# About the axioms for vector spaces. Can we derive the axiom which guarantees the existence of the zero vector from these $7$ axioms?

I am reading an applied mathematics book (in Japanese) by Kenichi Kanatani.

The author wrote as follows:

If non-empty set $$\mathcal{L}$$ along with an addition on $$\mathcal{L}$$ and a scalar multiplication on $$\mathcal{L}$$ has the following properties, then $$\mathcal{L}$$ is called a vector space.

1. For any $$u,v\in\mathcal{L}$$, $$u+v=v+u$$.
2. For any $$u,v,w\in\mathcal{L}$$, $$u+(v+w)=(u+v)+w$$.
3. For any $$u,v\in\mathcal{L}$$, there exists $$x\in\mathcal{L}$$ such that $$u+x=v$$. (We write $$x$$ as $$v-u$$.)
4. For any $$u,v\in\mathcal{L}$$ and any $$\lambda\in\mathbb{R}$$, $$\lambda(u+v)=\lambda v+\lambda u$$.
5. For any $$u\in\mathcal{L}$$ and any $$\lambda,\mu\in\mathbb{R}$$, $$(\lambda+\mu)u=\lambda u + \mu u$$.
6. For any $$u\in\mathcal{L}$$ and any $$\lambda,\mu\in\mathbb{R}$$, $$(\lambda\mu)u=\lambda(\mu u)$$.
7. For any $$u\in\mathcal{L}$$, $$1u=u$$.

And the author wrote as follows:

For any $$u\in\mathcal{L}$$, $$u-u\in\mathcal{L}$$. We write this as $$0$$. By definition, for any $$u\in\mathcal{L}$$, $$u+0=u$$.

By the way, the author is not a mathematician.

The author wrote "For any $$u,v\in\mathcal{L}$$, there exists $$x\in\mathcal{L}$$ such that $$u+x=v$$. (We write $$x$$ as $$v-u$$.)". But the author didn't prove that $$x$$ is unique.

The author didn't prove that for any $$u,v\in\mathcal{L}$$, $$u-u=v-v$$ held.

Can we really prove that $$u-u=v-v$$ for any $$u,v\in\mathcal{L}$$?

Suppose that we can prove that for any $$u,v\in\mathcal{L}$$, there exists $$x\in\mathcal{L}$$ such that $$u+x=v$$ uniquely.

Then, for any $$u,v\in\mathcal{L}$$, $$u+v+0u=(u+0u)+v=(1u+0u)+v=(1+0)u+v=1u+v=u+v,$$ and $$u+v+0v=u+(v+0v)=u+(1v+0v)=u+(1+0)v=u+1v=u+v.$$
So, by uniqueness, $$0u=0v.$$
Let $$z$$ be an arbitrary element of $$\mathcal{L}$$.
Let $$u$$ be an arbitrary element of $$\mathcal{L}$$.
Then, $$u+0z=u+0u=1u+0u=(1+0)u=1u=u.$$
So, $$0z$$ is zero vector.

• Good observation! I guess the way to prove it would be to first show that $u-v=u+(-1)v$, which shows that $u-u=0u$ and $v-v=0v$, and then use the various scalar-multiplication properties to show that $0u=0v$. Feb 21, 2021 at 3:48
• @tchappy ha, you may begin with the assumption $x=u-u$ and $y=v-v$ as well. Think about the consequences of what if $x≠y$ - are they in contradiction with properties (1) to (7). Feb 21, 2021 at 4:15
• @GregMartin Thank you very much for your comment. Feb 21, 2021 at 8:56
• @AweKumarJha Thank you very much for your comment. Feb 21, 2021 at 8:56
• See also math.stackexchange.com/a/2378914/589. It seems that $0u=0v$ is a requirement.
– lhf
Feb 21, 2021 at 12:46

Let $$v$$ be an arbitrary element of $$\mathcal{L}$$.
Let $$u$$ be an arbitrary element of $$\mathcal{L}$$.
There exists $$x\in\mathcal{L}$$ such that $$u+x=v$$.
$$v+0u=(u+x)+0u=(u+0u)+x=(1u+0u)+x=(1+0)u+x=1u+x=u+x=v$$.
So, $$0u$$ is zero vector.