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Linear algebra proof.

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data
    If A is an invertible idempotent matrix, then A must be the Identity matrix I_n.

    2. Relevant equations
    A^2==A ; A^2==AA; A^(-1); I==A^(-1)

    3. The attempt at a solution

    Suppose A is an nxn matrix =/= I_n.

    s.t. A^(2)==A

    so A^(2)==A ==> AA==A

    ==> A^(-1)AA==A^(-1)A ==> A==I==> A^(-1)A==A^(-1)I==>I==A^(-1)I==A^(-1)==A

    which yeilds a contradiction because we supposed our A =/= I_n.

    Therefore A==I_n

    Is this correct please help me understand where I have failed...
  2. jcsd
  3. Mar 8, 2010 #2


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    You should have stopped right here. You should have also stated that A is invertible.
    This is wrong. You don't know that the inverse of A is equal to A.
  4. Mar 8, 2010 #3
    Okay so if i stated A an invertible nxn matrix =/= I_n

    s.t A^(2)==A(idempotent)... truncating the last bit of foolishness. I was correct?
  5. Mar 8, 2010 #4
    another proof...
    problem statement.
    prove if A and B are idempotent and AB==BA then AB is idempotent.

    AB==BA ==> A^(-1), B^(-1) exist

    Since A and B are idempotent invertible matrices, from previously proven theorem, we know A=I and B=I. and since II==I ==> AB==I Therefore AB==BA and AB is Idempotent,
  6. Mar 8, 2010 #5


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    This isn't true.
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