Given that $A$ is a projection, $A$ is Hermitian if and only if $AA^\ast A=A$ In this question, $A^\ast$ is the conjugate transpose of $A$. I am asked to show that if $A$ is a projection matrix, that $A$ is Hermitian if and only if $A=AA^\ast A$. One direction is easy--if $A$ is Hermitian, the result is trivial. So what about the converse? The assumption that $A=AA^\ast A$ and that $A$ is a projection gives us some identities:
$$A^2=A,$$
$$A^\ast=A^\ast AA^\ast,$$
$$A=A(A^\ast)^2A,$$
etc.... But I am at a loss. How can I use these to prove the converse: If $A=AA^\ast A$ and $A$ is a projection, then $A$ is Hermitian?
 A: Using $A=AA^*A$, we can say that the restriction of $AA^*$ to the image of $A$ is the identity map. On the other hand, the kernel of $AA^*$ is the kernel of $A^*$, which is the orthogonal complement to the image of $A$. Conclude that $AA^*$ is the orthogonal projection onto the image of $A$.
Now, we may note that both $AA^*$ and $A$ are projections onto an $r$-dimensional subspace (where $r = rk(A)$).  It follows that $tr(A) = tr(AA^*) = r$. However, since $A$ satisfies $tr(AA^*) = tr(A)$, $A$ must be normal (this follows from Schur decomposition).  Since $A$ is normal with real eigenvalues, it must be Hermitian (this follows from the spectral theorem).
Thus, $A$ is Hermitian.
A: Consider the inner product $\langle C,D\rangle=tr(CD^*)$. 
Now $AA^*=AA^*AA^*$ and $A^*A=A^*AA^*A$. Thus, $A$, $A^*$, $AA^*$ and $A^*A$ are projections. So their traces are equal to their ranks, which  are also equal.
Finally, $\langle A-A^*,A-A^*\rangle=\langle A,A\rangle-\langle A,A^*\rangle-\langle A^* ,A\rangle+\langle A^*,A^*\rangle=$ 
$=tr(AA^*)-tr(AA)-tr(A^*A^*)+tr(A^*A)=0$. So $A=A^*$.
