Solution
= Solution
\b[Not unique in general.] For any invertible $r\times r$ <matrix> $S$,
$$
A=(BS)(S^{-1}C)
$$
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.