# What axiom or property of vector spaces implies that multiplication of a non-zero vector by 0 results in the zero vector?

I am trying to prove that a real vector space V not equal to the zero vector has infinitely many bases, using only the axioms of vector spaces. My strategy is assuming a valid basis $$B_1$$ = {$$v_1$$, ..., $$v_n$$}, multiplying every member of that basis by some number a $$\subset \Re$$ such that $$B_2$$ = {$$av_1$$, ... , $$av_n$$}, and showing that in general for every a $$\subset \Re$$, $$av$$ is a distinct vector.

My method is as follows:

1. Pick a vector V in v that is not the zero vector
2. Pick two numbers $$a_1$$ and $$a_2$$ in $$\Re$$ such that $$a_1 \neq a_2$$. Assume $$a_1v = a_2v$$.
3. $$a_1v - a_2v = a_2v - a_2v = 0$$, where $$0$$ is the zero vector.
4. $$(a_1 - a_2)v = 0$$, but $$v \neq 0$$, so
5. $$a_1 - a_2 = 0 \Rightarrow a_1 = a_2$$, which is a contradiction.

My problem is the jump from 4 to 5. I don't think any of the 8 basic axioms of a vector space imply that multiplication of a non-zero vector by $$0$$ results in the $$0$$ vector, so what property of real vector spaces allows for this? Or is my method completely wrong?

You're overcomplicating things. You can just multiply one of the vectors by a nonzero number $$a$$ (and one vector in the basis exists).
The set $$\{av_1,v_2,\dots,v_n\}$$ is easily seen to be linearly independent, hence a basis.
For every nonzero vector $$v\in V$$, the map $$f\colon\mathbb{R}\to V$$, $$f(a)=av$$ is linear and injective, so indeed the vectors you get are distinct.
Why does $$av=0$$, with $$v\ne0$$, implies $$a=0$$? Because if $$av=0$$ and $$a\ne0$$, we can do $$v=1v=(a^{-1}a)v=a^{-1}(av)=a^{-1}0=0$$ There would be another check to do, namely that for $$a\ne b$$ (both nonzero), the sets $$\{av_1,v_2,\dots,v_n\}$$ and $$\{bv_1,v_2,\dots,v_n\}$$ are distinct, but their equality means that either $$av_1=bv_1$$ or $$av_1=v_i$$ and $$bv_1=v_j$$, for some $$i,j>1$$. The latter possibility is excluded by linear independence, the former by the already established fact that the map $$f$$ is injective