Determinant of matrix exponential? 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))

 A: 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 $t=1$, $\det(e^{A})= e^{\text{tr}(A)}$.
A: You can reduce problem for diagonal matrix using the fact that every matrix can be approximated with any given precision and both functions: $e^X$ and $\det X$ are continuous. The problem with diagonal matrix is obvious.
P.S. It seems that lyj was faster than me.
A: If you're afraid of the density of diagonalizable matrices, simply triangularize $A$. You get $$A=P^{-1}UP,$$ with $U$ upper triangular and the eigenvalues $\{\lambda_j\}$ of $A$ on the diagonal.
Then
$$
\mbox{det}\;e^A=\mbox{det}(P^{-1}e^UP)=\mbox{det}\;e^U.
$$
Now observe that $e^U$ is upper triangular with $\{e^{\lambda_j}\}$ on the diagonal.
So
$$
\mbox{det} \;e^A=\mbox{det} \;e^U=e^{\lambda_1}\cdots e^{\lambda_n}=e^{\lambda_1+\ldots+\lambda_n}.
$$
A: You  just use the Jordan normal form, $A$ is your matrix, $D$ is the Jordan normal form of $A$ and $S$ is the transformation matrix, for the second equal just remember the definition of matrix exponential, and that 
$$(SAS^{-1})^n = S A  S^{-1}S A S^{-1} S A \dots S A S^{-1}=S A^n S^{-1}$$ 
\begin{align*}
\det(\exp(A))&=\det(\exp(S D S^{-1}))\\
&=\det(S \exp(D) S^{-1})\\
&=\det(S) \det(\exp(D)) \det (S^{-1})\\
&=\det(\exp (D))\\
&=\prod_{i=1}^n e^{d_{ii}}\\
&=e^{\sum_{i=1}^n{d_{ii}}}\\
&=e^{\text{tr}D}
\end{align*}
As the trace is invariant this works.
