SOLUTION: Hello: I am not sure how to answer this question. If A and B are nxn matrices, then show that there are no square matrices A and B such that AB-BA=I (where I is the Identity

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Question 25058: Hello:
I am not sure how to answer this question.
If A and B are nxn matrices, then show that there are no square matrices A and B such that AB-BA=I
(where I is the Identity).
I know that if AB = I and BA = I then AB-BA cannot equal I. Sorry I am confused.
Thanks
Answer by kev82(151) About Me  (Show Source):
You can put this solution on YOUR website!
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 then , 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 is

So the trace of is

Doing a similar calculation for we get

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.
Kev