Help with full rank factorization

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I've been tasked with proving the existence of a full rank factorization for an arbitrary m x n matrix, namely:

Let \textit{A} \in \textbf{R}^{m x n} with \textit{rank(A) = r} then there exist matrices \textit{B} \in \textbf{R}^{m x r} and \textit{C} \in \textbf{R}^{r x n} such that \textit{A = BC}. Furthermore \textit{rank(A) = rank(B) = r}.

I think I can prove the second property if I assume the first using \it{rank(AB)} \leq \it{rank(A)} and \it{rank(AB)} \leq \it{rank(B)}.

I'd appreciate a push in the right direction. Thanks.
 
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We have the following situation:
$$
V_n \stackrel{C}{\twoheadrightarrow} V_n/\operatorname{ker}A \cong V_r \cong \operatorname{im}A \stackrel{B}{\rightarrowtail} V_m
$$
which together ##BC## transform as ##A##. We have split the domain of ##A\, : \,V_n \longrightarrow V_m ## into ##V_n \cong \operatorname{ker}A \oplus V_n/\operatorname{ker}A## and the codomain in ##\operatorname{im}A \oplus V_{m-r}\,.##
 
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