Show that if ~'T and 11'T are contrasts and that if

~'11 = 0, then ~'T = 0 and r/T = 0 put orthogonal constraints on C(X, Z), i.e., the treatment sum of squares can be broken down with orthogonal contrasts in the usual way.

Hint: Let ~' = p~[X, Z] and 11' = p~[X, Z]. Show that p~(I - M)Z[Z'(I - M)Zr Z'(I - M)p2 = o.