Note 4.1.3.1:

If $AB=BA$, $e^{A+B}=e^A e^B; (e^A)^{-1} = e^{-A}; e^{0 \in M_{n \times n} (S)} = I$

If $A=diag(a_1, a_2, \cdots, a_n)$ then $e^A=diag(e^{a_1}, e^{a_2}, \cdots, e^{a_n})$
If $A=SDS^{-1}$ for a diagonal $D$ then $e^A=S e^D S^{-1}$.
if $A$ is non-diagonalizable, then cehck if there is $n$ s.t. $A^n=0$. If so, than a polynomial can be constructed from $e^A=\sum\limits_{i=0}^n \frac{A^i}{i!}$ $\left(=\sum\limits_{n=0}^{\infty} \frac{A^n}{n!}\right)$