A question about Span in Linear Algebra So I have this question about Span that I'm not sure how to do:
Let $\DeclareMathOperator{\Span}{Span}u_1 =\begin{bmatrix}
1\\
2\\
             3    
\end{bmatrix},
u_2 =\begin{bmatrix}
2\\
3\\
4
\end{bmatrix},
v_1 =\begin{bmatrix}
1\\
1\\
2
\end{bmatrix},
v_2 =\begin{bmatrix}
2\\
2\\
3
\end {bmatrix}$, in the real vector space $\mathbb{R}^3$.
Let $U=\Span(u_1,u_2)$ and $V=\Span(v_1, v_2)$.
Show that $U\cap V$ is equal to $\Span(v)$ where $v = \begin{bmatrix}
1\\
1\\
1
\end{bmatrix}$.
So I have already shown that all elements in $V$ can be written as $\begin{bmatrix}
\alpha +2\beta\\
\alpha + 2\beta\\
2\alpha + 3\beta
\end{bmatrix}$ (1) 
and that $\begin{bmatrix}
1 & 2\\
2 & 3\\
3 & 4
\end{bmatrix}
x = \begin{bmatrix}
\alpha +2\beta\\
\alpha + 2\beta\\
2\alpha + 3\beta
\end{bmatrix}$ only has a solution if $\alpha = \beta$ (2). I have also shown that $\begin{bmatrix}
\alpha +2\beta\\
\alpha + 2\beta\\
2\alpha + 3\beta
\end{bmatrix}$ is equal to $\beta\begin{bmatrix} 
1\\ 
1\\ 
1 
\end{bmatrix} = \beta v$ when $\alpha = \beta$. (3)
I know that I have to show that $U\cap V ⊆ \Span(v)$ and that $\Span(v) ⊆ U\cap V$.
My answer so far:
$U\cap V ⊆ \Span(v):$
Since we know that (3) and (2) is true then every element in $V$ can be written as a linear combination of $v$ which means that $V=\Span(v)$. Since $U\cap V⊆ V \Rightarrow U\cap V ⊆ \Span(v)$.
However I'm not sure how to show that $\Span(v) ⊆U\cap V$. Any help would be appreciated!
Edit:
With help from the comments I think I have found the answer:
Since $Span(v)=\{\alpha v|\alpha \in \mathbb{R}\}$ then $\alpha v=\alpha \cdot (−\gamma)\cdot v_1 + \beta \cdot v_2$, where $\beta=−\gamma$ and $\alpha \cdot v=\alpha \cdot (−\gamma)\cdot u_1 + \alpha \cdot \delta \cdot u_2$ where $\delta=−\gamma$ so $Span(v)⊆U\cap V$.
Are both these arguments correct?
 A: Hint: Can you show that $v \in U \cap V$?
A: Suppose a vector belongs to the intersection, so it can be written as
$$
\alpha\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}
+
\beta\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}
=
\gamma\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}
+
\delta\begin{bmatrix} 2 \\ 2 \\ 3 \end {bmatrix}
$$
Let's then look at the null space of the matrix
$$
\begin{bmatrix}
1 & 2 & -1 & -2 \\
2 & 3 & -1 & -2 \\
3 & 4 & -2 & -3
\end{bmatrix}
$$
A standard elimination yields
$$
\begin{bmatrix}
1 & 2 & -1 & -2 \\
2 & 3 & -1 & -2 \\
3 & 4 & -2 & -3
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & -1 & -2 \\
0 & -1 & 1 & 2 \\
0 & -2 & 1 & 3
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & -1 & -2 \\
0 & 1 & -1 & -2 \\
0 & 0 & -1 & -1
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & -1 & -2 \\
0 & 1 & -1 & -2 \\
0 & 0 & 1 & 1
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & 0 & -1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 1
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 1
\end{bmatrix}
$$
Thus we get $\alpha=-\delta$, $\beta=\delta$, $\gamma=-\delta$.
Choosing $\delta=1$ and $\gamma=-1$, we get the vector
$$
v=-u_1+u_2=-v_1+v_2=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\in U\cap V
$$
The equations imply that the dimension of $U\cap V$ is $1$, so $U\cap V=\operatorname{Span}\{v\}$. If you want a confirm of this, the elimination implies that the rank of $[u_1\ u_2\ v_1\ v_2]$, which equals the rank of $[u_1\ u_2\ {-v_1}\ {-v_2}]$ is $3$ and Grassmann's formula yields $\dim(U\cap V)=\dim U+\dim V-\dim(U+V)=2+2-3=1$.
