Finding trace and eigenvalues of P = I - X(X'X)^-1X'

[abiadams]

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]

do you know that for invertible matricies A,B that

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

 

[lanedance]

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]

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]

ok, yeah - misread that sorry

 

[lanedance]

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