A is a product of two self-adjoint matrices if and only if A is similar to adjoint of A? How can we prove that $A$ is a product of two self-adjoint matrices $X, Y$ if and only if $A$ is similar to $A^\ast$? I'm thinking about proving it but no useful techniques come to my mind. Thanks! 
 A: This is (part of) Theorem 1 from this paper of Radjavi and Williams; the proof is somewhat involved.  Here is a sketch of the proof.  If $A=BA^*B^{-1}$ for some invertible $B$, then by some simple manipulations you can get the equation $A(B+B^*)=(B+B^*)A^*$, so $A(B+B^*)$ is self-adjoint.  If $B+B^*$ were invertible, then we would be done, because we could take $X=A(B+B^*)$ and $Y=(B+B^*)^{-1}$.  But $B+B^*$ might not be invertible.  To fix this, note that we could have replaced $B$ by $\lambda B$ for any nonzero scalar $\lambda$, and you can show that for any invertible $B$, there exists a scalar $\lambda$ such that $\lambda B+\bar{\lambda}B^*$ is invertible.
Conversely, suppose $A=XY$ where $X$ and $Y$ are self-adjoint.  First note that the properties of being a product of two self-adjoint matrices and being conjugate to your own adjoint are both invariant under conjugation (for the first, use that $BXYB^{-1}=(BXB^*)((B^*)^{-1}YB^{-1})$).  So we may assume that $A$ is in Jordan normal form.  Splitting $A$ as a block diagonal matrix consisting of an invertible block and a nilpotent block, you can show that the diagonal blocks of $X$ and $Y$ corresponding to the invertible block of $A$ are invertible and conjugate that invertible block to its adjoint.  The nilpotent block is also similar to its adjoint (you can just explicitly show that a matrix in Jordan normal form with $0$s on the diagonal is similar to its adjoint).  Putting this together, you can conclude that $A$ is similar to its adjoint.
As a hint to why the proof is so complicated, Radjavi and Williams note that the result is not true (at least in the direction $A=XY\Rightarrow A\sim A^*$) for operators on an infinite-dimensional Hilbert spaces, so some special property of finite-dimensional spaces (such as the Jordan normal form used here) must be used.
