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Matrix construction

  1. Oct 21, 2009 #1
    Am really lost on this one:

    Suppose X is a full rank nxp matrix. Construct the matrix:

    P = I - X((X'X)^-1)X'

    Where X' is the transpose of X.

    I then have to find the trace and eigenvalues of P.

    Can anyone help?!
     
  2. jcsd
  3. Oct 21, 2009 #2

    lanedance

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    do you know that for invertible matricies A,B that

    [tex] (AB)^{-1} = B^{-1} A^{-1} [/tex]
     
  4. Oct 21, 2009 #3

    lanedance

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    easy to see this, say C is the inverse of (AB), then

    [tex](AB)C = ABC = I [/tex]

    [tex]A^{-1}ABC = BC = A^{-1} [/tex]

    [tex]B^{-1}BC = C = B^{-1}A^{-1} [/tex]
     
  5. Oct 21, 2009 #4

    statdad

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    lanedance's comment is true, but since your matrix need not be square it doesn't help. You can't simplify

    [tex]
    P = I - X(X'X)^{-1} X'
    [/tex]

    any more than it already is.

    This is the type of problem seen during a matrix introduction to multiple regression (although it may be in a different setting for your problem).

    Hint: You can show that [tex] P^2 = P [/tex] ([tex] P [/tex] is actually a projection matrix).
    1) What is its dimension?
    2) If you know the rank of [tex] X [/tex] you can find the rank of [tex] P [/tex]

    Those hints will (should, may) give you an idea of an attack to find the quantities you need.
     
  6. Oct 21, 2009 #5

    lanedance

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    ok, yeah - misread that sorry
     
  7. Oct 21, 2009 #6

    lanedance

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    along with statdads comments, the property of the trace being the sum of the eigenvalues
    [tex] tr(A) = \sum_i a_ii = \sum_i \lambda_i [/tex]

    and
    [tex] tr(A^k) = \sum_i \lambda_i^k [/tex]
    could help
     
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