Recall in this chapter we introduced the concept of the matrix exponential $e^M$ for a square matrix $M$. For an $n \times n$ matrix $M$, the $n \times n$ matrix exponential $e^M$ is the infinite series
$$e^M = \sum_{k=0}^{\infty} \frac{1}{k!} M^k = {\rm{I}}_n + M + \frac{1}{2!}M^2 + \frac{1}{3!}M^3 + \dots.$$
For example, for $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ we find $e^A = \begin{bmatrix} e&e\\0&e\end{bmatrix}$. More generally, let $M = \begin{bmatrix} t & x \\ 0 & t \end{bmatrix}$ for any $t$ and $x$. Then its matrix exponential is $e^M = \begin{bmatrix} e^t & xe^t\\0 & e^t \end{bmatrix}$.
One thing to be careful about is whether or not $e^{A+B} = e^A e^B$.
Since $e^{A+B}$ is the sum of terms $(1/k!)(A+B)^k$, the complication is that the "binomial formula"
for $(A+B)^k$ breaks down if $AB \ne BA$ (the analogy with how $e^x$ behaves for scalars $x$
breaks down in some ways). Already for $k=2$ we see a problem because
$(A+B)(A+B) = A^2 + AB+BA + B^2$ is not the same as $A^2+2AB+B^2$ when $AB \ne BA$.
The good news is that if
$AB=BA$ then the binomial formula does hold for $(A+B)^k$, and we have
$$e^A e^B = e^{A+B}\,\,\,\mbox{ when }
AB=BA.$$
For instance, since $c_1 M$ and $c_2 M$ commute for any scalars $c_1, c_2$, we have
$$e^{c_1 M} e^{c_2 M} = e^{c_1 M + c_2 M} = e^{(c_1+c_2)M}, $$
$$\mbox{and iterating } e^{c_1 M} e^{c_2 M} \cdots e^{c_k M} = e^{(c_1+\dots+c_k)M}.$$
As a special case,
\begin{equation}
(e^M)^k = e^{kM} \mbox{ for every } k \ge 1\,\,\,
\mbox{(so } (e^{(1/k)M})^k = e^M).
\end{equation}
Taking $c_1 = 1$ and $c_2 = -1$,
since $M$ and $-M$ commute for any $M$, we
have
$$e^M e^{-M} = e^{0} = {\rm{I}}_n$$
for any $M$, so $e^M$ is always invertible with inverse $e^{-M}$.
Sadly, if $AB \ne BA$ then generally $e^A e^B \ne e^{A+B}$. The story of how to deal with this "great matrix exponential tragedy" is beyond the level of this course.
The power identity above can be used to show
$$e^M = \lim_{k \to \infty}({\rm{I}}_n + (1/k)M)^k.$$
Here is why the matrix exponential $e^M$ for $M = tA$ is so useful in the study of $\x' = A\x$: For an $n$-vector $\mathbf{b}$ and an $n \times n$ matrix $A$, the function $\y(t) = e^{tA}\mathbf{b}$ satisfies
$$\y'(t) = A(e^{tA}\mathbf{b}) = A \y(t),\,\,\, \y(0)=\mathbf{b}.$$
Here we visualize the matrix exponential $e^{tA}$ from $t=0$ to $t=1$ for a $2\times 2$ matrix $A = \begin{bmatrix}a & b \\ c & d \end{bmatrix}$ by showing the two curves $e^{tA}\mathbf{e}_1$ and $e^{tA}\mathbf{e}_2$ for $0\le t \le 1$. (Here $\mathbf{e}_1 = \begin{bmatrix} 1 \\0\end{bmatrix}$ and $\mathbf{e}_2 = \begin{bmatrix} 0 \\1 \end{bmatrix}$.) We also visualize the approximation of $e^A$. To play with the visulization yourself, pick values for constants $a,b,c,d$, and choose $k$.