On Fri, 8 Jul 2005, Keean Schupke wrote: > Henning Thielemann wrote: > > >On Fri, 8 Jul 2005, 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?
Today: +----------------+ | The Linear Map | +----------------+ If F is a field, V a set of vectors, and (F,V,+,*) a vectorspace then a function f of V -> V is linear if: for all x and y from V f (x+y) == f x + f y for all k from F and x from V f (k*x) == k * f x Lemma If f is a linear map, then f 0 = 0. Proof: For any v from V it is v+(-1)*v = 0. f 0 = f (v+(-1)*v) = f v + f ((-1)*v) = f v + (-1) * f v = 0 Theorem If (v+) is a linear map then v == 0. Proof: (v+0) == 0 (conclusion from the lemma) => v == 0 8-] > (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. But the vectors I can add to v have one component less than necessary for multiplication with this matrix. Indeed you can map all v's with three components to an affine sub-space of the 4-dimensional space, namely to those vectors with a 1 in the last component. But you are no longer working with three dimensional vectors but with four-dimensional ones. Again: isomorphy is not identity. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
