Specific proof: $ span\{a_1,a_2\} \cap span\{b_1,b_2,b_3\} \subseteq \{ \vec0 \} $ Prove that:
$ span\{a_1,a_2\} \cap span\{b_1,b_2,b_3\} \subseteq \{ \vec0 \} $
where:
$ a_1 =\left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 3 \end{array}\right] $
, $ a_2 =\left[\begin{array}{cc} 1 \\ 2 \\ 1 \\ 3 \end{array}\right] $
;
$ b_1 =\left[\begin{array}{cc} 1 \\ 1 \\ 2 \\ 2 \end{array}\right] $
, $ b_2 =\left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 1 \end{array}\right] $
, $ b_3 =\left[\begin{array}{cc} 2 \\ 2 \\ 3 \\ 3 \end{array}\right] $
My attempt:
Let $ v$ be arbitrary. Suppose $ v \in span\{a_1,a_2\} \cap span\{b_1,b_2,b_3\} $ be arbitrary.
Therefore there exists scalars $ \alpha_1,\alpha_2,\beta_1,\beta_2,\beta_3 \in  \mathbb{R} $, such that
v = $ \alpha_1 \left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 3 \end{array}\right] $+ $ \alpha_2 \left[\begin{array}{cc} 1 \\ 2 \\ 1 \\ 3 \end{array}\right] $
and     v= $  \beta_1 \left[\begin{array}{cc} 1 \\ 1 \\ 2 \\ 2 \end{array}\right] $+ $ 
 \beta_2 \left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 1 \end{array}\right] $+ $ \beta_3 \left[\begin{array}{cc} 2 \\ 2 \\ 3 \\ 3 \end{array}\right] $
eventually we get the following matrix which describes the above system of equations:
$ \begin{bmatrix}
1 & 1 & -1 &  -1 & -2  | 0 \\
1 & 2 & -1 &  -1 & -2  | 0 \\
1 & 1 & -2 &  -1 & -3  | 0 \\
3 & 3 & -2 &  -1 & -3  | 0 \\  \end{bmatrix} $
but there are four equations and five variables, so there is one free variable and thus $ v $ is not necessarily zero.  [ Stopped here ]
Now I am stuck because the statement to be proven is a true one but in my solution I saw that $ v $ isn't necessarily zero, so it looks like there is a contradiction ( if so, I'd rephrase the proof to disproof ) but I'm not convinced.
The solution reduces $ span\{b_1,b_2,b_3\}  $ to $ span\{b_1,b_2\}   $, that is because  $ b_3 = b_1 + b_2 $ and I understood it, but I want to prove the statement with my way of not reducing the span a-priori , and essentially how do I deal with the apparent contradiction that v is not always zero vector?
 A: For your system of equations, there are non-zero solutions like
$$\begin{bmatrix}\alpha_1\\\alpha_2\\\beta_1\\\beta_2\\\beta_3\end{bmatrix}
= k\begin{bmatrix}0\\0\\1\\1\\-1\end{bmatrix}$$
But the $\alpha$s are still zero, which means the intersection of the spans is still $\{0a_1+0a_2\}= \left\{\vec 0\right\}$ when calculated from $a_1, a_2$.
Or when calculated from vectors $b_1, b_2, b_3$, the required intersection is ($k\in\mathbb R$)
$$\{kb_1+kb_2-kb_3\} = \left\{k\left(\begin{bmatrix}1\\1\\2\\2\end{bmatrix} + \begin{bmatrix}1\\1\\1\\1\end{bmatrix} - \begin{bmatrix}2\\2\\3\\3\end{bmatrix}\right)\right\} = \left\{k\vec0\right\} = \left\{\vec 0\right\}.$$
A: Your method is perfectly fine, just continue. Your solution will give a non-trivial combination of $(\alpha_1,\alpha_2,\beta_1,\beta_2,\beta_3)$, but if you plug it back into $v$, you will still get $v = 0$.
A: Your mistake lies in deducing from the fact that the system has necessarily a non-zero solution that there is a non-zero vector $v$. In fact,, and since $b_3=b_1+b_2$, if you take $\alpha_1=\alpha_2=0$, $\beta_1=\beta_2=1$, and $\beta_3=-1$, then $(\alpha_1,\alpha_2,\beta_1,\beta_2,\beta_3)$ is a solution of your system. But the $v$ that you et then is$$v=\alpha_1a_1+\alpha_2a_2=\beta_1b_1+\beta_2b_2+\beta_3b_3=0.$$
