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.