I am currently studying Introduction to Tensor Analysis and the Calculus of Moving Surfaces by Pavel Grinfeld. Exercise 3 of Chapter 1 Why Tensor Calculus? proceeds as follows:
Exercise 3. Show that the expression
$$\nabla T = \dfrac{1}{\sqrt{\mathbf{i} \cdot \mathbf{i}}} \dfrac{\partial{F}}{\partial{x}} + \dfrac{1}{\sqrt{\mathbf{j} \cdot \mathbf{j}}} \dfrac{\partial{F}}{\partial{y}} \tag{1.6}$$
yields the same result for all rescalings of Cartesian coordinates.
The provided solution proceeds as follows:
Let $a, b \in \mathbb{R}$, and consider the re-scaled coordinate basis
$$\mathbf{i}' = (a, 0), \\ \mathbf{j}' = (0, b),$$
where each of the above vectors is taken to be with respect to the standard basis for $\mathbb{R}^2$.
In studying linear algebra, I never gained an understanding of what it meant to have "one vector with respect to another vector". I would greatly appreciate it if people would please take the time to explain it, within this context.
One must be careful to distinguish between a vector per se and its coordinates relative to some basis, which are the coefficients of the unique linear combination of basis vectors that equals the given vector. For a real $n$-dimensional vector space, the latter is always an element of $\mathbb R^n$ regardless of the type of object the vectors themselves are. If the vector space is also $\mathbb R^n$, there’s a tendency to blur this distinction (as well as the distinction between points and vectors) in the interest of brevity, which can cause confusion if you’re not solid on the underlying concepts.
With that in mind, read $\mathbf i'=(a,0)$ as “let $\mathbf i'$ be the vector with coordinates $(a,0)$ relative to the standard basis of $\mathbb R^2$, i.e., $\mathbf i'=a\mathbf i$.”