Question about a step of showing $\text{det}~e^B= e^{\text{tr}(B)}$?

Question

I am checking the proof of $$ \text{det}~e^B= e^{\text{tr}(B)}.\tag{1} $$ Using the Jacobi's formula and adjugate matrix, I now $$d~\text{det}~A=\text{det}(A)\text{tr}(A^{-1}d~A)\tag{2}$$ holds. But here we see that, letting $A=e^B$, we have $$d~\text{det}~e^B=\text{tr}(B)~\text{det}~e^B,\tag{3}$$ which means that $$\text{tr}((e^B)^{-1}de^B)=\text{tr}~dB.\tag{4}$$

Why is that true?

Note that you consider $e^{tB}$ and differentiate w.r.t. $t$ — Kal S., Nov 10 '17 at 16:17
See carefully the proof in the provided link from wikipedia. You differentiate $e^{tB}$ w.r.t. $t$. It follows by a direct calculation! — Kal S., Nov 10 '17 at 16:38
@Test123 for the special case, there is no need to introduce t, right? — Roger, Nov 10 '17 at 16:59

score 1 · Answer 1 · answered Nov 10 '17 at 21:11

Note that $$ e^{-B}\mathrm{d}e^{B} ~=~ \int_0^1\!ds~e^{-sB}(\mathrm{d}B)e^{sB} \tag{A},$$ cf. e.g. my Phys.SE answer here. Therefore by cyclicity of the trace, we derive OP's sought-for formula: $$ {\rm Tr}(e^{-B}\mathrm{d}e^{B}) ~\stackrel{(1)}{=}~ \int_0^1\!ds~{\rm Tr}(e^{-sB}(\mathrm{d}B)e^{sB}) ~=~ \int_0^1\!ds~{\rm Tr}(\mathrm{d}B)~=~ {\rm Tr}(\mathrm{d}B)\tag{B}.$$

score 0 · Answer 2 · answered Nov 10 '17 at 17:59

If $f$ is an entire function, by invoking the Jordan normal form we have $\text{Spec}(f(M)) = f\left(\text{Spec}(M)\right)$. On the other hand $\det M=\prod_{\lambda \in\text{Spec} M}\lambda.$
It follows that $$ \det \exp M = \!\!\!\!\prod_{\lambda\in\text{Spec}\left(\exp M\right)}\!\!\!\!\!\!\lambda =\prod_{\lambda\in\text{Spec }M}e^{\lambda}=\exp\!\!\!\sum_{\lambda\in\text{Spec } M}\!\!\!\lambda=\exp\operatorname{Tr} M.$$ For brevity, $\det\exp=\exp\operatorname{Tr}$. Here $\text{Spec}$ is intended as a multiset, in which every eigenvalue occurs a number of times equal to its multiplicity.

Question about a step of showing $\text{det}~e^B= e^{\text{tr}(B)}$?

2 Answers2