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 $\;$ $Ex$, $(I-E)x$ $\;$ satisfies \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 $E^*=E^*E = (E^*E)^*=E$.
Suppose $E$ is self-adjoint then $E=E^*$, So $EE^*=E^2=E^*E$. Now suppose conversely that $EE^*=E^*E$ then, $<EE^*(v),E^*E(v)> = <EE^*(v),E(v)>$ and also $$<E^*E(v),EE^*(v)> = <E(v),EE^*(v)> = <E^*E(v), E^*(v)>$$ and hence from these two equations that are true for all $v$ in $V$. We get $E^*(v)=E(v)$ for all $v$ belong to $V$ that is, $E$ is self adjoint.
