While looking up something else on stack exchange, I ran across this question

Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ cannot be closed under scalar multiplication by $a \ne 0,1$

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.


1 Answer 1


1) You can just apply the definition of linearity. Let $r$ and $s$ be two real numbers, with unique representations $r = \sum_{x \in B} r_x x$ and $s = \sum_{x \in B} s_x x$. Then $r + s = \sum_{x \in B} (r_x + s_x) x$. Note: $r_x + s_x$ is rational, and is supported on a finite set in $B$, namely the union of the support of $r$ and the support of $s$. So by uniqueness we have $(r+s)_x = r_x + s_x$. Therefore, \begin{equation*} \alpha(r+s) = \sum_{x \in B} (r_x + s_x) = (\sum_{x \in B} r_x) + (\sum_{x \in B} s_x) = \alpha(r) + \alpha(s). \end{equation*}

Similarly, you can show that $\alpha(\lambda r) = \lambda \alpha(r)$ for all $\lambda \in \mathbb{Q}$ (although you won't need it for the rest of the argument). To guide you, think about the following: if $r = \sum_{x \in B} r_x x$, then what is the unique representation of $\lambda r$ in the basis $B$ over $\mathbb{Q}$? What is $\alpha(\lambda r)$?

2) Suppose $r = \sum_{x \in B} r_x x$. Then $ar = \sum_{x \in B} r_x (ax)$. But $ax$ is just another element in the basis $B$, so this is the unique representation of $ar$ with respect to $B$ - multiplying by $a$ simply permutes the elements in the basis - and so $(ar)_{ax} = r_x$. Thus, $\alpha(r) = \sum_{x \in B} r_x = \sum_{x \in B} (ar)_{ax} = \alpha(ar)$.

Write back if anything is still unclear!

  • $\begingroup$ This was an excellent answer. It made everything really clear for me, and actually helped re-enforce my understanding of the concepts involved that I was struggling with. $\endgroup$
    – user115411
    Sep 17, 2014 at 4:42

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