In $ M_n(\mathbb{C}) $ , if two matrices commute, then the exponential of their sum is the product of their exponentials. This property invited me to reflect on the matrices $ A $ for which $ \exp(A+B) = \exp(A)\exp(B) $ is true for all complex matrix $ B $. I would like to show that such matrices are scalar (proportional to $ I_n $ ). What if $ \mathbb{C} $ is replaced by $ \mathbb{R} $?

  • 2
    $\begingroup$ I suspect that with a judicious use of the BCH-formula and by taking $B \to 0$, we can deduce that $A$ must be a scalar matrix as you say. $\endgroup$ – Ben Grossmann Jul 10 '17 at 22:24
  • $\begingroup$ Note: I've added the Lie groups/algebras tags because I suspect that a tool from those areas (such as the BCH formula) will make quick work of this problem. $\endgroup$ – Ben Grossmann Jul 10 '17 at 22:26
  • 2
    $\begingroup$ Isn't the formula valid iff $A$ and $B$ commute? It should follow from the series. In that case, the answer is the center of $M_n(k)$, which are scalar matrices. $\endgroup$ – Ennar Jul 10 '17 at 23:24
  • $\begingroup$ Agree with @Ennar --no need to use BCH. $\endgroup$ – JMJ Jul 11 '17 at 0:26

Let $R$ denote the ring $\text{Mat}_{n\times n}(\mathbb{C})$ with additive identity $0_R$. From the required property of $A$, we have $$\exp(-B)=\exp\big(A+(-A-B)\big)=\exp(A)\,\exp(-A-B)$$ for every $B\in R$. That is, $$\exp(-A-B)=\exp(-A)\,\exp(-B)\,,$$ so $$1_G=\exp(A+B)\,\exp(-A-B)=\exp(A)\,\exp(B)\,\exp(-A)\,\exp(-B)$$ for all $B\in R$. Here, $G$ is the group $\text{GL}_n(\mathbb{C})$ with identity $1_G$. That is, $\exp(A)$ is in the center of $G$, which means $\exp(A)$ is a scalar matrix. Consequently, $A$ is a scalar matrix.

Alternatively, from $\exp(A)\,\exp(tB)\,\exp(-A)\,\exp(-tB)=1_G$ for all $B\in R$ and $t\in\mathbb{R}$, we have $$0_R=\left.\frac{\text{d}}{\text{d}t}\right|_{t=0}\,\exp(A)\,\exp(tB)\,\exp(-A)\,\exp(-tB)=\exp(A)\,B\,\exp(-A)-B\,,$$ so that $\exp(A)$ is in the center of $R$. Again, this immediately implies that $\exp(A)$ is a scalar matrix, so $A$ is a scalar matrix.

EDIT: Due to comments below, the statement that, if $\exp(A)$ is a scalar matrix, then $A$ is a scalar matrix is false. However, we can deduce that, in some basis, $A$ is of the form $kI+J$ where $k$ is a complex constant, $I\in R$ is the identity matrix (well, $I=1_G$), and $J \in R$ is a diagonal matrix with diagonal entries of the form $2\pi r \text{i}$, where $r\in\mathbb{Z}$. Without loss of generality, suppose that $k=0$.

If there are two diagonal terms of $J$ that are not equal, then we can assume that $A=J$ is $2$-by-$2$ and takes the form $$A=\begin{bmatrix}2\pi p\text{i} &0\\0&2\pi q\text{i}\end{bmatrix}\,,$$ where $p$ and $q$ are distinct integers. In this basis, take $B$ to be the nilpotent matrix $$B=\begin{bmatrix}0&1\\0&0\end{bmatrix}\,,$$ so that $$\exp(A+B)=I\text{ but }\exp(A)\,\exp(B)=I+B\neq I\,.$$ This is a contradiction, so all diagonal entries of $J$ are equal, and the claim follows.

| cite | improve this answer | |
  • 4
    $\begingroup$ why $ \exp(A) $ is scalar implies $ A $ is scalar ? What do you think of $ A = $ Diag( $ 0, 2 i \pi $ ) ? I think it only implies that $ A $ is diagonalizable with a particular spectrum. Using that, I think you only have to solve the problem when $ n = 2 $ and $ A $ is a diagonal matrix. $\endgroup$ – MrMaths Jul 11 '17 at 4:25
  • 1
    $\begingroup$ @MrMaths In that case you only need to find a matrix $B$ such that $e^{A+B}\ne e^Ae^B$, such as $B=\pmatrix{0&-2\pi\\ 2\pi&0}$. $\endgroup$ – user1551 Jul 11 '17 at 10:32
  • $\begingroup$ @Batominovski Is the result still true on $ \mathbb{R} $? $\endgroup$ – MrMaths Jul 11 '17 at 11:04
  • 1
    $\begingroup$ It should be true over $\mathbb{R}$ because the matrix $A$ can be "diagonalized" into $2$-by-$2$ blocks and $1$-by-$1$ blocks over $\mathbb{R}$. You only need to verify that there are no $2$-by-$2$ blocks, using a similar argument. $\endgroup$ – Batominovski Jul 11 '17 at 11:27

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.