# Prove that idempotent operator $E$ is self-adjoint if and only if $EE^∗$ = $E^∗E$

Let $V$ be a finite-dimensional inner product space, and let $E$ be an idempotent linear operator on $V$, i.e., $E^2 = E$. Prove that E is self-adjoint if and only if $EE^* = E^*E$.

Are there any simpler answers to the question that the answers provided here Normal, idempotent operator implies self-adjointness. . Both answers seem to be correct but contain logical steps that I can't comprehend e.g $(I−E)Ex=0 \Rightarrow (I−E^∗)Ex=0$ and $v^\ast E^\ast Ev=0 \Rightarrow Ev=0$

Suppose $E^2=E$. Then $E$ has a basis of eigenvectors with eigenvalues $0$ and $1$ because every vector can be written as $$x = Ex + (I-E)x,$$ and $Ex$, $(I-E)x$ satisfy \begin{align} E(Ex)&=1\cdot Ex,\\ E(I-E)x &= 0 \cdot (I-E)x. \end{align} $E$ is selfadjoint iff these eigenspaces are mutually orthogonal, which is equivalent to the condition that $$\langle Ex,(I-E)y\rangle=0,\;\;\; \forall x,y, \\ \iff \langle x,E^*(I-E)y\rangle = 0, \;\;\; \forall x,y, \\ \iff E^*(I-E)y=0,\;\; \forall y \\ \iff E^*(I-E) = 0 \\ \iff E^* = E^*E$$ The last condition holds iff $E^*=E^*E = (E^*E)^*=E$.