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

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

I'd appreciate a push in the right direction. Thanks.

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

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

I'd appreciate a push in the right direction. Thanks.

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