Why study finite-dimensional vector spaces in the abstract if they are all isomorphic to R^n?

If you decided that you are only going to call "vector space" those of the form $\mathbb R^n$, then you find yourself in the position that now subspaces are no longer vector spaces.

The cheeky answer is that we would not know that all finite-dimensional vector spaces are isomorphic to $\mathbb{R}^n$, if we did not study finite-dimensional vector spaces in their own right. In mathematics, we generally like to use as few assumptions as possible and to isolate them in the form of axioms.

See https://en.wikipedia.org/wiki/Examples_of_vector_spaces for examples of vector spaces that seem very different from those found in the world of geometry. Function spaces are good examples; the space $X \rightarrow \mathbb{R}$ of all continuous functions from a given topological space $X$ to $\mathbb{R}$ is a natural example.

For any integer $k$, the set $M_k$ of complex-differentiable functions $f$ defined on the upper-half plane $\{x+iy: \, y > 0\}$ that satisfy the equations $$f(z+1) = f(z), \; \; f(-1/z) = z^k f(z)$$ and have limit $\lim_{y \rightarrow \infty} f(iy) = 0$ is a vector space over $\mathbb{C}$.

Two specific elements of $M_k$ include the functions $$E_4(z) = 1 + 240 \sum_{n=1}^{\infty} \sigma_3(n) e^{2\pi i n z} \in M_4$$ and $$E_8(z) = 1 + 480 \sum_{n=1}^{\infty} \sigma_7(n) e^{2\pi i nz} \in M_8.$$ Here, $\sigma_k(n)$ is the divisor sum $\sum_{d | n} d^k$.

Assuming that $E_4 \in M_4$ it is rather easy to show that $E_4^2 \in M_8.$

It can be proved that $M_8$ is one-dimensional, so $E_4^2$ is a multiple of $E_8$. Comparing constant coefficients tells you that they must be equal, and comparing the others gives you the formula $\sigma_7(n) = \sigma_3(n) + 120 \sum_{m=1}^{n-1} \sigma_3(m) \sigma_3(n-m).$

For example $$\sigma_7(2) = 1 + 2^7 = 1 + 2^3 + 120$$ and $$\sigma_7(3) = 1 + 3^7 = 1 + 3^3 + 120(1+2^3 + 1 + 2^3).$$

A lot of vector spaces like this show up in number theory. They are typically finite-dimensional but working out a basis is pretty hard (certainly harder than showing that they are finite-dimensional).

Consider an analogous question:

Why consider finite sets in the abstract if they're all isomorphic to $\{1,\ldots,n\}$ for some $n$?

• Because there could be names for the elements that are more natural for a given situation than $1,\ldots,n$, e.g. we may want to refer to $$\{\text{red},\text{green},\text{blue}\}$$ instead of $$\{1,2,3\}\text{ where we agree that 1 stands for red, 2 for green, 3 for blue}$$

• In general, names for elements are not always important

• There are many subsets of a set of the form $\{1,\ldots,n\}$ for some $n$ that are not themselves sets of the form $\{1,\ldots,n\}$ for some $n$

All Hilbert spaces are unitarily isomorphic to some $\ell^2(E)$, but that does not mean we don't study Hilbert spaces.

The important thing is finite-dimensional vector spaces may carry additional structure (such as the space of polynomials) which we are interested in.

Another thing is people occasionally encounter non-trivial subspaces of $\mathbb R^n$, such as a line or a plane. In this case the theory of finite-dimensional vector spaces becomes useful.

Consider $V$, the space of polynomials of degree less than $3$, which is isomorphic to $\mathbb{R}^3$ via $$ a_0+a_1x+a_2x^2\mapsto \begin{bmatrix} a_0-a_2\\ a_0-a_1-2a_2\\ a_2 \end{bmatrix} $$ Would you recognize what is the linear map $f\colon V\to V$ whose matrix with respect to this isomorphism is $$ \begin{bmatrix} 2 & -1 & 0 \\ 1 & 0 & -2 \\ 0 & 0 & 1 \end{bmatrix} $$ without looking at the spoiler below?

The linear map can be easily expressed as $p(x)\mapsto p(x)+p'(x)$.

There are lots of ways that viewing a finite dimensional vector space in its "intrinsic form" is useful, especially in calculus. Here's a simple example. Consider the space of $n \times n$ matrices with real coefficients $\mathbb{M}^n$. We could just regard this as the space $\mathbb{R}^{n^2}$ and do calculus as per usual in higher dimensions.

This does not work so well when we want to consider derivatives of simple maps. For instance, let's look at the map $f: \mathbb{M}^n \to \mathbb{M}^n$ given by $f(M) = M^2$. This is a smooth map, and we can compute that the first derivative satisfies $$ <Df(M),N> = MN + NM $$ for all $N \in \mathbb{M}^n$. In this context we view $Df(M) \in \mathcal{L}(\mathbb{M}^n)$, i.e. as a linear map from $\mathbb{M}^n$ to itself. If we insist on throwing away this abstract formulation then we only have $\mathbb{R}^{n^2}$ to work with, in which case there is no product structure. We will still find that this map $f$, viewed as a map from $\mathbb{R}^{n^2}$ to itself, is smooth, but the formula for the derivative will be significantly worse, and even challenging to write down.

I would claim we study them precisely because they are isomorphic to $\mathbb{R}^n$. What do I mean?
I mean that since we are already familiar with $\mathbb{R}^n$, we can use this intuition to understand vector spaces in general, and once we do that, we can generalize the concepts to other less-intuitive objects (such as infinite-dimensional vector spaces) while carrying over our understanding.

Even studying $\mathbb{R}^n$ in the abstract is useful; sweeping irrelevant details under the rug makes the theory a lot cleaner. You can see what's going on more easily if you're not bogged down in coordinates!

This can also be true for actual computation too. Vector calculus and matrix algebra are important tools for doing calculations; one must be able to form a concept of such things as mathematical objects in their own right, rather than merely being arrays of scalar quantities.