A doubt about projection operators in a Hilbert space Suppose $H$ is a Hilbert space and $B(H)$ is the space of bounded linear operators on $H$. Let $\{E_{\alpha}\}_{\alpha\in A}$ be an arbitrary collection of orthogonal projection operators. Then let $E$ be the projection onto the smallest closed subspace containing $\cup_\alpha Range(E_\alpha)$.
I am trying to understand what happens to $Ev$ for $v\in H$. Is there a nice description of this in terms of $E_\alpha$?
 A: Thinking a bit more about my comment, I can give the negative answer: There seems to be no nice general description. 
Consider the easier special case where $H = \mathbb{R}^n$ and all $E_\alpha$ are of rank $1$ and are thus associated with vectors $v_\alpha$ such that $\operatorname{range}(E_\alpha)=\operatorname{span} (v_\alpha)$. Then we arrive at the classic textbook problem of projecting a vector onto the subspace spanned by $(v_\alpha)_\alpha$. The solution to this always is to first turn the $v_\alpha$ into an orthogonal basis of the subspace, see for example this question. And if there would be a nicer solution to creating an orthogonal basis, than the Gram-Schmidt process, we would probably use that instead.
As the general problem is more complicated than this special case (try it with several projections onto two-dimensional spaces), the nicest solution will be at most as nice as the Gram-Schmidt process, which I would confidently say is universally recognized as not very nice. (As evidenced by the groans you will get if you tell students to carry it out on a set of vectors as a homework problem...)
However there are special cases. As mentioned, if the ranges $(\operatorname{range}(E_\alpha))_{\alpha\in A}$ are pairwise orthogonal then $E = \sum_{\alpha \in A} E_\alpha$.
Also since you want $E$ to be the projection on the specified subspace, I assume you meant $E$ to be orthogonal. If you just want $E$ to be a projection (that is $E^2 = E$), then there might be more possibilities.
