7
$\begingroup$

It is fairly easy to see that if $A$ and $B $, real square matrices, commute, then $A $ and $e^B $ commute. In fact,

$$Ae^B = \sum_{n=0}^\infty A\frac{B^n}{n!} = \sum_{n=0}^\infty \frac{B^n}{n!}A = e^BA $$

But is the reverse true as well? If $A $ and $e^B $ commute, do $A $ and $B $ commute as well? If not, can you help me finding a counter-example?

What if both $A $ and $B $ are not constant matrices but matricial functions of $t $?

$\endgroup$
16
$\begingroup$

You can cook up nontrivial real matrices $B$ with $\exp(B)=I$, for instance $$\pmatrix{0&2\pi\\-2\pi&0}.$$ There are matrices $A$ that don't commute with $B$, say $$A=\pmatrix{1&0\\0&0}.$$ But $A$ commutes with $\exp(B)=I$.

$\endgroup$
  • 3
    $\begingroup$ It might be worth adding a proof that choice for $B$ satisfies $\exp B=I$. $\endgroup$ – J.G. Oct 27 '18 at 7:04

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.