Is $e^{(rA)} = (e^{A})^r$ when $r \in \mathbb{R}$ and $A$ is an element of a Banach algebra?


Clearly if $n$ is an integer, then


$e^{(nA)} = e^{A+A \cdots +A} = e^{A}e^{A}\cdots e^{A} = (e^{A})^n$,


where the second equality follows from the Baker-Hausdorff lemma and the fact that [A,A]=0. On the other hand, I think the equality is not generally true when $r \in \mathbb{C}$. But what about the reals?


Many thanks for your thoughts!



Is exp(rA) = (exp(A))^r for real r and A in a Banach space?
A particular case of a Banach algebra is the one-dimensional case $\mathbb C$, right?  

And according to Pietro, if you can prove this in that case, you get the general case by saying "functional calculus". (At least for $A$ where it applies, such as normal $A$.)  

But is it true for $\mathbb C$? 


$A = 2\pi i$, $r=1/2$, $1 = \exp(2\pi i)$ and $1^{1/2} =^{?} \exp(\pi i) = -1$ .


