Tx = T(λ_{1}x_{1}+ λ_{2}x_{2}) = λ_{1}T(x_{1}) + λ_{2}T(x_{2})

= λ_{1}x_{1}- λ_{2}x_{2}= λ_{1}x_{1}+ λ_{2}x_{2}- 2λ_{2}x_{2}

= x - 2λ_{2}x_{2}(this is just simple algebra, to this point).

now (x,x_{2}) = (λ_{1}x_{1}+ λ_{2}x_{2},x_{2}) = λ_{1}(x_{1},x_{2}) + λ_{2}(x_{2},x_{2})

and since x_{1},x_{2}are orthogonal, (x_{1},x_{2}) = 0, therefore:

(x,x_{2}) = λ_{2}(x_{2},x_{2}) = λ_{2}|x_{2}|^{2}.

but x_{2}lies on the unit circle, therefore it has length (and thus length squared) of 1.

so (x,x_{2}) = λ_{2}, and we have:

Tx = x - 2λ_{2}x_{2}= x - 2(x,x_{2})x_{2}.

so why?

the quantity (x,x_{2})x_{2}is a little misleading (because we are dealing with UNIT vectors x_{1},x_{2}).

normally, this is written as: [(x,x_{2})/(x_{2},x_{2})]x_{2}which is known as:

the projection of the vector x in the direction of the vector x_{2}. geometrically this is: "the part of x that lies on the line generated by x_{2}".

normally, we project a vector onto the unit vectors e_{1}= (1,0) and e_{2}= (0,1).

that is: if x = (x_{1},x_{2}) = x_{1}e_{1}+ x_{2}e_{2}, then:

the projection of x in the direction of e_{1}is [(x.e_{1})/(e_{1}.e_{1})]e_{1}

= [(x_{1}*1 + x_{2}*0)/(1*1 + 0*0)]e_{1}= x_{1}e_{1}= (x_{1},0).