The explanation of "What's that extra 1 for?" in the column representation of 3-d coordinates (x y z 1) could benefit from mentioning that translation—moving things—is not a linear transformation (the origin is not mapped to itself) but an affine transformation. Therefore, you cannot represent translation in 3-d space with a 3x3 matrix. What you can do, though, is embed that 3-d space within a 4-d space fixed at some coordinate on its 4th dimension, typically w=1. Then, a translation in the original 3-d space can be represented as a linear transformation in the 4-d space and thus can also be represented by a 4x4 matrix multiplication. So the extra 1 is actually what allows all common 3-d operations, including translation, to be done via linear algebra and thereby harness the brutal power of matrix multiplication on modern computing devices.
nice intuition there, this comment prompted me to consider a simpler example, 2d embedded within 3d. does the 2d plane (embedded in 3d) go through the 3d origin (where 0 maps to 0) and is thus a linear transformation in 3d but a 2d affine transform in 2d? it feels like this is the case?
No, the 2d x-y plane in your example cannot pass through the 3d space’s origin because that would imply that you fixed the z coordinate at zero. The plane must be fixed at some nonzero z because you need to be able move x and y values by some scaled version of z to make translation happen. If z is zero, that scheme does not work.
Consider a transformation f where we wish to move x-y coordinates s units to the right. In 2d, we could express it as:
f(x, y) = (x + s, y)
But that transformation is affine not linear. There is no way to generate the value s as a linear combination of the inputs x and y. So, our workaround is to embed the x-y plane into 3d space at z=1. Then we can move (x,y,1) points in that plane s units to the right using this transformation:
f(x, y, z) = (x + s*z, y, z)
This new transformation is linear: it maps (0,0,0) to itself. But it maps our embedded 2d plane's origin (0,0,1) to (s,0,1), shifting it right by s units, as we want.
The matrix form of that transformation is:
[[1 0 s]
[0 1 0]
[0 0 1]]
The same scheme would work if we had embedded the plane at any fixed z=r for nonzero r. We would only have to rescale the s in the matrix to s/r. Again, however, if r=0, this scheme will not work, as 1/r has gone to infinity.
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Consider a transformation f where we wish to move x-y coordinates s units to the right. In 2d, we could express it as:
f(x, y) = (x + s, y)
But that transformation is affine not linear. There is no way to generate the value s as a linear combination of the inputs x and y. So, our workaround is to embed the x-y plane into 3d space at z=1. Then we can move (x,y,1) points in that plane s units to the right using this transformation:
f(x, y, z) = (x + s*z, y, z)
This new transformation is linear: it maps (0,0,0) to itself. But it maps our embedded 2d plane's origin (0,0,1) to (s,0,1), shifting it right by s units, as we want.
The matrix form of that transformation is:
The same scheme would work if we had embedded the plane at any fixed z=r for nonzero r. We would only have to rescale the s in the matrix to s/r. Again, however, if r=0, this scheme will not work, as 1/r has gone to infinity.[deleted] · · focus · HN ↗
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