... { (1, 1) } and x = (1, 2). The orthogonal projection xZisxZ= (1, 2) · (1, 1) (1, 1) · (1, 1) (1, 1) =74 (1, 1) = (7/4, 7/4). (17 .19 )✷ 17 .6 Diagrammatic Methods in Hilbert SpacesOne of ... (h 1 , h2). Substituting 17 .39 in 17 .37 and 17 .38 we obtain23= 1 3h 1 + 1 3(h 1 + h2), (17 .40) and 23= 1 3(h 1 + h2) + 1 3h2. (17 . 41) The solution is h 1 = h2= 2/3 which giveske=23,43,23. ... KERNELS and y 1 2= xZ 1 − xZ22+ y22. (17 .15 )Eqs. 17 .14 and 17 .15 imply that xZ 1 − xZ22= 0 (17 .16 )so, by the strict positivity of inner products, xZ 1 = xZ2.✷If...