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In Spivak's book Calculus on Manifolds, he uses some notation that I'm not entirely sure of. On pg 20 when discussing the derivative of a function $p:R^2 \rightarrow R$ and $p(x,y) = x \cdot y$, he writes

$$ Dp(a,b)(x,y) = bx + ay $$

and concludes that $p'(a,b) = (b,a)$.

This is where I am a little confused. At one point, he states that $Dp(a,b)$ represents the derivative of $p$ at the point $(a,b)$ (pg 16). He goes on to say that it's common to represent the derivative in this case as the Jacobian, which would be a $1x2$ matrix in this case, as indicated by $p'(a,b) = (b,a)$ (I mean, it's not exactly a matrix, but the point is there). However, when defining $Dp(a,b)(x,y)$, what returns is a scalar value. So I'm not sure how to interpret this.

In exercise 2-12 (pg 23), he asks us to show for a bilinear function $f:R^n \times R^m \rightarrow R^p$ that $Df(a,b)(x,y) = f(a,y) + f(x,b)$. Well, now here, the derivative is in $R^p$, so clearly not a scalar. How am I supposed to interpret this? If it is at the point $(a,b)$, then why are both $x$ and $y$ present on the right hand side? What is being expressed in this case?

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    $\begingroup$ $D_p(a,b)$ is a linear transformation from $\mathbb R^2$ to $\mathbb R$. It wants a vector in$\mathbb R^2$ as input. It gives you a scalar as output. $\endgroup$
    – littleO
    Commented Jan 24, 2017 at 5:09

3 Answers 3

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$\newcommand{\Dif}{\mathrm D}$Think of $\Dif p$ as a mapping from $\mathbb R^2$ to the space of linear maps $\operatorname{Hom}(\mathbb R^2,\mathbb R)$, so when you evaluate it at $(a,b)$, the result is a linear map $\Dif p(a,b) \in \operatorname{Hom}(\mathbb R^2,\mathbb R)$. Evaluating this linear map at $(x,y)$ gives a scalar $\Dif p(a,b)(x,y) \in \mathbb R$.

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  • $\begingroup$ So...I have no idea what the Hom function is. A quick search said something about module homomorphisms, and I'm already completely lost. You wouldn't happen to have another explanation would you? More or less, anything involving algebra can be safely assumed to be beyond my grasp, as I typically focus in more applied areas $\endgroup$
    – cnolte
    Commented Jan 24, 2017 at 4:56
  • $\begingroup$ @burnmfburn $\operatorname{Hom}$ refers to the space of linear maps. $\endgroup$ Commented Jan 24, 2017 at 4:58
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    $\begingroup$ (for some reason I can't tag you?) Ok, after doing some more reading, it's not as scary of a thing as I was thinking. This makes a lot more sense, I'm going to keep playing with it until I get it down. Thanks for your help $\endgroup$
    – cnolte
    Commented Jan 24, 2017 at 5:11
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By definition, $Dp(a,b)$ is a linear transformation from $R^2$ to $R$. (It is a linear approximation of $p$). Thus $Dp(a,b)(x,y)$ should be an element of $R$.

$p'$ is standard matrix representation of $Dp(a,b)$. Thus it is 1 by 2 matrix.

In exercises 2-12, $Df(a)$ should be a linear transformation from $R^n \times R^m$ to $R^p$. Thus $Df(a,b)$ is linear transformation from $R^n \times R^m$ to $R^p$ and $Df(a,b)(x,y)$ is an element of $R^p$.

(I assume you already understand this by now, But since there is no accepeted answer, so...)

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  • $\begingroup$ This is the best answer. Here you clarify the one thing that is overlooked which is, to every linear map $T: \mathbb{R}^n \to \mathbb{R}^m$ their is an associated matrix $A$. Hence we make no difference between $A,T$. Now we just let $T = Df(p)$ and $A = \textbf{Jacobian}(Df(p)) = f'$. $\endgroup$ Commented Mar 26, 2017 at 14:55
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By using the bilinearity of $f(x,y)$, one has \begin{eqnarray} Df(a,b)(x,y)&=&\lim_{t\to\infty}\frac{f(a+tx,b+ty)-f(a,b)}{t}\\ &=&\lim_{t\to\infty}\frac{[f(a+tx,b+ty)-f(a,b+ty)]+[f(a,b+ty)-f(a,b)]}{t}\\ &=&\lim_{t\to\infty}\frac{tf(x,b+ty)+tf(a,y)}{t}\\ &=&f(x,b)+f(a,y). \end{eqnarray}

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