Vector valued function How can I determine if a function lies within the span of a vector valued function? 
For an example does $\begin{bmatrix} 2\\ 1\end{bmatrix}$ lie in the span of: 
$$a(x) = \begin{bmatrix} 1\\ x\end{bmatrix}$$
$$b(x) = \begin{bmatrix} x\\ 1\end{bmatrix}$$
$$c(x) = \begin{bmatrix} x\\ 2x\end{bmatrix}$$
 A: You could approach this problem in an unorganized brute force sort of way like: Suppose $$[2\;1]^T = r[1\;x]^T+s[x\;1]^T+t[x\;2x]^T.$$ This leads to equations: $2=r+sx+tx$ and $1=rx+s+t2x$. Thus $2=r+(s+t)x$ and $1=s+(r+2t)x$. So (equating coefficients) one has $2=r$, $0=s+t$, $1=s$, and $0=r+2t$. Therefore, $r=2$, $s=1$, and $t=-1$. Thus "Yes" $[2\;1]^T=2a(x)+b(x)-c(x)$.
Instead a coordinate approach is a bit more organized (although a little clunkier). Such an approach would be more successful (than brute force) if dealing with a larger problem.
The collection of all functions $\{f:\mathbb{R}\to\mathbb{R}^{2\times 1}\}$ form an infinite dimensional vector space. But your functions lie in a finite dimensional subspace, for example: $W = \left\{ \begin{bmatrix} ax+b\\cx+d \end{bmatrix} \;:\; a,b,c,d \in \mathbb{R} \right\}$. By choosing a basis like $$\beta=\left\{\begin{bmatrix} 1\\0 \end{bmatrix},\begin{bmatrix} x\\0 \end{bmatrix},\begin{bmatrix} 0\\1 \end{bmatrix},\begin{bmatrix} 0\\x \end{bmatrix}\right\}$$
we can write everything in coordinates. So 
$$ \left[ \begin{bmatrix} 2\\1 \end{bmatrix} \right]_\beta = \begin{bmatrix} 2 \\ 0 \\ 1 \\ 0 \end{bmatrix} \quad
\left[ \begin{bmatrix} 1\\x \end{bmatrix} \right]_\beta = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} \quad
\left[ \begin{bmatrix} x\\1 \end{bmatrix} \right]_\beta = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix} \quad
\left[ \begin{bmatrix} x\\2x \end{bmatrix} \right]_\beta = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 2 \end{bmatrix}$$.
Then asking if $[2\;1]^T$ is in the span of $\{a(x),b(x),c(x)\}$ amounts to asking if the final column of the following matrix lies in the span of the first three columns (i.e. is the final column a non-pivot column?):
$$ \begin{bmatrix} 1 & 0 & 0 & : & 2 \\ 0 & 1 & 1 & : & 0 \\ 0 & 1 & 0 & : & 1 \\ 1 & 0 & 2 & : & 0 \end{bmatrix}$$
This matrix row reduces to 
$$ \begin{bmatrix} 1 & 0 & 0 & : & 2 \\ 0 & 1 & 0 & : & 1 \\ 0 & 0 & 1 & : & -1 \\ 0 & 0 & 0 & : & 0 \end{bmatrix}$$
So the final column is not a pivot column and thus $[1\;2]^T$ lies in the span of $\{a(x),b(x),c(x)\}$. Moreover, the RREF tells us that the desired linear combination is $2a(x)+b(x)-c(x)$ (as before).
A: Maybe you mean to ask,
Does the point $(2,1)$ fall on the line parametrized by $a(x),b(x)$ or $c(x)$ if any?
$$ a(x) = (2,1)=(1,x) \ \ \Rightarrow \ \ 2=1, 1=x $$
$$ b(x) = (2,1)=(x,1) \ \ \Rightarrow \ \ 2=x, 1=1 $$
$$ c(x) = (2,1)=(x,2x) \ \ \Rightarrow \ \ 2=x, 1=2x $$
Obviously $2=1$ is hard to solve, the equation for $b(x)$ shows $b(2)=(2,1)$ and the equation for $c(x)$ gives $x=2$ and $x=1/2$ which is a contradiction. Only $b$ contains $(2,1)$ in its image.
Following Bill's suggestion: you could find $c_1,c_2,c_3$ such that
$$ (2,1) = c_1(1,x)+c_2(x,1)+c_3(x,2x) $$
This equation has to hold for all $x$ if $(2,1)$ is in the span of $a,b,c$ as vector-valued functions.
