Can all Hermitian matrices $H$ be written as $H=A^* A$? All the matrices below are square, complex matrices.
1) Is it true that, for every Hermitian matrix $H$, there exists $A$, that $A^*A=H$?
2) For any $A$, does $A^*A$ always have a square root? If it's not, is there any simple presumption of $A$ that makes $A^*A$ always have a square root?

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*This property is only true for matrices $H$ such that $x^*Hx\geq 0$ for all $x$ (non-negative definite). To see that, note that if it's the case $x^*A^*Ax=\lVert Ax\rVert^2\geq 0$. Conversely, if $x^*Hx\geq 0$ for all $x$, then we can diagonalize $H$, and each eigenvalue is non-negative (hence admits a square root).  

*Yes, as $A^*A$ is non-negative definite and symmetric, we can find $P$ unitary and $D$ diagonal such that $A^*A=P^*DP$. As each diagonal entry of $D$ is non negative, take $D'$ diagonal such that $D'^2=D$. Then take $R:=P^*D'P$.
