$AA^*A=A$ 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?
 A: Here's a proof that works:
Every matrix is unitarily triangularizable.  So, there exists a unitary $U$ such that $A = UTU^*$, where
$$
T = \pmatrix{1&t\\0&0}
$$
If we show that $t$ is necessarily zero, then we may conclude that $A$ is unitarily diagonalizable (since $T$ would then be diagonal).
With that goal in mind, we note that
$$
AA^*A = A \implies\\
UTU^*UT^*U^*UTU^* = UTU^* \implies\\
TT^*T = T
$$
Moreover, we compute
$$
TT^*T = T \pmatrix{1&t\\ \bar t & |t|^2} = \pmatrix{1 + |t|^2 & t(1 + |t|^2)\\0 & 0}
$$
We see that $TT^*T = T$ indeed implies that $t = 0$, which means that $A$ is unitarily diagonalizable.

It is interesting to see how this proof generalizes to the $n \times n$ case:
If $A$ has eigenvalues $0$ and $1$, then it is unitarily similar to the block matrix
$$
T = \pmatrix{I & Q\\0&N}
$$
where $N$ is strictly upper triangular (it is indeed true that the upper-left entry must be $I$; we note that the block associated with the eigenvalue $1$ is an invertible matrix satisfying $M = MM^*M$). As before, we may conclude that
$$
AA^*A = A \implies TT^*T = T
$$
and finally, we compute (with block-matrix multiplication)
$$
TT^*T = \pmatrix{I & Q\\0&N} \pmatrix{I & Q\\Q^* & Q^*Q + N^*N} = 
\pmatrix{I + QQ^* & (I + QQ^*)Q + QN^*N\\NQ^*&N(Q^*Q + N^*N)}
$$
noting that $QQ^* = 0 \iff \operatorname{tr}(QQ^*) = 0 \iff Q = 0$, we see that $TT^*T = T$ implies that $Q = 0$.
However, we have failed to prove that $N = 0$; we merely know that $N$ is strictly upper triangular with $N = NN^*N$.  This does not imply that $N$ is zero, as seen with the counterexample
$$
N = \pmatrix{0&1\\0&0}
$$
A: 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.
