Raising $e$ to the power of a matrix Does there exist a definition for matrix exponentiation?
If we have, say, an integer, one can define $A^B$ as follows:
$$\prod_{n = 1}^B A$$
We can define exponentials of fractions as a power of a radical, and we even have the following definition of the exponential:
$$e^z = \sum_{n = 0}^\infty \frac{z^n}{n!}$$
which comes from a Taylor Series for the function $\exp(z)$. Now, a problem seems to arise when we attempt to calculate $\exp(A)$, where $A$ is an $n$ x $n$ (square) matrix. We cannot define it as multiplication a "matrix number of times" as this makes no sense. The only reasonable definition that could work is the latter definition (the infinite series):
$$e^A = 1 + A + \frac{AA}{2!} + \frac{AAA}{3!} + \frac{AAAA}{4!} + \cdots$$
where we can define matrix exponentiation to the power of an integer, which is all that is required here. We know that $e^x$ will converge absolutely for all complex numbers, but do we know that this is true for matrices?
Can this "matrix sum" diverge, and are there ways to test divergence/convergence when a matrix is applied?
Or is this concept of "matrix divergence" not well defined?
 A: Yes, the exponential of a matrix can be defined by that Taylor series, and it is a very useful thing.  See
e.g. Wikipedia
The proof that it converges is not difficult, using any sub-multiplicative matrix norm.  If $\| \cdot \|$ is such a norm, then  any  power series
$\sum_n c_n A^n$  converges
whenever the real series $\sum_n |c_n| \|A\|^n$ converges.  
A: My favourite definition - as I teach quite often ODEs - is using differential equations.
Let us first note that $x(t)=\mathrm{e}^{at}$ is the unique solution of the initial value problem:
$$
x'=ax, \quad x(0)=1.
$$
Generalise that a little bit: Let $A$ be a square matrix ($A\in\mathbb R^{n\times n}$). Define as $\mathrm{e}^{tA}$ the unique, and globally defined, solution of 
$$
X'=AX, \quad X(0)=I,
$$ 
where $X$ is an $n\times n$ matrix, and $I$ the identity matrix. Through this definition a lot of properties of the exponential of a matrix, (which in order to prove require messy calculations) are proved very elegantly if we use the above definition. 
A: For diagonalizable $A$, you can write $A = P \Lambda P^{-1}$, where the column vectors of $P$ are eigenvectors and $\Lambda$ is diagonal matrix with eigenvalue entries.
Then $$e^{A} = e^{P \Lambda P^{-1}} = PIP^{-1} + P \Lambda P^{-1} + \frac{1}{2!}(P \Lambda P^{-1})^2 + \ldots = Pe^{\Lambda}P^{-1}$$
And raising $e$ to a diagonal matrix is easy—just replace all the diagonal entries with those raised by $e$.
