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Since $A\in M_{m\times n}(F)$ is an $m\times n$ matrix of rank $k$. Can $A$ be expressed as $A=BC$ where $B$ is an $m\times k$ matrix of rank $k$, and $C$ is a $k\times n$ matrix of rank $k$?

I know that since $A$ has rank $k$, then $A=PJ_kQ$ where $$ J_k=\begin{bmatrix} I_k & 0_{k,n-k}\\ 0_{m-k,k} & 0_{m-k,n-k} \end{bmatrix} $$ and $P$ is an invertible $m\times m$ matrix, and $Q$ is an invertible $n\times n$ matrix. However, I can't seem to fiddle this into two matrices like $B$ and $C$. Thanks.

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up vote 1 down vote accepted

This is called the rank factorization of a matrix. The wikipedia page provdies a rather detailed examination of the factorization so I will not go into detail. I will simply summarize the main result

Theorem: Let $A$ be an $m\times n$ matrix of rank $r$. Then there exists $m \times r$ matrix $C$ and $r\times n$ matrix $F$, each of rank $r$, such that $A = CF$. Moreover, one such decomposition can be obtained by taking $C$ as the reduced row echelon form of $A$ with non-pivot columns removed and taking $F$ as the reduced row echelon form of $A$ with zero rows removed.

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Thanks EuYu. ${}{}$ – Noomi Holloway Dec 8 '12 at 9:10

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