first of all, we have a direct sum decomposition:

E = H_{α}⊕span(α).

so we can write any v in E uniquely as v = x + kα, for some x in H_{α}, and k in R.

therefore, s_{α}(v) = s_{α}(x+kα) = s_{α}(x) + ks_{α}(α) = = x - kα.

so now let's look a bit more carefully at what "k" is.

note that v.α = x.α + k(α.α). but by the very definition of H_{α}, x.α = 0. so

v.α = k(α.α), and therefore k = v.α/(α.α).

putting this all together, we get:

s_{α}(v) = x - kα = x + kα - 2kα = v - [2(v.α)/(α.α)]α, which is what (1) states.

so then we compute:

(s_{α}(x)).(s_{α}(y)) = (x - [2(x.α)/(α.α)]α).(y - [2(y.α)/(α.α)]α)

= x.y - 2(x.α)(α.y)/(α.α) - 2(y.α)(x.α)/(α.α) + 4(x.α)(y.α)(α.α)/[(α.α)^{2}]

= x.y - 4(x.α)(y.α)/(α.α) + 4(x.α)(y.α)/(α.α) = x.y, which is what (2) states.