Question 25058
Hi,

This is quite easy to prove by considering the trace of both sides. The Trace of a square matrix is simply the sum of it's diagonal elements. Hopefully you can see that if *[tex P=Q] then *[tex Tr(P)=Tr(Q)], and that *[tex Tr(P+Q)=Tr(P)+Tr(Q)]. If you can't see where these identities come from then please email back.

Ok, The matrix *[tex AB] is given by the formula

*[tex AB_{ij}=\sum_k^n A_{ik}B_{kj}]

and the trace of a matrix *[tex P] is 

*[tex \sum_t^n P_{tt}]

So the trace of *[tex AB] is

*[tex \sum_t^n \sum_k^n A_{tk}B_{kt}]

Doing a similar calculation for *[tex Tr(BA)] we get

*[tex \sum_t^n \sum_k^n B_{tk}A{kt}]

Now you can either acccept that these are the same expression ie *[tex Tr(AB)=Tr(BA)] or you can read my discussion in green of why we're assuming it is.

<font color="green">Now you've not told me what the elements of *[tex A] and *[tex B] are, if they were matrices(making A a matrix of matrices) then we would be in trouble, because we wouldn't have commutivity, but I'm gonna assume that we are working with some nice subfield of *[tex \mathbb{C}] hence *[tex A_{ab}B_{cd}=B_{cd}A_ab}]. I'm also assuming our matrices are finite which allows us to change the order of summation(else we need to worry about absolute convergence on compact subsets - what fun!) So now you are happy that *[tex Tr(AB)=Tr(BA)]</font>

Well this is it, we're done. The trace of the LHS must be zero, and the trace of the RHS (the trace of the identity) definitly isn't zero, it's *[tex n]. So unless *[tex n=0] (doesn't make sense) there are no matrices that can satisfy this.

Hope that helps.

Kev