Why is $s_{\alpha}$ a Euclidean reflection? Let $V$ be a finite dimensional real vector space, and $\langle-,-\rangle$ a symmetric, nondegenerate positive definite bilinear form on $V$.  If $v \in V$, the Euclidean reflection about $v$ is defined to be the unique linear transformation $\phi: V \rightarrow V$ which sends $v$ to $-v$ and fixes pointwise all the elements in the orthogonal complement of $v$.
Let $G$ be a linear algebraic group, $T$ a maximal torus with character group $X$ and Weyl group $W$.  View $X$ as an additive subgroup of $V = \mathbb{R} \otimes_{\mathbb{Z}} X$.  Then $W$ extends to a group of automorphisms of $V$.  Let $\langle - ,-\rangle$ be a bilinear form on $V$ as above which is fixed by the elements of the Weyl group.
Let $\alpha$ be a weight of $T$ such that $G_{\alpha} = Z_G((\textrm{Ker } \alpha)^0)$ is not solvable, so $W_{\alpha} = W(G_{\alpha},T)$ is a subgroup of $W$ with order $2$.  Let $x \in N_{G_{\alpha}}(T)/Z_{G_{\alpha}}(T)$, and let $s_{\alpha}$ be the image of $x$ in $W$.  Then $s_{\alpha}$ extends to an automorphism of $V$.
I'm trying to understand why $s_{\alpha}$ is the Euclidean reflection about $\alpha$.  This is what Springer claims in Linear Algebraic Groups, (7.1.6).  I just don't know how to use the condition that $(\alpha,\chi) = 0$, since I don't know much about the bilinear form.  It is not difficult to see that $s_{\alpha}$ has to send $\alpha$ to $-\alpha$ (edit: not really, I would love for you to explain this as well), but I also need to show that if $\chi$ is a character of $T$ with $\langle \alpha, \chi \rangle = 0$, then $s_{\alpha}(\chi) = \chi$, or in other words, $\chi(y) = \chi(xyx^{-1})$ for all $y \in T$.
 A: There exists $\lambda \in Y(T)$ (a morphism $G_m \to T$) such that $s_{\alpha} \lambda = - \lambda$. (since $Y=X^{\vee}$) 
$s_{\alpha}$ action on $Y(T)$ is defined by $s_{\alpha} \lambda(t) = n_{\alpha} \lambda(t) n_{\alpha}^{-1}$.
we can assume $<\alpha, \lambda> \neq 0$ if we know $s_{\alpha} \alpha = - \alpha$.
For $\chi \in X$ satisfy $<\chi, \lambda>=0$, 
$s_{\alpha} \chi (t) = \chi(t)$, if $t \in (\text{Ker } \alpha)^{o}$, ($t$ commutative with $n_\alpha$)
$s_{\alpha} \chi (t) =\chi(n_{\alpha}^{-1}\lambda(s)n_{\alpha}) = \chi(\lambda(s^{-1})) = 1 = \chi(t)$, if $t=\lambda(s)$.
$s_{\alpha} \chi =\chi$ on Im $\lambda$ $\cdot$ $(\text{Ker } \alpha)^{o}$ which is a closed connected subgroup of $T$, must equal to $T$ if Im $\lambda$ $\not\subset$ Ker $\alpha$ (i.e. $<\alpha, \lambda> \neq 0$).
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Proof for $s_{\alpha} \alpha = - \alpha$.
$s_{\alpha} \alpha = \alpha \circ \text{Int } n_{\alpha}^{-1}$, $\text{Int } n_{\alpha}^{-1}$ induce $T/(\text{Ker} \alpha)^o \to T/(\text{Ker} \alpha)^o$, an automorphism of $G_m$, and $\text{Aut } G_m = \mathbb{Z}/2\mathbb{Z}$.
$\text{Int } n_{\alpha}^{-1}(t) \equiv t^{-1} \pmod{(\text{Ker} \alpha)^o}$, since $n_{\alpha}$ is not in $Z_G(T)$, can't induce the identity morphism.
Hence $s_{\alpha} \alpha (t) = (-\alpha)(t)$.
