# Every Hilbert space operator is a combination of projections

I am reading a paper on Hilbert space operators, in which the authors used a surprising result

Every $X\in\mathcal{B}(\mathcal{H})$ is a finite linear combination of orthogonal projections.

The author referred to a 1967 paper by Fillmore, Sums of operators of square zero. However, this paper is not online.

I wonder whether someone has a hint on how this could be true since there are all kinds of operators while projections have such a regular and restricted form.

Thanks!

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The mathscinet article says summary that "(...)every operator on a separable infinite-dimensional Hilbert space is (1) a sum of 64 operators with square zero and (2) a linear combination of 257 projections.". I think it's probably important that it is a separable space. Makes you wonder how did they exactly count the 257 projections. ;) – tomasz Jun 24 '12 at 0:48
@tomasz: You can generalize from the separable case to the infinite dimensional case by writing $\mathcal{H}$ as an orthogonal sum of separable $X$-invariant subspaces. For self-adjoint operators, this won't increase the number of projections needed. As every operator is a sum of self-adjoint ones, this allows the separability condition to be removed. – George Lowther Jun 24 '12 at 1:58
The number 257 was improved to 10 by Matsumoto 30 years ago. See this paper by Laurent Marcoux: math.uwaterloo.ca/~lwmarcou/Preprints/Mar2009Projections.pdf – Martin Argerami Jun 24 '12 at 8:41