Can any positive semi-definite Hermitian matrix $H$ be written as$ A^\dagger A$? Question 1： Give $n\times n$ Hermitian matrix $H$, if $H$ is positive semi-definite, does there always exist $n\times n$ matrix $A$ and $B$ such that $H=A^\dagger A=  B B^\dagger$?
Question 2: If question 1 is right, can we require stronger, that is, give $n\times n$ Hermitian matrix $H$, if $H$ is positive semi-definite, does there always exist $n\times n$ normal matrix $A$   such that $H=A^\dagger A= A A^\dagger$?

Yes. Since $H$ is Hermitian, it is unitarily diagonalizable, $H=UDU^*$, with $D$ diagonal with real entries (this the Spectral Theorem, which follows for example from the Schur triangularization). 
Since $H\geq0$, it follows that $D\geq0$. Now take $D^{1/2}$ to be the diagonal matrix with diagonal entries $d_{kk}^{1/2}$ for each diagonal entry $d_{kk}$ of $D$. Then 
$$
H=(UD^{1/2}U^*)^2=(UD^{1/2}U^*)^*\,UD^{1/2}U^*.
$$
