Homework Help: Determining if vectors in R3 are linear subspaces

1. Dec 14, 2011

csc2iffy

1. The problem statement, all variables and given/known data
U={(x1,x2,x3)$\in$ℝ3 | x1+x2=0}
Is this a linear subspace of ℝ3?

2. Relevant equations
x1+x2=0

3. The attempt at a solution
I know that in order to be a linear subspace, it must be closed under addition and scalar multiplication. I'm just not really sure how to incorporate the x1+x2=0. This is what I've done:
x=(x1,x2,x3), y=(y1,y2,y3)
x+y=(x1,x2,x3)+(y1,y2,y3)=(x1+y1, x2+y2, x3+y3)
Edit:
Do I simply just show this?
(x1+y1)+(x2+y2)=

cx=c(x1,x2,x3)=(cx1,cx2,cx3)
(cx1+cx2)=
c(x1+x2)=c(0)=0 (closed under scalar multiplication)
So it is a subspace?

Last edited: Dec 14, 2011
2. Dec 14, 2011

Staff: Mentor

I'm sure this isn't the question. U is a set of vectors. The question is whether U is a linear subspace of R3.
The key determinant of whether any vector <x1, x2, x3> is an element of U is whether its first two components of the vector add to 0.

Start with your two vectors x and y, and assume that they are in U. What can you say for certain about these two vectors? What can you say about their sum?

What can you say about kx, where k is a scalar?

3. Dec 14, 2011

csc2iffy

I think I got it...
Do I simply just show this?
(x1+y1)+(x2+y2)=

cx=c(x1,x2,x3)=(cx1,cx2,cx3)
(cx1+cx2)=
c(x1+x2)=c(0)=0 (closed under scalar multiplication)