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.)
Best Answer
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 $E = E^*$.