It is a well known fact that if $A,B\in M_{n\times n}(\mathbb C)$ and $AB=BA$, then $e^Ae^B=e^Be^A.$

The converse does not hold. Horn and Johnson give the following example in their Topics in Matrix Analysis (page 435). Let $$A=\begin{pmatrix}0&0\\0&2\pi i\end{pmatrix},\qquad B=\begin{pmatrix}0&1\\0&2\pi i\end{pmatrix}.$$ Then $$AB=\begin{pmatrix}0&0\\0&-4\pi^2\end{pmatrix}\neq\begin{pmatrix}0&2\pi i\\0&-4\pi^2\end{pmatrix}=BA.$$ We have $$e^A=\sum_{k=0}^{\infty}\frac 1{k!}\begin{pmatrix}0&0\\0&2\pi i\end{pmatrix}^k=\sum_{k=0}^{\infty}\frac 1{k!}\begin{pmatrix}0^k&0\\0&(2\pi i)^k\end{pmatrix}=\begin{pmatrix}e^0&0\\0&e^{2\pi i}\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$

For $S=\begin{pmatrix}1&-\frac i{2\pi}\\0&1 \end{pmatrix},$ we have $$e^B=e^{SAS^{-1}}=Se^AS^{-1}=S\begin{pmatrix}1&0\\0&1\end{pmatrix}S^{-1}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$

Therefore, $A,B$ are such non-commuting matrices that $e^Ae^B=\begin{pmatrix}1&0\\0&1\end{pmatrix}=e^Be^A.$

It is clear that $\pi$ is important in this particular example. In fact, the authors say what follows.

It is known that if all entries of $A,B\in M_n$ are algebraic numbers and $n\geq 2,$ then $e^A\cdot e^B=e^B\cdot e^A$ if and only if $AB=BA.$

No proof is given. How does one go about proving that?

  • $\begingroup$ $\Leftarrow$ is obvious, because for commuting $A,B$ we have $e^Ae^B=e^{A+B}=e^{B+A}=e^Be^A$. $\endgroup$ – tomasz Jun 30 '12 at 22:11
  • $\begingroup$ I wonder what the $n\ge 2$ assumption is for. The claim is trivially true for $n=1$ (and likewise for $n=0$, to the extent one considers it to be meaningful at all). $\endgroup$ – hmakholm left over Monica Jun 30 '12 at 22:27
  • $\begingroup$ @HenningMakholm I wondered that too. I guess they just didn't want that trivial case in the theorem. $\endgroup$ – user23211 Jun 30 '12 at 22:28
  • $\begingroup$ The paper (pdf linked) "A remark on commuting operator exponentials" by Wermuth contains generalizations to bounded operators on a Banach space, and references. It includes a reference to a paper where this result is proved, "Two remarks on matrix exponentials" by the same author, but I do not have access to this paper. Some of the other references in the first article I linked to might also have proofs of this result. $\endgroup$ – Jonas Meyer Jul 1 '12 at 10:41
  • $\begingroup$ I don't think I have access to that either, at least if I do, I don't know how to make it work. I'll try in a library, but not today. $\endgroup$ – user23211 Jul 1 '12 at 10:57

The proof comes from Wermuth's "Two remarks on matrix exponential" (DOI 10.1016/0024-3795(89)90554-5)

It seems to me that it is enough to assume that no two distinct eigenvalues of $A$ or $B$ differ by an integer multiple of $2\pi i$ (which is true if, but not only if $\pi$ is transcendental with respect to their entries). This is of course true if $A$ and $B$ have algebraic entries.

The basic idea is to somehow reverse the exponential function, and express $A$ and $B$ as power series (in fact, polynomials) in their exponentials.

Let $m(\lambda)=\prod_j (\lambda-\lambda_j)^{\mu_j}$ be the minimal polynomial of $A$. Then by assumption $e^{\lambda_j}$ are all different, so by Hermite's interpolation theorem we can find a polynomial $f$ such that for $g=f\circ \exp$ we have that $g(\lambda_j)=\lambda_j$, and if $\mu_j>0$, then $g'(\lambda_j)=1$ and $g^{(l)}(\lambda_j)=0$ for $2\leq l< \mu_j$.

Then we have $g(A)=A$ (this part I' don't really see, but apparently it's common knowledge, it's probably a corollary of Jordan's theorem on normal form), but $g(A)=f(e^A)$, so $A=f(e^A)$, and similarly for some polynomial $h$ we have $B=h(e^B)$, so $AB=f(e^A)h(e^B)=h(e^B)f(e^A)=BA$.


A broad-brush approach is as follows. We want to prove that if $e^A$ and $A$ have the same number of eigenvalues, then $e^A$ is a polynomial in $A$. This is true for diagonal matrices and hence for diagonalizable matrices. Since the diagonalizable matrices are dense, it is true for all matrices.

We can avoid the density argument though. Any square matrix $A$ can be written in exactly one way as $A=S+N$ where $S$ and $N$ are polynomials in $A$ and $S$ is diagonalizable and $N$ is nilpotent. Using the series expansion for $e^A$ we find that $$ e^A = e^D(I+N+\cdots+N^{k-1}) = e^D + M $$ with $M=e^D(N+\cdots+N^{k-1})$. Note that $M^k=0$. As $e^D$ and $M$ are polynomials in $A$ and $N$, they are polynomials in $A$; they are also the diagonalizable and nilpotent parts in the decomposition of $e^A$.

Now suppose $e^A$ and $e^B$ commute. Using the above we have $$ e^A = e^{D_A}+M_A,\quad e^B = e^{D_B}+M_B. $$ Then any two of the six matrices here commute (because the first three matrices are polynomials in $e^A$ and the second three polynomials in $e^B$) and therefore $e^{D_A}$ and $e^{D_B}$ commute. Since $e^A=e^{D_A}(I-N_A)^{-1}$ etc we see that that $(I-N_A)^{-1}$ and $(I-N_B)^{-1}$ commute, which implies that $N_A$ and $N_B$ commute. So if $D_A$ and $D_B$ are polynomials in $e^{D_A}$ and $e^{D_B}$ respectively, then $AB=BA$. As $D_A$ and $D_B$ are diagonalizable, we are OK provided the number of distinct eigenvalue of $e^{D_A}$ is equal to the number of distinct eigenvalues of $D_A$ (and ditto for $e^{D_B}$ and $D_B$).


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