Exponential of a matrix always converges I am trying to show that the exponential of a matrix converges for any given square matrix of size $n\times n$:
$M\mapsto e^M$ e.g. $\displaystyle e^M = \sum_{n=0}^\infty \frac{M^n}{n!}$
Can I argue that: Since $n!$ necessarily grows faster than $k^n$ will, that this converges. This seems to be an obvious fact, since:
$$n!=1\times 2\times 3\times \cdots \times k\times (k+1)\times (k+2)\times \cdots$$
$$k^n=k\times k\times k \times\cdots\times k \times k\times \cdots$$

If we have some $q\times q$ matrix, with $a$'s in each position(which will grow as fast as we make our $a$ and $q$ large) we still only get increasing at a rate of $q^{n-1}\times a^n$

In light of the comments, I know that in this banach space, I need only show that $\displaystyle e^M = \sum_{n=0}^\infty \frac{||M||^n}{n!}$ converges. Now I have many matrix norms to choose from, and I can't seem to get a good argument going rigorously. Any ideas?

This topic is extraordinarily well explained in the book Naive Lie Theory. Here is an extract that will answer your question.


