Decomposition of a Matrix $A$ into $C\cdot\overline{C^{T}}$ form

Question

Let $A\in\mathbb{K}^{n×n}$ be hermitian and positive definite.

Show that there exists $C\in GL(n,\mathbb{K})$ for wich

$C\cdot\overline{C^{T}}=A$

While I proved the simpler statement of there being a "Squareroot" for $A$ if $A$ fullfills the given conditions, I have no idea how to tackle this problem. Therefore I would appreciate some hints.

Answer 1

The spectral theorem implies that $A$ can be written as $UD\overline{U^\top}$ where $D$ is diagonal with positive entries. Now take $C=UD^{1/2}$.