3
$\begingroup$

Link to the post: A problem regarding the "exponential" of a continous linear mapping

Using notations from this post of mine, I have trouble proving that $e^A \circ e^B = e^B \circ e^A = e^{A+B}$ where $A$ and $B$ sastifies $A \circ B = B \circ A$.

Were them be series in $\mathbb{R}$ or $\mathbb{C}$, I would have multiplied the series together to get the desired result.

Therefore, I'm wondering if I could build the "multiplication" of series in the same manner. However, I have no clue how this can be done. Or maybe I'm going a wrong direction?

Please give me a hint to a correct way. Any help is greatly appreciated. Thank you for reading.

$\endgroup$
1
$\begingroup$

By definition, $$e^{A+B}x=\sum_{n=0}^{\infty}\frac{(A+B)^nx}{n!}=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}C(n,k)A^kB^{n-k}x $$ Since this series is absolutely convergent, we may exchange the summation order, obtaining $$ \sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\frac{C(n,k)A^k B^{n-k}x}{n!}=\sum_{k=0}^{\infty}\frac{A^{k}}{k!}\sum_{n=k}^{\infty}\frac{B^{n-k}x}{(n-k)!}=\sum_{k=0}^{\infty}\frac{A^k}{k!}e^Bx=e^Ae^Bx$$ Where moving $A^k$ out of the inner summation is justified because $A^k$ is continuous and the inner summation is a convergent series . Moreover, when $A$ and $B$ commute (i.e. $AB=BA$) we have $e^Ae^Bx=e^Be^Ax$ since they are both equal to $e^{A+B}x=e^{B+A}x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.