# 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.

• I think you're on the right track. but not just a random vector, ALL vectors in span x1,x2,x3. So rigorously define A, the intersection of span(x1,2,3) and span(x2,3,4), then show that for any vector in span(x2,3), you have a vector in A (and then, show it the other way around, too!) – Mohammad Athar Mar 7 '17 at 17:11
• You were kind of on the right track; I suspect you got stuck because you didn't use the fact that the $x_{i}$ are assumed linearly independent. Moral of the story: it often helps to ask, "Have I used all of the information available to me?" – Will R Mar 7 '17 at 17:20

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.

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 $\operatorname{Span}(x_2,x_3)$.

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 $x \in \span\{x_2, x_3\}$.