Keean Schupke wrote:
So the linear operator is translation (ie: + v)... effectively 'plus' could be viewed as a function which takes a vector and returns a matrix (operator)
(+) :: Vector -> Matrix
Since a matrix _is_ not a linear map but only its representation, this would not make sense. As I said (v+) is not a linear map thus there is no matrix which represents it. A linear map f must fulfill f 0 == 0
But since v+0 == v the function (v+) is only a linear map if 'v' is zero.
I can't see how to fit in your vector extension by the 1-component.
Eh?
Translation is a linear operation no?
No. It's affine, but not linear. As Henning said, to be linear, it must map zero to zero.
Adding vectors translates the first by the second (or the second by the first - the two are isomorphic)... A translation can be represented by the matrix:
1 0 0 0 0 1 0 0 0 0 1 0 dx dy dz 1
So the result of "v+" is this matrix.
No. If the above matrix is M, then:
[x y z w].M = [x+w.dx y+w.dy z+w.dz w]
which isn't a translation.
In the specific case of homogeneous coordinates, where:
h [x y z] = [x y z 1]
h' [x y z w] = [x/w y/w z/w]
then \v -> h'(h(v).M) is a translation, but M isn't itself a
translation.
--
Glynn Clements