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Linear algebra:Prove that the only invertible nxn idempotent matrix is I

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that the only invertible nxn idempotent matrix is the identity matrix.

    2. Relevant equations
    idempotent: given an nxn matrix, A^2=A
    invertible: There exists a matrix B such that AB=BA=I
    I, or identity matrix: for all matrices A, AI=IA=A

    3. The attempt at a solution
    So far I haven't gotten very far:
    Suppose A is an nxn matrix.
    For A to be invertible there exists a matrix B such that AB=BA=I.
    For A to be idempotent A^2=A.
    Otherwise, I've kind of mulled it over and gotten this far:
    To be
    invertible it has to have a determinant != 0. So, I think that I have
    to show that the determinant of everything else that is indempotent is
    0, which would mean widely defining the criteria for a matrix to be
    indempotent and showing that the two are mutually exclusive. The
    problem is I'm not quite sure about the properties of indempotent
    matrices and, since we don't have a formula for the determinant of a
    given nxn matrix, I'm not quite sure what to do.
    So far, I have not covered and am not allowed to use rank or eigenvalues in my class.
    Thanks!
     
  2. jcsd
  3. Feb 21, 2010 #2

    vela

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    You're overthinking the problem. You have one equation, [itex]A^2=A[/itex], and you know [itex]A^{-1}[/itex] exists. Think of how to solve for A.
     
  4. Feb 21, 2010 #3
    I know that A^2=A
    If I multiply both sides by A^(-1) I get: A=I
    Is that it? Does that show that it is universally true or that that is just one possibility?
     
  5. Feb 21, 2010 #4

    vela

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    Yeah, that's it. It holds for all A that are idempotent and invertible. You're not doing anything weird or wrong that opens up the possibility of additional solutions.

    You're probably worried about something along the lines of solving x^2=x, right? If you divide both sides by x, you get x=1, and you lose the solution x=0. The reason you lose x=0 is because it doesn't have a multiplicative inverse, so you can't justify the step of dividing by x when x=0. If you were, however, told that x has a multiplicative inverse, you'd know that x=1 is the only allowed solution. Well, that's essentially what you're being told here. The matrix A is invertible, so its multiplicative inverse [itex]A^{-1}[/itex] exists and you're justified in "dividing by A" to solve the equation.
     
  6. Feb 21, 2010 #5
    Oh, good point. Thank you very much for the help.
     
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