Let [itex]B \in \mathbb{R}^{n\times n}. [/itex]Show that [itex]\det e^B = e^{tr B}[/itex] where tr B is the trace of of B.(adsbygoogle = window.adsbygoogle || []).push({});

Clearly [itex]e^{tr B}[/itex] is the product of the diagonal entries of [itex]e^{B}[/itex].

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

[itex]BP = PJ[/itex]

[itex]P^{-1}BP = J[/itex]

[itex]P^{-1}e^{B}P = e^{J}[/itex]

[itex]e^{B} = Pe^{J}P^{-1}[/itex]

[itex]\det e^{B} = \det P \det e^{J} \det P^{-1}[/itex]

[itex]\det e^{B} = \det e^{J}[/itex]

This is where I get stuck. Since [itex]e^{J}[/itex] 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 [itex]e^{B}[/itex]?

Thanks

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# Homework Help: Determinant of the exponential of a matrix

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