My question is very short and simple.
Given a set, V, which contains a vector $(x_1,x_2,x_3)$ which follows a specific rule –
If I show that the set is closed under both component wise addition of two vectors also in the set, and show that the set is also closed under scalar multiplication, have I proven that the set is a Vector Space?
My immediate thought is, no, due to the fact that if a set doesn't contain the zero vector, then it cant be defined as a vector space.
Do I need to show that the zero vector is a member of the set, or is this assumed by proving the two conditions regarding addition and scalar multiplication above.
Thanks for your help in advance
Best Answer
Here is what I think you are asking.
The answer to this question: yes, as long as you can show that $V$ is not empty, because if there is some $v \in V$ then using what you have already proved, $$ v + (-1)v = 0 \in V. $$
The problem with the question as you asked it is that you did not tell us that the vectors in $V$ were elements of some larger space - in this case probably $\mathbb{R}^3$. The "rule" is irrelevant in the question, though you need it to write an answer.