Is span $\{[1,0],[0,1]\}$ a vector space? I can't figure this out. I would think that it is a vector space because it has the zero vector, and it seems to me to be closed under addition and scalar multiplication. But $[1,0]+[0,1] = [1,1]$ which is definitely not in the set. Can someone clarify? Is the span a vector space, or not?
 A: $\mathrm{span}\left\{\begin{bmatrix}
1\\0
\end{bmatrix},\begin{bmatrix}
0\\1
\end{bmatrix}\right\}$ is the set of linear combinations of the two vectors $\begin{bmatrix}
1\\0
\end{bmatrix}$
and $\begin{bmatrix}
0\\1
\end{bmatrix}$.
In other words, $\mathrm{span}\left\{\begin{bmatrix}
1\\0
\end{bmatrix},\begin{bmatrix}
0\\1
\end{bmatrix}\right\}=\left\{a\begin{bmatrix}
1\\0
\end{bmatrix}+b\begin{bmatrix}
0\\1
\end{bmatrix}:a,b\in\mathbb{R}\right\}$.
Thus, $\begin{bmatrix}
1\\1
\end{bmatrix}$ is in the span.
In particular,  the two vectors $\begin{bmatrix}
1\\0
\end{bmatrix}$
and $\begin{bmatrix}
0\\1
\end{bmatrix}$ form the standard basis of the vector space $\mathbb{R}^2$. Thus, they span $\mathbb{R}^2$ and so do any two linearly independent vectors in $\mathbb{R}^2$.
A: Yes, the (linear) span is a vector space. By definition it is the smallest vector space that contains all the elements in the set. In particular it will contain all linear combinations of those elements (and will in fact contain exactly all linear combinations that can be formed with those elements).
So in your example, if we consider $\{[1,0],[0,1]\} \subset \mathbb{R}^{2}$ we will get $$span\{[1,0],[0,1]\} = \mathbb{R}^{2}$$
because for every $[x,y] \in \mathbb{R}$ we can find a linear combination of $[1,0],[0,1]$ that represents $[x,y]$: 
$$x\cdot[1,0] + y\cdot[0,1] = [x,y].$$
With similar arguments we get 
get $$span\{[1,0]\} = \{[x,0] \in \mathbb{R}^{2}: x\in\mathbb{R}\}.$$
Wikipedia is also a good starting point for more on linear spans.
