A normal, idempotent linear operator must be self-adjoint I have been trying to solve this problem for quite a while. I am still unsure of whether any of the avenues I have pursued have been of any use. Any advice will be much appreciated.
Question:
Let $V$ be a finite-dimensional inner product space, and let $E$ be an idempotent linear operator on $V$. Prove that if $EE^* = E^*E$, then $E$ is self-adjoint.
(This is essentially exercise 5(a) in sec. 80 on p.162 of Paul R. Halmos, Finite-Dimensional Vector Spaces, but Halmos didn't assume that the dimension of $V$ is finite.)
 A: If $E$ is normal, $\| E x \| = \|E^* x \|$ for all $x$.  Similarly, $I-E$ is normal,
so $\|(I-E) x\| = \|(I - E^*)x \|$.  In particular, since $(I-E)Ex = 0$, $(I-E^*)Ex = 0$, i.e. $E^* E = E$, and similarly since $E(I-E)x = 0$,  $E^*(I-E)x = 0$, i.e. $E^* E  = E^* $.  But those together say  $E = E^*$.
A: Update. As $E$ is normal and idempotent, direct calculation shows that
$$
(E-E^\ast E)^\ast(E-E^\ast E)=0=(E^\ast-E^\ast E)^\ast(E^\ast-E^\ast E).\tag{1}
$$
Note that for any linear operator $A$, if $A^\ast A=0$, then $\langle Ax,Ax\rangle=\langle x,A^\ast Ax\rangle=0$ for every vector $x$, so that $A=0$. It follows from $(1)$ that $E-E^\ast E=0=E^\ast-E^\ast E$. Hence $E=E^\ast E=E^\ast$.

**(Old answer:)**
The equality $EE^\ast=E^\ast E$ implies that $(E+E^\ast-I)(E-E^\ast)=0$. If you can show that $E+E^\ast-I$ is invertible, you are done.
Suppose $(E+E^\ast -I)v=0$. Left-mulitply the equation by $E^\ast$, we get $E^\ast Ev=0$. Hence show that $Ev=0$. Since $EE^\ast=E^\ast E$, similarly, show that $E^\ast v=0$. Now, consider the equation $(E+E^\ast -I)v=0$ again and show that $E+E^\ast -I$ is invertible.
