[proofplan] We first compute the effect of $s_\alpha$ on $\alpha$ directly from the reflection formula. Then we take a positive root $\beta \ne \alpha$, write it in the simple-root basis, and observe that applying $s_\alpha$ changes only the coefficient of $\alpha$. If the reflected root were negative, all non-$\alpha$ coefficients would have to vanish, forcing $\beta$ to lie on the line $\mathbb{R}\alpha$; the reduced root-system axiom then forces $\beta=\alpha$, a contradiction. Finally, since $s_\alpha$ is an involution, the positive-root preservation gives equality rather than only inclusion. [/proofplan] [step:Compute the reflection of the simple root $\alpha$] Fix $\alpha \in \Delta$. By definition, \begin{align*} s_\alpha(\alpha) &= \alpha - (\alpha,\alpha^\vee)\alpha \\ &= \alpha - \frac{2(\alpha,\alpha)}{(\alpha,\alpha)}\alpha \\ &= -\alpha . \end{align*} [/step] [step:Show that reflecting a different positive root cannot make it negative] Let $\beta \in \Phi^+ \setminus \{\alpha\}$. Since $\Delta$ is a base for $\Phi$, there are coefficients $n_\delta \in \mathbb{R}_{\ge 0}$, indexed by $\delta \in \Delta$, such that \begin{align*} \beta = \sum_{\delta \in \Delta} n_\delta \delta . \end{align*} Using the definition of $s_\alpha$, we obtain \begin{align*} s_\alpha(\beta) &= \beta - (\beta,\alpha^\vee)\alpha \\ &= \sum_{\delta \in \Delta \setminus \{\alpha\}} n_\delta \delta + \bigl(n_\alpha - (\beta,\alpha^\vee)\bigr)\alpha . \end{align*} Thus the coefficients of $s_\alpha(\beta)$ in every simple-root direction $\delta \ne \alpha$ are still $n_\delta$. Because $s_\alpha$ is a root-system reflection, $s_\alpha(\beta) \in \Phi$. Suppose, for contradiction, that $s_\alpha(\beta) \in -\Phi^+$. Then its expansion in the base $\Delta$ has all coefficients non-positive. Comparing the coefficients in the simple-root basis gives \begin{align*} n_\delta \le 0 \qquad \text{for every } \delta \in \Delta \setminus \{\alpha\}. \end{align*} Since each $n_\delta \ge 0$, we get $n_\delta = 0$ for every $\delta \ne \alpha$. Hence \begin{align*} \beta = n_\alpha \alpha . \end{align*} Since $\beta \in \Phi^+$, we have $n_\alpha > 0$. The reduced root-system axiom gives \begin{align*} \Phi \cap \mathbb{R}\alpha = \{\alpha,-\alpha\}. \end{align*} Therefore the positive root $\beta$ on the line $\mathbb{R}\alpha$ must be $\alpha$, contradicting $\beta \ne \alpha$. Hence $s_\alpha(\beta) \notin -\Phi^+$. Since every root is either positive or negative with respect to the base $\Delta$, and since $s_\alpha(\beta) \in \Phi$, it follows that \begin{align*} s_\alpha(\beta) \in \Phi^+ . \end{align*} [guided] Let $\beta \in \Phi^+ \setminus \{\alpha\}$. The point is to track the simple-root coefficients. Since $\Delta$ is a base, every positive root has a unique expansion with non-negative coefficients in the simple roots, so there are coefficients $n_\delta \in \mathbb{R}_{\ge 0}$ such that \begin{align*} \beta = \sum_{\delta \in \Delta} n_\delta \delta . \end{align*} Now apply the reflection formula. The only term subtracted from $\beta$ is a scalar multiple of $\alpha$, so no coefficient in any other simple-root direction can change: \begin{align*} s_\alpha(\beta) &= \beta - (\beta,\alpha^\vee)\alpha \\ &= \sum_{\delta \in \Delta \setminus \{\alpha\}} n_\delta \delta + \bigl(n_\alpha - (\beta,\alpha^\vee)\bigr)\alpha . \end{align*} Thus for every $\delta \in \Delta \setminus \{\alpha\}$, the coefficient of $\delta$ in $s_\alpha(\beta)$ is exactly $n_\delta$. Because reflections associated to roots preserve the root system, $s_\alpha(\beta)$ is again a root. Assume for contradiction that this root is negative. A negative root has all coefficients non-positive in its expansion in the base $\Delta$. Therefore, for each $\delta \ne \alpha$, the coefficient $n_\delta$ must satisfy \begin{align*} n_\delta \le 0 . \end{align*} But $\beta$ was positive, so each $n_\delta \ge 0$. Combining the two inequalities gives $n_\delta = 0$ for all $\delta \ne \alpha$. Hence \begin{align*} \beta = n_\alpha \alpha . \end{align*} Since $\beta$ is a positive root, $n_\alpha > 0$. The reduced root-system axiom says that the only roots on the line through $\alpha$ are $\alpha$ and $-\alpha$: \begin{align*} \Phi \cap \mathbb{R}\alpha = \{\alpha,-\alpha\}. \end{align*} The positive one is $\alpha$, so $\beta=\alpha$, contradicting the assumption that $\beta \ne \alpha$. Therefore $s_\alpha(\beta)$ is not negative. Since it is a root, it must be positive: \begin{align*} s_\alpha(\beta) \in \Phi^+ . \end{align*} [/guided] [/step] [step:Upgrade the inclusion to a permutation of $\Phi^+ \setminus \{\alpha\}$] The previous step proves \begin{align*} s_\alpha(\Phi^+ \setminus \{\alpha\}) \subseteq \Phi^+ . \end{align*} Also $s_\alpha(\beta) \ne \alpha$ for every $\beta \in \Phi^+ \setminus \{\alpha\}$, because otherwise applying $s_\alpha$ to both sides gives \begin{align*} \beta = s_\alpha(\alpha) = -\alpha, \end{align*} contradicting $\beta \in \Phi^+$. Hence \begin{align*} s_\alpha(\Phi^+ \setminus \{\alpha\}) \subseteq \Phi^+ \setminus \{\alpha\}. \end{align*} Finally, $s_\alpha$ is an involution: for every $v \in V$, \begin{align*} s_\alpha(s_\alpha(v)) = v . \end{align*} Applying the same inclusion to $s_\alpha(\gamma)$ for each $\gamma \in \Phi^+ \setminus \{\alpha\}$ shows that every such $\gamma$ lies in the image. Therefore \begin{align*} s_\alpha(\Phi^+ \setminus \{\alpha\}) = \Phi^+ \setminus \{\alpha\}. \end{align*} Together with $s_\alpha(\alpha)=-\alpha$, this proves the theorem. [/step]