Showing that $2$ Weylgroups are the same

Question

Let $\Delta=\{\alpha_1,\dots,\alpha_n\}$ be a basis of a root system $R$ and $W=\langle s_\alpha\mid \alpha \in \Delta\rangle$ the Weylgroup

Let $\lambda>0$ and $R'=\{\lambda\alpha\mid\alpha \in R\}$. Now I want to show that $R'$ has the same Weyl group as $R$.

I have the following idea: Let $\Delta'=\{\alpha'_1,\dots,\alpha'_n\}\subset R'$ be a basis and $W'=\langle s_{\alpha'}\mid\alpha' \in \Delta'\rangle$ Then for $\alpha' \in \Delta', \beta'\in R'$ it follows $$s_{\alpha'}(\beta')=\beta'-\frac{2(\beta',\alpha')}{(\alpha',\alpha')}\alpha'=\lambda\beta-\frac {2\lambda^2(\beta,\alpha)}{\lambda^2(\alpha,\alpha)} \lambda\alpha = \lambda(s_\alpha(\beta))$$

So according to my computation it breaks down to show $s_\alpha(\beta)=\lambda s_\alpha(\beta)$. Since $s_\alpha$ is a reflection I am not sure what the additional factor $\lambda$ mean. It could mean that it just means an additional stretching (if $\lambda >1$) or a compression (if $\lambda <1$) after the reflection. However, the reflection should be the same because it is not affected by this action.

I hope someone can tell me if this is right or where my mistake is. Thanks for your help

Answer 1

You have shown that $s_{\alpha'}(\beta')=\lambda s_{\alpha}(\beta)$. Note that since $s_{\alpha}$ is a linear operator, $$\lambda s_{\alpha}(\beta)=s_{\alpha}(\lambda \beta)=s_\alpha(\beta').$$ Since $\beta'$ is arbitrary, you've shown that $s_{\alpha'}=s_{\alpha}$ for every $\alpha'$. Therefore, $W=W'$.

Answer 2

I think everything is OK. Your calculation shows that for two "new" (stretched or compressed) roots $\alpha'$ and $\beta'$, $s_{\alpha'}(\beta')$ is the accordingly stretched or compressed version of $s_\alpha(\beta)$. Which is probably what you want to show, isn't it?