For $A \in \mathbb{R}^2$. Assume A + A = $0$. Prove that A = $0$ using only vector space axioms.

Question

For $A \in \mathbb{R}^2$. Assume A + A = $0$. Prove that A = $0$ using only vector space axioms:

A1 : A + B = B + A

A2 : A + (B + C) = (A + B) + C

A3 : $0$ + A = A + $0$ = A

A4 : A + (-A) = $0$

M1 : (r + s)A = rA + sA

M2 : r(A + B) = rA + sB

M3 : r(sA) = rsA

M4 : 1A = A

Here is my thinking about how to solve this problem. A + A = $0$ imples A + A = A + (-A) by A4. Thus A = -A. Then since $ 1 \neq -1$, 1A = -1A can only happen if A = $0$.

However I'm not sure if there is a more concrete way to use the axioms. I feel like my end argument is weak. Any suggestion on how to show this?

Answer 1

$A + A = 1A + 1A = (1+1)A = 2A = 0$

Now, can you show that $\frac{1}{2} \cdot 0 = 0$?

Then, $A = (\frac{1}{2} \cdot 2) A = \frac{1}{2}(2A) = \frac{1}{2} 0 = 0$