Question about orthogonal projections.

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SUMMARY

This discussion clarifies the distinction between orthogonal and oblique projections in vector spaces. It emphasizes that orthogonal projections maintain the identity operator property on the range subspace, while oblique projections do not. The conversation also highlights the importance of inner products in determining orthogonality, specifically noting that two vectors can be orthogonal even when they lie in the same plane, provided their inner product equals zero. Key properties of projections are outlined, including the decomposition of vectors into unique components within their respective subspaces.

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  • Understanding of vector spaces and subspaces
  • Familiarity with inner product concepts
  • Knowledge of orthogonal and oblique projections
  • Basic linear algebra principles
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  • Study the properties of orthogonal projections in detail
  • Learn about oblique projections and their applications
  • Explore the concept of direct sums in vector spaces
  • Investigate the role of inner products in determining vector relationships
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Students and professionals in mathematics, particularly those studying linear algebra, as well as anyone interested in understanding vector projections and their applications in various fields such as physics and engineering.

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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