Lucid Dreamer said:
Perhaps I should have been more clear in my question. I am looking at dR(w)/dw where w is a vector and R(w) is a scalar function of a vector variable. In this case, jambaugh post seems to make sense as if w is a row vector, than dR(w)/dw is a column vector. Thanks for all your help!
Let me further elaborate as to
why you get a dual vector. To generalize the idea of a derivative to vector calculus use differentials as local coordinates in the local linear approximation to a function.
Given y = f(x) then the local linear approximation is:
dy = f'(x)dx \quad\quad\text{ that is to say } y+dy = f(x)+f'(x)dx \approx f(x+dx)
Allow either x or y or both to be vectors here. You then have the derivative as a linear operator valued function of x, said linear operator maps dx type objects to dy type objects. We can define it as a limit of a difference quotient if we are careful to avoid what looks like division by a vector:
\mathbf{dy} = f'(\mathbf{x})\mathbf{dx} \equiv \lim_{h\to 0} \frac{ f(\mathbf{x}+h\mathbf{dx}) - f(\mathbf{x})}{h}
Here h is a real number and the difference quotient is well defined provided the range and domain of f are vector spaces (so we can add elements and multiply by scalars h and 1/h). That includes of course the case of 1-dimensional vectors we call scalars.
The nature of f'(x) then is as a linear operator and we have the following cases:
- dy,dx both scalars: the linear operator f' is just multiplication by a number.
- dy vector and dx scalar: the linear operator f' maps scalars to vectors and so is multiplication by a vector.
- dy scalar and dx vector: the linear operator f' maps vectors to scalars and so is a dual vector (linear functional).
- dy vector and dx vector:f' is a full blown linear operator representable by a matrix.
Now as I mentioned, I prefer to use column vectors as coordinate vectors and row vectors for dual vectors so that in matrix format the action of f' is left multiplication by a matrix.
e.g.
u = f(x,y);\quad du = f'(x,y)\left[\begin{array}{c}dx \\ dy \end{array}\right] = ( {\scriptsize{\frac{\partial u}{\partial x}\quad \frac{\partial u}{\partial y}}} ) \left[\begin{array}{c}dx \\ dy \end{array}\right]
\left[\begin{array}{c}u \\ v\end{array}\right] = \mathbf{F}(x,y,z) ; \quad \left[\begin{array}{c}du \\ dv \end{array}\right] <br />
=\mathbf{F}' (x,y,z)\left[\begin{array}{c}dx \\ dy \\ dz \end{array}\right]= \left[\begin{array}{c c c} \scriptsize{ \frac{\partial u}{\partial x}} &\scriptsize{ \frac{\partial u}{\partial y}} &\scriptsize{ \frac{\partial u}{\partial z}}\\<br />
\scriptsize{ \frac{\partial v}{\partial x}} &\scriptsize{ \frac{\partial v}{\partial y}} &\scriptsize{ \frac{\partial v}{\partial z}}\end{array}\right] \left[\begin{array}{c}dx \\ dy \\ dz \end{array}\right]
Note the derivative of a scalar valued function of vectors is just the gradient and in resolving change of variables one gets the correct form of the gradient by preserving the differential relationship: du = \nabla u \cdot \mathbf{dr}.
2nd Note: Here I'm using primed notation just to match up with single variable calc. notation. More traditionally one uses the \nabla operator. F' \to \nabla F, or one may use a Leibniz type notation.
3rd Note: Reversing my use of rows vs columns would allow one to better express directional derivatives and the differential operator, e.g.:
\mathbf{d} =\left( \begin{array}{ccc}dx & dy & dz \end{array}\right) \left[ \begin{array}{c} \partial_x \\ \partial_y \\ \partial_z\end{array}\right]
A final note. Here I am treating the differentials simply as local coordinates and not as differential forms per se and not as infinitesimals. In full blown differential geometry of manifolds we can't add points and differentials become cotangent vectors while the partial derivatives become tangent vectors. What we call "vector" and what we call "dual vector" is relative. The distinction between "tangent vector" and "co-tangent vector" is not.
