# Thread: How do I determine which space these vectors span?

1. ## How do I determine which space these vectors span?

Hi guys. I'm gonna right down two different problems. The first one I understand but not the second. Find the space the vectors span.

1. a=(3,5,0) b=(1,2,0) sp(a,b)= $\displaystyle (3\alpha+\beta, 5\alpha+2\beta, 0) \rightarrow sp(a,b) = (\lambda_1, \lambda_2, 0)$

2. c=(1,1,2) d=(1,0,1) sp(c,d) = $\displaystyle (\alpha+\beta, \alpha, 2\alpha+\beta) \rightarrow sp(c,d) = ???$

How do I continue to get an answer like the first question - something of the form $\displaystyle (\lambda_1, 0, \lambda_2)$...(I know that's not correct, just giving an example)

2. Originally Posted by jayshizwiz
Hi guys. I'm gonna right down two different problems. The first one I understand but not the second. Find the space the vectors span.

1. a=(3,5,0) b=(1,2,0) sp(a,b)= $\displaystyle (3\alpha+\beta, 5\alpha+2\beta, 0) \rightarrow sp(a,b) = (\lambda_1, \lambda_2, 0)$

2. c=(1,1,2) d=(1,0,1) sp(c,d) = $\displaystyle (\alpha+\beta, \alpha, 2\alpha+\beta) \rightarrow sp(c,d) = ???$

How do I continue to get an answer like the first question - something of the form $\displaystyle (\lambda_1, 0, \lambda_2)$...(I know that's not correct, just giving an example)
There is no "something of the form" in case (2), but note that if we write $\displaystyle (x,y,z)=(\alpha+\beta, \alpha, 2\alpha+\beta)$ , then

it is true that $\displaystyle x+y-z=0$ , and this last is the equation of a plane in the 3-dimensional space, so

$\displaystyle Span\{c,d\}=\{(x,y,z)\,;\,x+y-z=0\}$

By the way, also (1) can be written in this thrifty way: $\displaystyle span\{a,b\}=\{(x,y,z)\,;\,z=0\}=$ the xy-plane in the 3-dim. space.

Tonio