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In a Hilbert space, is the idempotence condition of an operator P sufficient to assert that the operator is a projection operator or is it necessary that the operator is also self-adjoint?

Mario
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    Self-adjointness is necessary to have an orthogonal projection. You may also be willing to consider arbitrary idempotents as not-necessarily-orthogonal projections. – Qiaochu Yuan Oct 13 '23 at 00:31
  • Do you want just a projection, or an orthogonal projection? – lisyarus Oct 13 '23 at 00:32
  • @lisyarus Just a projection. As QiaochuYuan has said, for the projection to be orthogonal it is necessary for the operator to also be self-adjoint. My doubt is, do non-orthogonal projections exist in a Hilbert space and therefore idempotence is a sufficient condition? I am sorry if my question is a bit silly but I have not worked much with Hilbert spaces and in all the cases I have seen the projection operators have been idempotent and self-adjoint. – Mario Oct 13 '23 at 00:47
  • @Mario $\mathbb{R}^n$ is a Hilbert space. – CyclotomicField Oct 13 '23 at 01:20
  • @Mario Yes, there are many projection operators in any Hilbert space (and in fact just a vector space), including non-orthogonal ones. Idempotency is essentially equivalent to being a projection operator, and if the operator isn't self-adjoint, the projection won't be orthogonal. – lisyarus Oct 13 '23 at 01:51
  • Thank to all. My doubts have been clarified. – Mario Oct 13 '23 at 10:23

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