Raul Miller wrote: > This implies, to me, that in the context of any linear map, > there must be some value that corresponds to 0, and some > other value which corresponds to 1. (If this is not the case, > then there must be some way of defining addition and > multiplication which does not conform to the peano postulates?) > > That wikipedia page also asserts: > Differentiation is a linear map from the space of all differentiable > functions to the space of all functions. > > I think I can use the constant zero function for 0, but I am having > a problem figuring out what 1 would be. Does anyone know? >
You are absolutely right that the zero element in the vector space V of linear maps is the zero function. The element corresponding to 1 is not in V, but in the field (R) over which V is a vector space. You have to be able to add in V, and be able to multiply something in R by something in V. Saying that differentiation is a linear map D:V->V then means things like D(f+g)=(Df)+(Dg), D(kf)=k D(f). It says nothing about multiplying functions together. Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
