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

Question

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} $?

Answer 1

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.

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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.