Basis-free, non-power series definition of the exponential of linear operator? Given an arbitrary linear operator $A$ (be it real, complex or whatever), how can the exponential of it ($e^A$) be defined naturally, without stuff like power series? The exponential for regular numbers can be defined in terms of nice (although not necessarily best for generalising) axioms like $\exp(a+b) = \exp(a)\cdot\exp(b)$ and $\exp'(0) = \exp(0)$, which can be interpreted easily (in that case, a structure preserving mapping between the additive and multiplicative groups with other nice properties).
I've heard that the answer lies somewhere in the relationship between lie algebras and lie groups, but while I have an intuitive understanding of what a lie group is the topic as a whole is a bit beyond me (currently), and most explanations I've heard are mired in dense jargon. If there is a simpler (although perhaps less general) explanation, what is it? And if we have to go deep into the weeds of lie algebras, can you try and explain the meanings of the terms?
 A: There are two different definitions of the exponential that can be defined geometrically. One is for Lie groups and the other for Riemannian manifolds. However, they can be related to each other. We start with the definition on a Riemannian manifold:
Riemannian manifold
Let $M$ be a Riemannian manifold with either a definite or indefinite metric. Pick a point $p$ on the manifold. 
The exponential map will be a map  $exp_p:T_pM \rightarrow M$ 
This means there is not just one exponential map, there are many of them - one based at every point of the manifold. The domain of exponential function (at that point) is then the tangent space of the manifold (at that point). 
In particular, when we pick a vector $v$ at the tangent space $T_pM$ at $p$, it chooses a geodesic $\gamma_v$ on the manifold and we follow it and where we end up after a unit of time defines the value of the exponential. That is:
$exp_p(v) :=\gamma_v(1)$
Choosing the zero vector means we do not move at all, thus $exp_p(0) = p$.
Lie groups
Given a Lie group $G$ then the exponential map is $exp: G' \rightarrow G$ where $G'$ is the Lie algebra. It is defined as follows: when we pick a vector $v\in G'$ we find that it gives a canonically defined subgroup $\gamma_v$; in fact it's a subgroup that's parametised by time, and is usually referred to as a 1-parameter subgroup. Again, we define the exponential of the vector $v$ by following this subgroup for a unit amount of time and defining its value as where we end up at. Hence $exp(v):=_\gamma(1)$.
Compatibility
The correspondance between these two definitions are obvious. Now, Lie groups are in particular manifolds, but they are not Riemannian manifolds; and Riemann manifolds do not have a group structure. When we have compatible structures though, we find the definitions match. 
In particular, when we given a Lie group with a metric that is compatible with its group structure - what's called a translation invariant metric - then we find the definitions agree. 
Now, standard examples of Lie groups are groups of matrices, for example, $GL(n)$ or $SU(n)$. However not all Lie groups can be expressed in this way; those that can be are called matrix Lie groups. We can write out functions defined on such groups as power series. The exponential is such a function and when we write it as a power series, we will find it matches the usual definition. 
