If B(H) is the algebra of bounded linear operators acting on an infinite dimensional complex Hilbert space, then which elements of B(H) that can't be written as a linear combination of orthogonal projections ?
I found something more interesting here https://arxiv.org/pdf/1608.04445.pdf
In a finite Dimensional Vector space, a Square Diagonalizable matrix can be expressed as a linear combination of rank one Oblique projectors, the corresponding coefficients being the respective eigenvalues, you can see whether such a result carries over to the infinite dimensional Vector spaces such as Hilbert space in a straightforward manner.
The dependent variables part can't be written as a linear combination of orthogonal projections.
Dear Zouheir Amara
I don’t understand the your question
I suppose you are talking about infinite sums of projections, so...infinite sums in which topology?
Not infinite sums, but the finite ones. More precisely, operators that can't be written as $\alpha_1 P_1 + ...+ \alpha_n P_n$ where $\alpha_i$ are complex numbers and $P_i$ are orthogonal projections.
I might be wrong, but it's not difficult to prove the following statement.
Let $A\in B(\mathcal H)$ such that
$$ A= \sum_{k=1}^n \alpha_k P_k \ , \qquad P_h \bot P_k \qquad \forall h\neq k$$
its spectrum $\sigma(A)$ is discrete, because
if $I=\sum_{k=1}^n P_k $ then $\sigma(A)=\left\{ \alpha_1 , \alpha_2 \ldots \alpha_n \right\} $
while if $I\neq\sum_{k=1}^n P_k $ then $\sigma(A)=\left\{ 0 , \alpha_1 , \alpha_2 \ldots \alpha_n \right\} $.
The projections that I was talking about are not necessarily mutually orthogonal.
There is an ambiguity in the question. What do you mean by "orthogonal projection"?
If you mean selfadjoint idempotent (P=P^2=P^*) then the answer is "none", that is every bounded operator is a linear combination of n projections. This was first proven by Peter Fillmore, with n= 257. Then n was subsequently greatly reduced. I think the best for selfadjoint operators is n= 4 (Nakamura 1984).
If you mean "linear combination of mutually orthogonal projections" the operator will be diagonalizable (hence normal) and with finite spectrum.
Dear Victor Kaftal , I don't really understand your answer. Do you mean that every bounded operator is a linear combination of orthogonal projections ? (by orthogonal projections, I mean selfadjoint idempotents)
Yes. This was proven by Fillmore, Sums of operators with square zero, Acta Sci. Math. (Szeged), 28 (1967) 285-288. And as I wrote in my previous message, there are also estimates on how many projections are needed.
While the result is obvious for finitely dimensional spaces (every selfadjoint is diagonalizable and every operator is a linear combination of selfadjoints) the estimates on how many projections are needed is not trivial even for finite space.
Here is a link to the paper by Fillmore mentioned by Victor Kaftal
http://pub.acta.hu/acta/showCustomerArticle.action?id=7809&dataObjectType=article&returnAction=showCustomerVolume&sessionDataSetId=1e481b36a45b064a&style=
I found something more interesting here https://arxiv.org/pdf/1608.04445.pdf
Refer the below link
https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.math.tamu.edu/~yvorobet/MATH423-2012A/Lect3-10web.pdf&ved=2ahUKEwjdyOPikcLoAhU0meYKHc1kCx0QFjAFegQICBAB&usg=AOvVaw07HPbOl-kUl-wHL9uOD93R&cshid=1585569373546
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