Let $F:\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$ be a linear function, i.e., $$F(\alpha x + \beta y) = \alpha F(x) + \beta F(y)$$
Suppose you are given a routine that returns $F(x)$ given any $x\in\mathbb{R}^{n}$. How would you use this routine to determine a matrix $A\in\mathbb{R}^{m\times n}$ such that $F(x) = Ax$ for all $x\in\mathbb{R}^{n}$?
If we have a standard basis $e_1,e_2,\ldots,e_n$ we have a unique $x$ where $$x = \sum_{i}^{n}e_ix_i \ \ \forall x\in\mathbb{R}^{n}$$ By linearity we have $$F(x) = F(\sum_{i}^{n}e_i x_i) = \sum_{i}^{n}F(e_i)x_i = \sum_{i}^{n}a_i x_i = Ax$$ where $F(e_i) = a_i = A e_i$ determines the $i$-th column of $A$
This course is called foundations of computational math, although the above math is more on the linear algebra spectrum. I am just curious to see whether this approach makes sense and whether anyone has a better idea of what the question is asking.
You will get the matrix as follow:
First specify the basis on each space, say, the standard one on each: $${\Bbb R}^n={\rm gen}\{e_1,e_2,...,e_n\}$$ and $${\Bbb R}^m={\rm gen}\{e'_1,e'_2,...,e'_m\}$$ Then, map each $e_i$ to get $Te_i$, but since $Te_i$ lives in ${\Bbb R}^m$ hence $$Te_1=a^1{}_1e'_1+a^2{}_1e'_2+\cdots +a^m{}_1e'_m,$$ $$Te_2=a^1{}_2e'_1+a^2{}_2e'_2+\cdots +a^m{}_2e'_m,$$ $$\cdots$$ $$Te_n=a^1{}_ne'_1+a^2{}_ne'_2+\cdots +a^m{}_ne'_m.$$
So the matrix you are seeking is $$\left(\begin{array}{ccccc} a^1{}_1 & a^1{}_2 &...&a^1{}_n\\ a^2{}_1 & a^2{}_2 &...&a^2{}_n\\ \vdots&&\ddots\\ a^m{}_1 & a^m{}_2 &...&a^m{}_n \end{array}\right).$$