disprove that AB-BA = I

[hamsterman]

The task is to prove that for no two matrices A and B, A*B - B*A = I, where I is the identity matrix.
I tried multiplying by the inverses of A or B, but that doesn't seem to lead to a more manageable form. The only way I see this could be done is by writing down all n*n (assuming n by n matrices) linear equations. It's easy to do when n = 2, but the same contradiction may not be as obvious for higher n.
I hope there is a more intelligent way to go about this.

[phyzguy]

What do you know about determinants?

[hamsterman]

I know that det(AB) = det(BA), but I don't know what are the properties when subtraction is involved. Except for the case when only one line is different.

[Borek]

I know that det(AB) = det(BA), but I don't know what are the properties when subtraction is involved.

Determinant is just a number, isn't it?

[hamsterman]

What I mean is that I don't know what is det(AB-BA) even if I do know det(AB) and det(BA).
I'm looking at Sylvester's determinant theorem (http://en.wikipedia.org/wiki/Determinant#Sylvester.27s_determinant_theorem) which looks related, but I still don't see a solution. Now I need to prove that for no M, det(M+I) = det(M), at least when M = AB.. (now that I think about it, there is probably no matrix that can't be written as a product of two others, is there?)

[Dick]

What I mean is that I don't know what is det(AB-BA) even if I do know det(AB) and det(BA).
I'm looking at Sylvester's determinant theorem (http://en.wikipedia.org/wiki/Determinant#Sylvester.27s_determinant_theorem) which looks related, but I still don't see a solution. Now I need to prove that for no M, det(M+I) = det(M), at least when M = AB.. (now that I think about it, there is probably no matrix that can't be written as a product of two others, is there?)

Try taking the trace.

[hamsterman]

So tr(AB-BA) = 0 ? Great. Thanks.