Let be a projector. We prove that , with equality if and only if P is an orthogonal projector.
I suppose we could use the formula
and use the fact that
But I am not sure how to proceed.
Geometrically, there is a good way to see this. Translating it into algebra is not as easy.
For any vector , you can write . This is the unique expression for x as the sum of a vector in the range of P (namely Px) and a vector in the kernel of P ( namely x–Px). The projection P is orthogonal if and only if its range and kernel are orthogonal subspaces of .
If P is orthogonal then you can apply Pythagoras' theorem to deduce that . So (for all x), which tells you that .
If P is not orthogonal then you want to find a vector x such that . The way to do that is to take x to be a unit vector in the orthogonal complement of the kernel of P. That's the hard part of the problem, and I don't have time now to explain how the argument goes.