Why study finite-dimensional vector spaces in the abstract if they are all isomorphic to $R^n$? Timothy Gowers asks Why study finite-dimensional vector spaces in the abstract if they are all isomorphic to $R^n$? and lists some reasons. The most powerful of these is probably 

There are many important examples throughout mathematics of infinite-dimensional vector spaces. If one has understood finite-dimensional spaces in a coordinate-free way, then the relevant part of the theory carries over easily. If one has not, then it doesn't.

I mean sure, but what else? Does anyone know examples of specific vector spaces?
 A: 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.
A: A very concrete example of why we should not stop studying something once we have isomorphisms, although going away from vector spaces.
A lot of modern (asymmetric) cryptography is based on discrete logarithm problems in prime order groups. All these groups are isomorphic to $\mathbb{Z}/q\mathbb{Z}$ for some prime $q$. The two most common examples are probably


*

*A subgroup $G$ of the multiplicative subgroup $\mathbb{F}_p^*$ of a finite field $\mathbb{F}_p$, say of prime order $q$;

*The group of rational points $E(\mathbb{F}_p)$ on an elliptic curve $E$, also of prime order $q$.


Note that $G\cong \mathbb{Z}/q\mathbb{Z}\cong E(\mathbb{F}_p)$. As abstract groups, one could think of the discrete logarithm problem to be equivalent in all of them. If we forgot about any structure except for being a group, this is true, as one can show that we need at least $O(\sqrt{q})$ group operations. 
However, in practice there is additional structure. If we consider the set of integers modulo $q$, the discrete logarithm is quite trivial. In the group $G$ it is a bit harder, and if we set $\log_2{q}\approx 1000$ then it is currently infeasible (at least has never been publicly done). If we set $\log_2{q}\approx 256$, then the discrete logarithm problem in $G$ becomes feasible, while it is still very hard in $E(\mathbb{F}_p)$.
Groups can carry a lot of additional structure which we do not want to forget about, and I imagine the same to be true for vector spaces. For example, a finite field is a vector space over some prime order field, but is much more than that.
A: Some examples of finite-dimensional vector spaces that are useful in my field of practice, convex optimization:


*

*real or complex scalars (yes, $\mathbb{R}^1/\mathbb{C}^1$)

*$\mathbb{C}^n$, the space of complex vectors

*the space of $m\times n$ real or complex matrices

*$\mathcal{S}^n$, the space of $n\times n$ symmetric matrices

*$\mathcal{H}^n$, the space of $n\times n$ Hermitian matrices

*other spaces of matrices with specific structure; e.g., Toeplitz matrices, Hankel matrices

*Cartesian products of two or more of the above. For example, a tuple consisting of a symmetric matrix, a vector, and a scalar.


Knowing that these vector spaces are isomorphic to $\mathbb{R}^n$ simplifies the study of convex analysis and convex optimization tremendously, because most of results indeed deriving only once---with appeals to the isomorphism used to quickly apply those results to arbitrary finite-dimensional spaces.
How inconvenient would it be for us to avoid discussion of these other vector spaces? Well, let's consider a simple example. An important function in semidefinite programming is the maximum eigenvalue function on symmetric matrices:
$$f:\mathcal{S}^n\rightarrow\mathbb{R}, \qquad f(X) = \lambda_{\max}(X) = \max_{v\neq 0} (v^TXv)/(v^Tv)$$
This a convex function of its input $X$, and as a result we can build practically useful and tractable optimization problems that incorporate it into objective functions or constraints. (Other matrix functions, such as the minimum eigenvalue, the logarithm of the determinant, etc., arise frequently as well.)
Now, because of the isomorphism, we know that there is another function $g:\mathbb{R}^{n(n+1)/2}\rightarrow\mathbb{R}$ that could be used to represent $f$. In other words, to compute $g(x)$, we'd take the $n(n+1)/2$ values of $x$, arrange them in the lower triangle of a matrix, copy the off-diagonal elements to the upper triangle to symmetrize things, then compute the eigenvalue of that matrix. 
Whew! Do I really have to do that every time I want to work with a function defined on a non-$\mathbb{R}^n$ vector space? I hope not. Sure, sometimes it might be easier to prove a particular technical result if I do; but no, most of the time I want to work in the natural vector space the problem suggests. Knowing how the principles of vector spaces apply to spaces beyond $\mathbb{R}^n$ enables me to avoid constant translation to and fro.
A: 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).
A: 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.
A: 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 $n$
A: 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.
A: 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.
A: 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 $p(x)\mapsto p(x)+p'(x)$.

A: 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
$$
\langle Df(M),N \rangle = MN + NM
$$
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.
A: 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.
