#### evilpostingmong

Aren't all projections orthogonal projections? What I mean is that lets 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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#### evilpostingmong

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?

#### tiny-tim

Homework Helper
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

v is in V, and 1. only applies to U.

#### evilpostingmong

Forget what I said. I was mistaken. But thanks for the link, tiny-tim!

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