Determinant of the exponential of a matrix

[jdstokes]

Let $B \in \mathbb{R}^{n\times n}.$Show that $\det e^B = e^{tr B}$ where tr B is the trace of of B.

Clearly $e^{tr B}$ is the product of the diagonal entries of $e^{B}$.

By the Jordan canonical for theorem, $\exists P,J \in \mathbb{C}^{n \times n}$ where P is invertible and J is a diagonal sum of Jordan blocks, such that

$BP = PJ$
$P^{-1}BP = J$
$P^{-1}e^{B}P = e^{J}$
$e^{B} = Pe^{J}P^{-1}$
$\det e^{B} = \det P \det e^{J} \det P^{-1}$
$\det e^{B} = \det e^{J}$

This is where I get stuck. Since $e^{J}$ is upper triangular, its determinant is the product of its diagonal entries, but why should this be equal to the product of the diagonal entries of $e^{B}$?

Thanks

 

[StatusX]

Can you compute $\det e^J$ explicitly (in terms of the eigenvalues of B)? If so, then you can use the fact that the traces of similar matrices are equal.