I know that $A$ and $e^{At}$ conmute, as well as $A$ and $A^{1/2}$. So I am wondering if there is any chance that we can write: $$A^{1/2} A A^{1/2}-A^{1/2} A e^{At} A^{1/2}= A^2(1-e^{At})$$
I have been trying to show this, but I'm stuck determining if $A^{1/2}e^{At}A^{1/2}=Ae^{At}$. If so, why or why not? and which characteristics do I need for $A$?
It is possible to show this by using the definition of $e^{At}= \sum_{k=0}^\infty \frac{t^k}{k!} A^k$.
Since we have that $AA^{1/2}=A^{1/2}A$, then by induction one can show that $A^{k}A^{1/2}=A^{1/2}A^k$. Then you have:
$$A^{1/2}e^{At}=A^{1/2} \sum_{k=0}^\infty \frac{t^k}{k!} A^k = \sum_{k=0}^\infty \frac{t^k}{k!} A^{1/2}A^k=\sum_{k=0}^\infty \frac{t^k}{k!} A^k A^{1/2}=e^{At}A^{1/2}$$ as wanted.