What is an interpretation of the matrix exponential?

Consider a complex diagonalizable $n \times n$ matrix. If $X = A D A^{-1}$ where $A$ is invertible and $D$ is diagonal, then it's easy to see that $$e^X = A e^D A^{-1}$$ and $$e^D = \mathrm{diag}(e^{d_{11}}, \ldots, e^{d_{nn}})$$

Thus, for diagonalizable matrices, it corresponds to exponentiating each eigenvalue.

There is also a general interpretation but it is less intuitive. Every complex $n \times n$ matrix can be written in Jordan canonical form. Since a matrix in Jordan canonical form is block diagonal with blocks of the form $$J(\lambda) = \begin{pmatrix}\lambda & 1 & & \\ & \lambda & \ddots & \\ & & \ddots & 1 \\ & & & \lambda \end{pmatrix}$$ where $\lambda$ is an eigenvalue of $X$, $e^X$ is similar to a block diagonal matrix consisting of the blocks $$e^{J(\lambda)} = \begin{pmatrix}e^\lambda & \frac{e^\lambda}{1!} & \frac{e^\lambda}{2!} & \cdots & \frac{e^\lambda}{(k - 2)!} & \frac{e^\lambda}{(k - 1)!} \\ & e^\lambda & \frac{e^\lambda}{1!} & \frac{e^\lambda}{2!} & \cdots & \frac{e^\lambda}{(k - 2)!} \\ & & \ddots & \ddots & \ddots & \vdots \\ & & & e^{\lambda} & \frac{e^{\lambda}}{1!} & \frac{e^{\lambda}}{2!} \\ & & & & e^{\lambda} & \frac{e^\lambda}{1!} \\ & & & & & e^{\lambda}\end{pmatrix}$$ where $J(\lambda)$ is $k \times k$.

Thus, in general, exponentiating a matrix corresponds to exponentiating each of its Jordan blocks.

In fact, this interpretation also holds for any analytic function $f$ applied to a matrix and not just $e^X$. In general, $f(J(\lambda))$ involves the derivatives of $f$. See this question and Wikipedia for more details.

In ODEs, it's quite simple. For a single ODE $$y' = ay$$ We have that it's solution is $y(t) = y(0)e^{at}$. We can actually define the exponential function this way (as a solution to an ODE). If we let $\vec{y}(t) = (y_1(t),\dots,y_n(t))$, then the differential equation: $$\vec{y}' = Ay$$ where $A$ is now a matrix, is given by: $$e^{At}y(0)$$ So, it may be useful to think of the matrix exponential as the "Solution to the System of ODEs".

It's also used in Lie theory, as the connection between a Lie Algebra/Group, but if you don't already know what this is it's probably not useful to build intuition.

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.