Projection Operator Self-Adjointness

Question

Is it right to say that projection operator is not self-adjoint(Hermitian) if it is not orthogonal projection operator?

Answer 1

A projection operator $P$ is by definition orthogonal if and only if

$\langle Px, y \rangle = \langle x, Py \rangle \tag 1$

for all vectors $x$, $y$; if (1) binds then

$\langle Px, y \rangle = \langle x, Py \rangle = \langle P^\dagger x, y \rangle, \tag 2$

whence

$Px = P^\dagger x, \tag 3$

or

$P = P^\dagger, \tag 4$

that is, $P$ is self-adjoint. The argument may be reversed: (4) clearly implies (3); thus

$\langle Px, y \rangle = \langle P^\dagger x, y \rangle = \langle x, Py \rangle, \tag 5$

so $P$ is orthogonal.

Since we have shown that a projection operator is orthogonal if and only if it is self-adjoint, we conclude that if $P$ is not orthognonal, then $P$ is not self-adjoint.