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 $v^\ast E^\ast Ev=0 \Rightarrow Ev=0$
 A: 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 $E^*=E^*E = (E^*E)^*=E$.
A: 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.
