It is known that if two matrices $A,B \in M_n(\mathbb{C})$ commute, then $e^A$ and $e^B$ commute. Is the converse true?

If $e^A$ and $e^B$ commute, do $A$ and $B$ commute?

Edit: Addionally, what happens in $M_n(\mathbb{R})$?

Nota Bene: As a corollary of the counterexamples below, we deduce that if $A$ is not diagonal then $e^A$ may be diagonal.


5 Answers 5


No. Let $$A=\begin{pmatrix}2\pi i&0\\0&0\end{pmatrix}$$ and note that $e^A=I$. Let $B$ be any matrix that does not commute with $A$.

  • $\begingroup$ How does one see that $e^A = I$? Just curious how one would actually go about doing the computation. $\endgroup$
    – Stahl
    Apr 2, 2013 at 15:48
  • 7
    $\begingroup$ @Stahl: Since the matrix is diagonal, just take the exponential of every diagonal element. To wit: $\exp\operatorname{diag}(\lambda_1,\lambda_2,\ldots)= \operatorname{diag}(e^{\lambda_1},e^{\lambda_2},\ldots)$. $\endgroup$ Apr 2, 2013 at 15:48
  • 1
    $\begingroup$ Ah ok! I've seen the exponential map before, but never any explicit computations using it. +1 $\endgroup$
    – Stahl
    Apr 2, 2013 at 15:49
  • 1
    $\begingroup$ (Can someone explain why one of the \lambdas in my comment above does not work? I edited it several times, ensuring there is no space or other unwanted character in there. Well, actually, there seems to be two characters: U+200C ZERO WIDTH NON-JOINER followed by U+200B ZERO WIDTH SPACE, but I didn't put them there!) $\endgroup$ Apr 2, 2013 at 15:53
  • 2
    $\begingroup$ @CameronBuie The answer seems to be here $\endgroup$ Apr 2, 2013 at 16:02

Here's an example over $\mathbb{R}$, modeled after Harald's answer: let $$A=\pmatrix{0&-2\pi\\ 2\pi&0}.$$ Again, $e^A=I$. Now choose any $B$ that doesn't commute with $A$.

  • 13
    $\begingroup$ +1 We can even pick $B$ be to conjugate to $A$, e.g. $$B=\pmatrix{0&-\pi\cr4\pi&0\cr},$$ so then we have both $e^A=I_2$ and $e^B=I_2$. $\endgroup$ Apr 2, 2013 at 17:23
  • $\begingroup$ @JyrkiLahtonen That's dramatic! ;-D $\endgroup$
    – user1551
    Apr 2, 2013 at 17:32
  • $\begingroup$ Why all the examples?? $\endgroup$
    – Squirtle
    Apr 2, 2013 at 20:21
  • 4
    $\begingroup$ @dustanalysis: Because the statement that OP posted is not true, and these examples are the counterexamples of OP's statement. $\endgroup$ Apr 2, 2013 at 20:34

Another example: $$A=\pmatrix{0&-2\pi\\ 2\pi&0}, \textrm{ }B=\pmatrix{0&-2\pi\\49\cdot 2\pi&0}$$ This is a counterexample of the following statement: $$e^{A+B}=e^Ae^B\Longrightarrow AB=BA$$ since $$e^A=e^B=e^{A+B}=I$$ but $$AB=4\pi^2 \pmatrix{-49&0\\ 0&-1},$$ $$BA=4\pi^2 \pmatrix{-1&0\\0&-49}.$$


I just want to point out a general(maybe not the most general) way of constructing infinitely many counterexamples in $M_2(\mathbb{R})$. It is not hard to prove for a $2\times2$ traceless matrix $X$, we have

$$e^X=\cos(\sqrt{\det X})I+\frac{\sin\sqrt{\det X}}{\sqrt{\det X}}X,$$

where $\frac{\sin 0}{0}$ is to be understood as 1 for the second term. Thus to have $e^X=I$ we just need some traceless $X$ having $\det X=(2n\pi)^2,n\in\mathbb{Z/\{0\}}$, and this is a fairly easy job. Most of the matrices you construct in this way won't commute with each other. All the previously given $M_2(\mathbb{R})$ counterexamples fall into this category.


I would add that if $A$ and $B$ are Hermitian matrices, $[e^A,e^B] = 0$ does imply $[A,B] = 0$. You can refer to this answer by user8675309.

To put it shortly, the eigenspaces of $e^A$ are exactly those of $A$ if $A$ is Hermitian. Since $[e^A, e^B] = 0$, $e^A$ and $e^B$ have a common eigenbasis. This basis serves as a common eigenbasis of $A$ and $B$ as well. Therefore $A$ and $B$ commute.


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.