1. ## [SOLVED] Vector space

I am sure this is easy but I have been trying all day and just don't understand the question.

Determine whether the set, together with the indicated operations, is a vector space.

1. The set {(x, -x) : x is a real number} with the standard operations.

2. The set {(x, y) : x, y are a real numbers such that x > y} with the standard operations.

Any help would be appreciated. Thanks

2. Originally Posted by Neoxl

1. The set {(x, -x) : x is a real number} with the standard operations.
Let $S = \{(x,-x)\}$. Now if $a\in S, \ b\in S$ then $a=(x,-x)$ and $b=(y,-y)$ so $a+b = (x+y,-(x+y))$ so $a+b$ in $S$. Similarly $ka$ in $S$ where $k\in \mathbb{R}$. So it is a vector space.

3. Originally Posted by ThePerfectHacker
Let $S = \{(x,-x)\}$. Now if $a\in S, \ b\in S$ then $a=(x,-x)$ and $b=(y,-y)$ so $a+b = (x+y,-(x+y))$ so $a+b$ in $S$. Similarly $ka$ in $S$ where $k\in \mathbb{R}$. So it is a vector space.
don't you have to verify a list of like 10 axioms or something. that's what we had to do in my linear algebra class. i guess if the standard operations hold, then verifying that it is closed under addition and scalar multiplication is fine, but i don't know

4. Originally Posted by Jhevon
don't you have to verify a list of like 10 axioms or something. that's what we had to do in my linear algebra class. i guess if the standard operations hold, then verifying that it is closed under addition and scalar multiplication is fine, but i don't know
You do not need to. For example, since $\mathbb{R}^2$ is commutative then certainly any subset! So it just comes down to checking the closure of addition and scalar multiplication.