Why do we generalise the definition of derivative as a linear transformation instead of as its associated matrix?

Question

The derivative of a function $f:\mathbb{R}^n \to \mathbb{R}^m$ at a point $a \in \mathbb{R}^n$ is defined as a linear transformation $T:\mathbb{R}^n \to \mathbb{R}^m$; the one that gives the best linear aporoximation to the function at that point. However, to me it seems more accurate to define it as its associated matrix, because otherwise we have two different definitions for the derivative of a differentiable function $f:\mathbb{R} \to \mathbb{R}$ at $a \in \mathbb{R}$.
Take for instance $f(x)=x^2$, what is the derivative of $f$ at $a$? Is it $2a$ or is it the linear functional that takes $x$ to $2ax$? Those are two different objects that live in two different spaces

Answer 1

Those may be to different objects, but the spaces (namely $\mathbf R$ and $L(\mathbf R, \mathbf R)$) are isomorphic. By the way, sticking to the matrices doesn't solve the problem, as then we have (sticking to your example) the number $2a \in \mathbf R$ and the map $\{1\} \times \{1\} \to \mathbf R$ (matrices are maps), which maps $(1,1)$ to $2a$. As thankfully, $$ \mathbf R \cong \mathbf R^{1 \times 1} \cong L(\mathbf R, \mathbf R)$$ we do not have a problem.

Sticking to the linear map has the advantage that we can easily generalize that to (e.g. infinite-dimensional) vector spaces that do not come with a canonical choice of a basis, so linear maps cannot be described by a canonical associated matrix.