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Fredrik

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If V is a finite-dimensional vector space and U is a subspace of V, every x in V can be uniquely expressed as x=y+z, with y in U, and z in the orthogonal complement of U. The map [itex]x\mapsto y[/itex] is the projection operator associated with the subspace U. It's linear, self-adjoint and idempotent (P

Let P be any linear, self-adjoint and idempotent operator. Its range W is a subspace. So every x in V can be uniquely expressed as x=y+z, with y in W and z in the orthogonal complement. Since the decomposition is unique, and x=Px+(1-P)x, we have y=Px and z=(1-P)x. So P is the projection operator associated with W.

This means that the two standard ways to define a projection operator are equivalent. So if you're using either of these definitions, there's only one projection operator associated with ker M. Are you using some other definition?

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Thank you for the reply. I think I was not sufficiently clear about the concept I am considering.

I want to consider a linear operator M on a vector space V whose image is linearly independent from its Kernel.

On a finite dimensional vector space, this implies that V = ker M + Img M.

In this case, there is a unique projector P such that ker P = Img M and Img P = ker M. It can be considered the natural projector onto the kernel of M. It is not necessarily an orthogonal projector---note that I have not specified any notion of inner product on V.

If the vector space is infinite dimensional, in general we do not have

V = ker M +Img M. But suppose that ker M is a closed subspace of V, so there is a projection P onto the kernel of M: ker M = Img P. If we in addition require P M =0, this is a natural infinite dimensional analogue of the projection operator defined in the last paragraph.

My question is twofold:

1) Is the projection operator defined above unique in the infinite dimensional case?

2) Is there a standard way to refer to this projection operator among mathematically knowledgable?

Thank you!

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