Let $A$ be a square matrix and $A=BC$ be its rank factorization. Show that rank $(A)$=rank $(A^2)$ if and only if $CB$ is non-singular.
please give a hint.
My approach : Suppose that rank $(A)$=rank $(A^2)$. This implies range $A$=range $A^2$. Now from this how to proceed.
Hint: What can we say about $\operatorname{range}(B) \cap \ker(C)$?
Alternatively, note that $A^2 = B[CB]C$.
One approach is as follows. Let $A$ be $n \times n$ with rank $r$, so that $B$ is $n \times r$ and $C$ is $r \times n$.
Suppose that $\operatorname{rank}(CB) < r$ (i.e. $CB$ is singular). It follows that $$ \operatorname{rank}(A^2) = \operatorname{rank}(B(CB)C) \leq \operatorname{rank}(CB) < r = \operatorname{rank}(A). $$ On the other hand, suppose that $CB$ is non-singular. It follows that $$ \operatorname{rank}((CB)C) = \operatorname{rank}(C) = r. $$ Now, using the fact that that $B$ has a trivial kernel, conclude that $$ \operatorname{rank}(A^2) = \operatorname{rank}(B[CBC]) = \operatorname{rank}(CBC) = r. $$