Basis of the intersection of two spaces

[Zero2Infinity]

Homework Statement

Consider two vector spaces $A=span\{(1,1,0),(0,2,0)\}$ and $B=\{(x,y,z)\in\mathbb{R}^3 s.t. x-y=0\}$. Find a basis of $A\cap B$.

I get the solution but I also inferred it without all the calculations. Is my reasoning correct

Homework Equations

linear dependence definition.

The Attempt at a Solution

First of all I look for a basis of $B$, which is $(1,1,0),(0,0,1)$.
Now, the complete method for finding the basis of $A\cap B$ is the following one. I set
\begin{equation}
a_{1}(1,1,0)+a_{2}(0,2,0)=b_{1}(1,1,0)+b_{2}(0,0,1)
\end{equation}
or, equivalently (the coefficients are unknown so the sign is not that important as I can name $-b_{i}=b_{i}$),
\begin{equation}
a_{1}(1,1,0)+a_{2}(0,2,0)+b_{1}(1,1,0)+b_{2}(0,0,1)=(0,0,0)
\end{equation}
By solving the linear system I get:
\begin{equation}
\begin{cases} a_{1}+0+b_{1}+0=0\\
a_{1}+2a_{2}+b_{1}+0=0 \\ 0+0+0+b_{2}=0 \end{cases}
\Leftrightarrow \begin{cases} a_{1}=-b_{1}\\
a_{1}+2a_{2}+b_{1}=0 \\ b_{2}=0 \end{cases}
\end{equation}
$a_{1}=-b_{1},a_{2}=b_{2}=0$. Thus, the generic vector $\textbf{v}\in A\cap B$ can be written as the linear combination of the basis of $A$ with $a_{1}$ and $a_{2}$ as coefficents.
\begin{equation}
\textbf{v}=-b_{1}(1,1,0)+0\cdot(0,2,0)=-b_{1}(1,1,0)
\end{equation}
The conclusion is that the searched basis is $(1,1,0)$.

Now, I know this is the correct procedure but is it correct to say that, since I know a basis of $A$ and a basis of $B$, the basis of $A\cap B$ are the common vectors to both basis, if there is at least one?
This way I could have immediately said that $\mathcal{B}_A:(1,1,0),(0,2,0), \mathcal{B}_B:(1,1,0),(0,0,1)\Rightarrow \mathcal{B}_{A\cap B}:(1,1,0)$.

 

[RUber]

I think the answer is yes...but you have to be careful. If you are able to write a basis vector for one space in terms of basis vectors from another, then clearly, those vectors will be in the intersection.
However, your space A is simply the xy-plane. It could have been written with other basis vectors that would not be so simple as [1,1,0] to pick out. Consider [1,0,0] and [0,1,0].
If you have two 2-D spaces, as above, your intersection is either all of the 2D space (if A = B) , spanned by a single vector, or empty (parallel planes).

 

[PeroK]

You could get the dimension of $A \cap B$ from considering the dimensions of the subspaces. If the second basis vector for $B$ is not in $A$, then the intersection must be the 1D subspace spanned by the common basis vector.

Geometrically, two planes through the origin, if not equal, intersect in a line through the origin.