If $PP^*=P^*P$ and $P^2=P$ then show that $P^*=P$ To show an idempotent normal matrix is an $\textbf{orthogonal}$ projection.
Proof: If we use, spectral decomposition of normal matrices we can write, $P=U\Lambda U^* \implies P^2=U\Lambda^2U^*$. Idempotence gives $\lambda(P)\in \{0,1\}$, so that  $P^*=U\Lambda^* U^*=U\Lambda U^*=P$.
Can we do this proof without using spectral demcomposition?
 A: Here is an elementary proof (which works in infinte dimesional spaces too): $\|P^{*}x-Px\|^{2}= \langle P^{*}x, P^{*}x \rangle-2\Re \langle P^{*}x, Px\rangle+\|Px\|^{2}=2\|Px\|^{2}-2\Re \langle P^{*}x, Px\rangle$ since the $P^{*}P=PP^{*}$. Now consider two cases: $x=Py$ for some $y$ and $x=(I-P)y$ for some $y$. In both of these cases it follows easily that $\|P^{*}x-Px\|^{2}=0$ so $P^{*}x=Px$. Now use the fact that $x =Px+(x-Px)$ so the $P^{*}$ and $P$ coincide at every point $x$.
[I have used the following calculation: $\langle P^{*}x, P^{*}x \rangle =\langle PP^{*}x, x \rangle = \langle P^{*}Px, x \rangle=\langle Px, Px \rangle=\|Px\|^{2}$].
A: The main argument supposes that you know that the trace of an idempotent matrix (in characteristic 0) gives you said matrix's rank-- this result does not require understanding of spectral theorem / normality or even eigenvalues (see post-script).
main argument
$(P^*P)^2 = P^*PP^*P= P^*P^*PP = (P^*P)$
hence $(P^*P)$ is idempotent.  Since we work over $\mathbb C$, we know
$\text{rank}\Big(P^*P\Big) =\text{rank}\Big(P\Big)\implies \text{trace}\Big(P^*P\Big) =\text{trace}\Big(P\Big)=\text{trace}\Big(P^2\Big)$
but
$\text{trace}\Big(P^*P\Big) \geq \text{trace}\Big(P^2\Big)$
by Cauchy-Schwarz (met with equality)  $\implies P^*=P$
optional post-script:
'eigenvalue-free' proof that an idempotent matrix has rank given by its trace (and that said matrix must be diagonalizable).
Suppose idempotent $A$ has rank r, then apply rank nullity, and build a basis for its kernel -- call these vectors $\big\{\mathbf q_{r+1},..., \mathbf q_{n}\big\}$.  Further we have $\text{rank}\big(A^2\big)=\text{rank}\big(A\big)$ so $V=\text{image}\big(A\big)\oplus \text{ker}\big(A\big)$.  Extend the $\mathbf q_j$ to a basis for the vector space by appending $r$ linearly independent vectors from the image of $A$, and collect these vectors in a matrix.
$Q:=\bigg[\begin{array}{c|c|c|c|c} 
\mathbf q_1  &\cdots & \mathbf q_{r} & \mathbf q_{r+1} &\cdots & \mathbf q_{n}\end{array}\bigg]$
$ AQ  = QD$
where $D=\begin{bmatrix} B_r &\mathbf {0}\\ * &\mathbf {0}\end{bmatrix}=\begin{bmatrix} I_r &\mathbf {0}\\ \mathbf {0} &\mathbf {0}\end{bmatrix}$
because
1.) $A$ sends vectors $\mathbf q_j$ for $r+1\leq j\leq n$ to zero
2.) $*$ is seen to be zero because e.g. $A\mathbf q_1 \in \text{image}\big(A\big)$ hence $A\mathbf q_1$ is uniquely written as a linear combination of $\mathbf q_k$ for $1\leq k\leq r$ (i.e. basis vectors for the image of $A$).
3.) $QD^2 = A^2Q =AQ=QD\implies B_r^2 = B_r\implies B_r =I_r$
because $\text{rank}\big(A\big)= r= \text{rank}\big(B_r\big)$ and the only invertible idempotent element is the identity (a one line proof).  Since $Q$ is invertible we have  $A=QDQ^{-1}$
$\implies \text{rank}\Big(A\Big)=r = \text{trace}\Big(D\Big)= \text{trace}\Big(QDQ^{-1}\Big)= \text{trace}\Big(A\Big)$
