How to extract a subspace


by weetabixharry
Tags: extract, subspace
weetabixharry
weetabixharry is offline
#1
Nov22-12, 04:04 PM
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.
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
weetabixharry
weetabixharry is offline
#2
Nov23-12, 09:14 AM
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.
weetabixharry
weetabixharry is offline
#3
Nov24-12, 12:39 PM
P: 96
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%
} \\
&=&\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 re-ask the question in Linear Algebra.


Register to reply

Related Discussions
Prove: sum of a finite dim. subspace with a subspace is closed Calculus & Beyond Homework 1
Dimension of an intersection between a random subspace and a fixed subspace Set Theory, Logic, Probability, Statistics 4
extract h General Math 6
Finding a subspace (possibly intersection of subspace?) Calculus & Beyond Homework 7
a subspace has finite codimension n iff it has a complementary subspace of dim nu Calculus & Beyond Homework 3