# If matrices $A$ and $AB$ have full column rank, how do I prove that $P_A - P_{AB}$ is positive semidefinite?

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?

## 1 Answer

$$P_A$$ 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.

• 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 $(P_A - P_{AB})^2 = P_A - P_{AB}$. – Maarten Meijering Dec 21 '18 at 16:02