Intersection of two vector subspaces Let $F$ and $G$ be vector subspaces: $F=\operatorname{Span}\{(0,1,0)\}$ and $G=\operatorname{Span}\{(1,1,1)\}$.
What is the intersection between $F$ and $G$? In other words, what is $F\cap G=?$
I thought it would be $\{0\}$ but WolframAlpha tells me otherwise. I try to find the intersection in the following way:


*

*$a(0,1,0)=b(1,1,1)$

*I get the following system.
$a=0$
$a=b$
$0=b$

*The result is
$a=0$ and $b=0$. Thus, $0(0,1,0)=0(1,1,1)=\{0\}$.   
It seems that I am wrong. Please correct me.
P.S. If $F=\operatorname{Span}\{(x,0,z)\}$ and $G=\operatorname{Span}\{(0,0,z)\}$, the intersection would be the $z$ axis that traverses the $xz$ plane of $F$ subspace. But I can't prove that the way I do with the system. How do I prove it?
 A: We've already taken care of your first question in the comments. Here's an easy way to answer your second question: since $G\subseteq F$, it follows immediately that $F\cap G=G$. The reason for $G\subseteq F$: any element $\mathbf{v}\in G$ is of the form $\mathbf{v}=(0,0,z)$ for some $z\in\mathbb{R}$; but then $\mathbf{v}\in F$, because we can represent it as $\mathbf{v}=(0,0,z)=(x,0,z)$, where $x=0$.
A note about your notation in the second question. Although it's not really wrong, but it's very unlikely that you would be given something like $F=\operatorname{Span}\{(x,0,z)\}$. It's more likely that the same subspace would be described as
$$F=\operatorname{Span}\{(1,0,0),(0,0,1)\} \quad \text{or} \quad F=\{(x,0,z)\mid x,z\in\mathbb{R}\}.$$
If $x$ and $z$ run over all reals, the set of all vectors of the form $(x,0,z)$ already describes a vector subspace, so there's no need to take its span. Again, it's not wrong at all to take its span, as it's a meaningful concept for any given set of vectors; but since
$$\operatorname{Span}\{(x,0,z)\mid x,z\in\mathbb{R}\}=\{(x,0,z)\mid x,z\in\mathbb{R}\},$$
it's redundant to say "Span" there, so it's unlikely that a textbook or an instructor would do that.
