The task is to prove that for an Hermitian matrix $A$ exists one and only matrix $S\in \text{GL}(\mathbb{C},n)$ upper triangular with positive entries on diagonal such that $A = S^T \bar{S}$?
As I know, for each Hermitian matrix $A$ we can find matrix $S\in\text{GL}(\mathbb{C},n)$ such that $A=S^TES = S^TS$, but sure $S$ mustn't an upper triangular. As a hint for a task I am given that if you use Gram-Schmidt orthonormalization, then transform matrix between bases is upper triangular with positive entries on the diagonal (this fact is actually easy to show), so I can assume that this matrix $S$ I am looking for is a transform matrix for two bases such that one was generated by other by Gram-Schmidt orthonormalization. But I still have no idea what these bases could be.
EDIT: Sorry, I sure meant "upper triangular" not "upper diagonal" matrix $S$. Thank you in advance!
Hint: Let $P=S^T$. Then you are looking for a matrix $P$ with positive diagonal such that $A=PP^\ast$. Suppose $A=PP^\ast$ for some $P$ that does not necessarily have a positive diagonal. Then you may consider $(PD)(PD)^\ast$, where $D=\operatorname{diag}(e^{i\theta_1},e^{i\theta_2},\ldots,e^{i\theta_n})$ for some appropriate angles $\theta_1,\theta_2,\ldots,\theta_n\in\mathbb{R}$.