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The full question is:

Let $M \in \mathbb{R}^{n\times m}$ and $r=\operatorname{rank}(M)$. Show there exists $A\in\mathbb{R}^{n\times r}$ and $B\in\mathbb{R}^{r\times m}$ such that $M=AB$.

So the way I tried it was that there are $r$ number of basis vectors $v_1,\dots, v_r$. And $M$ can be expressed as $$ \begin{bmatrix} | & | & & |\\ m_{1,1}v_1+\cdots+m_{n,1}v_r & m_{1,2}v_1+\cdots+m_{n,2}v_r & \dots & m_{1,m}v_1+\cdots+m_{n,m}v_r \\ | & | & & | \end{bmatrix} $$ So all the columns can be expressed as the linear combinations of linearly independent columns in $M$, but I'm not sure where to go from here to prove that $M=AB$.

Could anyone help please?

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First all there exist invertible matrices $P_{n\times n}, Q_{m\times m}$ such that $$ PMQ =\left(\begin{matrix}I_{r\times r}&N_{(n-r)\times (m-r)}\\0_{(n-r)\times r}&0_{(n-r)\times(m-r)} \end{matrix}\right). $$ Then $$ M=P^{-1}\left(\begin{matrix}I&N\\0&0 \end{matrix}\right)Q^{-1}=P^{-1}\binom{I}{0}(I, N)Q^{-1}. $$ Let $$ A=P^{-1}\binom{I}{0}, B=(I, N)Q^{-1} $$ and then $$ M=AB. $$ It is easy to check that $\text{rank}{A}=\text{rank}{B}=r$.

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  • $\begingroup$ What are the notations of matrices inside the matrix mean? $\endgroup$ – donnyan Sep 24 '19 at 19:28
  • $\begingroup$ $I_{r\times r}$ is the $r\times r$ identity matrix and so on. $\endgroup$ – xpaul Sep 24 '19 at 19:32
  • $\begingroup$ I meant the I, N, 0, 0 in what it seems like 2x2 matrix. What does that mean? $\endgroup$ – donnyan Sep 24 '19 at 20:04
  • $\begingroup$ Also, is PMQ basically rref matrix that reduced all linear dependent so that only r number of rows have nonzero entries? $\endgroup$ – donnyan Sep 24 '19 at 22:07
  • $\begingroup$ Yes, you are right. $\endgroup$ – xpaul Sep 25 '19 at 13:39

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