$A^2=A^*A$. Why is matrix $A$ Hermitian? Let $A$ be $n \times n$ matrix and $A^2=A^*A$.  Why is $A$ a Hermitian matrix?
 A: Let us first do a calculation (where we use the assumption)
$$
(A-A^*)^*(A-A^*)=(A^*-A)(A-A^*)=A^*A-(A^*)^2-A^2+AA^*=AA^*-(A^*)^2.
$$
Now
$$
((A^*)^2)^*=A^2=A^*A,
$$
so
$$
(A^*)^2=(A^*A)^*=A^*A.
$$
Inserting this into the calculation above,
$$
(A-A^*)^*(A-A^*)=AA^*-A^*A
$$
But then the trace of that matrix on the left-hand side is zero, since (here we use that the trace is linear and that $\text{tr}\,(AB)=\text{tr}\,(BA)$)
$$
\text{tr}\,(AA^*-A^*A)=\text{tr}\,AA^*-\text{tr}\,A^*A=0.
$$
Hence 
$$\text{tr}\,(A-A^*)^*(A-A^*)=0.
$$ 
Thus, the square of the Frobenius norm of $A-A^*$ is zero. It follows that $A-A^*=0$, and thus that $A$ is Hermitian.
A: This question is the complex counterpart of

if matrix such $AA^T=A^2$ then $A$ is symmetric?

With the inner product $\langle X,Y\rangle=\operatorname{Re}\operatorname{tr}(XY^\ast)$ defined on the real linear space $M_n(\mathbb C)$, Hermitian matrices are orthogonal to skew-Hermitian matrices. Now, if we denote the Hermitian and skew-Hermitian parts of $A$ by respectively $H$ and $K$, the condition $AA^\ast=A^2$ implies that $\langle K,K\rangle=\langle K,H\rangle=0$. Therefore $K=0$ and $A$ is Hermitian.
