Understanding Orthogonal Projection in Linear Operators

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    Orthogonal Projection
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Discussion Overview

The discussion revolves around the concept of orthogonal projection in the context of idempotent linear operators on finite-dimensional inner product spaces. Participants explore the definitions, implications, and properties of such projections, particularly focusing on the relationship between the operator T, its image, and its kernel.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the meaning of T being "the orthogonal projection onto its image," prompting further exploration of the concept.
  • Another participant suggests that any vector v can be expressed as a sum of components from the image of T and its kernel, leading to the assertion that T(e + f) = e.
  • A participant seeks clarification on whether T(e) = e, and how this relates to the notion of projections.
  • There is a discussion about the nature of projections, with one participant asserting that a projection onto a subspace is an idempotent linear map, particularly in the presence of an inner product.
  • One participant raises a question about the identity transformation, arguing that it is not an orthogonal projection, while another counters that it is indeed a projection onto the subspace itself.
  • Further inquiries are made regarding the equivalence of T(v) to the projection of v onto the image of T, and whether the knowledge of the image of T is necessary a priori.
  • A detailed explanation is provided about the properties of projections, including the uniqueness of vector representation in terms of complementary subspaces and the definition of orthogonal projections.
  • A participant concludes that their understanding aligns with the assumption that "orthogonal projection onto its image" implies T(v) = Proj_{im(T)}(v).

Areas of Agreement / Disagreement

Participants express differing views on the nature of the identity transformation as an orthogonal projection, and there is no consensus on the necessity of knowing the image of T a priori. The discussion remains unresolved regarding some of the nuances of the definitions and implications of projections.

Contextual Notes

Participants highlight the importance of verifying statements about projections and their properties, indicating potential limitations in their assumptions and definitions. The discussion reflects varying interpretations of the concepts involved.

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Let T in L(V) be an idempotent linear operator on a finite dimensional inner product space. What does it mean for T to be "the orthogonal projection onto its image"?
 
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Every element in v is a combination e+f where e is in the image of T and f in the kernel and T(e+f)=e
 
So T(e)=e?

What does it mean in terms of projections?
 
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Eh? What do you think a projection onto a subspace is? To me it *is* an idempotent linear map. Then nice thing about having an inner product (non-degenerat) around is that there is an obvious choice of complementary subspace
 
What about the identity transformation? It's not an orthogonal projection of anything.
 
Yes it is, onto the subspace itself.
 
Just to be certain, is it equivalent to saying

[tex]T(v)=Proj_{im(T)}(v)[/tex] for all v? And not knowing what the image of T is a priori does not matter? And how does it follow that each vector in V is a combination of some vector in its image and its nullspace?
 
Last edited:
a projection is amap onto a subspace, that sends every vector to a vector in the subspace, and leaves vectorsd that are already in the subspace where they are.

so if V is a vector space and X is a subspace we want to project on, and if

Y is any complementary subspace, i.e. X and Y together generate V, and X and Y have only the zewro vector in common, then every vector in V can be written uniquely in the form x+y where x is in X and y is in Y.

Then the map f sending x+y to x, is a projection onto X, "along" Y.

Notice that f(f(v)) = v for all v, since once v gets into X it stays put. And also Y = ker(f), since vectors in Y go to zero.


Indeed any linear map f such that f^2 = f is asuch a projection.


f is called an "orthogonal" projection if Y = ker(f) is orthogonal to X = im(f).\

at least i think so, you should of course verify everything by proving all these either trivial or false statements.
 
I figured it out. Of course I assumed that what is meant by "orthogonal projection onto its image" is that [tex]T(v)=Proj_{im(T)}(v)[/tex]. Thanks for the help.
 

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