This is Example II-2.3 of https://www.scribd.com/read/282634061/Advanced-Calculus-of-Several-Variables?mode=standard
If $F:\mathbb{R}^n\to\mathbb{R}^m$ is linear, then $F$ is differentiable everywhere, and $$dF_\mathbf{a}=F\text{ for all } \mathbf{a}\in\mathbb{R}^n,$$ In short, a linear mapping is its own differential, because $$\lim_{\mathbf{h}\to\mathbf{0}}\frac{F\left(\mathbf{a}+\mathbf{h}\right)-F\left(\mathbf{a}\right)-F\left(\mathbf{h}\right)}{\left|\mathbf{h}\right|}=\lim_{\mathbf{h}\to\mathbf{0}}\frac{\mathbf{0}}{\left|\mathbf{h}\right|}=\mathbf{0}$$ by linearity of $F$.
When I encountered that, I was baffled by it. It's typically difficult to explain confusion after the fact, but here is an attempt. Consider the linear mapping
$$F_{\mathbf{x}}\left(\mathbf{x}\right)=\vec{m}\cdot\mathbf{x}.$$
I was thinking, well, what about something like
$$F_{\mathbf{x}}\left(\mathbf{x}\right)^{\prime}=\left(\vec{m}\cdot\mathbf{x}\right)^{\prime}=\vec{m}^{\prime}\cdot\mathbf{x}+\vec{m}?$$
It took me a long time to sort out what Edwards was intending by the above example.
Here's what I believe to be a more complicated instance of the same difficulty:
https://physics.stackexchange.com/q/528210/117014
How might Edwards's example be better stated, without being too sophisticated?
There is certainly a chance that I missed something obvious in his material leading up to the example, or that my thinking wasn't logical. But I doubt I am the only person to stumble on this or similar situations.
Perhaps the following will better explain the source of my confusion. This is the context I am accustomed to when considering linear mappings at a point. Note that the mapping itself is a function of position. https://www.amazon.com/Theoretical-Physics-Dover-Books/dp/0486652270
And to clear up any misconceptions regarding my respect for Edwards's book, this is my copy.
If $f : \mathbb{R}^n \to \mathbb{R}^m$ is a differentiable map, then its differential is a map: \begin{align} \mathrm{d}f : \mathbb{R}^n &\to L\left(\mathbb{R}^n,\mathbb{R}^m\right) \\ x & \mapsto \mathrm{d}_xf \end{align} Hence, $\mathrm{d}f$ is a map from $\mathbb{R}^n$ to $L(\mathbb{R}^n,\mathbb{R}^m)$. It does not have sense to say "$f$ is equal to its own differential $\mathrm{d}f$": these two functions do not live in the same space.
But if $f$ is linear, what is true is that, for all $x \in \mathbb{R}^n$, then $\mathrm{d}_xf = f$. This is because: $$ \forall x \in \mathbb{R}^n, \forall h \in \mathbb{R}^n, f(x+h) = f(x) + f(h) $$ and consequently $$ f(x+h) = f(x) + f(h) + o(\|h\|) $$ and the very definition of the differential of $f$ at $x$ is the only linear map $\mathrm{d}_xf$ such that: $$ f(x+h) = f(x)+ \mathrm{d}_xf(h) + o(\|h\|) $$ Hence, what makes sense is saying "$\mathrm{d}f$ is a constant map and at each point $x\in\mathbb{R}^n$, $\mathrm{d}_xf = f$". An analogy can be made with linear maps on $\mathbb{R}$, that is if $f(x) = ax$. Then $f'$ is a constant map, with $f'(x)=a$.