For orthogonal projection matrices $A, B \in \mathbb R^{n\times n}$, we have that
$A^\top = A$ and $A^2 = A$
$B^\top = B$ and $B^2 = B$
where $A^\top$ is the transpose of $A$, and $A \succeq B $ says that $A-B$ is a positive semidefinite matrix.
I would like to prove the following proposition
$$A \succeq B \Leftrightarrow AB = B$$
For showing $A \succeq B \Rightarrow AB = B$, we have that
$$ \begin{aligned} A-B \succeq 0 &\Rightarrow x^\top (A-B)x \ge 0 \; \forall x \\ &\Rightarrow y^\top(I-A)^\top(A-B)(I-A)y \ge 0 \; \text{for } x = (I-A)y \; \forall y\\ &\Rightarrow y^\top(I-A)^\top B(I-A)y \le 0 \; \forall y \; \text{ as } A(I-A)=0 \\ &\Rightarrow y^\top(I-A)^\top B(I-A)y = 0 \; \forall y \; \text{ as } B \succeq 0\\ &\Rightarrow y^\top(I-A)^\top B^\top B(I-A)y = 0 \; \forall y \; \text{ as } B=B^2=B^\top B \\ &\Rightarrow B(I-A) = 0 \\ &\Rightarrow A^\top B^\top = B^\top \\ &\Rightarrow AB = B \end{aligned} $$
Not sure how to show $A-B\succeq 0 \Leftarrow AB = B$.
For the $\Leftarrow$ direction: Suppose that $AB=B$. Observe that $BA=B^{T}A^{T}=(AB)^{T}=B^{T}=B=AB$. That is, $A$ and B commutes. Also, $(I-B)^{2}=I-B$. Let $x$ be an arbitrary vector. Consider \begin{eqnarray*} & & \langle(A-B)x,x\rangle\\ & = & \langle A(I-B)x,x\rangle\\ & = & \langle A(I-B)^{2}x,x\rangle\\ & = & \langle(I-B)A(I-B)x,x\rangle\\ & = & \langle A(I-B)x,(I-B)^{T}x\rangle\\ & = & \langle A(I-B)x,(I-B)x\rangle\\ & \geq & 0. \end{eqnarray*} (In the above, $\langle\cdot,\cdot\rangle$ denotes the usual inner product.) Obviously $A-B$ is symmetric. Hence, $A-B$ is positive semi-definite.