Projection associated to the decomposition $H=M⊕N$ Let $H$ be a Hilbert space and let $M$ and $N$ be two closed subspaces in $H$ such that $H=M⊕N$.
I'm trying to find a formula giving $P_{M,N}$ (the projection onto $M$ with respect to $N$) in terms of $P_{M,M^{\bot }}$ (the orthogonal projection onto $M$) and $P_{N,N^{\bot }}$ (the orthogonal projection onto $N$).
Thank you !
 A: We are going to write $\,Q=P_{M,M^\perp}$ and $\,R=P_{N,N^\perp}$ for the orthogonal projections.  
Loosely speaking, $P_{M,N}$ and $Q$ have eigenvalue multiplicities which coincide. Moreover, they possess the same supporting subspace
$\,M=\operatorname{im}P_{M,N}=\operatorname{im}Q=\,$ the eigenspace for the eigenvalue $1$. So let's look out for a similarity transformation $T$ intertwining them, i.e.
$$TP_{M,N}=Q\,T\tag{1}$$ 
with $T$ invertible.
Necessarily, $T$ has to map $N$ to $\ker Q = M^\perp$,
and this is fulfilled by choosing $T=\mathbb 1-QR\,$:
$$Tn \:=\: (\mathbb 1-QR)n \:=\: n-Qn \:=\: (\mathbb 1-Q)n\:\in M^\perp\quad\forall n\in N,$$
hence $(1)$ is satisfied when restricted to $N$.
For $m\in M$ the RHS evaluates to $$QTm \:=\: (Q-QR)m \:=\: m-QRm \:=\: Tm\,,$$
and $(1)$ is again satisfied, hence it's satisfied in general.
$T=\mathbb 1-QR\,$ is invertible because $\|QR\|<1$, which in turn follows(*) from
the central assumption $H=M\oplus N=\operatorname{im}Q\oplus\operatorname{im}R\,$.
Thus,
$$P_{M,N}\:=\:(\mathbb 1-QR)^{-1}Q\,(\mathbb 1-QR)\tag{2}$$ 
is a formula as asked for.
* Norm estimate for a product of two orthogonal projectors 
