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Given a square matrix $A$, form the Lie series of it, which is defined by: $$ e^A = I + A + \frac{1}{2} A^2 + \frac{1}{3!} A^3 + \cdots + \frac{1}{n!} A^n = \sum_{k=0}^\infty \frac{1}{k!} A^k $$ Is there a simple expression for the determinant of $\,e^A$ ?

This question is motivated by:

Where I've proved for the 2-D case: $$ S = e^{\large t \begin{bmatrix} 0 & p \\ p & 0 \end{bmatrix}} = \begin{bmatrix} \cosh(p t) & \sinh(p t) \\ \sinh(p t) & \cosh(p t) \end{bmatrix} \quad \Longrightarrow \quad \det(S) = 1 $$ I'd like to know if the determinant equals one for the 3-D case as well, where $$\,S = \exp\left(t \begin{bmatrix} 0 & p & q \\ p & 0 & r \\ q & r & 0 \end{bmatrix}\right)$$
Han de Bruijn
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1 Answers1

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It turns out $\det(e^{A}) = e^{\text{tr}(A)}$. I leave it as an exercise to prove.

Indeed, your last matrix also has determinant one.