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

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 $$M=AB$$.

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 $$\text{rank}{A}=\text{rank}{B}=r$$.
• $I_{r\times r}$ is the $r\times r$ identity matrix and so on. – xpaul Sep 24 '19 at 19:32