Orthogonal Projections in $U\subset V$ two subspaces in a Hilbert space Let $U,V\subset H$ be closed subspaces of a Hilbert space $H$, and let $P_U$ and $P_V$ the respective orthogonal projections. Show:
$U\subset V \Longleftrightarrow P_U=P_UP_V=P_VP_U$ 
Trying to prove $\Rightarrow$, it's easy to see that $P_U=P_VP_U$ since for every $x\in H$ we have that $P_U(x)\in U\subset V$ and therefore $P_VP_U(x)=P_U(x)$, but I cannot prove why it should be also $P_U=P_UP_V$.
The proof of $\Leftarrow$ is easier; if $x\in U$ we have $P_U(x)=x$ and, since $P_VP_U=P_U$ we have $P_V(x)=P_VP_U(x)=P_U(x)=x$, and therefore $x\in V$.
Can anyone help me with the second part of $\Rightarrow$? Thanks!

In a pedestrian way: If f is contained in $U$ then the equality holds for f in $U$ and acting with $P_U$, ${P_U}{P_V}$ or ${P_V}{P_U}$ on f reproduces f. If, on the other hand f is contained in $V$ but not in $U$ then the equality also holds but the result is $0$ since ${P_V}f=f$ and ${P_U}f=0$.
The equality is sometimes encountered in connection with the spectral measure associated with the spectral decomposition of a self-adjoint operator.
