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} $?

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    $\begingroup$ 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. $\endgroup$ – Omnomnomnom 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. $\endgroup$ – Omnomnomnom Jul 10 '17 at 22:26
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    $\begingroup$ 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. $\endgroup$ – SZN 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.

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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

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