Prove that if $A^*+A=AA^*$ then $A$ is normal

Question

Prove that if $A^*+A=AA^*$ then $A$ is normal. I've tried some basic algebraic operations with no luck.

Answer 1

\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*}

Answer 2

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
$\text{trace}\big((A^*A)^2\big) = \text{trace}\big((AA^*)(A^*A)\big)\longrightarrow A \text{ is normal}$