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Jan 28, 2012 at 9:43 comment added Qmechanic ...where the answer(v1) has implicitly used at one point that $A$ is antisymmetric, either in the definition of the Pfaffian, or in the very last equality...
Jan 27, 2012 at 22:11 vote accept Lagerbaer
Jan 27, 2012 at 22:01 comment added Lagerbaer Ah, I think I already did that, I just had no clue what a Pfaffian was, and got stuck at proving Pfaff(A)^2 = det(A).
Jan 27, 2012 at 21:57 history answered user1504 CC BY-SA 3.0