Let $X_{n\times p}$ and $Z_{n\times q}$ be two matrices such that $n>>p,q$. Prove $$(I-P_X)(I-P_Z)(I-P_X)\ge (I-P_{X:Z})$$ in Loewner order where $P_Y=Y(Y^TY)^-Y^T$ (Note the pseudo inverse)
I was able to reduce the problem to proving $$Z_1 ((Z_1^T Z_1)^- - (Z_1^T Z_1 + Z_2^T Z_2)^-) Z_1^T$$ is Positive Semi Definite where $Z_1=(I-P_X)Z$ and $Z_2= P_XZ$
Spectral decomposition seems like the way to go but I can't really proceed.
Let $\mathcal Y$ be the column space of the augmented matrix $Y=[X|Z]$. Every vector in $\mathbb R^n$ can be written as $y+p$ for some $y\in\mathcal Y$ and $p\in\mathcal Y^\perp$. Since $(I-P_X)p=(I-P_Y)p=(I-P_Z)p=p$ and $(I-P_Y)y=0$, we have \begin{aligned} &\big\langle\left[(I-P_X)(I-P_Z)(I-P_X)-(I-P_Y)\right](y+p),\ y+p\big\rangle\\ &=\big\langle\left[(I-P_X)(I-P_Z)(I-P_X)-(I-P_Y)\right]y,\ y+p\big\rangle\\ &=\big\langle y,\,\left[(I-P_X)(I-P_Z)(I-P_X)-(I-P_Y)\right](y+p)\big\rangle\\ &=\big\langle y,\,\left[(I-P_X)(I-P_Z)(I-P_X)-(I-P_Y)\right]y\big\rangle\\ &=\big\langle y,\,(I-P_X)(I-P_Z)(I-P_X)y\big\rangle\\ &=\big\langle y,\,(I-P_X)(I-P_Z)^2(I-P_X)y\big\rangle\\ &=\big\langle (I-P_Z)(I-P_X)y,\,(I-P_Z)(I-P_X)y\big\rangle\\ &\ge0. \end{aligned} Hence $(I-P_X)(I-P_Z)(I-P_X)\ge(I-P_Y)$.