Question about orthogonal projections.

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    Orthogonal Projections
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Discussion Overview

The discussion revolves around the concept of orthogonal projections in vector spaces, particularly in the context of projecting vectors from 3D to 2D space. Participants explore the definitions and properties of orthogonal and oblique projections, as well as the implications of these concepts on vector relationships.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether all projections are orthogonal projections, using an example of a vector in 3D space projected onto a 2D plane and discussing the implications of the null space.
  • Another participant references external resources to clarify the distinction between orthogonal and oblique projections.
  • A participant outlines properties of projections, including the identity operator on the range and the direct sum decomposition of the vector space, suggesting that oblique projections map vectors back to themselves.
  • A later reply expresses confusion regarding the application of the identity operator to the null space, indicating a misunderstanding of the properties discussed.
  • One participant acknowledges a mistake in their earlier statement and thanks another for providing clarification.

Areas of Agreement / Disagreement

Participants express differing views on the nature of projections, with some asserting that not all projections are orthogonal while others provide definitions and properties that may imply otherwise. The discussion remains unresolved regarding the classification of projections.

Contextual Notes

There are limitations in the discussion regarding the definitions of orthogonal and oblique projections, as well as the assumptions made about vector spaces and their properties. Some mathematical steps and implications are not fully explored.

evilpostingmong
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Aren't all projections orthogonal projections? What I mean is that let's say there
is a vector in 3d space and it gets projected to 2d space. So [1 2 3]--->[1 2 0]
Within the null space is [0 0 3], which is perpendicular to every vector in the x-y plane,
not to mention the inner product of [0 0 3] and (column)[1 2 0] is 0, which shows
that [0 0 3] is perpendicular to [1 2 0]. What also gets me is that I have seen
a picture of two vectors and one vector projected (orthogonally) onto the
other vector, but both were on the x-y plane, yet orthogonality is shown
by taking the inner product between two vectors and getting 0. I can't see
how 0 can be obtained when both vectors are on the same plane (note that they
were in the positive x positive y quadrant).
 
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Let W be an underlying vector space. Suppose the subspaces U and V are the range and null space of P respectively. Then we have these basic properties:

1. P is the identity operator I on U: \forall x \in U: Px = x.
2. We have a direct sum W = U ⊕ V. This means that every vector x may be decomposed uniquely in the manner x = u + v, where u is in U and v is in V. The decomposition is given by u = Px,\ v = x - Px.

So by 1., oblique projections map the vector back to itself Pv=v right?
 
evilpostingmong said:
Let W be an underlying vector space. Suppose the subspaces U and V are the range and null space of P respectively. Then we have these basic properties:

1. P is the identity operator I on U: \forall x \in U: Px = x.
2. We have a direct sum W = U ⊕ V. This means that every vector x may be decomposed uniquely in the manner x = u + v, where u is in U and v is in V. The decomposition is given by u = Px,\ v = x - Px.

So by 1., oblique projections map the vector back to itself Pv=v right?

(just got up … :zzz:)

Sorry, not following you :redface: …

v is in V, and 1. only applies to U. :confused:
 
Forget what I said. I was mistaken. But thanks for the link, tiny-tim!
 

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