linear system of equations and the intersection of two spans

Question

Given the vectors:

$u_1=\begin{bmatrix} 1 \\ 2 \\3 \end{bmatrix},u_2=\begin{bmatrix} 2 \\ 3 \\4 \end{bmatrix},v_1=\begin{bmatrix} 1 \\ 1 \\2 \end{bmatrix},v_2=\begin{bmatrix} 2 \\ 2 \\3 \end{bmatrix}$

Let $U=span(u_1,u_2)$ and $V=span(v_1,v_2)$

Let $\alpha,\beta$ be real numbers. We are informed that the linear system

$\begin{bmatrix} 1 & 2 \\ 2 & 3 \\ 3 & 4 \end{bmatrix}x=\begin{bmatrix} \alpha & 2\beta \\ \alpha & 2\beta \\ 2\alpha & 3\beta \end{bmatrix}$

only has solutions if $\alpha+\beta=0$

And lastly we are informed that the vector

$v=\begin{bmatrix} \alpha & 2\beta \\ \alpha & 2\beta \\ 2\alpha & 3\beta \end{bmatrix}$ can be written as $v=\beta\begin{bmatrix} 1 \\ 1 \\1 \end{bmatrix}$

Task: Conclude that the intersection of U and V equals $span(v)$, when $v=\begin{bmatrix} 1 \\ 1 \\1 \end{bmatrix}$

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What I've tried:

I'm not 100% certain how to properly explain it (or if I've even understood it properly).

As I understand it, the information regarding the linear system of equations and solutions only being there if $\alpha+\beta=0$ is the intersection between U and V. And as $\alpha+\beta=0$ you can rewrite the right side to a column vector of only beta's. But if $v$ is a column vector of 1's then that is the same as $\beta=1$ so the right side is equal to the vector $v$. And the left side is all combinations that give, what is now equal to $v$ which the definition of the span.

So I think what I am uncertain of is, what exactly the intersection of U and V in this task is? And if it is the linear system we are given, how to explain that.

Answer 1

Alternative method to find the intersection:

In $3$ dimension, the span of $2$ vectors is a plane $u_1 \times u_2$ would give you the normal direction of the plane spanned by $u_1 \times u_2$.

Similarly for $v_1 \times v_2$.

To find the direction of intersection, just compute $(u_1 \times u_2) \times (v_1 \times v_2)$.

Remark: Your matrix dimension is strange, on the LHS, it seems like a column vector but on the RHS, it seems like a matrix that has more than one column.

I believe you intended to meant

$$\begin{bmatrix} 1 & 2 \\ 2 & 3 \\ 3 & 4 \end{bmatrix}x=\begin{bmatrix} \alpha + 2\beta \\ \alpha + 2\beta \\ 2\alpha+ 3\beta \end{bmatrix}$$

only if $\alpha+\beta=0$. which is basically,

$$\sum x_iu_i = \alpha v_1+\beta v_2$$ only has solution if $\alpha + \beta =0$. The equation describe the interesection of the two planes and they meet when $\alpha+\beta = 0$ of which the right hand side can be simplified (by letting $\alpha = -\beta)$ to show it is a multiple of the all-one vector.

Answer 2

Note that to find the intersection we need to find a subspace which is spanned by both $U$ and $V$, that is we need to solve

$$au_1+bu_2=cv_1+dv_2$$

from here we obtain a system (which correspond to your first) and the solution reveals what subspace the intersection is (in your case the condition $\alpha=-\beta$ lead to a subspace with dimension $1$).