Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Definition clarifications would be appreciated:

How do I interpret the following ? For $f: R^n\to R^m$, $Df(\vec\xi)(\vec{x})$ in differentiation of a vector function. I know it as a function that maps $\vec x$ to some other vector and this function's definition depends on the point of evaluation $\vec\xi$. But I am still slightly confused. Is $Df(\vec\xi)$ some sort of gradient or something? Because I was told that $Df(\vec\xi)(\vec x) \circ \vec x=0$, where $\circ$ is the dot product.

Also, how does it act on $\vec x$? Just multiplying by $\vec x$? Since surely $Df(\vec\xi)$, having a given point of evaluation takes a "value". And what happens if $Df(\vec\xi)$ contains a vector? Then do we automatically use the dot product to define its "acting on" $\vec x$?

share|improve this question
    
Please clarify your question. In particular, what is $f$? Is it a mapping $f:{\mathbb R}^n \to {\mathbb R}^m$? –  Jeff Nov 22 '11 at 21:06
    
@Jeff: Yes, you're right. Edited. –  vino Nov 22 '11 at 21:12
    
Welcome to MSE vino =) I hope our answers satisfy you. If you appreciate answers you should "upvote" them by clicking the up-arrow on the left of the answer. If you wish to consider an answer as the one you were looking for just click on the check button under those up-down-arrows. Users like upvotes a lot. For a new user your way of asking questions is perfect =) Have fun here! –  Patrick Da Silva Nov 22 '11 at 21:16

2 Answers 2

up vote 1 down vote accepted

Think about differentiation in one variable for a second. The derivative of the one-dimensional vector function (from $\mathbb R$ to $\mathbb R$) gives you the slope at each point. Another way to look at it is that at each point $x \in \mathbb R$ you associate to it a linear transform, $f'(x) : \mathbb R \to \mathbb R$ which for $y$ gives you the number $f'(x)y$ (here this is just standard number multiplication). The transform way is the right way to see things in $\mathbb R^n \to \mathbb R^m$ and furthermore from $V \to W$ where $V$ and $W$ are arbitrary normed vector spaces.

Now when you think of the gradient, say $f : \mathbb R^n \to \mathbb R$, you have a differentiation operator $\nabla : \mathbb R^n \to \mathrm{Hom}(\mathbb R^n, \mathbb R)$ (the latter means all linear applications from $\mathbb R^n$ to $\mathbb R$) which maps every $\mathbb x \in \mathbb R^n$ to $\nabla f(x)$, but instead of seeing $\nabla f(x)$ as a vector, think of it as a linear transform : $\nabla f(x) : \mathbb R^n \to \mathbb R$ and $\nabla f(x)(y) = \nabla f(x) \cdot y$, where on the left you apply the function $\nabla f(x)$ on $y$ and on the right this is the scalar product of both vectors.

When $f : \mathbb R^n \to \mathbb R^m$, exactly the same thing happens : you have a differentation operator $D : \mathbb R^n \to \mathrm{Hom}(\mathbb R^n, \mathbb R^m)$ and $x \in \mathbb R^n$ gets mapped to $Df(x)$ (here the old $\nabla$ became the more general $D$), the matrix with the partial derivatives in it. When you compute $Df(x)(y)$, this is just matrix multiplication : $Df(x)$ is the $m \times n$ matrix and $y$ is a $n \times 1$ column vector.

Note that it isn't true that $Df(\vec{\xi})(\vec x) \cdot \vec x = 0$ in general : think of the case where $f$ is the identity function, thus $Df(\vec{\xi})$ is the identity matrix for every $\vec{x}$, hence $Df(\vec{\xi})(\vec x) \cdot \vec x = \vec x \neq 0$ as long as $\vec x \neq 0$. Also note that the $\circ$ you had put up there is not the scalar product but rather matrix multiplication when $f : \mathbb R^n \to \mathbb R^m$.

Hope that helps,

share|improve this answer
    
A good way to feel this a little better is to do some drawings in the one-dimensional case : I know it helped me a lot at first. –  Patrick Da Silva Nov 22 '11 at 21:17
1  
Thanks, Patrick! –  vino Nov 22 '11 at 21:31

Here I assume that $f$ is a map from some normed vector space $X$ to another normed vector space $Y$.

$Df(y)$ is a linear map from $X$ to $Y$, depending on the point $y \in X$. In the case where $X=R$, it is the map $x \rightarrow f'(y) x$. In the case where $Y=R$, it is a linear form on $X$ and therefore may be represented as $x \rightarrow (u,x)$ for some vector $u$, if $X$ has a scalar product. Then $u$ is the gradient of $f$ at $y$. But in general it is just a linear map, taking the vector $x \in X$ as an argument.

Actually, it is much better to think of $y$ as a point (a "reference position") and of $x$ as a vector (a "perturbation", or a "movement"): you can have for example $f$ defined only on an open subset $U$ (and then $y$ must belong to $U$) but $Df(y)$ must always be defined on the whole $X$.

Also this way of thinking will make it easier if you ever learn about manifolds ;)

share|improve this answer
    
Thanks, Glougloubarbaki! –  vino Nov 22 '11 at 21:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.