Specific product of projection matrices gives one of the factors I'm trying to give a proof of the following statement:  

Let $X=[x_1 \; X_{-1}]$ be an $n\times k$ matrix, with $X_{-1}$ being $X$ without the first column $x_1$. Furthermore, let $H=X(X^TX)^{-1}X^T$ and $H_{-1}=X_{-1}(X_{-1}^TX_{-1})^{-1}X_{-1}^T$ be the corresponding projection matrices for $X$ and $X_{-1}$, respectively. Then $H_{-1}H=H_{-1}$.  

Geometrically, the result makes sense. If a vector $v\in\mathbb{R}^3$ is projected onto the $xy$-plane, and then onto the $x$-axis, the same result would have followed from just projecting $v$ directly onto the $x$-axis. However, I'm struggling with an algebraic proof.
 A: If $A\in\mathbb R^{m\times n}$ with $m\ge n$ has full rank, then $A(A^TA)^{-1}A^T$ is the orthogonal projection onto the column space of $A$. So, $H$ projects onto the column space of $X$ and $H_{-1}$ projects onto the column space of $X_{-1}$, which is contained in that of $X$. That's why $H_{-1}H = H_{-1}$.

I see from your question that you already know that these are orthogonal projections. So, assume that we have two subspaces $M\subset N$ of $\mathbb R^n$. We want to prove that $P_MP_N = P_M$. For this, let $\oplus$ denote the orthogonal sum of subspcaes. We have $N = M\oplus Z$ with a subspace $Z$. Moreover, $\mathbb R^n = N\oplus N^\perp$. So, $\mathbb R^n = M\oplus Z\oplus N^\perp$. From here we see that $M^\perp = Z\oplus N^\perp$.
Let $x\in\mathbb R^n$ and decompose $x = m+z+n$ according to the decomposition of $\mathbb R^n$. Then $P_Mx = m$. Moreover, $P_Nx = m+z = m+(z+0)$ and so $P_MP_Nx = m = P_Mx$.
Another proof: Since for $x\in N$ we have $P_Nx = x$, it easily follows that $P_NP_M = P_M$. As orthogonal projections are symmetric, we conclude $P_MP_N = (P_NP_M)^T = P_M$.
