$AA^*A=A$ with eigenvalues $1$ and $0$, prove that $A$ is unitarily diagonalizable. 
Let $A$ be a $2$ by $2$ complex matrix such that $AA^*A=A$, and has eigenvalues of $1$ and $0$. Prove that $A$ is unitarily diagonalizable. 

Well, if a matrix is Hermitian, then it is unitary diagonalizable. It also has real eigenvalues which is another property of Hermitian matrices. Hence, I want to prove Hermitian. 
I want to prove it by showing that $\langle Az , w\rangle = \langle z, Aw\rangle$ but I am unable to find a clear way to showing this given the information. Is this the right approach?

The problem statement can be generalised as follows:

Let $A\in M_n(\mathbb C)$. If both $A$ and $AA^\ast$ are idempotent,  $A$ must be Hermitian.

Proof. By unitary triangularisation, we may assume that
$$
A=\pmatrix{P&X\\ 0&N}
$$
where all eigenvalues of $P$ are $1$ and all eigenvalues of $N$ are zero. As $A$ is idempotent, $P$ and $N$ must be idempotent too. Thus $P=I,\ N=0$ and
$$
AA^\ast = \pmatrix{I+XX^\ast&0\\ 0&0}.
$$
Since $I+XX^\ast\succeq I$, in order that $AA^\ast$ is idempotent, $X$ must be zero. Thus $A$ is unitarily similar to $I\oplus0$ and it is Hermitian.
