You can put this solution on YOUR website!
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
, and that
. If you can't see where these identities come from then please email back.
Ok, The matrix
is given by the formula
and the trace of a matrix
So the trace of
Doing a similar calculation for
Now you can either acccept that these are the same expression ie
or you can read my discussion in green of why we're assuming it is.
Now you've not told me what the elements of and 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 hence . 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
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
. So unless
(doesn't make sense) there are no matrices that can satisfy this.
Hope that helps.