Quick Fact Check if $A$ and $B$ Commute, $\exp((A+B)t)= \exp(At)\cdot\exp(Bt)$?

Question

If $A$ and $B$ Commute, $\exp((A+B)t)= \exp(At)\cdot\exp(Bt)$? is this statement true?

see http://math.stackexchange.com/questions/449354/commuting-exponential-matrices?rq=1 — James S. Cook, Feb 05 '14 at 05:05
Do you mean exp(At+Bt)=exp(At)exp(Bt)? (If so, the t is unnecessary.) — anon, Feb 05 '14 at 05:06
Yes, the statement is true. Note: as a matter of fact, this is true if and only if $A$ and $B$ commute. — Ben Grossmann, Feb 05 '14 at 05:10
@Omnomnomnom: Can you explain why the converse is true, i.e. why $\exp(t(A + B)) = \exp(tA)\exp(tB)$ for all $t \geq 0$ implies that $A$ and $B$ commute? — el_tenedor, Oct 30 '16 at 12:39
@Omnomnomnom: if so, i would appreciate a comment or answer to my post: http://math.stackexchange.com/questions/1991471/if-expta-b-expta-exptb-for-all-t-geq-0-then-a-b-commute — el_tenedor, Oct 30 '16 at 12:58

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

1

It is in fact the case. For a thorough explanation, please see my answer to this question. It's only a mouse click away!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

edited Apr 13 '17 at 12:21

Community

answered Feb 05 '14 at 05:29

Robert Lewis

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

0

if $A\cdot B=B\cdot A$

then $e^{A+B}=e^{A}\cdot e^{B}$

now multiply both side by $t$

we have

$e^{At+Bt}$,

you can apply same rule for this too

there is question related to matrix expoenntial

edited Apr 13 '17 at 12:20

Community

answered Feb 05 '14 at 05:05

dato datuashvili

Do you mean multiply $A$ and $B$ by $t$? – anon Feb 05 '14 at 05:07
he has done like this,$t$ is on the exponential part right? – dato datuashvili Feb 05 '14 at 05:09
huh? ${}{}{}{}$ – anon Feb 05 '14 at 05:10
@anon is my answer mistake? – dato datuashvili Feb 05 '14 at 05:11
1

No, I just think instead of saying "multiply both side[s] by $t$" (which would result in $e^{A+B}t=e^A*e^Bt$) you should say "multiply $A$ and $B$ by $t$." Although there is the issue that you are simply stating the desired conclusion without proof. – anon Feb 05 '14 at 05:11
yes yes sorry,English is not my native language – dato datuashvili Feb 05 '14 at 05:13
i have shown about commute matrix exponential – dato datuashvili Feb 05 '14 at 05:18

2 Answers2