6 ms·
Shearings cannot be represented in this way.
by wolfgke 8y ago
Shearings cannot be represented in this way.
- munchbunny 8y agoThey can if you add a dimension to the space. That's one of the reasons 3d graphics use 4d vectors and matrices.
- bonoboTP 8y agoYou're talking about translation.
- munchbunny 8y agoNo, I wasn't, but I did confuse the terms. Shear can be done without the extra dimension. Skew transforms require the extra dimension, as does translation.
- bonoboTP 8y agoWhat do you mean by skew? A perspective transformation (homography)? I'm not sure it's standard terminology.
- rocqua 8y agoNotably though, shearings are very 'rare'. Any pertubation will make a shearing no longer a shearing. At least, if I remember correctly.
- blt 8y agoSame for the the orthogonal matrices, or the diagonal matrices, or the symmetric matrices, or the unit determinant matrices, or the singular matrices ... They are all sets of Lebesgue measure zero.
- rocqua 8y agoOrthogonal, diagonal, symmetric, and unit-determinant matrices are all sub-groups though, which makes them 'more special' then all shearing matrices. Singular matrices are special in the sense that they keep the matrix monoid from being a group. My category theory isn't strong enough to characterize it, but this probably also has a name. Edit: I think the singular matrices are the 'kernel' of the right adjoint of the forgetful functor from the category of groups to the category of monoids. Though I must admit a lot of that sentence is my stringing together words I only vaguely know.
- throwawaymath 8y agoYes they can. This follows from singular value decomposition. Let S be the matrix representation of a shear transformation. There exist rotation matrices R, B and a diagonal matrix D such that S = RDC, where C is the transpose of B. D is the matrix representation of a scaling transformation and R, B are the matrix representations of rotation transformations. Since S is a product of rotation and scaling matrices, its corresponding linear transformation is a composition of rotations and scalings. It would ordinarily be weird to represent shear transformations using rotations and scalings because shear matrices are elementary. But it checks out.
- wolfgke 8y agoOK, point taken. I considered "scaling" in a less general sense (scalar multiple of the unit matrix), while you want to allow arbitrary diagonal entries. My definition is to my knowledge the common one in linear algebra textbooks because in yours, the feasible maps depend on the chosen basis. EDIT: To state my point more clearly: in textbooks, "scaling" is the linear map that is induced by the "scalar multiplication" in the definition of the vector space (that is why both terms start with "scal").
- magicalhippo 8y agoReminds me of the old "hack" to use three shear transformations to rotate an image. The idea being that a shear is relatively much faster on weaker CPUs, relative to doing a "proper" (reverse mapping) rotation. A nice write-up can be found here: https://www.ocf.berkeley.edu/~fricke/projects/israel/paeth/rotation_by_shearing.html https://www.ocf.berkeley.edu/~fricke/projects/israel/paeth/r...