Same-Matrix Product of Matrix Exponentials

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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:

\(\ds \map {\Phi'} t\) \(=\) \(\ds \mathbf A e^{\mathbf A t} e^{\mathbf A s} - \mathbf A e^{\mathbf A \paren {t + s} }\)
\(\ds \) \(=\) \(\ds \mathbf A \paren {e^{\mathbf A t} e^{\mathbf A s} - e^{\mathbf A \paren {t + s} } }\)
\(\ds \) \(=\) \(\ds \mathbf A \map \Phi t\)


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.

$\blacksquare$