Why is it useful to write a vector as a finite (or infinite) linear combination of basis vectors? I'm working on a project in an applied mathematics course and a professor asked me a basic question:
What is useful about writing any element of a vector space in terms of a possibly infinite linear combination of basis vectors?
I didn't really have an answer. So I'm wondering what's so useful about it? Why can't we just come up with an alternate expression for any arbitrary element and use that?
 A: There's quite a few useful properties,  but probably the #1 nice thing about a basis is that you can define your linear operators only on what they do to basis elements.    Then the fact that they are linear means you immediately know what they do to any arbitrary element of the set (We normally don't do infinite sums of basis elements,  except in a hilbert space  since if $$x=\sum _{i=1}^nm_ie_i$$ where the $e_i$'s are your basis elements and your $m_i's$ come from the field,  linearity gives us 
$$f(x)=f(\sum _{i=1}^nm_ie_i)=\sum _{i=1}^nf(m_ie_i)=\sum _{i=1}^nm_if(e_i)$$
thus,  determining what the values of $f$ are on each $e_i$ uniquely determines what $f$ does to arbitrary elements.
Note this requires $f$ to be a LINEAR operator, but that's mostly what we deal with in vector spaces
A: A lot of this way of doing things started with solving partial differential equations. For example, suppose you want to solve the heat equation
$$
                     u_{t}(t,x) = c^{2}u_{xx}(t,x), \\
                         0 \le t < \infty,\;\;\; 0 \le x \le 1 \\
                       u(t,0) = 0,\;\; u(t,1) = 0, \\
                       u(0,x) = h(x).
$$
The operator $Lf = f''$ on the right has a basis of eigenfunctions
$$
                           \{ \sin(n\pi x) : n=1,2,3,\cdots \}
$$
This is an orthogonal basis. So you can write the solution $u(t,x)$ as
$$
                        u(t,x) = \sum_{n=1}^{\infty}a_n\sin(n\pi x).
$$
You can think of fixing $t$ on the left, expanding in $x$ and obtaining coefficients $a_n$ for each fixed $t$. So the coefficients $a_n$ will depend on $t$, but that's okay. When you plug this form back into the equation, you get
$$
                     \sum_{n=1}^{\infty}a_{n}'(t)\sin(n\pi x) = -c^{2}\sum_{n=1}^{\infty}a_{n}(t)n^{2}\pi^{2}\sin(n\pi x)
$$
Because the $\sin$ terms are a basis, then you conclude that the coefficients of the sin functions on the left must match those on the right, one for one:
$$
                         a_{n}'(t) = -c^{2}n^{2}\pi^{2}a_{n}(t).
$$
Now you have a simpler equation which is an ODE in $t$:
$$
                           a_{n}(t) = a_{n}(0)e^{-c^{2}n^{2}\pi^{2}t}.
$$
The final solution has the form
$$
        u(t,x) = \sum_{n=1}^{\infty}a_{n}(0)e^{-c^{2}n^{2}\pi^{2}t}\sin(n\pi x).
$$
The $a_{n}(0)$ is determined by matching the initial condition $u(0,x)=h(x)$, which is the initial temperature distribution:
$$
                         h(x) = \sum_{n=1}^{\infty}a_{n}(0)\sin(n\pi x).
$$
Because the $\sin$ functions on the right are an orthogonal basis, you can multiply by $\sin(m\pi x)$ and integrate over $[0,1]$ to isolate each coefficient:
$$
           \int_{0}^{1}h(x)\sin(m\pi x)dx = a_{m}(0)\int_{0}^{1}\sin(m\pi x)^{2}dx = \frac{1}{2}a_m(0).
$$
Now you get the complete solution:
$$
           u(t,x) = \sum_{n=1}^{\infty}\left(2\int_{0}^{1}h(x)\sin(m\pi x)dx\right)e^{-c^{2}n^{2}\pi^{2} t}\sin(nx).
$$
The expressions in parentheses on the right are just constants determined by the initial temperature data.
You needed a basis in $x$ to do all of this. But the key is that you are using a basis that diagonalizes the operator on the right so that operating on the basis with $\frac{\partial^{2}}{\partial x^{2}}$ becomes a scalar multiplication problem by the eigenvalue:
$$
        \frac{d^{2}}{dx^{2}}\sin(n\pi x) = -n^{2}\pi^{2}\sin(n\pi x).
$$
That's key. You don't want just any old basis; you want one that simplifies the operator on the right. That way you end up with coefficient equations for the expansion where the operator on the right has been reduced to mulitiplication by a scalar. You end up with infinitely many coefficient equations, but they're ODEs that are much simpler to solve.
A: A short answer: the basis vectors acts like a coordinate system that can describe the entire vectorial space.
A: Sometimes there is a linear transformation that you want to understand. If you can find a basis such that the linear transformation has a simple effect on the basis vectors (like it simply scales them, for example) then this helps a lot to understand what the linear transformation does to an arbitrary vector (which can be written as a linear combination of the special basis vectors).
