Projection matrices identity: $P_{M_ZX}=P_{M_Zx}+P_{M_{[x\; Z]}X_2}$ For any real matrix $Y$ with $n$ rows and full column rank, we define the orthogonal projection matrices
$$
\underbrace{P_Y}_{n\times n}=Y(Y'Y)^{-1}Y',\quad \underbrace{M_Y}_{n\times n}=I_n-P_Y.
$$
Among many other properties, $P_Y$ fixes the column space of $Y$ whereas $M_Y$ sends each vector in that space to $0$.

A paper (reference available if you are interested) I'm reading cites the following identity
  $$
P_{M_ZX}=P_{M_Zx}+P_{M_{[x\; Z]}X_2}.\tag{$\star$}
$$
  (The subscript of the first term on the RHS is $M_Z\cdot x$.) The dimensions of various things are
  $$
Z:n\times l;\quad x:n\times1;\quad X_2:n\times k;\quad X=[x\; X_2]:n\times(1+k).
$$

I've spent some time trying to prove ($\star$) suspecting it is probably clever algebraic manipulation but I can't figure it out. A better approach probably involves looking at the RHS of ($\star$) and recognize some relationship between $M_Zx$ and $M_{[x\;Z]}X_2$. I don't have enough knowledge at the moment for this latter approach but I'm interested in learning more about projection matrices. This book will probably aid me but I don't have a copy yet. 
Can someone help please? Thank you very much.
 A: First of all, if $P_1,P_2$ are orthogonal projections then $P_1$ and $P_2$ are positive semidefinite and $\Im(P_1+P_2)=\Im(P_1)+\Im(P_2)$. 
Now, if $P=P_1+P_2$ and $P,P_1,P_2$ are orthogonal projections then $\Im(P)=\Im(P_1)+\Im(P_2)$ . Thus, $P_1=P_1P=P_1^2+P_1P_2=P_1+P_1P_2$. Therefore, $P_1P_2=0$ and $\Im(P_1)\perp\Im(P_2)$.
Notice that the converse is also true if $P_1,P_2$ are orthogonal projections and $\Im(P_1)\perp\Im(P_2)$ then $P=P_1+P_2$ is an orthogonal projection.
Thus, we must prove that $\Im(P_{M_ZX})=\Im(P_{M_Zx})+\Im(P_{M_{[x,Z]}X_2})$ and $\Im(P_{M_Zx})\perp\Im(P_{M_{[x,Z]}X_2})$.
Now $\Im(P_{M_Z X})=\Im(M_Z X)$, $\Im(P_{M_Z x})=\Im(M_Z x)$ and $\Im(P_{M_{[x,Z]}X_2})=\Im(M_{[x,Z]}X_2)$. $\hspace{2cm}(1)$
Notice that $\Im([x,Z])=\Im(Z)\oplus\Im(M_Z x)$ (because the unique column of $M_Z x$ is the projection of $x$ on the subspace $\Im(Z)^{\perp}$).
Thus, $\Im([x,Z])^{\perp}=\Im(Z)^{\perp}\cap\Im(M_Z x)^{\perp}$ and $\Im(Z)^{\perp}=\Im(Z)^{\perp}\cap\Im(M_Z x)^{\perp}\oplus \Im(M_Z x)$. $\hspace{1.2cm}(2)$
Since $M_Z$ is the projection on the subspace $\Im(Z)^{\perp}$, $M_{[x,Z]}$ is the projection on the subspace $[x,Z]^{\perp}$ and $P_{M_Z x}$ is the projection on the subspace $\Im(M_Z x)$
then $M_Z=M_{[x,Z]}+P_{M_Z x}$ and $M_ZX_2=M_{[x,Z]}X_2+P_{M_Z x}X_2$. 
Therefore, $\Im(M_ZX_2)\subset \Im(M_{[x,Z]}X_2)+\Im(M_Z x)\subset \Im(M_ZX_2)+\Im(M_Z x)$.
So $\Im(M_ZX_2)+\Im(M_Z x)=\Im(M_{[x,Z]}X_2)+\Im(M_Z x)$ and, by $(2)$,  $\Im(M_{[x,Z]}X_2)\perp\Im(M_Z x)$.   $\hspace{0.5cm}(3)$
Notice also that $M_ZX=[M_Z x,M_Z X_2]$. Thus, $\Im(M_ZX)=\Im(M_Z x)+\Im(M_Z X_2)$.  By $(3)$, $\Im(M_ZX)=\Im(M_Z x)+\Im(M_{[x,Z]}X_2)$.$\hspace{10cm}(4)$  
Finally, by $(4)$ and $(1)$, $\Im(P_{M_Z X})=\Im(P_{M_Z x})+\Im(P_{M_{[x,Z]}X_2})$ and $\Im(P_{M_Z x})\perp\Im(P_{M_{[x,Z]}X_2})$.
