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Matrix Proof: Idempotent

  1. Oct 7, 2008 #1
    A matrix P is called idempotent if P^2 = P. If P is idempotent and P =/= I show that det(P)=0.

    I don't really know where to go with this but i have a feeling that it involves taking the det of each side.

    det(P^2) = det(P)
    det(P)det(P) = det(P)

    where to from here if thats even the right step/method to take, or if its even right at all >_>

    Thanks :)
     
  2. jcsd
  3. Oct 7, 2008 #2

    gabbagabbahey

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    looks fine to me; now for what values of det(P) does that equation hold?
     
  4. Oct 7, 2008 #3
    det(P) = 1 or 0
     
  5. Oct 7, 2008 #4

    gabbagabbahey

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    Okay, if detP=1, and P^2=P, what matrix must P be?
     
    Last edited: Oct 7, 2008
  6. Oct 7, 2008 #5

    gabbagabbahey

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    Hint: use the fact that if [itex]det(P) \neq 0[/itex], then P is invertible. Multiply [itex]P^2=P[/itex] by [itex]P^{-1}[/itex].
     
  7. Oct 27, 2008 #6
    but it says det(P)=/=1. How do you show that det(P)=0???
     
  8. Oct 27, 2008 #7

    rock.freak667

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    det(P2) = det(P)

    => det(P)^2-det(P)=0

    This is the same as t^2-t=0 where t=det(P). Factorise and use that fact that P=/= I
     
  9. Sep 13, 2009 #8
    I have a questions:
    if A=I-X(X'X)^-1X'
    is it A idempotent?
     
  10. Sep 13, 2009 #9

    statdad

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    kendarto: don't jump into another poster's thread.

    try to calculate [tex] A^2 [/tex] and answer this for yourself.
     
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