# Matrices Commuting with Matrix Exponential

• I
Gold Member

## Summary:

Does ##\left[ A, e^B \right]=0## imply ##[A,B]=0##?

## Main Question or Discussion Point

The summary pretty much explains my question. I know that ##\left[ A, e^B \right]=0## if ##[A,B]=0## (and can prove it), but I can't figure out how to prove if it is or is not an "if and only if" statement.

• etotheipi

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fresh_42
Mentor
I would look for a counterexample. A nilpotent matrix ##B## of degree three or four perhaps. You need at least the square for a counterexample.

mathman
I haven't much background in matrix theory. However wouldn't ##[A,B]=0## imply ##[A,e^B]=A##? Expand ##e^B## in power series and the first term ends up as ##[A,I]##.

• PeroK
Gold Member
I haven't much background in matrix theory. However wouldn't ##[A,B]=0## imply ##[A,e^B]=A##? Expand ##e^B## in power series and the first term ends up as ##[A,I]##.
Right, but all matrices commute with the identity, so ##[A,I]=0##.

Gold Member
I would look for a counterexample. A nilpotent matrix ##B## of degree three or four perhaps. You need at least the square for a counterexample.
Ok, I'm working on that. So I am assuming that my statement is not "if and only if." What conditions must be present for ##\left[ A, e^B \right]=0## to imply ##[A,B]=0##? Also, while a general answer is nice, I'm really only interested in skew-Hermitian matrices, if that makes it any easier.

fresh_42
Mentor
Well, ##e^B## is unitary then, but I don't see how this helps. It only makes the search for a counterexample more complicated since you cannot just use upper triangular matrices for ##B##.

Where did you get the element ##[\mathfrak{g},G]## from?

fresh_42
Mentor
I would start with ##n=2## and ##B## diagonal with different eigenvalues. Something like ##e^B=\operatorname{diag}(e^{\pi i},e^{3\pi i})## and then look for an ##A## which does not commute with ##\operatorname{diag}(\pi i, 3\pi i).##

Gold Member
I would start with ##n=2## and ##B## diagonal with different eigenvalues. Something like ##e^B=\operatorname{diag}(e^{\pi i},e^{3\pi i})## and then look for an ##A## which does not commute with ##\operatorname{diag}(\pi i, 3\pi i).##
Everything that I can think of that does commute with ##e^B=\operatorname{diag}(e^{\pi i},e^{3\pi i})## also commutes with ##\operatorname{diag}(\pi i, 3\pi i).##

Well, eB is unitary then, but I don't see how this helps. It only makes the search for a counterexample more complicated since you cannot just use upper triangular matrices for B.
I was just wondering if there is a general rule about when ##\left[ A, e^B \right]=0## does imply that ##[A,B]=0## given that A and B are skew-Hermitian (or just in general for two matrices A and B).

StoneTemplePython
Gold Member
2019 Award
I was just wondering if there is a general rule about when ##\left[ A, e^B \right]=0## does imply that ##[A,B]=0## given that A and B are skew-Hermitian (or just in general for two matrices A and B).
skew hermitian is an interesting place for this -- such matrices are unitarily diagonalizable and have purely imaginary eigenvalues... so check the kernel of the complex scalar exponential map:
##\exp\big(2i\pi\cdot n\big) = 1##
where ##n## is any integer

The problem you have is for some skew ##S## with spectrum given by ##\{2i\pi\cdot n\}## you have
##e^S = I## and the Identity matrix commutes with everything, but ##S## need not commute with everything

you should be able to engineer an countexample from here -- I'd probably use a 3x3 matrices for it, where ##S## is diagonal with ##[0, 2i\pi ,4 i\pi ]## on the diagonal, and ##A## some skew hermitian matrix that has no zeros anywhere on it. ##SA## has all zeros on its first row but not its first column, and ##AS## has all zeros on first column but not first row, so they do not commute, yet
##\left[ A, e^S \right] = \left[ A, I \right]=0##

• PeroK, etotheipi and Isaac0427
Gold Member
skew hermitian is an interesting place for this -- such matrices are unitarily diagonalizable and have purely imaginary eigenvalues... so check the kernel of the complex scalar exponential map:
##\exp\big(2i\pi\cdot n\big) = 1##
where ##n## is any integer

The problem you have is for some skew ##S## with spectrum given by ##\{2i\pi\cdot n\}## you have
##e^S = I## and the Identity matrix commutes with everything, but ##S## need not commute with everything

you should be able to engineer an countexample from here -- I'd probably use a 3x3 matrices for it, where ##S## is diagonal with ##[0, 2i\pi ,4 i\pi ]## on the diagonal, and ##A## some skew hermitian matrix that has no zeros anywhere on it. ##SA## has all zeros on its first row but not its first column, and ##AS## has all zeros on first column but not first row, so they do not commute, yet
##\left[ A, e^S \right] = \left[ A, I \right]=0##
Ah, thank you. As an easy example,
$$\begin{pmatrix} 0 & 0\\ 0 & 2\pi i \end{pmatrix}$$
and
$$\begin{pmatrix} 1 & i\\ i & 2 \end{pmatrix}.$$

Still curious if anyone has an idea about what conditions in addition to ##[A,e^B]=0## are necessary to know that ##[A,B]=0##.

