What are the matrices $ A $ such as $ \exp(A+B) = \exp(A)\exp(B) $ for all $ B $ 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} $? 
 A: 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.
