2 dimensional plane of a 3 dimensional space

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Discussion Overview

The discussion revolves around the properties of projection operators in the context of a 2-dimensional plane within a 3-dimensional space. Participants explore the nature of these operators, their identity characteristics, and uniqueness related to subspaces.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant suggests that the projection operator onto a subspace acts as an identity operator for that subspace, while also questioning the nature of the identity operator (I) in this context.
  • Another participant agrees that the projection operator is indeed an identity on its image and states that the projection operator is uniquely determined by its image, citing definitions.
  • A later reply expresses gratitude for the response and indicates a willingness to attempt proofs before asking for them.
  • Subsequently, a participant retracts their earlier statement, claiming that the projection operator is not uniquely determined by the image due to the nature of the complementary subspace, apologizing for any confusion caused.

Areas of Agreement / Disagreement

Participants express differing views on the uniqueness of the projection operator, with one asserting uniqueness based on definitions and another countering that it is not uniquely determined. The discussion remains unresolved regarding this aspect.

Contextual Notes

The discussion highlights the dependence on definitions and the implications of complementary subspaces in the context of projection operators.

Gear300
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For a subspace of Rn, such as a 2 dimensional plane of a 3 dimensional space, wouldn't the projection operator onto the subspace act as an identity operator for that subspace, while at the same time, (I) would also be a (trivial) identity operator?

Furthermore, is a projection operator for a subspace unique?
 
Last edited:
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Yes, by the definition of projection operator, it's an identity on its image. Undoubtedly, I is also an identity on this subspace.
[STRIKE]Yes, a projection operator is uniquely determined by its image. It follows (easily) from the definition as well.[/STRIKE]
Hope that helps. Should you need proofs, just say.
 
Last edited:


Thanks for the reply. Before asking, I'll give the proofs a try.
 


I apologise, the projection is NOT uniquely determined by the image, because complementary subspace is not. Sorry if I wasted your time
:blushing:
 

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