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?

1. It is false, because $A^{\ast}A$ has non-negative eigenvalues, but $H$ only needs to have real eigenvalues. However, if we instead require that $H$ be positive semi-definite (i.e. non-negative eigenvalues), then the result is true - this is the _Cholesky decomposition_. However, such a decomposition is not unique.

2. This is true. $A^{\ast} A$ is a self-adjoint, positive semidefinite matrix, so it can be diagonalized as $A^{\ast}A = P^{-1}\Lambda P$ with $\Lambda$ a diagonal matrix with non-negative entries. Then we can simply square-root the diagonals of $\Lambda$, which I will denote by $\sqrt{\Lambda}$, and then the matrix $S = P^{-1} \sqrt{\Lambda}P$ serves as a square root of $A^{\ast}A$.