I Matrices Commuting with Matrix Exponential

[Isaac0427]

TL;DR 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.

Thanks in advance!

 

[fresh_42]

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 $[A,I]$.

 

[Isaac0427]

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 $[A,I]=0$.

 

[Isaac0427]

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]

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]

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 $\operatorname{diag}(\pi i, 3\pi i).$

 

[Isaac0427]

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]

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
$\left[ A, e^S \right] = \left[ A, I \right]=0$

 

[Isaac0427]

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]

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]

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.

 

[HomogenousCow]

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]

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.

Thanks in advance!

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.

 

[mathman]

Right, but all matrices commute with the identity, so $[A,I]=0$.

$[A,I]=A$ for any matrix.

 

[fresh_42]

$[A,I]=A$ 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.

 

[Isaac0427]

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

 

[zinq]

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

 

[fresh_42]

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.

 

[PeroK]

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?

 

[zinq]

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. )

 

[zinq]

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

And are you sure that e^tB = e^te^B ?

 

[Isaac0427]

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]

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.

There's a relationship between Hermitian and Unitary operators, in that $U$ is unitary iff $U = e^{iH}$ for some Hermitian operator $H$.

 

[Isaac0427]

There's a relationship between Hermitian and Unitary operators, in that $U$ is unitary iff $U = e^{iH}$ for some Hermitian operator $H$.

Right, I have that part. I've been trying to figure out if this correspondence between Hermitian operators and unitary operators has anything to do with the correspondence between conserved quantities and symmetries (as, if I am not mistaken, a symmetry can be described by a Hamiltonian's invariance under the transformation $U^\dagger H U$).

 

[StoneTemplePython]

Fair enough. (I still suspect [A exp(tB)] = 0 for t in (-ε, ε) implies [A,B] = 0

For skew hermitian matrices this does hold-- as long as you may select small enough $\epsilon$ after knowing $B$. I said as much in post #12.

 

[fresh_42]

For skew hermitian matrices this does hold-- as long as you may select small enough $\epsilon$ after knowing $B$. I said as much in post #12.

It does not hold. Only up to $O(\varepsilon^2)B^2$. But mathematically it's wrong.

 

[StoneTemplePython]

It does not hold. Only up to O(ε2)B2. But mathematically it's wrong.

once again for skew hermitian matrices $A$, $B$ only

outline of proof:
1.) exponential function is injective when the domain is constrained to purely imaginary $z\in (−i\pi,i\pi)$. If desired, we can shrink this to, say $z\in (\frac{−i\pi}{2},\frac{i\pi}{2})$.
2.) B is unitarily diagonalizable with purely imaginary eigenvalues, and for real $t$ sufficiently close to zero, i.e. all $t\in (-\epsilon, \epsilon)$, then $tB$ has eigenvalues solely in the above interval. Call this $B':=tB$.
3.) $[A,e^{B^′}]=0\longrightarrow A$ and $e^{B'}$ may be simultaneously diagonalized
4.) $e^{B^′}$ has the same eigenvectors/ eigenspace as $B^{′}$, i.e.
$\ker\big(e^{B^{'}} - \lambda I \big) =\ker\big(B' - \log(\lambda) I \big)$ . This is the standard principal argument and is unique because of (1).
5.) $A$ and $B^′$ may be simultaneously diagonalized by $U$
6.) $[A,B^{'}]=0$
- - - -
Looking at this in terms of a power series really misses the mark as we have powerful diagonalization theorems that we may lean on.

PS:
after writing this up and editing (it is late at night so gaps can creep into my posts) the PF server bizarely overwrote all my LaTeX into some mixture of blank spaces and plain text; I think it has something to do with the 'preview' function... I tried to rewrite everything and fix this, but who knows.

 

[PeroK]

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

And are you sure that e^tB = e^te^B ?

No, sorry, I was tired and what I posted was nonsense. Apologies.

 

[fresh_42]

Thanks.

Looking at this in terms of a power series really misses the mark as we have powerful diagonalization theorems that we may lean on.

And if you want to kill the last "by-foot" step, you could substitute 4) by the adjoint representations and that exp is a local diffeo around 0.