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Disprove that AB-BA = I

  1. Jan 6, 2012 #1
    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.
  2. jcsd
  3. Jan 6, 2012 #2


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    What do you know about determinants?
  4. Jan 6, 2012 #3
    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.
  5. Jan 6, 2012 #4


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    Staff: Mentor

    Determinant is just a number, isn't it?
  6. Jan 6, 2012 #5
    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 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)[STRIKE], at least when M = AB..[/STRIKE] (now that I think about it, there is probably no matrix that can't be written as a product of two others, is there?)
  7. Jan 6, 2012 #6


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    Try taking the trace.
  8. Jan 6, 2012 #7
    So tr(AB-BA) = 0 ? Great. Thanks.
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