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

Question

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 $u,v\in\mathcal{L}$?

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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.

Answer 1

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.