mathbb{R}), I)$. In this way you can prove that $\gamma$ is a one-parameter subgroup.
