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

  • 3
    $\begingroup$ You may want to first prove that $A^n$ commutes with $B$ for all $n>0$. $\endgroup$
    – Somos
    May 5, 2022 at 20:00
  • 4
    $\begingroup$ 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?) $\endgroup$ May 5, 2022 at 20:00
  • $\begingroup$ @HagenvonEitzen: Do you really need to worry about convergence? ananta's answer looks good to me. $\endgroup$
    – TonyK
    May 5, 2022 at 20:29

2 Answers 2


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.


Possibly overkill, but a technique that is worth knowing:

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 $$[B, e^A]=U(1)=e^{A}U(0)=0$$


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