I am reading the book "Simple groups of Lie type" by R.Carter, and stuck with the following lemma:
Let $r \in \Pi$. Then $w_r$ transforms $r$ into $-r$ but every other positive root into a positive root.
And the proof is something wibbly-wobbly: he takes the first coefficient $\alpha_i > 0$ of $r_i$ in the expansion of some positive root $s$, which is different to that of $r$, and then says "the coefficient of $r_i$ in $w_r(s)$ is therefore also positive". Could you please explain this to me, since neither I do understand the proof nor do I believe in the statement of this lemma?
Here $\Pi$ denotes the fundamental system of roots (which is contained in the positive root system $\Phi^+$) and $w_r$ denotes the reflection in the hyperplane orthogonal to $r$.
Let $\Phi$ be the root system. I assume the $\Pi$ in your question is the set of simple roots (otherwise the statement is false). Then any root $\alpha$ has a unique expression $\alpha=\sum_{r\in \Pi} a_r r$ with $a_r\in \mathbb Z$ such that at least one $a_r\neq 0$, and either all $a_r\geq 0$ (in which case we say $\alpha\in \Phi^+$, or that it is a positive root), or all $a_r\leq 0$ ($\alpha\in\Phi^-$, negative root). This is essentially the definition/construction of simple/positive roots.
Now let $r\in \Pi$. We define the reflection $w_r$ on the ambient Euclidean space of $\Phi$ by $w_r(v)=v-2\frac{(r,v)}{(r,r)}r$. Now clearly $w_r(r)=-r$. On the other hand, let $\alpha$ be a positive root different from $r$. Then $\alpha=\sum_{s\in \Pi} a_s s$ with $a_s\geq 0$ for all $s\in \Pi$ and $a_t\neq 0$ for some $t\neq r\in \Pi$ (otherwise, $\alpha=0$ which is impossible, or $\alpha=r$).
Now apply $w_r$ to $\alpha$. Since $w_r(\alpha)\in \Phi$, $w_r(\alpha)=\beta$ for some $\beta\in\Phi$ hence either $\beta\in\Phi^+$, the set of positive roots, or $\beta\in\Phi^-$, the set of negative roots. We can tell which is which by checking the coefficient of some simple root. This is easy using the definition of $w_r$:
$$\beta=w_r(\alpha)=\alpha-2\frac{(r,\alpha)}{(r,r)} r=\sum_{s\neq r} a_s s + (a_r-2\frac{(r,\alpha)}{(r,r)})r$$
Aha, this is the unique expression of $\beta$ in terms of simple roots, hence the coefficients are eiher all positive integers or all negative integers. The coefficient of $r$ isn't obviously positive or negative, true. However, clearly the coefficient $a_t$ of $t$ is the same, and we assumed $a_t>0$ since $\alpha\neq r$. Since any root must have all positive or all negative coefficients, they must all be positive hence $\beta\in \Phi^+$.