While looking up something else on stack exchange, I ran across this question
and it intrigued me. The answer was provided by Jonathan Golan (https://math.stackexchange.com/users/32745/jonathan-golan) and the proof seemed accessible to my level, but I've not been able to fully follow his argument. Admittedly, my linear algebra background is a weak (and I'm currently starting to study a text to understand it better) - but perhaps someone could be kind enough to help me with some issues I'm having comprehending the proof. If the difficulty is above what I'm capable of, I'll understand though.
Here I reprint Dr. Golan's answer from the above link for convenience:
Let $B$ be a Hamel basis. Then any real number $r$ can we written uniquely as $\Sigma_{x \in B} {r_x}x$ where the $r_x$ are rational numbers only finitely many of which are nonzero. The function $\alpha : r \to \Sigma_{x \in B} r_x$ is a linear transformation of vector spaces over the rational numbers. Now suppose that $a \ne 1$ and $ax \in B$ for all $x \in B$. Then $\alpha(ar)=\alpha(r)$ for all real numbers $r$. In particular, if $x \in B$ and if $r = x(a-1)^{-1}$ then $1 = \alpha(x) = \alpha([a-1]r) = \alpha(ar) - \alpha(r) = 0$. Contradiction!
1) My first issue is how do we know this map is a linear one? I tried using the basic definition of linearity - e.g. $\alpha (x+y) = \alpha(x) + \alpha(y)$ etc. - but I was unable to come up with anything sensible.
2) How is he able to conclude $\alpha(ar)=\alpha(r )$ ? I have no thoughts on this step.
Thanks to anyone willing to help with my probably silly questions in advance. I thought some explanations might help me understand these things better.