Show that:
\begin{equation} \dv t e^{tA} = e^{tA}A \end{equation}
We can apply the result we shown in eigenvalue:
\begin{equation} \dv t \qty(e^{tA}) = \dv t \qty(I + \sum_{k=1}^{\infty} \frac{t^{k}}{k!}A^{k}) = \qty(\sum_{k=1}^{\infty }\frac{1}{k!}kt^{k-1}A^{k-1})A \end{equation}
We do this separation because \(k=0\) would’t make sense to raise \(A\) (\(k-1=-1\)) to as we are unsure about the invertability of \(A\). Obviously \(\frac{1}{k!}k = \frac{1}{(k-1)!}\). Therefore, we can shift our index back yet again:
\begin{equation} \qty(\sum_{k=1}^{\infty }\frac{1}{k!}kt^{k-1}A^{k-1})A = \qty(\sum_{j=0}^{\infty }\frac{1}{j!}t^{j}A^{j})A \end{equation}
Awesome. So now we have the taylor series in \(e^{tA}\) back, times \(A\).
So therefore:
\begin{equation} \qty(\sum_{j=0}^{\infty }\frac{1}{j!}t^{j}A^{j})A = e^{tA}A \end{equation}
Be forewarned:
\begin{equation} e^{A}e^{B} \neq e^{A+B} \end{equation}
mostly because matrix multiplication is not commutative..