# Thread: Set is Subspace

1. ## Set is Subspace

I need to determine whether all vectors (a,b,c) for which a+b+c=0 are a subspace of R^3.

I believe it is under scalar multiplication:
k(a,b,c)= (ka, kb, kc)
(ka + kb +kc) = 0
let a= 1, b=2, c=-3, k=5
(5(1) + 5(2) + 5(-3)) = 0
0 = 0
So therefore this is closed under scalar multiplication.

I think I'm correct in that, but how do I find out addition?

2. Originally Posted by Hellreaver
I need to determine whether all vectors (a,b,c) for which a+b+c=0 are a subspace of R^3.

I believe it is under scalar multiplication:
k(a,b,c)= (ka, kb, kc)
(ka + kb +kc) = 0
let a= 1, b=2, c=-3, k=5
(5(1) + 5(2) + 5(-3)) = 0
0 = 0
So therefore this is closed under scalar multiplication.

I think I'm correct in that, but how do I find out addition?
You just need to check two things. First, it is closed under vector addition. Second, it is closed under scalar multiplication. You were correct in showing it is closed under scalar multiplication. Now let $\bold{a} = (a_1,a_2,a_3)$ and $\bold{b} = (b_1,b_2,b_3)$. Therefore, $a_1+a_2+a_3=0$ and $b_1+b_2+b_3 = 0$. Then $\bold{a}+\bold{b} = (a_1+b_1,a_2+b_2,a_3+b_3)$. And $(a_1+b_1)+(a_2+b_2) + (a_3+b_3)$ can be rearranged as $(a_1+a_2+a_3)+(b_1+b_2+b_3) = 0 + 0 = 0$. Thus, it is closed under vector addition.

3. Awesome, thanks. I was really unsure about the addition, and I'm happy I have the scalar multiplication part. Thanks a bunch.