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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: Additionally, 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.

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5 Answers 5

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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 $A$.

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  • $\begingroup$ How does one see that $e^A = I$? Just curious how one would actually go about doing the computation. $\endgroup$
    – Stahl
    Commented Apr 2, 2013 at 15:48
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    $\begingroup$ @Stahl: Since the matrix is diagonal, just take the exponential of every diagonal element. To wit: $\exp\operatorname{diag}(\lambda_1,\lambda_2,\ldots)= \operatorname{diag}(e^{\lambda_1},e^{\lambda_2},\ldots)$. $\endgroup$ Commented Apr 2, 2013 at 15:48
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    $\begingroup$ Ah ok! I've seen the exponential map before, but never any explicit computations using it. +1 $\endgroup$
    – Stahl
    Commented Apr 2, 2013 at 15:49
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    $\begingroup$ (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!) $\endgroup$ Commented Apr 2, 2013 at 15:53
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    $\begingroup$ @CameronBuie The answer seems to be here $\endgroup$ Commented Apr 2, 2013 at 16:02
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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 $A$.

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    $\begingroup$ +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 $e^B=I_2$. $\endgroup$ Commented Apr 2, 2013 at 17:23
  • $\begingroup$ @JyrkiLahtonen That's dramatic! ;-D $\endgroup$
    – user1551
    Commented Apr 2, 2013 at 17:32
  • $\begingroup$ Why all the examples?? $\endgroup$
    – Squirtle
    Commented Apr 2, 2013 at 20:21
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    $\begingroup$ @dustanalysis: Because the statement that OP posted is not true, and these examples are the counterexamples of OP's statement. $\endgroup$ Commented Apr 2, 2013 at 20:34
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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 $$AB=4\pi^2 \pmatrix{-49&0\\ 0&-1},$$ $$BA=4\pi^2 \pmatrix{-1&0\\0&-49}.$$

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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.

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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.

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