TITLE: Determinant of matrix exponential?
QUESTION [13 upvotes]: Suppose $A$ is a $n \times n$ constant matrix. How can I prove
$\det(e^A) = e^{\displaystyle \sum_{\lambda_i\in\sigma(A)} \lambda_i}$,
where $\sigma(A)$ is the multiset of eigenvalues of $A$?
The following matlab code shows this is true in $n = 3$ case:
A = rand(3)
detA = exp(sum(eig(A)))
detmA = det(expm(A))

REPLY [9 votes]: For an analytic method, using differential equations:
Let $f(t)= \det(e^{tA})$. Then $f'(t)=D \det(e^{tA}) \cdot Ae^{tA}=\text{tr} \left(^t \text{com}(e^{tA})Ae^{tA} \right)$. But $A$ and $e^{tA}$ commute, and $^t\text{com}(e^{tA})e^{tA}=\det(e^{tA}) \operatorname{I}_n$. Therefore, $f'(t)=\text{tr}(A)f(t)$ and $f(0)=1$, hence $f(t)=e^{\text{tr}(A)t}$. For $t=1$, $\det(e^{A})= e^{\text{tr}(A)}$.