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 $?


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 $\exp(B)=I$.

  • 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

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