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?

The spectral theorem implies that any Hermitian matrix $H$ is diagonizable using an unitary base change $U$, so $U^* H U = D$, where $D$ is diagonal with real entries. For $D$ proposition 2) is true, since $\mathbb{C}$ is closed under squareroots, so $D = R^2$ for some diagonal $R$. Therefore, $H = U R^2 U^* = (U R U^*)^2$.
For any matrix $A$, the matrix $AA^*$ is Hermitian.
However, if you go to dimension $1$, $H = -1$ is Hermitian, but for any $A \in \mathbb{C}$ you have $AA^* = \lvert A \rvert ^2 > 0$, so 1) is false.
As pointed out by others, it is true for positive-semi-definite matrices $H$, let me add a construction: You can run through the argument above, noting $R^* = R$, and take $A$ to be $UR$. So $AA^* = UR (UR)^* = UR^2 U^* = H$.
