I'm having trouble understanding the following in a review textbook I'm using, particularly the long equality with inverses and transposes, and the subsequent conclusion regarding the rank and trace of Qx. !

Could someone please explain it, or direct me to a link or freely available document that explains what is going on step by step?

Thank you!


It may help to call the matrix $X(X'X)^{-1}X'$ ,$\:$ "$H$" (for "hat", since it's usually called the 'hat-matrix', because it puts the hat on $y$): $\quad\hat{y}=Hy$.

Note that $H^2=H$ (expand it out, and cancel an adjacent pair of terms that is a matrix and its inverse).

Then $Q=I-H$ and $Q^2=(I-H)^2 = I^2 -HI - IH +H^2$. Using known results, simplify this down to $Q$.

Then note that in the page you copied, above the marked section it says "the rank of an idempotent matrix equals its trace (see below for a proof)" -- so look after that point for a proof of that fact. So now we know that rank(Q)=trace(Q).

Now you would want to relate X and Q. (If you don't see what to do there, you can always fall back on something like this)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.