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abiadams
Oct21-09, 06:47 PM
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?!

lanedance
Oct21-09, 07:25 PM
do you know that for invertible matricies A,B that

(AB)^{-1} = B^{-1} A^{-1}

lanedance
Oct21-09, 07:29 PM
easy to see this, say C is the inverse of (AB), then

(AB)C = ABC = I

A^{-1}ABC = BC = A^{-1}

B^{-1}BC = C = B^{-1}A^{-1}

statdad
Oct21-09, 08:16 PM
lanedance's comment is true, but since your matrix need not be square it doesn't help. You can't simplify


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


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 P^2 = P ( P is actually a projection matrix).
1) What is its dimension?
2) If you know the rank of X you can find the rank of P

Those hints will (should, may) give you an idea of an attack to find the quantities you need.

lanedance
Oct21-09, 09:02 PM
ok, yeah - misread that sorry

lanedance
Oct21-09, 09:07 PM
along with statdads comments, the property of the trace being the sum of the eigenvalues
tr(A) = \sum_i a_ii = \sum_i \lambda_i

and
tr(A^k) = \sum_i \lambda_i^k
could help