I'm asked to verify if:
$B=(x,y,z) \in \mathbb{R}^3 : ||(x,y,z)|| \leq 1$
is a subspace. I know I have to check for the zero vector, addition and scalar multiplication. Here lies my question however.
If I use $(0,0,0)$ to check if the zero vector exists, it clearly does. However, if I multiply $(0,0,1)$ by some scalar, say, $50$, scalar multiplication does not hold so this isn't a subspace.
Here lies my problem. I thought because the $0$ vector exists, this set MUST be a subspace and hence, scalar multiplication and addition should hold.
Am I wrong in assuming that if $0$ vector exists, all the other axioms must hold? Or is it possible that some axioms work and some fail?
Best Answer
You have the idea somewhat reversed. If the space is closed under addition and scalar multiplication, then you need to have the zero vector since $$0\cdot (x,y,z)=(0,0,0)$$ and $$(x,y,z)-(x,y,z)=(0,0,0)$$
So in fact the crucial axioms are addition and scalar multiplication. The zero vector comes for free
Edit thanks to comment: The zero vector comes for free after checking the set is non-empty. Just so happens that it is often easy to see that the zero vector is in the space, hence non-empty.So this whole discussion kind of chicken and egg