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Suppose a 3D configuration of points is given, $X\in\mathbb{R}^{n\times 3}$, and a non-zero matrix $Q\in\mathbb{3\times 2}$, with orthonormal columns. Now, suppose a mapping to 2D is obtained as $$Y=XQ.$$ I'm interested in the intuitive interpretation of the result $Y\in\mathbb{R}^{n\times 2}$. Suppose a plane $P$ is given; does the above mapping imply the result of overlaying the configuration with plane $P$ ? I introduced the plane since I'm not sure about the notion of the viewpoint in this mapping. Does the viewpoint enter the above? I'm also interested in the orthonormality condition of the direction vectors. What would happen if they are not orthogonal?

For instance, choosing $Q$ to contain the first two columns of a $3\times 3$ identity matrix would yield a solution where each point is mapped to a point that is obtained at the intersection of a line passing through that point and $xy$ plane, such that the the line is orhogonal to the $xy$ plane. Given such an interpretation, how could I cast the above?

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The way you composed your question makes me think that you have a) a very good mathematical intuition b) a maybe not so broad theoretical background. This makes it a bit hard to choose the right words for an answer. I'll try it below, but if you have any questions feel free to ask. –  Simon Markett May 17 '12 at 15:30
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Let us start with your example: $Q$ has the first two canoncial base vectors as columns. What you obtain is the projection to the $xy$-plane as you have correctly identified in your question: You take a perpendicular of any point to the $xy$-plane and your image is the point where the perpendicular hits this plane. If you want you can picture this as follows: You sit in the three dimensional space such that you look straight at the $xy$-planen and all you see is the shadow of your configuration on this plane.

What happens if you choose other orthonormal vectors? Maybe I start with a bit earlier. When you write a linear transformation $A$ of $\mathbb R^3$ to itself as a matrix (I'll come to your case in a second), then the three columns are precisely the image of the three canoncial base vectors $e_i$ ($1$ at the $i^{th}$ position and $0$ on the other two positions, $1\leq i\leq 3$). When you want to compute the image of an arbitrary vector you represent it in terms of the three base vectors $(x,y,z)=xe_1+ye_2+ze_3$, take the image of the base vectors again and recombine: $A(x,y,z)=xAe_1+yAe_2+zAe_3$. What happens now if you have an orthonormal matrix, i.e. a $3\times 3$ matrix where all three columns are orthonormal? The images of your three base vectors (which have length one and are orthonormal) are again of length one and orthonormal. So your transformation is just a rotation (and possibly a reflection). Another way to picture this is that you rotate your coordinate system (where you might mess up the orientation).

Now you want a linear map $Q:\mathbb R^3\to\mathbb R^2$ such that the two columns are orthonormal. Note that you can add a third vector at the end such that all three are orthonormal. Let us call this matrix $Q'$. Now our interpretation from above is valid and $Q'$ is just a rotation (+ maybe reflection). And afterwards you just forget the third coordinate, or you project to the plane spanned by the images of the first two base vectors - precisely as in the initial example.

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In other words, if I fix the coordinate system (so, I do not rotate the coordinate system), but actually rotate (and maybe reflect the configuration), and then take the perpendicilar of each point to the $xy$ plane. The points where the perpendiculars hit the $xy$ plane correspond to the configuration $Y$. Am I right? (I'm deliberately repeating, because I want it to be consistent with the initial intuition). –  user506901 May 17 '12 at 15:57
    
Nearly: You project to the plane given by the two vectors in the matrix Q. Alternatively you can first project to the $xy$-plane and then rotate this plane. –  Simon Markett May 17 '12 at 16:01
    
Well, how should I then interpret your "So your transformation is just a rotation (and possibly a reflection). Another way to picture this is that you rotate your coordinate system (where you might mess up the orientation)." As I now see it, the matrix Q, if one more column is added, corresponds to a rotation matrix, which is orthonormal. Or? Maybe you misinterpreted my comment above. I meant that $Y$ corresponds to a orthogonal projection of rotated/reflected $X$ onto $xy$ plane. –  user506901 May 17 '12 at 19:26
    
I think we just talk past each other. Yes, $Q$ which one extra column is a rotation/reflection whcih is the same ar being orthonormal. But: after applying thir rotation/reflection you project on the plane spanned by the first two columns, which is not necessarily the $xy$-plane. –  Simon Markett May 18 '12 at 10:16
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