Why is $\det(e^X)=e^{\operatorname{tr}(X)}$? I've seen on Wikipedia that for a complex matrix $X$, $\det(e^X)=e^{\operatorname{tr}(X)}$.
It is clearly true for a diagonal matrix. What about other matrices ?
The series-based definition of exp is useless here.

A alternative to doing this by normal forms which perhaps assumes more but is much more natural to me is (as suggested in the comment on $\det(\exp X)=e^{\mathrm{Tr}\, X}$ for 2 dimensional matrices) to note that it clearly holds for diagonalizable matrices (see the duplicate How to prove $\det(e^A) = e^{\operatorname{tr}(A)}$? ...), and by


*

*the continuity of $\det, \mathrm{tr}$ and $\exp$

*the density of diagonalizable matrices in the space of all complex matrices (Diagonalizable matrices with complex values are dense in set of $n\times n$ complex matrices.)


we have the result more generally for all matrices.
