Prove one projection matrix is larger than another

Question

Please give me some hints for the right direction, but don't give me a full answer.

We define $\mathbf{P_X}$ as projection matrices: $\mathbf{P_X = X(X'X)^{-1}X'}$. My exercise reads: Prove if both $\mathbf{A}$ and $\mathbf{AB}$ are full column rank, then: $\mathbf{P_A - P_{AB}}\geq 0$.

First of all, I'm pretty sure the zero should be boldfaced, as I expect to end up with a matrix, not a scalar. Furthermore, I know $\mathbf{P_X} \in \mathcal{P}$ implies $\mathbf{I-P_X} \in \mathcal{P}$. Hence, we have $\mathbf{0 \leq P_A \leq I}$ and $\mathbf{0 \leq P_{AB} \leq I}$. However, this proof requires me to show $\mathbf{P_A \geq P_{AB}}$, and I'm not sure how to approach it.

Does it have somehting to do with the ranks of matrices $\mathbf{A}$ and $\mathbf{AB}$? If so, how does it relate to the "size" of the projection matrices?

Answer 1

The hint that eventually solved it for me was that if you indeed had to show $\mathbf{P_A - P_{AB}} \geq \mathbf{0}$, you could also show $\mathbf{P_A - P_{AB}}$ was a projection matrix itself. (Thus: $\mathbf{P_A - P_{AB}} \in \mathcal{P}$)