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Oct 13, 2017 at 8:41 comment added Vincent @IlyaBogdanov Ah thank you! I missed the $\det(A) = \det(B) = 1$ condition. It all makes sense now.
Oct 13, 2017 at 7:56 comment added Ilya Bogdanov @Vincent: IAs for your second question --- no, I do not know how to assume that.
Oct 13, 2017 at 7:55 comment added Ilya Bogdanov @Vincent: We know that $\lambda_3=\overline{\lambda_2}$, so the relation holds for it. Next, $\lambda_1\lambda_2\lambda_3=1=\mu_1\mu_2\mu_3$, whence the relation for $\lambda_1$. (I could use the basic Galois theory technique, but I needed the fixed correspondence between the roots.)
Oct 13, 2017 at 7:31 comment added Vincent Perhaps a stupid question, but how do you go from $\lambda_2^k = \mu_2^l$ to $\lambda_1^k = \mu_1^l$ in the last line? Can we somehow assume wlog that $\lambda_1 = \mu_1 = 1?$
Sep 24, 2017 at 5:31 comment added Misha Thank you, Ilya! I was hoping for a positive answer, but alas...
Sep 24, 2017 at 5:29 vote accept Misha
Sep 23, 2017 at 19:13 history answered Ilya Bogdanov CC BY-SA 3.0