Exponential of a sum of commuting matrices

Statement

Let $\mathbb{K}$ be $\R$ or $\C$, and let $A$ and $B$ be $n \times n$ matrices over $\mathbb{K}$ with $AB = BA$. Then $$e^{A + B} = e^A e^B,$$ where $e^A$ is the matrix exponential.

Remarks

The proof multiplies the two series and uses the binomial formula, which needs $AB = BA$. It gives $e^A e^{-A} = I_n$, so $e^A$ is always invertible. Without $AB = BA$, the identity can fail, for example for $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$.

Explanations

L1 human

For numbers, $e^{a + b} = e^a e^b$. For matrices, this rule holds only when the two matrices commute. In particular, $e^{sA} e^{tA} = e^{(s + t)A}$ for all scalars $s$ and $t$.