TITLE: Show that the additive inverse condition can be replaced by $0v = v$ for all $v \in V$
QUESTION [8 upvotes]: In the definition of a vector space, the additive inverse condition
  requires that for every $v \in V$ (where $V$ is a vector space over
  $\mathbf{F} = \mathbb{R}$ or $\mathbb{C}$), there exists $w \in V$
  such that $$v + w = 0$$  Show that this condition can be replaced with
  the condition that 
$$0v = 0$$
for all $v \in V$. The $0$ on the left side is the number $0$ in
  $\mathbf{F}$ and the $0$ on the right side is the additive identity of
  $V$.
(The phrase "a condition can be replaced" in a definition means that the >collection of objects satisfying the definition is unchanged if the >original condition is replaced with a the new condition)
(Taken from "Linear Algebra Done Right (3rd edition), by S. Axler)

I am new to abstract algebra and have close to no clue how to tackle this problem. Here is my futile "attempt" nonetheless:
I tried some substitution and got
$$v + w = 0v$$
but am unable to make any sense out of this, much less determine if I am headed towards the right direction.

REPLY [10 votes]: Call the first condition C1 and the second condition C2. Then we have to prove two things:

The axioms of a vector space (with C1) imply C2
The axioms of a vector space (without C1) together with C2 imply C1

This means that, within the other vector space axioms, C1 and C2 are equivalent. This is already an important first step.
Now let's prove both directions.

Suppose that every $v\in V$ has an inverse. Using the axioms of a vector space, we see that: $$0\cdot v = (0+0)\cdot v = 0\cdot v+0\cdot v$$ Now, because we assumed that C1 holds, we know that $0\cdot v$, which is a vector, has an inverse $w\in V$. Now add this inverse to both sides to get that $$ 0\cdot v +w = 0\cdot v+0\cdot v+w \Leftrightarrow 0 = 0\cdot v+0 = 0\cdot v$$
So C2 is proven.
Suppose that C2 holds, so $0\cdot v = 0$ for every vector $v\in V$. Again using only vector space axioms (but not using C1!!) we find that
$$0 = 0\cdot v = (1+(-1))\cdot v = 1\cdot v+(-1)\cdot v = v+(-1)\cdot v $$
So $(-1)\cdot v$ is an inverse of $v$ for every $v\in V$. So C1 is proven.