TITLE: Why a subspace of a vector space is useful
QUESTION [25 upvotes]: I'm in a linear algebra class and am having a hard time wrapping my head around what subspaces of a vector space are useful for (among many other things!). My understanding of a vector space is that, simplistically, it defines a coordinate plane that you can plot points on and figure out some useful things about the relationship between vectors/points.
I think what I'm curious about is more application of some of these ideas. Such as, is a subspace useful for a reason other than you don't have to look at the entire space something exists in (I guess one way I've been thinking about it is if you want to make a map of a city, you don't necessarily need to make a map of the state it's in) or am I even wrong about that much? Also, even though I feel like I should know this at this point, is if the subspace is linearly independent, is it still a subspace? If it is, what exactly does that describe and/or why is that still useful? If it's not, is it still useful for something?
I think the most difficult part of this for me is I'm having a hard time being able to visualize what exactly we're talking about and I have a hard time thinking that abstractly. I know one or two examples of this might be too specific and doesn't generalize the concept enough, but I think if I have some example to relate back to when applying the idea to new things it might be helpful.

REPLY [2 votes]: One particular application of subspaces that I find illuminating is that they reveal the structure of solutions of linear inhomogeneous equations. Say, you have the following equation:
$$L(x)=f(x)\tag{1}$$
where $x\in X$; $X$ being some vector space and  $L:X\to X$ is a linear operator. Then the most general solution of $(1)$ has the form $x = l+v$ where $l\in U \subset X$, a linear subspace of $X$ spanned by the fundamental solutions of the corresponding homogeneous equation:
$$L(x)=0 \tag{2}$$
and $v\in X \setminus V$. For example, if $L=\cfrac{d^2}{dt^2}+\omega^2$, $U$ would be spanned by $\{e^{-\omega t},e^{\omega t}\}$
Once the general structure is clear, if say, you are lucky to know the fundamental system of $(2)$ then you only have to find, perhaps, guess, just one more solution not spanned by the fundamental system to construct the general solution.
This theory is most frequently illustrated on the examples of linear ordinary differential equations, however, I like this example from the domain of the recurrence equations.
The problem of expressing function $f(x)=x^2$ in terms of cubic splines on the entire real axis leads to the following equation:
$$\lambda_{j}+4\lambda_{j+1}+\lambda_{j+2}=24(j+3)^2$$
Here $L=I+4E+E^2$ where $I$ is the identity and $E$ is the shift operator. The general solution of the corresponding homogeneous relation
$$\lambda_{j}+4\lambda_{j+1}+\lambda_{j+2}=0$$
is given by the following expression:
$$\lambda^h_j=\alpha\left(-2-\sqrt{3}\right)^{j}+\beta\left(-2+\sqrt{3}\right)^{j}$$
Which is a linear subspace spanned by $\{\left(-2-\sqrt{3}\right)^{j},\left(-2+\sqrt{3}\right)^{j}\}$
It can be found, for example, using generating functions or via the characteristic equation using ansatz $\lambda_j=r^j$. Not surprisingly it depends on 2 arbitrary constants, as it takes 2 initial terms, $\lambda_0$ and $\lambda_{-1}$ to define the sequence from the three-term recurrence, so the subspace is 2-dimensional. 
Applying the general ideas from the linear systems we deduce that in order to obtain the general solution of the inhomogeneous recurrence we have to add a particular solution to the expression above.
Since the RHS is the quadratic polynomial it makes sense to look for the particular solution in the form:
$$\lambda^p_j=aj^2+bj+c$$
Substituting this into the original recurrence and gathering together the powers of $j$ and equating them on both sides we obtain the general solution:
$$\lambda_j=\alpha(-2-\sqrt{3})^{j}+\beta(-2+\sqrt{3})^{j}+4j^2+16j+\frac{44}{3}$$
Using initial data $\lambda_0=\frac{44}{3}$ and $\lambda_{-1}=\frac{8}{3}$ to determine the unknown constants we finally get
$$\lambda_j=4j^2+16j+\frac{44}{3}$$
which is the solution of the original problem.