Let M and N are two closed subspaces of Hilbert space H such that $N\subset M$. Also $P_M$ and $P_N$ are orthogonal projections on M and N respectively. It is clear that $P_M-P_N$ is again an orthogonal projection. My question is that $P_M-P_N$ on which subspace of H is orthogonal projection?
thanks for your help
It projects onto the orthogonal complement of $N$ in $M$, i.e. $\mathcal{R}(P_M - P_N) = N^\perp \cap M$. To see the inclusion "$\subset$", let $x \in N$. Then $\langle x, (P_M - P_N) y \rangle = \langle P_M x - P_N x, y \rangle = \langle x - x, y \rangle = 0$ for all $y \in H$, so $\mathcal{R}(P_M - P_N) \subset N^\perp$. The inclusion $\mathcal{R}(P_M - P_N) \subset M$ is clear. Finally, for the "$\supset$" direction, let $x \in N^\perp \cap M$. We need to find $y \in H$ with $(P_M - P_N) y = x$. In fact, $y = x$ does this, since $P_M x - P_N x = P_M x = x$.
It's easy to verify that $P_{M}-P_{N}$ is an orthogonal projection because $N\subset M$ gives $P_{M}P_{N}=P_{N}$ and, by taking adjoints, also $P_{N}=P_{N}P_{M}$. As with all orthogonal projections, $P_{M}-P_{N}$ is the orthogonal projection onto its range, where it is equal to $I$. Because $$ P_{M}-P_{N} = (I-P_{N})P_{M}= P_{M}(I-P_{N}) $$ it's fairly easy to check that the range of $P_{M}-P_{N}$ is $$ \mathcal{R}(I-P_{N})\cap\mathcal{R}(P_{M})= N^{\perp}\cap M. $$
