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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 $r>1$.

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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 $B$).

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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.

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