TITLE: What is an interpretation of the matrix exponential?
QUESTION [21 upvotes]: I just read about the existence of the "matrix exponential"
$$e^X := \sum_{k = 0}^\infty\frac1{k!}X^k$$
Is there a simple way to interpret this? I understand the analog between  real number exponentials as infinite Taylor expansions. However, I have no easy way of interpreting in the case of a matrix.
I've read that it relates to linear ODE's

REPLY [4 votes]: Think about $\exp$ as a function which translates a (relative) infinitesimal additive change into a finite multiplicative change after one unit of time (this view comes from Lie groups/algebras). Sounds strange, and the connection between the infinitesimal change and the resulting change is not always obvious:

If you add nothing, your infinitesimal additive change is zero ($x\mapsto x+0x$). The corresponding finite multiplicative change is $\exp 0=1$.
If your infinitesimal change is a multiple of the current value, i.e. $x\mapsto x+\lambda x$, then your finite change after one unit of time is $\exp\lambda$.

This can be applied to matrices (and even more general constructs):

If your infinitesimal change is just adding a scaled version of the current vector, i.e. $x\mapsto x+(\lambda\Bbb I)x$ (wheere $\Bbb I$ is the identity matrix), then after one unit of time this corresponds to scaling $x$ by a factor $e^\lambda$, which corresponds to the map $x\mapsto(e^\lambda\Bbb I)x$. You see $\exp(\lambda\Bbb I)=e^\lambda \Bbb I$.
If your infinitesimal additive change does not point in the direction of the current vector $x$, but perpendicular to it, i.e. $x\mapsto x+\lambda Rx$ with 
$$R:=\begin{pmatrix}0&-1\\1&0\end{pmatrix},$$
then the corresponding multiplicative change is a rotation with angle $\lambda$ (in radians). If we express the rotation matrices by $R_\theta$, then we see $\exp(\lambda R)=R_{\lambda}$.

All this needs some intuition and quite a lot of it can be delivered from differential equations as decscribed by @Mark.