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.)

Solution 1:

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 $E = E^*$.

Solution 2:

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 $E=E^\ast E=E^\ast$.

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.