Prove that if $A^*+A=AA^*$ then $A$ is normal Prove that if $A^*+A=AA^*$ then $A$ is normal.
I've tried some basic algebraic operations with no luck.
 A: \begin{align*}
&A^*+A=AA^*\\
\implies &(A-I)(A^*-I)=I\\
\implies &A-I=(A^*-I)^{-1}\\
\implies &(A^*-I)(A-I)=I\\
\implies &A^*A=A+A^*=AA^*\tag*{$\blacksquare$}
\end{align*}
A: for arbitrary $A \in \mathbb C^\text{n x x}$  if you compute
$\big\Vert A^*A - AA^*\big \Vert_F^2 $
and recall the positive definiteness of a norm, you get a well known trace inequality
$\text{trace}\big((A^*A)^2\big)\geq \text{trace}\big((A^2)^*A^2\big)=\text{trace}\big((A^*A)(AA^*)\big)$
with equality iff $A$ is normal.
(i) for this problem,  multiplying the original equation on the right by $AA^*$ and taking traces
$\text{trace}\big((AA^*)^2\big)= \text{trace}\big(A^*AA^*\big)+\text{trace}\big(AAA^*\big) = \text{trace}\big((A^*)^2A\big)+\text{trace}\big(A^2A^*\big)$
note that $\text{trace}\big((AA^*)^2\big)= \text{trace}\big((A^*A)^2\big)$
ii) multiplying the original equation on the right by $A^*A$ and taking traces
$\text{trace}\big((AA^*)(A^*A)\big)=  \text{trace}\big(A^* A^*A\big)+\text{trace}\big(AA^*A\big)=  \text{trace}\big((A^*)^2A\big)+\text{trace}\big(A^2A^*\big)$
thus
$\text{trace}\big((A^*A)^2\big) = \text{trace}\big((AA^*)(A^*A)\big)\longrightarrow A \text{ is normal}$
