Has the additive identity to be same for all elements of a set V which is a vector space over a field F??

Question

I am solving a problem in which it's given that V = $\mathbb R$ and F = $\mathbb R$ and the definition of vector addition and scalar multiplication is given. The problem is to check which axioms of vector space are satisfied and which are not.

I am getting unique additive identity as $ \frac{-v}{2}$ $\forall$ $v\in V$.

My doubt is that has the additive identity to be same for $\forall$ $v\in V$, like the one in the usual addition (which is 0) or it can be different but unique for different elements of V like the additive inverse in usual addition (which is $-v$ $\forall$ $v\in V$).

Answer 1

Indeed the additive identity has to be the same for all elements of the vector space. By definition, the additive identity is an element $e\in V$ such that $$e + v = v + e = v$$ for all $v\in V$.

This definition is what it means for an element to be an additive identity, and a natural question is: Can there only be one additive identity or can there be more?

To answer this, assume that $e_1$ and $e_2$ are two vectors that are both additive identities. Since $e_1$ is an additive identity, we have for all $v\in V$ that $e_1 + v = v$. Since this is true for all $v$, I can put $v=e_2$, so $e_1+e_2=e_2$. On the other hand, since $e_2$ is an additive identity, we have for all $v$ that $v+e_2 = v$. In particular, we can put $v=e_1$, so $e_1+e_2=e_1$.

Ok, but now $e_1+e_2 = e_2$, and also $e_1+e_2=e_1$, so it turns out that $e_2=e_1$. In conclusion, there can only be one additive identity.

Thus for all vectors $v\in V$, the additive identity will be the same element (which is usually called "$0$" for vector spaces).