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Sep 2, 2012 at 21:37 history edited AlexArvanitakis CC BY-SA 3.0
Added the material about multilinearity
Aug 31, 2012 at 2:35 comment added AlexArvanitakis You know, I never thought about explicitly proving that. I think I can show it formally, in the sense that if one writes the operator $S$ (as in the wikipedia page) as a sum of terms $|v_i\rangle\langle i|$ where the $\langle i|$ form a basis, then the functional determinant is linear wrt every $|v_i\rangle$ --that is, the columns. I'm going to sleep on it.
Aug 30, 2012 at 19:41 comment added Matthieu Romagny One point I'm wondering about is whether these determinants are indeed multilinear in any sense.
Aug 30, 2012 at 16:44 history answered AlexArvanitakis CC BY-SA 3.0