 Solution

The perpendicular decomposition in part (a) gives, by the Pythagorean theorem,Thus an orthogonal projection cannot lengthen a vector and $\boxed{\Vert P{\Vert }_2\le 1}$. Algebraically the cross term vanishes because $P^T(I-P)=P-P^2=0$, giving the same identity directly from the hypotheses. If $P\ne 0$, take a nonzero $x$ in its range; then $Px=x$ and $\Vert P{\Vert }_2=1$. For $P=0$ the norm is zero.