Is the derivative a 'mapping' or a 'value'?

The derivative (or 'total derivative') a function $f : \mathbb{R}^m \rightarrow \mathbb{R}^n$ at some point $a \in \mathbb{R}^m$ is usually defined as a linear mapping $Df_a : \mathbb{R}^m \rightarrow \mathbb{R}^n$ which obeys $$\lim_{h \to 0} \frac{f(a+h) - f(a) -Df_a(h)}{\Vert h \Vert_2} = 0.$$ This can be generalized to functions between Banach spaces. In the special case $m = n = 1$ this would mean that the derivative of $f$ at $a$ is a mapping from $\mathbb{R}$ to $\mathbb{R}$.

Now some authors (e.g. Munkres, Analysis on Manifolds) define the derivative as the $n \times m$ matrix $J_a(f)$ which obeys $$\lim_{h \to 0} \frac{f(a+h) - f(a) -J_a(f) h}{\Vert h \Vert_2} = 0.$$

My question, therefore, is as follows:

What is the 'correct' definition of the derivative of $f$ at $a$? Is it the mapping $Df_a$ or the matrix $J_a(f)$?

You will agree that a linear map and its matrix (wrt some bases) are not the same thing. If $f : \mathbb{R} \rightarrow \mathbb{R}$, then the definition of the derivative as taught in elementary calculus and real analysis is Munkres' definition ($f'(a)$ is a $1 \times 1$ matrix). However if we have to use the first definition above, then we must define the derivative as the linear map $y \mapsto f'(a)y$.

Related question:

Are derivatives linear maps?

Derivative as a linear transformation

• "You will agree that a linear map and its matrix (wrt some bases) are not the same thing." > With a basis fixed, there's a canonical correspondence between a linear map and its corresponding matrix. There's no reason to pretend that $\lambda$ and the map $x \to \lambda x$ are fundamentally different things. In fancier terms, the tangent bundle over $\mathbb{R}$ is a trivial line bundle. – anomaly May 31 '17 at 7:38
• @anomaly Your comment led me to an interesting discovery (of course it might be known already). If I understand correctly, you are saying that there is a bijective map, say $g$, from $\mathbb{R}$ to the set of all linear maps from $\mathbb{R}$ to $\mathbb{R}$ (let us call this set $H$) which preserves all usual operations defined on $\mathbb{R}$. Since $\mathbb{R}$ is a field, $g$ must be a 'field isomorphism', so that $H$ is also a field. Indeed I verified that $H$ is a field. Interesting because I had never known fields which were not subsets of $\mathbb{R}$ or $\mathbb{C}$. – EpsilonDelta May 31 '17 at 10:11
• @anomaly By the way I do not know what tangent bundles are, but will revisit your point when I study them. – EpsilonDelta May 31 '17 at 10:12
• Incidentally, this is where the determinant comes from. For a vector space $V$, the space of linear maps $\operatorname{End}(V)$ acts on the wedge product $\Lambda^n V$ by $f(v_1 \wedge \cdots \wedge v_n) = f(v_1) \wedge \cdots \wedge f(v_n)$. For $n = \dim V$, the space $\Lambda^n V$ has dimension $1$, and by (a) definition, $f$ acts on it by multiplication by $\det f$. In particular, this gives you the fact that $\det (fg) = (\det f)(\det g)$ for free. – anomaly May 31 '17 at 17:51

• But the matrix of a linear map is not unique and depends on the choice of the bases. So if I use a non-standard basis for $\mathbb{R}^m$ or $\mathbb{R}^n$ the matrix of the derivative map will change. However the underlying mapping will remain unaffected. So wouldn't it make more sense to always define the derivative as a mapping? (As opposed to in real analysis, where we are taught to think of the derivative at a point as just a real number.) – EpsilonDelta May 31 '17 at 9:03
• @EpsilonDelta What do you mean? Even you choose different bases to represent $[D_a f]_B$, you never change the vectors $D_a f(e_k)$, $k=1,2,..m$, where $e_k$ are standard vectors. When you represent $D_a f (h)=Ah$, then $A$ is uniquely determined, and it is independent from choice of base. Two definitions coincide and $Ja(f)=A$. Note that for linear transformation $T :R^m \rightarrow R^n$, the only matrix representation (i.e., $T(h)=Ah$) is when $A=[T]_B$, where $B$ is the ordered base consist of standard vectors. – Red shoes May 31 '17 at 16:04
• By Matrix-Representation I mean $T(h)=Ah$, and we identify $T$ by $A$ and vice versa. – Red shoes May 31 '17 at 16:23