# Linear algebra- matrix of rank $r$ as product of two matrices of rank $r$

Let $$A$$ be $$m\times n$$ matrix of rank $$r$$ then show that $$A$$ can be written as product of matrices of order $$m\times r$$ and $$r\times n$$.

I know how to prove the above relation if $$r=1$$ but don't know how to proceed for $$r>1$$.

By definition the span of the columns of $$A$$ has dimension $$r$$, so this subspace has a basis of size $$r$$.

Let these basis elements be the columns of the $$m \times r$$ matrix $$B$$.

Each column of $$A$$ can be written as a linear combination of these basis elements, so this means there exists some $$r \times n$$ matrix $$C$$ such that $$A=BC$$. Specifically, the $$i$$th column of $$C$$ contains the coefficients that appear when expressing the $$i$$th column of $$A$$ as a linear combination of the basis (columns of $$B$$).

Adding up on previous responses, any $$m\times n$$ matrix $$A$$ of rank $$r$$ is equivalent to: $$J_{r,m,n} = \begin{pmatrix} 1 & 0 & 0 & & \cdots & & 0 \\ 0 & 1 & 0 & & \cdots & & 0 \\ 0 & 0 & \ddots & & & & 0\\ \vdots & & & 1 & & & \vdots \\ & & & & 0 & & \\ & & & & & \ddots & \\ 0 & & & \cdots & & & 0 \end{pmatrix} \in \mathcal{M}_{m,n}(\mathbb{K})$$

with $$r$$ ones on the diagonal.

So there exists two invertible matrices $$P,Q \in \mathcal{GL}_m(\mathbb{K}) \times \mathcal{GL}_n(\mathbb{K})$$ so that $$A = PJ_{r,m,n}Q$$. Since $$J_{r,m,n}$$ can easily be made into the desired product (since $$r\leq m,n$$): $$J_{r,m,n} = \underset{= J_{r,m,r}}{ \begin{pmatrix} I_r \\ \hline 0 \end{pmatrix}} \times \underset{= J_{r,r,n}}{\begin{pmatrix} I_r | 0 \end{pmatrix}}$$ Now notice how: $$A = (PJ_{r,m,r})(J_{r,r,n}Q)$$ yields the desired result because both $$J_{r,m,r}$$ and $$J_{r,r,n}$$ are of rank $$r$$ and $$P$$ and $$Q$$ have full rank.