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Thread: Have I done anything false here?

  1. #1
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    Default Have I done anything false here?

    The statement is:
    Show that if a matrix, A, is idempotent, then 2A-I is invertible and it is its own inverse.

    I did
    (2A-I)*I
    =(2A-I)*A*A^-1
    =(2A^2-A)*A^-1
    =(2A-A)*A^-1
    =A*A^-1 = I
    and I is invertible, thus 2A-I is invertible.

    Have I done something that is not always true here? or is 2A-I really equal to I? meaning A = I
    (Idempotent means that A^n = A for any matrix A and n being an element of Z+

    Just seems too neat

  2. #2
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    Default

    I - A is idempotent

    ( 2A - I )( 2A - I ) = 4A² - 2AI - 2IA + I² = 4A - 2A - 2A + I = I, which is the identity

  3. #3
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    Default

    hmm thats pretty neat, almost TOO neat. I think I have done something wrong, or A=I

    Now moving on to getting bloody vandermonde determinantes through row operations... :P

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