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.
$\blacksquare$