On Mon, 21 Jan 2013 21:47:11 -0800 (PST), email@example.com wrote:
>Hi, All: > >I'm just curious about wether orthogonalprojection generalizes to cases such as >this: > >Say we have a 1-D subspace L (i.e., a line thru the origin) in R^3 , and >let q=(x,y,z) be a point in R^3 which is not on the line. Then I don't see >how to project q orthogonally onto L; I can see how to project q into a >2-D subspace P : the projection of q into P is the ortho complement, and >every vector in P is in the orthogonal complement of the ortho projected >line. But, the same is not the case with q and L. Sorry for the rambling; >my question is then actually: > > If L is a 1-D subspace of R^3, and q=(x,y,z) is a point not on L. Can we >define the orthogonal projection of q into L, or do we need to have a plane >P (as subspace) , to define an ortho projection of q?
Yes. The projection is p, defined by two conditions:
(i) p lies on L
(ii) q - p is orthhogonal to L.
Or: If V is any subspace of R^n, let W be the orthogonal complement: W is the space of all w such that
v.w = 0 for all v in V
(or in another standard notation, <v, w> = 0).
Then every x in R^n has a uniquue decomposition
x = v + w
where v is in V and w is in W; now v is the orthogonal projection of x onto V.
Yes, you could also define this in terms of certain planes, but that's sort of missing the point.