Sorry for the delayed response, which has now led to two separate
threads. See below.
> From: Henning Thielemann
> 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 [...]
> I think matrices and derivatives are very different issues. [...]
Of course. My suggestion is to use linear maps instead of vectors or
matrices when the vectors or matrices serve as representations of
linear maps.
> I have often seen that the first derivative is considered as vector,
> and the second derivative is considered as matrix.
I'm guessing you mean for derivatives of functions in R^n->R.
> 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.
What does work I think, for all degrees of derivatives and all
dimensions of vector spaces (and well beyond R^n), is keeping a clear
distinction between linear maps and representations of linear maps.
Linear maps get composed and applied, but certainly not multiplied.
> 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.
This formulation is not much more general than R^n (i.e., {1,..,n}->R).
The vectors are still restricted to "tuples" (indexed by i) of elements
of
the same type, right?
> 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.
I prefer something like the following instead, where "VS s u" means that
u
is a vector space over the scalar field s.
> derive :: (VS s u, VS s v) => (u -> v) -> (u -> LMap u v)
Since VS s (LMap u v), the result of derive may be given back to derive.
Second derivatives then have type u -> LMap u (LMap u v), where as we'd
expect, LMap u (LMap u v) is isomorphic to the type of bilinear maps
from
u to v. By using LMap instead of Matrix, we're not tempted to confuse
(for instance) linear and bilinear maps, just as you pointed out.
- Conal
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