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

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 $$-v$$ $$\forall$$ $$v\in V$$).

• The additive identity must be unique. Aug 9, 2021 at 7:23
• @Berci what I have understood from uniqueness of identity element is that there can't exist more than one identity element for any element of the set V. Is it correct? As in the question stated above, there exists only one identity element for all elements of V. So we can't say that a unique additive identity element exists for each element of V ? Aug 9, 2021 at 7:26
• @Berci I am new to linear algebra course. Please let me know if I have understood the concept wrongly. Aug 9, 2021 at 7:31
• I mean there should be exactly one element $e$ of $V$ which satisfies $e+x=x$ for all $x\in V$, with the given meaning of '+'. Aug 9, 2021 at 7:31
• @Berci Thanks. I understood !! Aug 9, 2021 at 7:37

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 $$v\in V$$.
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).