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.
ive been thinking -- if i have a 3x3 matrix [1 0 q; 0 1 r; 0 0 1], when dotted with x, those rows are planes in 3d with normals n1=(1 0 q), n2=(0 1 r) and n3=(0 0 1). plane 3 is parallel to the x-y plane and 1 unit up. plane 1 is tilted by q and parallel to the y-axis, plane 2 is tilted by r and parallel to the x-axis. the intersection of those planes (i.e. the solution x) when calculating Ax=b gives an output vector b sitting in 3d space at b=(x+qz, y+rz, 1*z). since we always specify z=1 we have b=(x+q, y+r, 1). in 3d this is a linear shear because we are translating proportionally by z but because z always equals 1 in this case we effectively get a translation in 2d.
so as the other 2 helpful commenters also just said: 3d shears using linear algebra degenerate to 2d affine transformations when z=1 (or w in 4d)
re-read this a day later and ofc it still makes sense to me but really most of it is barking up the wrong tree, focusing on solutions (which we already know) rather than outputs. if i could delete it i would but oh well.
tmoertel · · focus · HN ↗
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[0]: Here’s a visual: <a href="https://gunn-gatm.github.io/textbook/gatm.pdf#page=28" rel="nofollow">https://gunn-gatm.github.io/textbook/gatm.pdf#page=28
tmoertel · · focus · HN ↗
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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so as the other 2 helpful commenters also just said: 3d shears using linear algebra degenerate to 2d affine transformations when z=1 (or w in 4d)
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