1. ## Projection

Let $H$ be a Hilbert Space and $P$ and $Q$ be the projections on closed linear subspaces $M$ and $N$ of $H$.
Prove that if $Q-P$ is a projection,then its range is $N\cap M^\perp$.

2. Originally Posted by problem
Let $H$ be a Hilbert Space and $P$ and $Q$ be the projections on closed linear subspaces $M$ and $N$ of $H$.
Prove that if $Q-P$ is a projection,then its range is $N\cap M^\perp$.
For a start, $I-P$ is a projection, with range $M^\perp$. So if $x\in N\cap M^\perp$ then $(I-P)x=x$, $Px=0$, $Qx=x$ and hence $(P-Q)x=x$. Thus if $P-Q$ is a projection then its range contains $N\cap M^\perp$.

For the converse inclusion, if $P-Q$ is a projection then
$(1)\quad (Q-P)^2 = Q-P$,
$(2)\quad Q - PQ - QP + P = Q-P$ (since $P^2=P$ and $Q^2 = Q$),
$(3)\quad PQ+QP=2P$.
Multiply both sides of (3) on the left to get
$(4)\quad QPQ=QP$.
Multiply both sides of (3) on the right to get
$(5)\quad QPQ=PQ$.
From (4), (5) and (3),
$(6)\quad PQ=QP=P$.
Therefore
$(7)\quad (I-P)Q=Q(I-P)=Q-P$.
It follows from (7) that the range of $Q-P$ is contained in the range of $(I-P)$ and in the range of $Q$, and thus in $N\cap M^\perp$.