StoneTemplePython
Gold Member
2019 Award
Ah, thank you. As an easy example,
$$\begin{pmatrix} 0 & 0\\ 0 & 2\pi i \end{pmatrix}$$
and
$$\begin{pmatrix} 1 & i\\ i & 2 \end{pmatrix}.$$

Still curious if anyone has an idea about what conditions in addition to ##[A,e^B]=0## are necessary to know that ##[A,B]=0##.
note:

$$\begin{pmatrix} 1 & i\\ i & 2 \end{pmatrix}.$$
is not skew hermitian (check the diagonals)

StoneTemplePython
Gold Member
2019 Award
Still curious if anyone has an idea about what conditions in addition to ##[A,e^B]=0## are necessary to know that ##[A,B]=0##.
For this, again with skew hermitian matrices in mind, one approach is to restrict the eigenvalues some how so that the exponential function is injective. E.g. constraining so all eigenvalues to have modulus ##\lt \pi## should do it.

The result is that whatever diagonalizes ##e^B## diagonalizes ##B## and if ##e^B## and ##A## commute then they are both simultanouesly diagonalizable, by ##U##, which in turn simultaneously diagonalizes ##B## and ##A## which implies ##B## and ##A## commute.

Ah, thank you. As an easy example,
$$\begin{pmatrix} 0 & 0\\ 0 & 2\pi i \end{pmatrix}$$
and
$$\begin{pmatrix} 1 & i\\ i & 2 \end{pmatrix}.$$

Still curious if anyone has an idea about what conditions in addition to ##[A,e^B]=0## are necessary to know that ##[A,B]=0##.
I would look into matrix logairthims

PeroK
Homework Helper
Gold Member
Summary:: Does ##\left[ A, e^B \right]=0## imply ##[A,B]=0##?

The summary pretty much explains my question. I know that ##\left[ A, e^B \right]=0## if ##[A,B]=0## (and can prove it), but I can't figure out how to prove if it is or is not an "if and only if" statement.

Here's a counterexample using Pauli spin matrices. We know that:
$$\exp(-i2\pi \sigma_k) = I$$
commutes with everything, but the ##\sigma_k## do not.

• • Abhishek11235, etotheipi and DrClaude
mathman
Right, but all matrices commute with the identity, so ##[A,I]=0##.
##[A,I]=A## for any matrix.

fresh_42
Mentor
##[A,I]=A## for any matrix.
No. ##[X,Y]## is short for ##XY-YX## if ##X,Y## are algebra elements, and ##X^{-1}Y^{-1}XY## if ##X,Y## are group elements. So the result is either ##[A,I]=0## or ##[A,I]=I##.

mathman
No. ##[X,Y]## is short for ##XY-YX## if ##X,Y## are algebra elements, and ##X^{-1}Y^{-1}XY## if ##X,Y## are group elements. So the result is either ##[A,I]=0## or ##[A,I]=I##.
It looks like I misunderstood the symbolism. I thought it ([...]) meant product.

Gold Member
note:

$$\begin{pmatrix} 1 & i\\ i & 2 \end{pmatrix}.$$
is not skew hermitian (check the diagonals)
For some reason...I just forgot that the diagonals existed. Let’s change that second matrix to
$$\begin{pmatrix} i & 3\\ -3 & 2i \end{pmatrix}.$$ Whoops

If A commutes with exp(tB) for real t in some interval about 0, then I think A and B commute.

fresh_42
Mentor
If A commutes with exp(tB) for real t in some interval about 0, then I think A and B commute.
Only if you cut the quadratic terms onwards. Will say: no, but good enough for physicists.

• Isaac0427
PeroK
Homework Helper
Gold Member
If A commutes with exp(tB) for real t in some interval about 0, then I think A and B commute.
And what do you think about all the countereaxmples above?

Fair enough. (I still suspect [A exp(tB)] = 0 for t in (-ε, ε) implies [A,B] = 0 for *almost all B*, but I'm not sure I can identify the exact exceptions. )

Sorry, I missed where it was "proven otherwise".

And are you sure that etB = eteB ???

Last edited:
Gold Member
Only if you cut the quadratic terms onwards. Will say: no, but good enough for physicists.
Speaking of physics, while this was really a math question, this was the context that I was asking it in (just in case someone may have any thoughts about the specific application of my question):

In quantum mechanics, say you have a Hermitian operator Q that commutes with the Hamiltonian. Thus the observable quantity corresponding to Q is conserved, which implies a symmetry by Noether's theorem. At least for the cases of momentum and energy conservation, this symmetry is an invariance of the energy under some unitary transformation ##U=e^{iQ}## (i.e. translational symmetry for momentum conservation and time-translation symmetry for energy conservation). In these cases, both [Q,H]=0 and ##[e^{iQ},H]=0##. I was curious if the symmetry corresponding to the "conservation of Q" could in general be associated with some unitary transformation ##e^{iQ}## and vice versa, but it would seem like that is a no, given that it is possible for ##e^{iQ}## to commute with H without Q commuting with H.

I'm hoping this isn't nonsense. Also, I apologize if this should just be a new thread-- I can do that if I need to.

PeroK