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JasonRox

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http://www.math.ucdavis.edu/~daddel...plications/Determinant/Determinant/node3.html

It doesn't seem like the definition assume commutativity. So, I guess it would work.

I'm totally not sure about this. I'm just throwing this thought in the air.

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Hurkyl

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Hurkyl

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Your algebra A is a finite dimensional algebra over the reals -- I think that implies that it can be represented as an algebra of real matrices. So you can view your matrix over A as a matrix over R, and take the (real-valued!) determinant of that.

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However, that is not that I am working with. I am playing with Clifford algebra and it is known that every element of Cl(p+1, q+1) can be represented by M(2, Cl(p,q)) and M(n, T) where T = R, C or H. Of course the definition of determinant is trivial when T = R or C, but I am not sure if the Laplace expansion still works on quaternionic matrices. If I can compute the determinant in that case, it will suggest the norm function for Cl(p+1, q+1).

Your algebra A is a finite dimensional algebra over the reals -- I think that implies that it can be represented as an algebra of real matrices. So you can view your matrix over A as a matrix over R, and take the (real-valued!) determinant of that.

Thank you for your suggestion by the way. I searched with the keyword pseudodeterminant while the suggested keyword 'noncommutative determinant' is more obvious.

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