Can the composite of two projections really fail to be a projection? Let $H$ denote a Hilbert space. For any closed subspace $C \subseteq H$, write $P_C$ for the orthogonal projection onto $C$. Then according to wikipedia, the composite $P_U \circ P_V$ needn't be a projection. That is, it needn't be the case that $P_U \circ P_V = P_{U \cap V}.$ Is this really true?

It is true, which can be readily shown in ${\mathbb{R}^2}$.
Let $U = \left\{ {\left( {x,0} \right):x \in \mathbb{R}} \right\}$ and $V = \left\{ {\left( {x,y} \right):x - y = 0} \right\}$.
Let ${P_U}\left( {x,y} \right) = \left( {x + y,0} \right)$ and ${P_V}\left( {x,y} \right) = \left( {x,x} \right)$. Note that both are projections, since $P_U^2 = {P_U}$ and $P_V^2 = {P_V}$.
Then ${\left( {{P_V}{P_U}} \right)^2} \ne {P_V}{P_U}$, since ${P_U}\left( {1,1} \right) = \left( {2,0} \right),{P_V}\left( {2,0} \right) = \left( {2,2} \right),{P_U}\left( {2,2} \right) = \left( {4,0} \right),{P_V}\left( {4,0} \right) = \left( {4,4} \right)$, so ${{P_V}{P_U}}$ isn't a projection, although both ${P_V}$ and ${P_U}$ are.
EDIT: Note that the notions of projection and orthogonal projection are not the same. Definition of projection is that it is a linear map which is idempotent. Definition of orthogonal projection is that it is a projection whose image and kernel are orthogonal. It can be shown that the orthogonality condition is equivalent to the condition that ${P^*} = P$. So, for projections we have ${P^2} = P$, while for orthogonal projections we have ${P^2} = P = {P^*}$ (${P^*}$ is the adjoined map to $P$).
One can wonder, since the statement doesn't hold for projections in general, is it true if we restrict ourselves to orthogonal projections?
The answer remains no, since ${P_U}\left( {x,y} \right) = \left( {x,0} \right)$ and ${P_V}\left( {x,y} \right) = \left( {\frac{{x + y}}{2},\frac{{x + y}}{2}} \right)$ are orthogonal projections ($P_U^2 = {P_U}$, $P_V^2 = {P_V}$, $\operatorname{Im} {P_U} = \left\{ {\left( {x,0} \right):x \in \mathbb{R}} \right\} \bot \left\{ {\left( {0,y} \right):y \in \mathbb{R}} \right\} = \ker {P_U}$ and $\operatorname{Im} {P_V} = \left\{ {\left( {x,x} \right):x \in \mathbb{R}} \right\} \bot \left\{ {\left( { - x,x} \right):x \in \mathbb{R}} \right\} = \ker {P_V}$), but ${P_U}{P_V}{P_U}{P_V}\left( {1,1} \right) = {P_U}{P_V}{P_U}\left( {1,1} \right) = {P_U}{P_V}\left( {1,0} \right) = {P_U}\left( {\frac{1}{2},\frac{1}{2}} \right) = \left( {\frac{1}{2},0} \right)$, so ${P_U}{P_V}\left( {1,1} \right) = \left( {1,0} \right) \ne \left( {\frac{1}{2},0} \right) = {\left( {{P_U}{P_V}} \right)^2}\left( {1,1} \right)$.
We conclude that even orthogonality doesn't guarantee that ${{P_U}{P_V}}$ remains a projection, let alone an orthogonal projection.
