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!
 A: First, for every $x\in H$, $P_Ux\in U\subset V$, and hence
$$
P_VP_Ux=P_Ux.
$$
Next, we shall use the following fact.
Fact. Let $M\subset H$ be a closed subspace of $H$ and
$$
M^\perp=\{x\in H: \langle x,y\rangle=0\,\,\text{for all $y\in M$}\},
$$
the normal subspace of $M$. Then every $x\in H$ is expressed uniquely as
$$
x=x_M+x_{M^\perp}=P_Mx+P_{M^\perp}x,
$$
i.e. $P_{M^\perp}=I-P_M$ and $P_{M^\perp}P_M=P_MP_{M^\perp}=0$.
In order to show that $P_UP_V=P_U$ it suffices to show that $P_U(I-P_V)=0$ or $P_UP_{V^\perp}=0$ or for every $x\in H$
$$
P_UP_{V^\perp}x=P_Ux_{V^\perp}=0.
$$
But $x_{V^\perp}$ has the property the $\langle x_{V^\perp},y\rangle=0$, for every $y\in V$, and hence for every $y\in U$, and thus $x_{V^\perp}\in U^\perp$ and therefore $P_Ux{V^\perp}=0$.
A: 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.
