# On the commutativity of matrices and their exponentials

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