On Wed, 29 Jun 2005, Conal Elliott wrote:
On row & column vectors, do you really want to think of them as {1,...,n)->R? They often represent linear maps from R^n to R or R to R^n, which are very different types. Similarly, instead of working with matrices, how about linear maps from R^n to R^m? In this view, column and row vectors, matrices, and often scalars are useful as representations of linear maps.
We should not identify things which can be mapped bijectively. "1" and 1
are very different, although they may mean the same in a given context.
Yes it is possible to describe linear maps with vectors but vectors are
not linear maps. Namely, x |->
I've played around some with this idea of working with linear maps instead of the common representations, especially in the context of derivatives (including higher-dimensional and higher-order), where it is the view of calculus on manifolds. It's a lovely, unifying approach and combines all of the various chain rules into a single one. I'd love to explore it more thoroughly with one or more collaborators.
I think matrices and derivatives are very different issues. I have often seen that the first derivative is considered as vector, and the second derivative is considered as matrix. In this spirit it is used like x^T * (D2 f)(x) * x but this is only abuse of the common multiplication definitions. A good interpretation and notation should seamless extend to higher derivatives. But the interpretation above does not work in higher dimensions. I like the following type for derivation. derive :: ((i -> a) -> b) -> ((i -> a) -> (i -> b)) Here i is the index type, (i -> a) is the vector type, b is the type the vector function maps to. Its derivative has the same type of argument, but the result is a vector with indices of type i. You see that it is easy to repeat the application of 'derive', just replace b by say i->b. The second derivative yields vectors of type (i -> i -> b). This can be interpreted as matrix because it has two indices. But this is certainly not a matrix which represents a linear mapping as usual, but it is a matrix representing a bilinear form. The only thing we need is a multiplication to reduce one level of indices. mul :: (i -> c) -> (i -> b) -> b Though, what we still need is a general (overloaded?) definition of the scaling of b by c and a sum of b.