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

  • $\begingroup$ "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. $\endgroup$ – anomaly May 31 '17 at 7:38
  • $\begingroup$ @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 $\mathbb{C}$. $\endgroup$ – EpsilonDelta May 31 '17 at 10:11
  • $\begingroup$ @anomaly By the way I do not know what tangent bundles are, but will revisit your point when I study them. $\endgroup$ – EpsilonDelta May 31 '17 at 10:12
  • $\begingroup$ 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. $\endgroup$ – anomaly May 31 '17 at 17:51

Both are correct and equivalent ( as long as we identify any linear map via its matrix-representation in base of standard vectors).

First one is more general, it can be used in any normed vector space, but you cannot use second one in infinite dimension vector space, since the size of a matrix can't be infinite.

  • $\begingroup$ 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.) $\endgroup$ – EpsilonDelta May 31 '17 at 9:03
  • $\begingroup$ @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. $\endgroup$ – Red shoes May 31 '17 at 16:04
  • $\begingroup$ By Matrix-Representation I mean $T(h)=Ah$, and we identify $T$ by $A$ and vice versa. $\endgroup$ – Red shoes May 31 '17 at 16:23

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