Proving the 'additive inverse' and 'zero vector' axiom for vector space.

Question

I'm confused about the axiom for the additive inverse and zero vector. If I define a vector space with normal addition but scalar multiplication like so: k(x, y) = (kx, 0).

One axiom says: There is a special object in V, denoted 0 and called the zero vector, such that for all u in V we have u + 0 = u. Since vector addition is normal, 0 in this case is (0, 0).

However another axiom says: For every u in V there is another object in V, denoted -u and called the negative of u, such that u + (-1)u = 0. However, u + (-1)u = (x, y) + (-x, 0) = (0, y) which doesn't equal the zero vector.

Does this mean this axiom fails? The textbook I read this from does not say this axiom fails. Why?

Answer 1

You are wrong. What the “other axiom” says is that for each $u\in V$, there is a $u'\in V$ such that $u+u'=0$. It does not say that $u'=(-1)u$.

Answer 2

Vector spaces axioms also require $1u=u$, which is in general not satisfied by your scalar multiplication. So your structure is not a vector space.