# Show that if $A$ and $B$ commute, then $B$ commutes with $e^A$

Show that if $$A$$ and $$B$$ commute, then $$B$$ commutes with $$e^A$$

For the first one I have $$e^A = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} +\dots$$

I believe I can pull out $$I + A + \frac{A^2}{2!} + \frac{A^3}{3!} +\dots$$ and multiply by $$B$$ on either side but not sure the validity of this proving this

• You may want to first prove that $A^n$ commutes with $B$ for all $n>0$. May 5, 2022 at 20:00
• That's the correct path. You can show that $B$ commutes with every finite partial sum. And then you need to deal with convergence. (E.g., do you know that multiplication is continuous?) May 5, 2022 at 20:00
• @HagenvonEitzen: Do you really need to worry about convergence? ananta's answer looks good to me. May 5, 2022 at 20:29

Your approach to the problem is nice. Expanding the exponent $$e^A$$ is a key step in proving that if $$A$$ and $$B$$ commute then $$e^A$$ and $$B$$ also commute.
It is given to us that $$A$$ and $$B$$ commute and thus we can write: $$\left[A,B\right] = \left[B,A\right]$$ Note:If $$A$$ and $$B$$ are matrices this also means that $$A$$ and $$B$$ are both square matrices.
Now, consider an arbitrary exponent of $$A$$ i.e. $$A^k$$. Let us multiply this by $$B$$: $$A^kB = (AAA\cdots k\text{ times}) B$$ Now, since $$A$$ and $$B$$ commute, we can interchange the position of the last $$A$$ in the RHS of the previous product and $$B$$. $$\implies A^kB = (AAA\cdots k-1\text{ times}) BA$$ This can be done repeatedly and $$B$$ can be moved to an arbitrary position within the product: $$\implies A^kB = (AAA\cdots k-n\text{ times})B(AAA\cdots n\text{ times})$$ In particular we can place $$B$$ in the beginning of the product:$$\implies A^kB = B(AAA\cdots k\text{ times})$$ Now, Let us prove that $$e^A$$ and $$B$$ commute. As you correctly mentioned: \begin{align} &\boxed{e^A = \Sigma_{i=0}^{\infty}\dfrac{A^i}{i!}}\\ \implies \left[ e^A,B\right] = e^AB &= \left[ \Sigma_{i=0}^{\infty}\dfrac{A^i}{i!}\right ]B\\ \implies e^AB &= \Sigma_{i=0}^{\infty}\dfrac{A^i}{i!}B\\ &= \Sigma_{i=0}^{\infty}\dfrac{A^iB}{i!}\\ &=\Sigma_{i=0}^{\infty}\dfrac{BA^i}{i!}\\ &=\Sigma_{i=0}^{\infty}B\dfrac{A^i}{i!}\\ &=B \left[ \Sigma_{i=0}^{\infty}\dfrac{A^i}{i!}\right] = \left[ B,e^A\right] \end{align}\\ Hence proved that $$e^A$$ and $$B$$ also commute.
Define $$U(x)=[B, e^{x A}]$$, then $${d U \over dx} = [B, Ae^{x A}]$$. Since $$[A,B]=0$$, we have $${d U \over dx} = BAe^{x A} - A e^{x A}B = A[B, e^{x A}] = A U(x)$$ Solving this: $$U(x)=e^{x A} U(0)$$ and since $$U(0)=[B, e^{0A}]=[B,I]=0$$ we deduce that $$[B, e^A]=U(1)=e^{A}U(0)=0$$