Re: Projection in non-othogonal space

On 01.10.2006 20:55, c186282 wrote:
On Sun, 01 Oct 2006 09:14:53 -0700, Tonico wrote:

c186282 wrote:
I have a space spanned by a set of non-orthogonal (not even
normalized) basis vectors (e_{i}). I do have the metric for the space
(\eta) that I have successfully used for a variety of
calculations. Now I need a non-orthogonal projection operator. Does
something like that exist? What should I look for? It should have the
property that if I have a vector V=\sum a_i e_i on which I apply the
projection operator P_{j} it returns just a_j e_j. The normal
orthogonal projection operator gives me some other junk along with
what I want. I'm currently looking into if I can write this other junk
in the form of an operator and subtract it form my orthogonal
projection operator.

Thank you
Well, you want projection...on what?? You just specify a subspace to
project on.
You say you want your "projection" to return the j-th component of a
given vector written according to the basis you have, times the vector
e_j...well, then define P to be so on that basis and extend by
linearity. P then will be the projection on Span<e_j>.
Regards
Tonio

What you have described is the standard orthogonal projection
operator. If I use the standard orthogonal projection operator to
project one of my basis vectors say e_{1} on to one of the other
basis vectors say e_{2} I get a non-zero result because my metric is
not diagonal. I need a non-orthogonal projection operator which would
give me a zero result.

If I take my vector V and project out the components of V that lays
along the basis vectors e_{i} then add up all of my projections I
should recover V. This is obvious in an orthogonal space with an
orthogonal projection operator. But I need a non-orthogonal projection
operator for my non-orthogonal space.

In matrix notation (w.r.t. the basis e_1, ..., e_n) let P_i be a square
matrix with every element 0 but the element a_[ii}=1. It represents a
projection on U_i=span{e_i} along V_i=span{e_j|j<>i}. Since, for any i,
the direct sum U_i (+) V_i is equal to the original vector space, every
vector x has a unique representation a e_i + v with a scalar a and v in
V_i; in this case the projection of x on U_i along V_i is a e_i.

Note that given a projection there are two sub vector spaces U and V
such that the vector space decomposes into a direct sum of U and V: one
on which is projected (U) and the other which is exactly the kernel of
the projection (V). Given U and V as before then there is a unique
projection on U along V. For orthogonal projections U determines V
uniquely as the orthogonal complement of U, hence only U is needed in
this case.
.

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