Proof that $-v = (-1)*v$ I need to prove that for every Vector Space this is valid:
$$
-v = (-1)*v
$$
-v = inverse element of addition
-1 a real number
$*$ the multiplication by real number of the Vector Space
My teacher said that $-v$ is just a notation for the inverse element of addition. I'd like to prove that $-v = (-1)*v$.
So I came up with the following solution and I'd like to know if it's correct:
\begin{align}
v + (-1)*v = u \\
(1)*v + (-1)*v = u \\
(1-1)*v = u \\
0*v = u \\
o = u\\
\end{align}
Since $v + -v = o$
$v + (-1)*v = o = v + -v$
adding -v to both sides
v + (-1)*v -v = v + -v + -v
o + (-1)*v = o + -v
(-1)*v = -v
Did I commit any mistakes? Did I make any assumptions that may not be valid for EVERY Vector Space?
Edit: As S. Sheng said I have not proved that (0)*v = o. I'll try to prove that and come back later with a proof of that.
I also haven't proved that (1)*v = v
Oh my... I'm starting to think this is beyond my abilities..
 A: One way to deduce it is
$$
\vec0=0\vec u\\
\vec0=(1+(-1))\vec u\\
\vec0=1\vec u+(-1)\vec u\\
\vec 0+(-\vec u)=\vec u +(-1)\vec u+(-\vec u)\\
-\vec u=(\vec u +(-\vec u))+(-1)\vec u\\
-\vec u=(-1)\vec u
$$
A: The u is extraneous for your proof. But in any case the edited version is now correct. I will provide a rewrite so that others can use this thread for future reference.
$$v+(-1)v=(1)v+(-1)v=(1-1)v=(0)v=0$$
Adding $-v$ to both sides yields $(-1)v=-v$
Note that in our proof we assume $(0)v=0$. I am assuming that you have already proved that yourself or are allowed to assume it.
A: See the Axiom of the vector spaces
Axiom (4),(6) give the desired result.
one more note:-In a group, the additive inverse is unique!
A: $$(-1)\vec{u}+(\vec{u}) =(-1)\vec{u}+(1)(\vec{u})=$$
$$ (-1+1)(\vec{u})= 0(\vec{u})=0$$
That makes $$(-1)\vec{u}=-\vec{u}$$
A: Instead of saying "Since $v +-v = 0$", you might want to say "Hence $v + (-1)v = 0$". And from this (depending on your definition, it follows immediately that $-v + (-1)v$.
Also, be sure to follow your teachers notation. He/she might want you to write $\vec{v}$ for vectors. That way we can distinguish between the number $0$ and the vector $\vec{0}$.
Also, if you insist on having the $u$ in your proof, then I would say in the beginning of the proof "Let $u = v + (-1)v$".
A: In any vector space for any vector $\vec u$ exists an inverse element of addition $\vec v$ such that
$$\vec u+\vec v=\vec 0$$
that is precisely $\vec v=-\vec u=(-1)\vec u.$
A: Let $v \in V$, where is $V$ is a vector space.
$(-1)v = (-1)v + 0 = (-1)v + (v+(-v)) = (-1)v + v + (-v) = (-1+1)v + (-v) = -v$
You don't need to prove that $1v=v$ because it's an property of vector space.
But you need to prove that $0v=0$, where is $0$ on the left side is scalar and vector is on the right side.
$0v=(0+0)v = 0v+0v$
$0v+(-0v) = 0v+0v+(-0v)$ 
Thus we got that
$0v = 0$.
In both proofs i used only properties of vector space.
