 Solution

Not unique in general. For any invertible $r\times r$ matrix $S$,is another rank factorization, with the same full ranks. This is a change of basis in the intermediate $r$-dimensional space. For example, even $[1]=[1][1]=[2][1/2]$ has different full-rank factors. Once a particular ordered basis $B$ is fixed, its coordinate matrix $C$ is unique; it is the basis choice that creates the freedom.