A is a Hermitian projection if and only if it is an orthogonal projection I need to figure out this property of Hermitian / Orthogonal projections
"A is a Hermitian projection if and only if it is an orthogonal projection"
Your assistance will be highly appreciated.
Thank You
 A: 
If $A$ is a Hermitian projection, then $A$ is an orthogonal projection.

Let $u\in\mathcal{R}(A)$ and $v\in\mathcal{N}(A)=\mathcal{R}(I-A)$, that is, $u=Ax$ and $v=y-Ay$ for some $x$ and $y$. Since $A$ is Hermitian, we have $\langle u,v\rangle=\langle Ax,y-Ay\rangle=\langle x,A(y-Ay)\rangle=\langle x,Ay-Ay\rangle=0$. Hence $\mathcal{R}(A)\perp\mathcal{N}(A)$.

If $A$ is an orthogonal projection, then $A$ is a Hermitian.

We show that if $A$ is not Hermitian, then $A$ is not orthogonal. Assume $A$ is not Hermitian. Then $\langle Ax,y\rangle -\langle x,Ay\rangle\neq 0$ for some $x$ and $y$. Hence
$$
\begin{split}
0&\neq \langle Ax,y\rangle -\langle x,Ay\rangle
=
\langle Ax,y\rangle -\langle Ax+(I-A)x,Ay\rangle \\
&=
\langle Ax,(I-A)y\rangle-\langle (I-A)x,Ay\rangle.
\end{split}
$$
Since the sum of the two inner products is nonzero, it means that at least one of them is nonzero (sum of two zeros cannot be nonzero, right?). Hence there is a $u\in\mathcal{R}(A)$ and $v\in\mathcal{R}(I-A)=\mathcal{N}(A)$ such that $\langle u,v\rangle\neq 0$ and, consequently, $A$ is not orthogonal.

It is easy to see that $\mathcal{R}(I-A)=\mathcal{N}(A)$. If $u\in\mathcal{R}(I-A)$, then $u=x-Ax$ for some $x$ and $Au=Ax-A^2x=Ax-Ax=0$. Hence $\mathcal{R}(I-A)\subset\mathcal{N}(A)$. Let $v\in\mathcal{N}(A)$ and consider $v=Av+(I-A)v=(I-A)v$ since $Av=0$. Hence $\mathcal{N}(A)\subset\mathcal{R}(I-A)$.
