Show that $Span(x_1, x_2, x_3) ∩ Span(x_2, x_3, x_4) = Span(x_2, x_3)$. Hi I'm having trouble with this homework question:
Let $V$ be a vector space over a field $F$. Let $x_1, x_2, x_3, x_4$ be four linearly independent vectors in $V$ . Show that $Span(x_1, x_2, x_3) ∩ Span(x_2, x_3, x_4) = Span(x_2, x_3)$.
So I know that since the $4$ vectors are linearly independent then none of them can be expressed as a linear combination of the other. This would also mean that $x_1, x_2, x_3$ can't be expressed as a linear combination of the others and the same for $x_2, x_3, x_4$.To solve this I guessed that I would take a random vector in the $Span(x_2, x_3, x_4)$ and set it equal to the other random vector chosen in the $Span(x_1, x_2, x_3)$. Then I'd solve these two equations to show that it reduces down to a vector that's equal to a linear combination of $x_2$ and $x_3$ but this didn't get me anywhere.
Any help is very much appreciated. 
 A: An alternative way.
The inclusion $\supseteq $ is obvious. Since $\dim\operatorname{Span}(x_1,x_2,x_3,x_4)$ is finite, you can use Grassman's formula: $$\dim(X+Y)+\dim(X\cap Y)=\dim X+\dim Y$$
with $X=\operatorname{Span}(x_1,x_2,x_3),\ Y=\operatorname{Span}(x_4,x_2,x_3)$ and $X+Y=\operatorname{Span}(x_1,x_2,x_3,x_4)$ (easily proved to be this).
It yields that $X\cap Y$ is a subspace of dimension $2$ which contains $\operatorname{Span}(x_2,x_3)$.
A: For convenience, set $U = \operatorname{span}{(x_{1},x_{2},x_{3})}$ and $W = \operatorname{span}{(x_{2},x_{3},x_{4})}.$ Clearly $\operatorname{span}{(x_{2},x_{3})}\subseteq U\cap W,$ since $x_{2},x_{3}\in U\cap W.$ Therefore, if we can prove that $U\cap W\subseteq \operatorname{span}{(x_{2},x_{3})},$ then we'll be done.
Let $v\in U\cap W.$ Then, since $v\in U,$ there exist $a_{1},a_{2},a_{3}\in F$ such that
$$v = a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3};$$
and since $v\in V,$ there exist $b_{2},b_{3},b_{4}\in F$ such that
$$v = b_{2}x_{2}+b_{3}x_{3}+b_{3}x_{4}.$$
Hence we have $a_{1},a_{2},a_{3},b_{2},b_{3},b_{4}\in F$ such that
$$a_{1}x_{1} + (a_{2}-b_{2})x_{2} + (a_{3}-b_{3})x_{3}-b_{4}x_{4}=0.$$
But we know that $x_{1},x_{2},x_{3},x_{4}$ are linearly independent! Hence $a_{1}=0,$ $a_{2}=b_{2},$ $a_{3}=b_{3}$ and $b_{4}=0.$ Therefore $v\in \operatorname{span}{(x_{2},x_{3})},$ and this completes the proof.
A: To show equality between two sets $A$ and $B$, one has to show that both $A \subseteq B$ and $B \subseteq A$ are true. $\DeclareMathOperator{\span}{\mathrm{span}}$
Let $x \in \span\{x_2, x_3\}$. Then $x = \beta x_2 + \gamma x_3$ for some $\beta,\gamma \in \mathbb R$. It follows right away that $x \in \span\{x_1, x_2, x_3\} \cap \span\{x_2,x_3,x_4\}$: it suffices to choose $0$ as coefficients for $x_1$ in the first case and $x_4$ in the second one. We proved that
$$\span\{x_2, x_3\} \subseteq \span\{x_1, x_2, x_3\} \cap \span\{x_2,x_3,x_4\}.$$
Now, take $x \in \span\{x_1, x_2, x_3\} \cap \span\{x_2,x_3,x_4\}$. Then we have
$$x = \alpha x_1 + \beta x_2 + \gamma x_3 = \beta' x_2 + \gamma' x_3 + \delta x_4,\qquad\alpha,\beta,\beta',\gamma,\gamma',\delta \in \mathbb R$$
It follows that
$$0 = x - x = \alpha x_1 + (\beta - \beta')x_2 + (\gamma - \gamma')x_3 - \delta x_4$$
and since $\{x_1,x_2,x_3,x_4\}$ are linearly independent, $\alpha$ and $\delta$ have to be zero, giving that $x \in \span\{x_2, x_3\}$.
