# How to extract a subspace

by weetabixharry
Tags: extract, subspace
 P: 96 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.
 P: 96 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.
P: 96
 Quote by weetabixharry A2D = E2+A which seems to give the desired 2D representation.
Of course, this can be simplified as:$$\begin{eqnarray*} \mathbf{A}_{2D} &=&\mathbf{E}_{2}^{+}\mathbf{A} \\ &=&\left( \mathbf{E}_{2}^{T}\mathbf{E}_{2}\right)^{-1} \mathbf{E}_{2}^{T}\mathbf{A% } \\ &=&\mathbf{E}_{2}^{T}\mathbf{A} \end{eqnarray*}$$
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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