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Inverse of a linear operator

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that if an operator A satisfies A2 - A + I = 0 then A has an inverse. Express A-1 as a simple polynomial of A.


    2. Relevant equations

    I'm not sure that this is relevant, but A-1=1/(detA)TrC where TrC is the transpose of the matrix of cofactors. Also:
    If detA = 0 then the matrix has no inverse

    3. The attempt at a solution
    So I notice immediately that adding by the identity matrix in this equation will result in a matrix with its diagonal having numbers (real or complex) and the rest being zero, as I can be expressed as the kronecker delta. And if the determinant must be nonzero in order to have an inverse, there has to be a way to relate the diagonal of an n dimensional matrix with its determinant. I'm just stuck as to how to do that. Any help greatly appreciated, thank you.
    *Edit:
    I've been thinking more about this problem, and it seems like there should be a way to use the secular equation to solve it. We went over it briefly in class (the class is quantum and I haven't had linear algebra yet, so it's kind of a chore), but not in enough detail that I would be able to use it in a proof.
     
    Last edited: Feb 2, 2009
  2. jcsd
  3. Feb 2, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi seek! Welcome to PF! :smile:
    oooh … so complicated:cry:

    Try writing it A2 - A = -I :wink:
     
  4. Feb 3, 2009 #3
    My oversight is to my pride as a cold slap to the visage. Thanks so much for the help, maybe next time I'll be able to use the skills I learned in 5th grade.
     
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