Matrix exponentiation intuition. What does $x^A$ intuitively mean if $x \in \mathbb{C}$ and $A$ is any matrix? Also, what if we had $x$ being a matrix too? Last but not least, what happens if we have a complex $x$ raised to a quaternion(answering the first question probably generalizes this, but I include it in case only this case makes sense)? What does $x^a, x \in \mathbb{C}, a \in \mathbb{H}$ intuitively mean? What if $x \in \mathbb{H}$ too? 
I know that matrix exponentiation with base $e$ is $e^A=\displaystyle \sum_{k=0}^{\infty} \frac{1}{k!} A^k$ but apart from symbols in a "paper", I see nothing else.
 A: This is only a partial answer to your question. 
Some intuition as to what
$e^A$ means:
In the finite dimensional case consider the following.
For any $A \in M_n(\mathbb{R})$ and $t\in \mathbb{R_+}$ the matrix exponential $e^{tA}$ can be defined by,
$$  e^{tA}= \sum_{k=0}^\infty \frac {t^kA^k}{k!}$$
Since we have defined our 'matrix exponential' via an infinite sum we need to make sense of this first. More precisely we need to be sure that this converges. 
Indeed, taking any norm on $\mathbb{R^n}$ (note that any norm in finite dimensions are equivalent) and the corresponding matrix norm on $M_n(\mathbb{R})$ we can show that the partial sums of the above series form a Cauchy sequence. Since $\mathbb{R^n}$ is complete it follows that the the above infinite series converges. So the definition does make sense.
Now to the question of why would anybody care about this?. Well consider the following one dimensional initial value problem.
$$ \dot x =ax, \quad x(0)=x_0 $$
where $ x\in \mathbb R$ and $a \in \mathbb{R_+}$ is some constant. Now notice that the solution to this differential equation is given by 
$$ x=x_0e^{at}$$
Now, consider the following  IVP given by the autonomous system of differential equations,
$$ \dot x=Ax, \quad x(0) =x_0 $$
where $x\in \mathbb{R^n}$ and $A\in M_n(\mathbb{R}$). Suppose now we want to solve this system. Then if we can make sense of the matrix exponential, using the one dimensional case as inspiration we can write the solution to this system as
$$ x=x_0e^{tA}$$
In the infinite dimensional case where $A$ is some operator then we need to do a lot more work to make sense of this 'matrix exponentiation'. But at least in the finite dimensional case it is quite easy to see 'why' anybody would want to consider something like this.
There is a whole industry called 'Theory of operator semigroups' that generalizes this notion to infinite dimensions.
Some examples.
If $A$ is diagonal say,  $A= \begin{pmatrix} a_{11} & 0 \\ 0 & a_{22} \end{pmatrix}$ then $ e^A= \begin{pmatrix} e^{a_{11}} & 0 \\ 0 & e^{a_{22}}\end{pmatrix}$. In the diagonal case finding $e^A$ is pretty easy. In the general case it would be quite difficult. You will have to consider the Jordan form of your matrix to find the matrix exponential.
A: *

*Let $x\in\mathbb{C}^*$ and $A\in M_n(\mathbb{C})$. We can put $x^A=e^{\log(x)A}=\sum_k 1/k!(\log(x))^kA^k$ but this formula depends on the definition of $\log(x)$. For instance the formula has a sense if $x\notin \mathbb{R}^-$ and $\log$ is the principal logarithm. Unfortunately, in general $(xy)^A\not= x^Ay^A$ (choose $x=-1-\epsilon i,y=-1+\epsilon i$ where $\epsilon >0$). Yet, we obtain $(xy)^A= x^Ay^A$ if $Re(x)>0,Re(y)>0$.

*Take for example $x=1+i$ and $Q$ the quaternion $Q=\begin{pmatrix}1-2i&-3+i\\3+i&1+2i\end{pmatrix}$.Then $x^Q\approx \begin{pmatrix}18-10i&-8.4-7.6i\\2.15+11i&5.4-3i\end{pmatrix}$.

*If $x$ is a quaternion that has no eigenvalues in $\mathbb{R}^-$, then the equality $e^{\log(x)A}=\sum_k 1/k!(\log(x))^kA^k$ is no more valid.

*"What does $x^a,x∈C,a∈H$ intuitively mean ?" Another question: what does $\pi ^e$ intuitively mean ? Perhaps the volume of a cube in $\mathbb{R}^e$ with side length $\pi$.
