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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 $E_\alpha$?

Landon Carter
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    If the ranges are all orthogonal to each other, $E$ is simply the sum of all $E_\alpha$. If they are not orthogonal, then you could probably try to mimic Gram-Schmidt, but I would not call this a nice description... – mlk Apr 28 '18 at 08:08

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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.

mlk
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  • Thanks for your answer Milk! I was wondering what difference in visualisation would occur when I am working in an infinite dimensional space. Like in finite dimensional vector space I know every projection is associated with a subspace on which it projects, then the whole space can be written as a finite sum of such subspaces, etc. Do these still hold in the infinite dimensional setup? – Landon Carter Apr 28 '18 at 16:49
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    In general Hilbert spaces behave rather similar to Euclidean finite dimensional spaces. The big difference is probably that instead of finite sums, you generally have infinite square summable series instead. So in your example, if you have orthogonal subspaces $(S_\alpha){\alpha\in A}$ which span $H$ (for example in the sense that any nonzero vector has a nonzero projection on one of them), then $H$ is not a direct sum but the set of all $v=\sum{\alpha\in A} v_\alpha$ where $\sum_{\alpha\in A} |v_\alpha|^2 < \infty$. – mlk Apr 29 '18 at 21:20
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    For me a good example to visualize and understand those things have always been Fourier series. Just think of them as rewriting functions in the Hilbert space of periodic square integrable functions in another orthogonal basis. In that interpretation, calculating the Fourier coefficient of $\cos(kx)$ is exactly finding the orthogonal projection of your function onto the subspace spanned by $\cos(kx)$. Also many of the theorems have geometric meaning, for example Parseval's theorem is actually a variant of Pythagoras and so on. – mlk Apr 29 '18 at 21:28