Same-Matrix Product of Matrix Exponentials

Theorem
Let $\mathbf A$ be a square matrix.

Let $s, t \in \R$ be real numbers.

Let $e^{\mathbf A t}$ denote the matrix exponential of $\mathbf A$.

Then:
 * $e^{\mathbf A t} e^{\mathbf A s} = e^{\mathbf A \paren {t + s} }$

Proof
Let
 * $\map \Phi t = e^{\mathbf A t} e^{\mathbf A s} - e^{\mathbf A \paren {t + s} }$

for some fixed $s \in \R$.

From Derivative of Matrix Exponential:

Since $\map \Phi 0 = e^{\mathbf A s} - e^{\mathbf A s} = 0$, it follows that:
 * $\map \Phi t = e^{\mathbf A t} \map \Phi 0 = 0$

independent of $s$.

Hence the result.