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.