Checking if a set of vectors spans a specific subspace 
Is it possible that $\{(1,2,0),(2,0,3)\}$ can span the subspace $U={(r,s,0) | r}$ and $s\in\mathbb{R}$?

I'm trying to wrap my head around vector spaces, but I think yes.
My reasoning is that the span of $\{(x_1, x_2,x_3...x_n)\}=\{k_1x_1,k_2x_2,k_3x_3...k_nx_n\}$ for any set of scalars $k$. You can get $U$ with $1 (1,2,0) + 0(2,0,3)$, so it spans the subspace $U$. Am I correct? Additionally, does the set span $\mathbb{R}^2$ as well, since the matrix's rank is 2?
 A: First question,  no.  For instance $(1,1,0)$ is not in the span.  One way to see this is to put the vectors in a matrix and compute the determinant. 
Second question,  the vectors are linearly independent so span a $2$-dimensional subspace. Thus they span a space isomorphic to $\Bbb R^2$.
A: The span of $(1,2,0)$ and $(2,0,3)$ is $a(1,2,0)+b(2,0,3)=(a+2b, 2a, 3b)$ where $a,b\in\mathbb R$.
If we restrict the span to $U=(r,s,0)$ then $3b=0\implies b=0$.
Therefore, the span becomes $(a,2a,0)=a(1,2,0)$ which is a one dimensional subspace of $U$.
Therefore, $U$ is not spanned by the two vectors. 
A: Vectors that span a space must perforce be elements of that space. The vector $(2,0,3)$ doesn’t lie in $U$—its last coordinate isn’t $0$—so we have to discard it. You could also reject $(2,0,3)$ by considering how you might form $(r,s,0)$ by combining the two vectors: you can’t cancel out that final $3$ with a multiple of $(1,2,0)$, so $(2,0,3)$ can’t be used in a linear combination that produces $(r,s,0)\ne0$.  
That leaves multiples of $(1,2,0)$ to work with, but $(1,0,0)\in U$ isn’t a multiple of this vector, so the answer is no.
A: This problem can be answered by thinking about $U$ geometrically.  Since $U$ is the set of all vectors $(r, s, 0): r,s\ \epsilon\ \mathbb{R}$, we can think about $U$ as being the set of all vectors in $\mathbb{R}^3$ that lie in the $xy$ plane.  Now consider your spanning set, $\{(1,2,0),\ (2,0,3)\}$.  $(1,2,0)$ and all of its scalar multiples definitely lie in the $xy$ plane, because its 3rd element is zero while its first and second elements are nonzero.  So far so good.  But look at the second vector in your spanning set: $(2,0,3)$ and its scalar multiples do not lie in the $xy$ plane, but rather in the $xz$ plane.
Most of the answers that you see here are based on this observation, and rely on selecting a point in the $xy$ plane that is not on the line parametrically defined by $x=t;\ y=2t, t\ \epsilon\ \mathbb{R}$.
We can come to this same conclusion in a more roundabout way by setting up an augmented matrix and taking it down to reduced row echelon form:
$$\begin{bmatrix}
1&2&r\\
2&0&s\\
0&3&0
\end{bmatrix} \sim
\begin{bmatrix}
1&0&r-\frac{2r-s}{2}\\
0&1&\frac{2r-s}{4}\\
0&0&\frac{s-2r}{4}
\end{bmatrix}$$
Since we only want to deal with consistent solutions (we are considering the span of two vectors, after all), then we need $\frac{s-2r}{4} = 0$.  This tells us that $s=2r$ for all of our consistent solutions.  If we subtitute in $2r$ for $s$, then we find that the only place where our two vectors actually generate solutions that lie in $U$ are when the coefficient of $(2,0,3)$ is $0$ and the coefficient of $(1,2,0)$ is some $r\ \epsilon\ \mathbb{R}$.  $U$ generates a plane, while the subspace generated by $(1,2,0)$ is a line.  You can't span a plane with a line, so your spanning set doesn't span $U$.
Your spanning set is linearly independent, however, and so it does generate a plane in $\mathbb{R}^3$, which we can treat like $\mathbb{R}^2$, but it is not $\mathbb{R}^2$ or $U$.
