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How to extract a subspace

by weetabixharry
Tags: extract, subspace
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Nov22-12, 04:04 PM
P: 108
I have a (3 x N) matrix of column rank 2. If each column is treated as a point in 3-space, then connecting the points draws out some planar shape.

What operation can I apply such that this planar shape is transformed onto the x-y axis, so that the shape is exactly the same, but is now described fully by x-y coordinates in a (2 x N) matrix?

I feel like there should be a (2 x 3) matrix that would do this, but I can't figure out what it should be. (I have a hunch that I'm looking for a mapping that is isometric and conformal... some kind of rotation?). Also, I'd like to be able to generalise to higher dimensions.
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Nov23-12, 09:14 AM
P: 108
I think I may have found a solution, but would appreciate any further discussion... since my understanding is rather weak. I basically started thinking about pseudoinverses and figured I wanted a pseudoinverse of something orthonormal, to avoid distorting my shape (?).

Let's call my (3 x N) matrix A. To get an orthonormal basis spanning the (2D) column space, I eigendecompose AAT and take the eigenvectors associated with the 2 largest eigenvalues, denoted by the (3 x 2) matrix E2.

Finally, I left-multiply A by the pseudoinverse of E2:

A2D = E2+A

which seems to give the desired 2D representation.
Nov24-12, 12:39 PM
P: 108
Quote Quote by weetabixharry View Post
A2D = E2+A

which seems to give the desired 2D representation.
Of course, this can be simplified as:[tex]\begin{eqnarray*}
\mathbf{A}_{2D} &=&\mathbf{E}_{2}^{+}\mathbf{A} \\
&=&\left( \mathbf{E}_{2}^{T}\mathbf{E}_{2}\right)^{-1} \mathbf{E}_{2}^{T}\mathbf{A%
} \\
However, I still don't really have an intuitive idea for why this works. Perhaps I should re-ask the question in Linear Algebra.

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