# Prove that span{x,y}=span{y,z}

• September 22nd 2010, 07:28 AM
Runty
Prove that span{x,y}=span{y,z}
I have never liked span questions, so I could use some help with this.

First, suppose that $V$ is a vector space over a field $F$.

Now suppose $x,y,z\in V$ satisfies $x+y+z=0$. Prove that $span\{x,y\}=span\{y,z\}$.
• September 22nd 2010, 07:46 AM
Swlabr
Take x over to the other side, y+z=-x, to see that $x \in Span(y, z) \Rightarrow Span(x, y) \subseteq Span(y, z)$ (can you see why?)

Then, do it the other way!
• September 23rd 2010, 04:41 AM
HallsofIvy
Saying that v is in the span of x and y means v= ax+ by for some scalars a and b.

Since x+ y+ z= 0, x= -y- z so v= a(-y- z)+ by.
• September 23rd 2010, 10:43 AM
Runty
Quote:

Originally Posted by HallsofIvy
Saying that v is in the span of x and y means v= ax+ by for some scalars a and b.

Since x+ y+ z= 0, x= -y- z so v= a(-y- z)+ by.

This means that $v=a(-y-z)+by$ implies that $v\in span(y,z)$, right? Or am I misinterpreting something?
• September 23rd 2010, 11:36 AM
Raoh
Hi :)
..it also means,
$v=(b-a)y-az\Rightarrow v\in Span\left \{ y,z \right \}$
$v,y$ and $z$ are vectors.
• September 24th 2010, 05:17 AM
HallsofIvy
Quote:

Originally Posted by Runty
This means that $v=a(-y-z)+by$ implies that $v\in span(y,z)$, right? Or am I misinterpreting something?

Well, v= a(-y- z)+ by= (-a+ b)y+ (-a)z so that v is a linear combination of y and z and therefore is in span(y,z).

You will also need to prove the other way: if u is in span(y, z), then it is in span(x, y).