Question on Notation from Spivak's Calculus on Manifolds

Question

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?

Answer 1

$\newcommand{\Dif}{\mathrm D}$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 $\Dif p(a,b)(x,y) \in \mathbb R$.

Answer 2

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}

Answer 3

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...)