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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!

Ninja7777
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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)

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Glen_b
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