Showing determinant of product is product of dets for linear operators

Hakkinen
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Homework Statement


Assume A and B are normal linear operators [A,A^{t}]=0 (where A^t is the adjoint)

show that det AB = detAdetB


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The Attempt at a Solution


Well I know that since the operators commute with their adjoint the eigenbases form orthonormal sets. I'm not really sure how to proceed though so any assistance would be appreciated.
 
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I think I've got a solution for this but I'm not sure if it's general enough.

So since each operators eigenbasis forms an orthonormal set you can choose any convenient orthonormal basis. If I pick the standard cartesian basis (i,j,k) then each operator can be represented by the identity matrix. From here it is straight forward to show the identity.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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