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How to extract a subspace 
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#1
Nov2212, 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 3space, then connecting the points draws out some planar shape.
What operation can I apply such that this planar shape is transformed onto the xy axis, so that the shape is exactly the same, but is now described fully by xy 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. 


#2
Nov2312, 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 AA^{T} and take the eigenvectors associated with the 2 largest eigenvalues, denoted by the (3 x 2) matrix E_{2}. Finally, I leftmultiply A by the pseudoinverse of E_{2}: A_{2D} = E_{2}^{+}A which seems to give the desired 2D representation. 


#3
Nov2412, 12:39 PM

P: 108

\mathbf{A}_{2D} &=&\mathbf{E}_{2}^{+}\mathbf{A} \\ &=&\left( \mathbf{E}_{2}^{T}\mathbf{E}_{2}\right)^{1} \mathbf{E}_{2}^{T}\mathbf{A% } \\ &=&\mathbf{E}_{2}^{T}\mathbf{A} \end{eqnarray*}[/tex] However, I still don't really have an intuitive idea for why this works. Perhaps I should reask the question in Linear Algebra. 


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