Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
Thanks EuYu. ${}{}$ –  Noomi Holloway Dec 8 '12 at 9:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.