Why is $\langle Px,y\rangle=\langle x,P^*y\rangle$

Question

This property is at the heart of a Wikipedia article on projections: this one.

I have not been able to make any headway in my understanding of the statement “A projection matrix $P$ is orthogonal if and only if it is Hermitian”. The article proves this by saying (I paraphrase): “well obviously $\langle Px,y\rangle=\langle x,P^*y\rangle\implies P=P^*$”.

Any answers or hints would be greatly appreciated. As well as an explanation of the property $\langle Px,y\rangle=\langle x,P^*y\rangle$, if there are alternative explanations/intuitions for why an orthogonal projection must be Hermitian/self-adjoint I’d appreciate that too.

Answer 1

The line you quote is part of the proof of the direction "if a projection $P$ is orthogonal, then it is self-adjoint."

• If a projection $P$ is orthogonal, then $\langle x, Py \rangle = \langle Px, y\rangle$. (This is proved a few lines above by rearranging the orthogonality condition $\langle Px, y-Py\rangle = \langle x-Px, Py \rangle = 0$.)

• Next, the equation $\langle Px, y \rangle = \langle x, P^* y\rangle$ holds by the definition of the adjoint. If it helps, think about matrix transposes and note that $(Px)^* y = x^* P^* y = x^* (P^* y)$.

Combining the two steps above yields $\langle x, Py \rangle = \langle Px, y\rangle = \langle x, P^* y\rangle$. This holds for all $x$ and $y$. In general if linear operators $A$ and $B$ satisfy $\langle x, A y \rangle = \langle x, B y\rangle$ for all $x,y$, then $A=B$. (One way to think about this is to consider $x,y$ being standard basis elements.) Thus $P=P^*$.

Answer 2

If $P$ is orthogonal then $\langle Px, y \rangle = \langle Px, Py \rangle + \langle Px, y- Py \rangle = \langle Px, Py \rangle $. Switching $x,y$, we obtain $\langle x, Py \rangle = \langle Px, Py \rangle $. Thus, $\langle Px, y \rangle = \langle x, Py \rangle$ which implies $\langle Px, y \rangle = \langle P^*x, y \rangle$ and further implies $\forall x,y\, \langle (P-P^*)x, y \rangle =0$. Taking $y = (P-P^*)x$, we find $\| (P-P^*)x\| = 0$ for all $x$ which implies $P-P^* = 0$ and therefore $P = P^*$.

If $P^* = P$ then $$\langle Px, y- Py \rangle = \langle x, P^*y- P^*Py \rangle = \langle x, Py- P^2y \rangle = \langle x, Py- Py \rangle =0,$$ since $P^2 = P$.