Decomposition of a hermitian symmetric matrix.

Question

Lets say we have a $N$ ranked hermitian positive semidefinite matix $\textbf{X}$. Is there an algorithm or a way to decompose it to a $M \leq N$ ranked matrix $\textbf{F}$ such that $$\textbf{X} = \textbf{F} \textbf{F}^H$$

Answer 1

Note that ${\rm rank}\{\mathbf F \mathbf F^H\} = {\rm rank}\{\mathbf F\}$. So for the decomposition you seek, $\mathbf X$ and $\mathbf F$ always have the same rank. You write that $\mathbf X$ has rank $N$ and $\mathbf F$ has rank $M\leq N$, this would only work if $M=N$.

What you might mean is the decomposition of a given matrix $\mathbf X$ of size $N \times N$ with rank $M \leq N$ into $\mathbf F \mathbf F^H$ where $\mathbf F$ is of size $N \times M$. This is possible, e.g., via the Cholesky decomposition, as Daniel was pointing out.