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First of all, the projection matrix $P_A$ is given by $P_A = A(A'A)^{-1}A'$. Similarly, $P_{AB} = AB(B'A'AB)^{-1}B'A'$.

I have tried proving that $P_A - P_{AB}$ is itself a projection matrix, then it would follow that it is positive semidefinite. This didn't work out for me, I don't think this is the way to prove it. Premultiplying by x' and postmultiplying by x also hasn't helped me. Can anybody point me in the right direction?

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$P_A$ is the orthogonal projection on $\text{Ran}(A)$ (the range, aka column space, of $A$). $P_{AB}$ is the orthogonal projection on $\text{Ran}(AB)$, which is a linear subspace of $\text{Ran}(A)$. Therefore $P_A - P_{AB}$ is the orthogonal projection on the orthogonal complement of $\text{Ran}(AB)$ in $\text{Ran}(A)$. Being an orthogonal projection, it is positive semidefinite.

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  • $\begingroup$ Thank you very much! After reading this I realized I could write out $P_AP_{AB}$ and then $(A'A)^{-1} A'A$ cancels out, so $P_AP_{AB} = P_{AB}$. I can use this in proving $(P_A - P_{AB})^2 = P_A - P_{AB}$. $\endgroup$ – Maarten Meijering Dec 21 '18 at 16:02

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